١
ﻨﻅﺭﻴﺔ ﺍﻟﺯﻤﺭ
J.S. Milne
١
، ﺃﻴﻠﻭل
٧٠٠٢
ﻤﻘﺩﻤﺔ
:
ﻜﺘﺒﺕ ﺍﻟﻨﺴﺨﺔ ﺍﻷﻭﻟﻰ ﻤﻥ ﻫﺫﻩ ﺍﻟﻤ
ﻼﺤﻅ
ﺎﺕ ﻟ
ﻁﻼﺏ ﺍﻟﺴﻨﺔ ﺍﻷﻭﻟﻰ ﻤﻥ ﺨﺭﻴﺠﻲ ﺍﻟﺠﺒ
ﺭ
.
ﻭـﻫ ﺎـﻤﻜ
ﺍﻟﺤﺎل ﻓﻲ ﻤﻌﻅﻡ ﻫﺫﻩ ﺍﻟﺩﻭﺭ
ﻭﻜﺎﻨﺕ ﻤﺘﺭﻜﺯﺓ ﻋﻠﻰ ﺍﻟﺯﻤﺭ ﺍﻟ،ﺕﺍ
ﻤﺠﺭﺩ
ﻰـﻠﻋ ،ﺹﺎـﺨ لﻜﺸﺒﻭ ،ﺓ
ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
ﺒﻜل ﺍﻷﺤﻭﺍل
ﺇﻥ ﺍﻟﺯﻤﺭ،
ﺍﻟ
ﻤﺠﺭﺩ
ﺓ
ﻟﻴﺴﺕ ﻫﻲ ﺍﻟﻤﺸﻜﻠ
ﺔ ﺍﻟﻭﺤﻴﺩﺓ ﺍﻟﺘﻲ ﺘﻭﺍﺠﻪ ﻤﻌﻅﻡ
ﻋﻠﻤﺎﺀ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻭﺇﻨﻤ
ﺍﻟﺯﻤﺭ،ﺔﻴﺭﺒﺠﻟﺍ ﺍﻟﺯﻤﺭﺎ
ﺍﻟ
ﻁﺒﻭﻟﻭﺠﻴ
ﺔ
ﺎﻟﺯﻤﺭـﺒ ﻡﺘـﻬﻨ ﻻﻭ ،ﻲﻟ ﺭﻤﺯ ،ﻭﺃ ،
ﺒﺤﺩ ﺫﺍﺘﻬﺎ ﻓﻘﻁ ﻭﺇﻨﻤﺎ ﺒﺘﻤﺜﻴﻼﺘﻬﺎ ﺍﻟﺨﻁﻴﺔ ﺃﻴﻀﺎﹰ
.
ﻴﺘﺭﻜﺯ
ﺍﻫﺘﻤﺎﻤﻲ
)
ﻓﻲ
ﺴﺘﻘﺒلـﻤﻟﺍ
(
ﺫﻩـﻫ ﻊﻴـﺴﻭﺘﺒ
ﺍﻟﻤ
ﻼﺤﻅ
ﺎﺕ
ﻟﺘﻌﻁﻲ ﻭﺘﻨﺘﺞ
ﻤﺠﻠﺩﺍﹰ ﻴﺒﻠﻎ ﺤﺠﻤﻪ
)
٠٠٢
ﺼﻔﺤﺔ
(
ﻭ،
ﺍﻟﺫﻱ ﺴﻴﺒﺭﻫﻥ
،ﺭـﺜﻜﺃ ﺔﻴﻟﻭﻤﺸﺒ ،
ﻭﺍﻟﻤﺠﺎﻻﺕ ﺍﻟﻤﺭﺘﺒﻁﺔ ﺒﻬﺎ،ﺀﺎﻴﺯﻴﻔﻟﺍ ،ﺕﺎﻴﻀﺎﻴﺭﻟﺍ ﻲﺠﻴﺭﺨﻟ ﺭﻤﺯﻟﺍ ﺔﻴﺭﻅﻨ ﻰﻟﺇ ﺔﻤﺩﻘﻤ ﻡﻴﻫﺎﻔﻤ
.
ﻴﺭﺠﻰ ﺇﺭﺴﺎل ﺍﻟﻤﻼﺤﻅﺎﺕ ﻭﺍﻟﺘﺼﻭﻴﺒﺎﺕ ﻋﻠﻰ ﺍﻟﻤﻭﻗﻊ
math0 at jmilne.org
.
V2.01
)
١٢
، ﺁﺏ
٦٩٩١
.(
؛ﺍﻟﻨﺴﺨﺔ ﺍﻷﻭﻟﻰ ﻋﻠﻰ ﺸﺒﻜﺔ ﺍﻹﻨﺘﺭﻨﺕ
٧٥
ﺼﻔﺤﺔ
.
V2.11
)
٩٢
، ﺁﺏ
٣٠٠٢
.(
ﺤﺩﺩﺕ ﺍ
ـ ﺍﻟﺘ؛ﺓﺭﻴﻐﺼﻟﺍ ﺀﺎﻁﺨﻷﺍ ﻥﻤ ﺩﻴﺩﻌﻟ
ﺭ
؛ﺭـﻴﻐﺘﻴ ﻡـﻟ ﻡﻴﻗ
٥٨
ﺼﻔﺤﺔ
.
V3.00
)
١
، ﺃﻴﻠﻭل
٧٠٠٢
.(
؛ﻤﻊ ﺍﻟﺘﻨﻘﻴﺢ ﻭﺍﻟﺘﻭﺴﻴﻊ
١٢١
ﺼﻔﺤﺔ
.
ﺤﻘﻭﻕ ﺍﻟﺘﺄﻟﻴﻑ ﻭﺍﻟﻨﺸﺭ
.© 1996, 2002, 2003, 2007, J.S. Milne
ﻥـﻤ ﺡﺭﺼـﻤ ﻥﺫﺇ ﻥﻭﺩﺒ ﻱﺭﺎﺠﺘﻟﺍ ﺭﻴﻏ ﻲﺼﺨﺸﻟﺍ لﺎﻤﻌﺘﺴﻼﻟ ﺓﺩﺤﺍﻭ ﺓﺭﻤ لﻤﻌﻟﺍ ﺍﺫﻫ ﺦﺴﻨ ﻥﻜﻤﻴ
ﺼﺎﺤﺏ ﺤﻘﻭﻕ ﺍﻟﻁ
ﺒﺎﻋﺔ ﻭﺍﻟﻨﺸﺭ
.
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٢
ﻴﻤﻜﻥ ﺍﻟﻘﻭل ﺒﺄﻥ ﻨﻅﺭﻴﺔ ﺍﻟﺯﻤﺭ ﺫﺍﺕ ﺍﻟﺭﺘﺏ ﺍﻟﻤﻨﺘﻬﻴﺔ
ﺘﻌﻭﺩ
ﻲـﺸﻭﻜ ﻥـﻤﺯ ﻰﻟﺇ
.
ﻟﻓﺈ
ـﻴ
ﻪ
ﻊـﺠﺭﺘ
ﺍﻟﻤﺤﺎﻭﻻﺕ ﺍﻷﻭﻟﻰ ﻓﻲ ﺍﻟﺘﺼﻨﻴﻑ ﻤﻊ ﻤﺨﻁﻁ ﻟﺘﺸﻜﻴل
ﻨﻅﺭﻴﺔ
ﻤﻥ ﻋﺩﺩ ﻤﻥ ﺍﻟﺤ
ﺔـﻟﺯﻌﻨﻤﻟﺍ ﻕﺌﺎـﻘ
.
ﺃﺩﺨل ﻏﺎﻟﻭ
ﺍ
ﺇﻟﻰ
ﻫﺫﻩ
ﺍﻟﻨﻅﺭﻴﺔ
ﻓﻜﺭﺓ
ﻓﻲ
ﻏﺎﻴﺔ
ﻤﻥ
ﺔـﻴﺌﺯﺠﻟﺍ ﺓﺭـﻤﺯﻟﺍ ﻥﻋ ﺔﻴﻤﻫﻷﺍ
)
ﺔـﻴﻤﻅﺎﻨﻟﺍ
(
،
ﻭﺘﻘﺴﻴﻡ ﺍﻟ
ﺯﻤﺭ ﺇﻟﻰ ﺒﺴﻴﻁﺔ ﻭﻤﺭﻜﺒﺔ
.
ﻋﻼﻭﺓ ﻋﻠﻰ ﺫﻟﻙ
،
ﻭ
ﺒﻤﺎ
ﺔـﻴﻬﺘﻨﻤ ﺔـﺠﺭﺩ ﻥﻤ ﺔﻟﺩﺎﻌﻤ لﻜ ﻥﺃ
ﺘﻘﺎﺒل ﺯﻤﺭﺓ ﻤﻥ ﺭﺘﺒﺔ ﻤﻨﺘﻬﻴﺔ ﺒﺤﻴﺙ ﺘﻌﺘﻤﺩ ﻋﻠﻴ
ﺒﻴﻥ ﻏﺎﻟﻭ،ﺔﻟﺩﺎﻌﻤﻟﺍ ﺹﺍﻭﺨ ﻊﻴﻤﺠ ﺎﻬ
ﺍ
ﺒﺄﻥ ﺍﻟﻭﺼﻭل
ﺇﻟﻰ ﺘﻁﺒﻴﻘﺎﺕ
ﺍﻟﻨﻅﺭﻴﺔ
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻹﺴﻬﺎﻡ،ﹰﺍﺩﻴﻌﺒ ﻥﻭﻜﻴ ﻥﺃ ﻥﻜﻤﻴ
ﺴﻴﻜﻭﻥ ﺃﻜﺒﺭ
ﺭـﻴﻏ لﻜﺸـﺒ ،
ﻟﺘﻁﻭﻴﺭ ﻤﺘﺘﺎﻟ،ﺭﺸﺎﺒﻤ
ﻴﺎﺘﻬﺎ ﺍﻟﺠﺯﺌﻴﺔ
.
ﻼلـﺨ ﻥـﻤ ،ﻥﻴﻴﺴﻨﺭﻔﻟﺍ ﺕﺎﻴﻀﺎﻴﺭﻟﺍ ﺀﺎﻤﻠﻋ لﺒﻗ ﻥﻤ ﺩﻴﺩﺤﺘﻟﺎﺒﻭ ،ﺕﺎﻓﺎﻀﻹﺍ ﻥﻤ ﺩﻴﺩﻌﻟﺍ ﺕﻤﺩﻗ
ﻤﻨﺘﺼﻑ ﺍﻟﻘﺭﻥ
)
ﺍﻟﺘﺎﺴﻊ ﻋﺸﺭ
.(
لـﺒﻗ ﻥﻤ ﺔﺜﻟﺎﺜﻟﺍ ﺔﺨﺴﻨﻟﺍ ﻲﻓ ﺔﻨﻫﺭﺒﻤﻠﻟ ﻲﻠﻴﺼﻔﺘ ﺽﺭﻋ لﻭﺃ ﻡﺩﻗ
ﺴﻴﺭﺕ
“Cours d’Alge`bre Supe´rieure,”
ﺍﻟﺘﻲ ﻁﺒﻌﺕ ﻓﻲ ﻋﺎﻡ
٦٦٨١
.
ﻙـﻟﺫ ﺩﻌﺒ ﺽﺭﻋ
ﺍﻟﺴﻴﺩ
ﺎﻡـﻋ ﻲـﻓ ﻥﺍﺩﺭﻭـﺠ
٠٧٨١
ﺴﺨﺔـﻨﻟﺍ
“Traite des substitutions et des e´quations
alge´briques.”
ﻭﻗﺩ
ﻜﺭﺱ ﺍﻟﺠﺯﺀ
ﺍﻷﻫﻡ
ﻤﻥ ﺒﺤﺙ
ﺎﻟﻭـﻏ ﺭﺎﻜﻓﺃ ﺭﻴﻭﻁﺘﺒ ﻥﺍﺩﺭﻭﺠ
ﺍ
ﺎـﻬﺘﺎﻘﻴﺒﻁﺘﻭ
ﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻌﺎﺩﻻﺕ
.
ﻜﺄﺠﺯﺍﺀ ﻤﺘ،ﺔﻴﺭﻅﻨﻟﺍ ﻲﻓ ﻅﻭﺤﻠﻤ ﻡﺩﻘﺘ ﺙﺩﺤﻴ ﻡﻟ
ﻔﺭ
ﺫﻜﺭﺍﺕـﻤ ﺭﻭـﻬﻅ ﻰﺘﺤ ،ﺎﻬﺘﺎﻘﻴﺒﻁﺘ ﻥﻤ ﺔﻗ
ﻫﻴﺭ ﺴﻴﻠﻭ
“The´ore`mes sur les groupes de substitutions”
ﻲـﻓ
ﺯﺀـﺠﻟﺍ
ﻥـﻤ ﺱﻤﺎـﺨﻟﺍ
Mathematische Annalen
.
ﺘﻘﺩﻤﺕ،ﺓﺭﻴﺨﻷﺍ ﺕﺍﻭﻨﺴﻟﺍ ﻲﻓ ﺔﺼﺎﺨﻭ ،ﺕﺍﺭﻜﺫﻤﻟﺍ ﻩﺫﻫ ﺦﻴﺭﺎﺘ ﺫﻨﻤ
ﺍﻟﻨﻅﺭﻴﺔ ﺒﺸﻜل
ﻤﺴﺘﻤﺭ
.
W. Burnside, Theory of Groups of Finite Order, 1897.
ﺃﺩﺨل ﻏﺎﻟﻭ
ﺍ
ﻤﻔﻬﻭﻡ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨ
ﺎﻅﻤﻴﺔ ﻓﻲ ﻋﺎﻡ
٢٣٨١
ﻭ ﺃﺸﺎﺭ ﻜﺎﻤﻴل ﺠﻭﺭﺩﺍﻥ ﻓﻲ ﻤﻘﺩﻤﺔ،
Traite´
ﻓﻲ ﻋﺎﻡ
٠٧٨١
ﻡـﻅﻌﻤ ﻲـﻓ ﺎـﻤﻜ ﺔـﺒﻜﺭﻤﻟﺍ ﺭﻤﺯﻟﺍﻭ ﺔﻁﻴﺴﺒﻟﺍ ﺭﻤﺯﻟﺍ ﻥﻴﺒ ﺯﻴﻴﻤﺘﻟﺍ ﻰﻟﺇ
ﺍﻟﺘﻔﺭﻋﺎﺕ ﺍﻟﻬﺎﻤﺔ ﻓﻲ ﻨﻅﺭﻴﺔ
ﺯﻤﺭ
ﺍﻟﺘﺒﺩﻴﻼﺕ
.
ﺒﻌﺩ ﺫﻟﻙ
ﺒﺩﺃ ﺠﻭﺭﺩﺍﻥ،
Traite´
ﺎﺕـﻨﺎﻴﺒ ﺓﺩﻋﺎﻗ ﺀﺎﻨﺒﺒ
ﻟﻠﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
–
ﺍﻟﺯﻤﺭ ﺍﻟﻤﺘﻨﺎﻭﺒﺔ ﻤﻥ
ﺭـﻤﺯﻟﺍ ﻥﻤ ﺭﻴﺜﻜﻟﺍﻭ لﻗﻷﺍ ﻰﻠﻋ ﺔﺴﻤﺎﺨﻟﺍ ﺔﺠﺭﺩﻟﺍ
ﺍﻟﺨﻁﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺍﻹﺴﻘﺎﻁﻴﺔ ﻋﻠﻰ ﺤﻘﻭل ﺫﺍﺕ ﺩﻟﻴل ﺃﻭﻟﻲ
.
ﻪـﺘﺎﻨﻫﺭﺒﻤ ﻭﻠﻴـﺴ ﻎﻴﻨﻭﺩﻭﻟ ﻊﺒﻁ ،ﹰﺍﺭﻴﺨﺃ
ﺍﻟﺸﻬﻴﺭﺓ ﻋﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺘﻲ ﺭﺘﺒﻬﺎ ﻗﻭﻯ ﻷﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ ﻓﻲ ﻋﺎﻡ
٢٧٨١
.
R. Solomon, Bull. Amer. Math. Soc., 2001.
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٣
ﺍﻟﻔﻬﺭﺱ
ﻤﻘﺩﻤﺔ
...
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١
ﺍﻟﻔﻬﺭﺱ
..................................................................................
٣
ﻤﻼﺤﻅﺎﺕ
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...
٦
ﺍﻟﻔﺼل ﺍﻷﻭل
...........................................................................
٧
ﺘﻌﺭﻴﻔﺎﺕ ﻭ ﻨﺘﺎﺌﺞ ﺃﺴﺎﺴﻴﺔ
...............................................................
٧
ﺍﻟﻔﺼل ﺍﻟﺜﺎﻨﻲ
.............................................................
..............
٦٣
؛ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ ﻭﺍﻟﺘﻘﺩﻴﻤﺎﺕ
ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ
.................................................
٦٣
ﺃﻨﺼﺎﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ
................................................................
٦٣
ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ
..............................................
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٧٣
ﺍﻟﻤﻭﻟﺩﺍﺕ ﻭﺍﻟﻌﻼﻗﺎﺕ
................................................................
١٤
ﺍﻟﺘﻘﺩﻴﻡ ﺍﻟﻤﻨﺘﻬﻲ
ﻟﻠﺯﻤﺭ
................................................................
٣٤
ﺯﻤﺭ
ﻜﻭﻜﺴﺘﻴﺭ
.............................
.........................................
٥٤
ﺘﻤﺎﺭﻴﻥ
.............................................................................
٩٤
ﺍﻟﻔﺼل ﺍﻟﺜﺎﻟﺙ
...........................................................................
١٥
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻭﺍﻟﺘﻤﺩﻴﺩﺍﺕ
.
...........................................................
١٥
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭ
..............................................................
١٥
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ
..............................................................
٤٥
ﺍﻟﺠﺩﺍﺀ
ﺍﺕ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭﺓ
.............................................................
٥٥
ﺘﻤﺩﻴﺩﺍﺕ ﺍﻟﺯﻤﺭ
.....................................................................
٠٦
ﻤﺨﻁﻁ ﻫﻭﻟﺩﺭ
......................................................................
٢٦
ﺘﻤﺎﺭﻴﻥ
.............................................................................
٤٦
ﺍﻟﻔﺼل ﺍﻟﺭﺍﺒﻊ
..........................................................................
٦٦
ﺘﺄﺜﻴﺭ ﺯﻤﺭ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺎﺕ
...................................................
...........
٦٦
ﺘﻌﺭﻴﻑ ﻭﺃﻤﺜﻠﺔ
......................................................................
٦٦
ﺯﻤﺭ ﺍﻟﺘﺒﺩﻴﻼﺕ
......................................................................
٦٧
ﺨﻭﺍﺭﺯﻤﻴﺔ
Coxeter
Tood –
...............................
....................
٤٨
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٤
ﺍﻟﺘﺄﺜﻴﺭﺍﺕ ﺍﻻﺒﺘﺩﺍﺌﻴﺔ
..................................................................
٦٨
ﺘﻤﺎﺭﻴﻥ
.............................................................................
٨٨
ﺍﻟﻔﺼل ﺍﻟﺨﺎﻤﺱ
...................................
.....................................
١٩
؛ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
ﺘﻁﺒﻴﻘﺎﺕ
..............................................................
١٩
ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
.....................................................................
١٩
ﺍﻟﺘﻘﺭﻴﺒﺎﺕ ﺍﻟﺒﺩﻴﻠﺔ ﻟﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
....
................................................
٦٩
ﺃﻤﺜﻠﺔ
...............................................................................
٧٩
ﺘﻤﺎﺭﻴﻥ
.............................................................................
٢٠١
ﺍﻟﻔﺼل ﺍﻟﺴﺎﺩﺱ
............
...........................................................
٣٠١
؛ﺍﻟﻤﺘﺴﻠﺴﻼﺕ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ
ﺍﻟﺯﻤﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل ﻭﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
........................
٣٠١
ﺍﻟﻤﺘﺴﻠﺴﻼﺕ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ
........................................................
٣٠١
ﺍﻟﺯﻤﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠ
ﺤل
.................................................................
٦٠١
ﺍﻟﺯﻤﺭ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
...............................................................
١١١
ﺍﻟﺯﻤﺭ ﺒﻤﺅﺜﺭﺍﺕ
...................................................................
٥١١
ﻤﺒ
ﺭﻫﻨﺔ ﻜﺭﻭل
ـ
ﺴﺸﻤﻴﺩﺕ
........................................................
٨١١
ﺘﻤﺎﺭﻴﻥ
............................................................................
٩١١
ﺍﻟﻔﺼل ﺍﻟﺴﺎﺒﻊ
........................................................................
٠٢١
ﺘﻤﺜﻴﻼﺕ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
...............................................................
٠٢١
ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﻤﺼﻔﻭﻓﻴﺔ
..............................................................
٠٢١
ﺠﺫﻭﺭ
1
ﻓﻲ ﺍﻟﺤﻘﻭل
....................................................
...........
١٢١
ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺨﻁﻴﺔ
..................................................................
٢٢١
ﻤﺒﺭﻫﻨﺔ
Maschke
...............................................................
٣٢١
؛ ﺠﺒﺭ ﺍﻟﺯﻤﺭ
ﻨﺼﻑ ﺍﻟﺒﺴﻴﻁﺔ
.....................................
..................
٥٢١
ﺍﻟﻤﻭﺩﻭﻻﺕ ﻨﺼﻑ ﺍﻟﺒﺴﻴﻁﺔ
.........................................................
٦٢١
F
-
ﺍﻟﺠﺒﻭﺭ ﺍﻟﺒﺴﻴﻁﺔ ﻭ ﻤﻭﺩﻭﻻﺘﻬﺎ
.................................................
٧٢١
F
-
ﺍﻟ
ﺠﺒﻭﺭ ﻨﺼﻑ ﺍﻟﺒﺴﻴﻁﺔ ﻭﻤﻭﺩﻭﻻﺘﻬﺎ
..........................................
٤٣١
ﺘﻤﺜﻴﻼﺕ
G
......................................................................
٦٣١
ﻤﻴﺯﺍﺕ
G
..............................
.........................................
٨٣١
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٥
ﻗﺎﺌﻤﺔ ﺨﻭﺍﺹ ﺍﻟﺯﻤﺭﺓ
...............................................................
٢٤١
ﻟﻠﻜﺘﺎﺒﺔ
............................................................................
٢٤١
ﺘﻤﺎﺭﻴﻥ ﻤﺤﻠﻭﻟﺔ
......
..
.........
....
.....................
..
...........................
٤٤١
ﻤﺴﺎﺌل
ﻏﻴﺭ ﻤﺤﻠﻭﻟﺔ
...................................................................
٣٥١
ﺍﻤﺘﺤﺎﻥ ﻟﺴﺎﻋﺘﻴﻥ
.....................................................................
٠٦١
ﺍﻟﻤﺭﺍﺠﻊ
..............
................................................................
٣٦١
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٦
ﻤﻼﺤﻅﺎﺕ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻟﺭﻤﻭﺯ
ﺍﻵﺘﻴﺔ
:
{
}
0 , 1 , 2 , ...
,
=
ﺤﻠﻘﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺢ
=
Z=
ﺤﻘل ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ
=
ﺤﻘل ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ
=
ﺍﻟﺤﻘل ﺍﻟﻤﻜﻭﻥ ﻤﻥ
p
ﺤﻴﺙ،ﺭﺼﻨﻋ
p
ﻋﺩﺩ ﺃﻭﻟ
ﻲ
p
p
=
=
F
Z
Z
ﻟﻜل ﻋﺩﺩﻴﻥ ﺼﺤﻴﺤﻴﻥ
n
m ,
،
n
m
ﺘﻌﻨﻲ ﺒﺄﻥ
m
ﻴﻘﺴﻡ
n
، ﺃﻱ ﺃﻥ،
Ζ
∈
m
n
.
ﻨﻌﺘﺒﺭ
p
، ﺃﻱ ﺃﻥ،ﹰﺎﻴﻟﻭﺃ ﹰﺍﺩﺩﻋ ،ﺕﺎﻅﺤﻼﻤﻟﺍ ﻊﻴﻤﺠ ﻲﻓ
...
,
1000000007
,
...
,
11
,
7
,
5
,
3
,
2
=
p
.
ﻟﺘﻜﻥ ﻟﺩﻴﻨﺎ ﻋﻼﻗﺔ
ﺍﻟ
،ﺘﻜﺎﻓﺅ
[ ]
∗
ﺍ
ﻰـﻠﻋ ﻱﻭـﺤﻴ ﻱﺫﻟﺍ ﺅﻓﺎﻜﺘﻟﺍ ﻑﺼﻟ ﺯﻤﺭﺘ ﻲﺘﻟ
⋅∗
ﺯـﻤﺭﻨ
ﻟﻠﻤﺠﻤﻭﻋﺔ ﺍﻟﺨﺎﻟﻴﺔ ﺒﺎﻟﺭﻤﺯ
φ
.
ﺃﻤﺎ ﻗﺩﺭﺓ ﺍﻟﻤﺠﻤﻭﻋﺔ
S
ﻓﻨﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ
S
)
ﺈﻥـﻓ ﻙﻟﺫـﻟ
S
ﻫﻲ ﻋﺩﺩ ﻋﻨﺎﺼﺭ ﺍﻟﻤﺠﻤﻭﻋﺔ
S
ﻭﻥـﻜﺘ ﺎﻤﺩـﻨﻋ ﻙـﻟﺫﻭ ،
S
ﺔـﻴﻬﺘﻨﻤ
.(
ﺘﻜﻥـﻟ
I
,
A
ﻨﺭﻤﺯ ﻟﻤﺠﻤﻭﻋﺔ ﻋﻨﺎﺼﺭ،ﻥﻴﺘﻋﻭﻤﺠﻤ
A
ﺔـﻋﻭﻤﺠﻤﻟﺍ ﻥﻤ ﺎﻬﺘﻟﺩﺃ ﻲﺘﻟﺍ
I
ﺎﻟﺭﻤﺯـﺒ ،
( )
I
i
i
a
∈
،
ﻭﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﺘﻁﺒﻴﻕ ﻤﻥ ﺍ
ﻟﺸﻜل
١
:
i
i
a
I
A
→
a
.
ﻴﺘﻁﻠﺏ ﻭﺠﻭﺩ ﺍﻟﺤﻠﻘﺎﺕ ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎﻴﺩ
1
ﻭﺘﺸﺎﻜﻼﺕ ﺍﻟﺤﻠﻘﺎﺕ ﻟﻭﺠﻭﺩ،
ﺘﻁﺒﻴﻕ ﺘﻘﺎﺒل
.
ﻴﻜﻭﻥ ﺍﻟﻌﻨﺼﺭ
a
ﻤﻥ ﺤﻠﻘﺔ ﻤﺎ ﻋﻨﺼﺭ ﻭﺤﺩﺓ ﺇﺫﺍ ﻜﺎﻥ ﻟﻪ ﻤﻌﻜﻭﺱ
)
ﻋﻨﺼﺭ
b
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،ﹰﻼﺜﻤ
ba
ab
=
=
1
.(
Y
X
⊂
،
X
ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻤﻥ
Y
)
ﻟﻴﺱ ﺒﺎﻟﻀﺭﻭﺭﺓ ﻤﺠﻤﻭﻋﺔ ﻓﻌﻠﻴﺔ
(؛
def
=
X
Y
،
X
ﻑ ﻟﺘﺴﺎﻭﻱﺭﻌﺘ
Y
ﺃﻭ ﺘﺴﺎﻭﻱ،
Y
ﺒ
؛ﺎﻟﺘﻌﺭﻴﻑ
Y
X
≈
،
X
ﺘﻤﺎﺜل
Y
؛
X
Y
،
X
ﻭ
Y
ﻤﺘﻤﺎﺜﻼﻥ ﻗﺎﻨﻭﻨﻴﺎﹰ
)
ﺃﻭ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﻤﺎﺜل ﻭﺤﻴﺩ
ﺍﹰ
.(
1
A
ﺃﺴﺭﺓ
ﻴﺠﺏ ﺘﻤﻴﻴﺯﻫﺎ
ﻋﻠﻰ
ﻤﺠﻤﻭﻋﺔ
ﻤﺎ
.
ﺇ،ﹰﻼﺜﻤ
ﺫﺍ ﻜﺎﻨﺕ
f
ﺩﺍﻟﺔ ﻤﻥ ﺍﻟﺸﻜل
3
Ζ → Ζ Ζ
لـﺴﺭﺘ ﻲﺘﻟﺍ
ﻋﻨﺩﺌﺫ،ﻪﺌﻓﺎﻜﺘ ﻑﺼ ﻰﻟﺇ ﺢﻴﺤﺼﻟﺍ ﺩﺩﻌﻟﺍ
( )
{
}
Ζ
∈
i
i
f
,
، ﻤﺠﻤﻭﻋﺔ ﻤﻜﻭﻨﺔ ﻤﻥ ﺜﻼﺜﺔ ﻋﻨﺎﺼﺭ
ﺤﻴﺙ ﺇﻥ
( )
( )
Ζ
∈
i
i
f
ﺃﺴﺭﺓ ﻤﻊ ﻤﺠﻤﻭﻋﺔ
ﺒ
ﺩﻟﻴل ﻏﻴﺭ ﻤﻨﺘﻬﻲ
.
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٧
ﺍﻟﻔﺼل ﺍﻷﻭل
ﺘﻌ
ﺭﻴﻔ
ﺎﺕ
ﻭ ﻨﺘﺎﺌﺞ ﺃﺴﺎﺴﻴ
ﺔ
ﺘﻌ
ﺭﻴﻔ
ﺎﺕ
ﻭ ﺃﻤﺜﻠﺔ
ﺘﻌﺭﻴﻑ
1.1
ﺍﻟﺯﻤﺭﺓ
(group)
ﻫﻲ ﻜل ﻤﺠﻤﻭﻋﺔ
G
ﻤﻊ ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﺜﻨﺎﺌﻴﺔ
(
)
,
:
∗
× →
a
a b
a b G G
G
ﺘﺤﻘﻕ ﺍﻟﺸﺭﻭﻁ
ﺍﻵﺘﻴﺔ
:
.G
1
)
ﺍﻟﺨﺎﺼﺔ ﺍﻟﺘﺠﻤﻴﻌﻴﺔ
(
ﻟﻜل
G
c
b
a
∈
,
,
ﻴﻜﻭﻥ
(
)
(
)
;
∗
∗ = ∗
∗
a
b
c
a
b
c
.G
2
)
ﻴﻭﺠﺩ ﻋﻨﺼﺭ ﺤﻴﺎﺩﻱ
(
ﻴﻭﺠﺩ ﻋﻨﺼﺭ
G
e
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
( )
1
∗ = ∗ =
a
e
e
a
a
ﻟﻜل
G
a
∈
.G
3
)
ﻴﻭﺠﺩ ﻤﻌﻜﻭﺱ
(
ﻟﻜل
G
a
∈
ﻴﻭﺠﺩ،
G
a
∈
′
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
.
a
a
e
a
a
∗
′
=
=
′
∗
ﻨﺨﺘﺼﺭ ﻋﺎﺩﺓﹰ ﺍﻟﻜﺘﺎﺒﺔ
( )
∗
,
G
ﺇﻟﻰ
G
.
ﻨﻜﺘﺏ ﺃﻴﻀﺎﹰ
ab
ﺒﺩﻻﹰ ﻤﻥ
b
a
∗
ﻭ
1
ﻥـﻤ ﹰﻻﺩﺒ
e
،
ﺃﻭ ﻨﻜﺘﺏ
b
a
+
ﺒﺩﻻﹰ ﻤﻥ
b
a
∗
ﻭ
0
ﺒﺩﻻﹰ ﻤﻥ
e
.
ﺭﺓـﻤﺯﻟﺍ ﻥﺄـﺒ لﺎﻘﻴ ،ﻰﻟﻭﻷﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
ﻀﺭﺒﻴﺔ
(multiplicative)
ﻭﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻴﻘﺎل ﺒﺄﻨﻬﺎ ﺠﻤﻌﻴﺔ،
(additive)
.
2.1
ﻓﻴﻤﺎ ﻴﻠ
ﻲ
،
...
,
, b
a
ﺘﺭﻤﺯ ﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
.
)
a
(
ﺇﻥ ﺍﻟﻌﻨﺼﺭ ﺍﻟﺫﻱ ﻴﺤﻘﻕ ﺍﻟﻌﻼﻗﺔ
(1)
ﻴﺩﻋﻰ ﺒﺎﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎﻴﺩ
(neutral element)
.
ﻓﺈﺫﺍ ﻜﺎﻥ
e
′
ﻋﻨﺼﺭﺍﹰ
ﻋﻨﺩﺌﺫ،ﺭﺨﺁ ﹰﺍﺩﻴﺎﺤﻤ
e
e
e
e
=
′
∗
=
′
.
ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﺇﻥ
e
ﻭـﻫ
ﺍﻟﻌﻨﺼﺭ ﺍﻟﻭﺤﻴﺩ ﻤﻥ
G
ﺍﻟﺫﻱ ﻴﺤ،
ﻘﻕ
e
e
e
=
∗
)
ﺒﺘﻁﺒﻴﻕ
G
1
.(
)
b
(
ﺇﺫﺍ ﻜﺎﻥ
e
a
b
=
∗
ﻭ
e
c
a
=
∗
ﻋﻨﺩﺌﺫ،
(
) (
)
c
c
e
c
a
b
c
a
b
e
b
b
=
∗
=
∗
∗
=
∗
∗
=
∗
=
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻟﻌﻨﺼﺭ
a
′
ﻓﻲ
(G
3
)
ﺔـﻟﻻﺩﺒ ﺩﻴﺤﻭ لﻜﺸﺒ ﻑﺭﻌﻤ
a
.
ﺫﺍـﻫ ﻰﻤﺴـﻴ
ﺍﻟﻌﻨﺼﺭ ﻤﻌ
ﻜﻭﺱ
(inverse)
ﺍﻟﻌﻨﺼﺭ
a
ﻭﻴﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ
1
−
a
)
ﺭـﻴﻅﻨ ﻰﻤﺴﻴ ﻭﺃ
(negative)
ﺍﻟﻌﻨﺼﺭ
a
ﻭ
ﺭﻤﺯﻩ
a
−
.(
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٨
)
c
(
ﻨﻼﺤﻅ ﻤﻥ
(G
1
)
ﺃﻥ ﺠﺩﺍﺀ
ﺜﻼﺜﺔ ﻋﻨﺎﺼﺭ
3
2
1
,
,
a
a
a
ﻤﻥ
G
ﻴﻜﻭﻥ ﻤﻌﺭﻓﺎﹰ ﺒﺸﻜل
ﻓﻴﻤﺎ ﺇﺫﺍ ﺸﻜﻠﻨﺎ،ﺢﻀﺍﻭ
ﺃﻭﻻﹰ
2
1
a
a
ﻭﻤﻥ ﺜﻡ ﻨﺸﻜل
(
)
3
2
1
a
a
a
ﺸﻜلـﻨ ﻭﺃ ،
3
2
a
a
ﺃﻭﻻﹰ ﻭ
ﺒﻌﺩ ﺫﻟﻙ ﻨﺸﻜل
(
)
3
2
1
a
a
a
،
ﻓ
ﺎﻟﻨﺘﻴﺠﺔ ﺴﺘ
ﻜﻭﻥ ﻨﻔﺴﻬﺎ
.
ﺭﻫﻥـﺒﻴ ،ﺔـﻘﻴﻘﺤﻟﺍ ﻲﻓ
(G
1
)
ﺃﻥ ﺍﻟﺠﺩﺍﺀ ﻷﻱ
n
-
ﻋﺩﺩ
n
a
a
a
,
...
,
,
2
1
ﻤﻥ ﻋﻨﺎﺼﺭ
G
ﺸﻜلـﺒ ﻑﺭﻌﻤ
ﻭﺍﻀﺢ
.
ﻨﺒﺭﻫﻥ ﻋﻠﻰ ﺫﻟﻙ ﺒﺎﻻﺴﺘﻘﺭﺍﺀ ﺍﻟﺭﻴﺎﻀﻲ ﻋﻠﻰ
n
.
ﺇﻥ ﺃﺤﺩ ﺍﻟﺠﺩﺍﺀﺍﺕ
ﻗﺩ ﻴﻨﺘﻬﻲ
ﺇﻟﻰ
( )
2
(
)(
)
1
1
...
, ,
+
i
i
n
a
a
a
a
ﻜ
ﺒﻴﻨﻤﺎ ﻴﻭﺠﺩ ﺠﺩﺍﺀ ﺁﺨﺭ ﻤﻥ ﺍﻟﺸﻜل،ﻲﺌﺎﻬﻨ ﺀﺍﺩﺠ
( )
3
(
)(
)
1
1
...
, ,
+
j
j
n
a
a
a
a
ﻭﺫﻟﻙ ﺤﺴﺏ ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭ،ﺩﻴﺠ لﻜﺸﺒ ﻥﺎﻓﺭﻌﻤ ﻥﻴﻘﺒﺎﺴﻟﺍ ﻥﻴﺭﻴﺒﻌﺘﻟﺍ ﻥﺃ ﻅﺤﻼﻨ
ﺍﺌﻲ
.
ﺇﺫﺍ ﻜﺎﻥ،ﻙﻟﺫﻟ
j
i
=
،
ﻓﺈﻥ
(2)
ﻭ
(3)
ﻤﺘﻜﺎﻓﺌﺎﻥ
.
ﺎﻥـﻜ ﺍﺫﺇ ﺎﻤﺃ
j
i
≠
ﺫـﺌﺩﻨﻋ ،
ﻴﻤﻜﻥ ﺃﻥ ﻨﻔﺭﺽ ﺒﺄﻥ
j
i
<
.
ﻭﺒﺎﻟﺘﺎﻟﻲ
(
)(
) (
)
(
)(
)
(
)
(
)(
)
(
)
(
)
(
)
(
)
1
1
1
1
1
1
1
1
1
1
...
...
...
...
...
...
...
...
...
...
+
+
+
+
+
+
=
=
i
i
n
i
i
j
j
n
j
j
n
i
i
j
j
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
ﺇﻥ ﺍﻟﺘﻌﺒﻴﺭﺍﻥ ﻓﻲ ﺍﻟﺠﻬﺔ ﺍﻟﻴﻤﻨﻰ ﻤﺘﺴﺎﻭﻴﺎﻥ ﻭﺫﻟﻙ ﻤﻥ
(G
1
)
.
)
d
(
ﺇﻥ ﻤ
ﻌﻜﻭﺱ
n
a
a
a
...
2
1
ﻫﻭ
1
1
1
1
2
...
−
−
−
a
a
a
n
ﺴﺎﻭﻱـﻴ ﺀﺍﺩـﺠﻟﺍ ﺱﻭﻜﻌﻤ ﻥﺃ ﻱﺃ ،
ﺠﺩﺍﺀ
،ﺍﻟﻤﻌﻜﻭﺴﺎﺕ
ﻭﻟﻜﻥ
ﺒﺘﺭﺘﻴﺏ ﻤﻌﺎﻜﺱ
.
)
e
(
ﺇﻥ
)
(G
3
ﻴﺒﻴﻥ ﺒﺄﻥ ﻗﺎﻨﻭﻥ ﺍﻟﺤﺫﻑ
(cancellation)
ﺃﻱ ﺃﻥ،ﺭﻤﺯﻟﺍ ﻲﻓ ﻕﻘﺤﻤ
)
ﺒﺎﻟﺠﺩﺍﺀ ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﺒﺎﻟﻌﺩﺩ
1
−
a
(
ﻭ
ﺇﺫﺍ ﻜﺎﻨﺕ،ﺱﻜﻌﻟﺎﺒ
G
ﻓﺈﻥ ﻗﺎﻨﻭﻥﺫﺌﺩﻨﻋ ،ﺔﻴﻬﺘﻨﻤ
ﺍﻟﺤﺫﻑ ﻤﺤﻘﻕ ﻓﻲ
(G
3
)
،
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ
G
G
ax
x
→
:
a
ﻭـﻬﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ ،ﻥﻴﺎﺒﺘﻤ
ﺇﺫﺍ ﻜﺎﻥ،ﺹﺎﺨ لﻜﺸﺒ ،لﺒﺎﻘﺘ
e
ﺫـﺌﺩﻨﻋ ،ﻕﻴﺒﻁﺘﻟﺍ ﺓﺭﻭﺼ ﻲﻓ
ﻭﻥـﻜﻴ
ﺼﺭـﻨﻌﻠﻟ
a
ﻤﻌ
ﻴﻜﻭﻥ ﻟﻪ ﻤﻌﻜﻭﺱ،ﻪﺒﺎﺸﻤ لﻜﺸﺒﻭ ،ﻲﻨﻴﻤﻴ ﺱﻭﻜ
ﻥـﻤﻭ ،ﻱﺭﺎﺴﻴ
)
b
(
ﺄﻥـﺒ ﺩـﺠﻨ
ﺍﻟﻤﻌﻜﻭﺴﻴﻥ ﻤﺘﺴﺎﻭﻴﺎﻥ
.
ﺎﻥــﺘﺭﻤﺯﻟﺍ ﻥﻭــﻜﺘ
(
) (
)
∗′
′
∗
,
,
,
G
G
لــﺒﺎﻘﺘ ﻕــﻴﺒﻁﺘ ﺩــﺠﻭ ﺍﺫﺇ ﻥﻴﺘﻠﺜﺎــﻤﺘﻤ
G
G
a
a
′
↔
′
↔
,
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
(
)
b
a
b
a
′
∗′
′
=
′
∗
ﻟﻜل
G
b
a
∈
,
.
c
b
ca
ba
c
b
ac
ab
=
⇒
=
=
⇒
=
,
PDF created with pdfFactory Pro trial version
٩
ﺇﻥ
G
ﺭﺘﺒﺔ
(order)
ﺍﻟﺯﻤﺭﺓ
G
ﻫﻲ ﻗﺩﺭﺓ ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ
.
ﺘﺴﻤﻰ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻲ
ﺭﺘﺒﺘﻬﺎ ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
p
ـ ﺒ
p
-
ﺯﻤﺭﺓ
.
ﻟﻜل ﻋﻨﺼﺭ
a
ﻤﻥ
G
ﻑﺭﻌﻨ ،
<
=
>
=
−
−
−
0
...
0
0
...
1
1
1
n
a
a
a
n
e
n
a
aa
a
n
ﺇﻥ ﺍﻟﻨﺘﺎﺌﺞ
ﺍﻵﺘﻴﺔ
ﻤﺤﻘﻘﺔ
( )
4
( )
,
,
+
=
=
n
n
m
n m
m
mn
a a
a
a
a
ﻟﻜل
Ζ
∈
n
m,
ﻨﺠﺩ ﻤﻥ
(4)
ﺒﺄﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
e
a
n
n
=
Ζ
∈
ﺘﺸﻜل ﺇﻴﺩﻴﺎﻻﹰ ﻓﻲ
Ζ
ﻭﻟﺫﻟﻙ ﻓﻬﻲ ﺘﺴﺎﻭﻱ،
Ζ
m
ﺼﺤﻴﺤﺔـﻟﺍ ﺩﺍﺩﻋﻷﺍ ﺽﻌﺒﻟ
0
≥
m
.
ﺎﻥـﻜ ﺍﺫﺇ
m
0
=
،
n
a
e
≠
ﺒﺎﺴﺘﺜﻨﺎﺀ
0
=
n
ﺔـﺒﺘﺭ ﻥﺄﺒ لﺎﻘﻴ ﺫﺌﺩﻨﻋ ،
a
ﺔـﻴﻬﺘﻨﻤ ﺭـﻴﻏ
.
ﺎﻥـﻜ ﺍﺫﺇﻭ
0
≠
m
ﻴﻜﻭﻥ ﺃ،
ﺼﻐﺭ ﻋﺩﺩ ﺼﺤﻴﺢ
0
>
m
ﺒﺤﻴﺙ ﻴﺤﻘﻕ،
e
a
m
=
ﺔـﺒﺘﺭ ﻥﺄﺒ لﺎﻘﻴ ﺫﺌﺩﻨﻋ ،
a
ﻤﻨﺘﻬﻴﺔ
.
،ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
1
1
−
−
=
m
a
a
ﻭ
n
m
e
a
n
⇔
=
ﺃﻤﺜﻠﺔ
3.1
ﻟﺘﻜﻥ
∞
C
ﺍﻟﺯﻤﺭﺓ
( )
+
,
Z
ﺤﻴﺢـﺼ ﺩﺩﻋ لﻜﻟ ،ﻭ ،
1
≥
m
ﺘﻜﻥـﻟ ،
m
C
ﺭﺓـﻤﺯﻟﺍ
(
)
+
Ζ
Ζ
,
m
.
4.1
ﺩﻴﻼﺕــﺒﺘﻟﺍ ﺓﺭــﻤﺯ
(permutation groups)
.
ﺘﻜﻥــﻟ
S
ﺘﻜﻥــﻟﻭ ﺔــﻋﻭﻤﺠﻤ
( )
Sym S
ﺎﺒﻼﺕـﻘﺘﻟﺍ لﻜ ﺔﻋﻭﻤﺠﻤ
S
S
→
:
α
.
ﻥـﻤ ﻥﻴﺭﺼـﻨﻋ ﺀﺍﺩـﺠ ﻑﺭـﻌﻨ
( )
Sym S
ﺒﺘﺭﻜﻴﺒﻬﺎ ﻋﻠﻰ ﺍﻟﺸﻜل
β
α
αβ
o
=
ﻟﻜل
( )
S
Sym
∈
γ
β
α ,
,
ﻭ
S
s
∈
( )
5
(
)
(
)
( )
( )
(
)
(
)
(
)
(
)
( )
,
α β
γ
α β γ
α
β γ
=
=
s
s
s
o
o
o
o
ﻭﺒﺎﻟﺘﺎﻟﻲ
ﻓﺈﻥ
ﻗﺎﻨﻭﻥ ﺍﻟﺘﺠﻤ
ﻴﻌﻲ ﻤﺤﻘﻕ
.
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻤﺤﺎﻴﺩ
s
s a
ﺩـﻴﺎﺤﻤﻟﺍ ﺭﺼـﻨﻌﻟﺍ ﻭﻫ
ﻓﻲ
( )
Sym S
ﻟﻜﻲ ﺘﺸﻜل،ﺱﻭﻜﻌﻤ ﺩﺠﻭﻴﻭ ،
( )
Sym S
ﺘﻁﺒﻴﻘﺎﺕ ﺘﻘﺎﺒل
.
ﺈﻥـﻓ ﻲﻟﺎـﺘﻟﺎﺒﻭ
( )
S
Sym
ـ ﺘﺩﻋﻰ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﻟ،ﺓﺭﻤﺯ لﻜﺸﺘ
S
.
ﺇﻥ،ﹰﻼﺜﻤ
n
S
ﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻼﺕ
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٠١
ﻤﻥ
n
ﺤﺭﻑ ﻤﻌﺭﻓﺔ ﻟﺘﺸﻜل ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﻟﻠﻤﺠﻤﻭﻋﺔ
{
}
n
,
...
,
2
,
1
ﻥـﻤ ﻲـﻫﻭ
ﺍﻟﺭﺘﺒﺔ
!
n
.
5.1
ﻟﺘﻜﻥ
H
G ,
ﻟﻨﺒﻨﻲ ﺍﻟﺯﻤﺭﺓ،ﻥﻴﺘﺭﻤﺯ
H
G
×
ــﺭ ﻟـﺸﺎﺒﻤﻟﺍ ﺀﺍﺩﺠﻟﺎﺒ ﻰﻤﺴﺘ ﻲﺘﻟﺍ ،
H
G ,
.
ـ ﺘﺴﻤﻰ ﺒﺎﻟﺠﺩﺍﺀ ﺍﻟﺩﻴﻜﺎﺭﺘﻲ ﻟ،ﺔﻋﻭﻤﺠﻤﻜ ﺎﻤﺃ
H
G ,
ﻀﺭﺏـﻟﺍ ﺔﻴﻠﻤﻋ ﻑﺭﻌﺘﻭ ،
ﻜﻤﺎ
ﻴﻠﻲ
:
(
)(
) (
)
h
h
g
g
h
g
h
g
′
′
=
′
′
,
,
,
6.1
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺘﺒﺩﻴﻠﻴﺔ
)
ﺃﻭ ﺁﺒﻠﻴﺔ
٢
(
ﺇﺫﺍ ﻜﺎﻥ
:
ba
ab
=
ﻟﻜل،
G
b
a
∈
,
ﻓﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ
.
ﻴﻜﻭﻥ ﺍﻟﺠﺩﺍﺀ ﺍﻟﻤﻨﺘﻬ
ﻲ
ﻷﻱ ﻤﺤﻤﻭﻋﺔ
S
ﻥ ﺍﻟﻌﻨﺎـﻤ ﺔﻴﻬﺘﻨﻤ
ﺭـﺼ
)
ﻴﺱـﻟ
ﺒﺎﻟﻀﺭﻭﺭﺓ ﺃﻥ ﺘﻜﻭﻥ ﻤﺭﺘﺒﺔ
(
ﺍﻟﺠﺩﺍﺀ ﺍﻟﺨﺎﻟﻲ ﻫﻭ،ﹰﺍﺩﻴﺠ ﹰﺎﻓﺭﻌﻤ
e
.
ﺭـﻤﺯﻟﺍ ﻊـﻤ لـﻤﺎﻌﺘﻨ ،ﹰﺓﺩﺎﻋ
ﺍﻟﺘ
ﺒﺩﻴﻠﻴﺔ ﻋﻠﻰ ﺃﻨﻬﺎ ﺯﻤﺭ ﺠﻤﻌﻴﺔ
.
ﺘﺼﺒﺢ ﺍﻟﻤﻌﺎﺩﻟﺔ
(4)
:
(
)
( )
mna
na
m
na
ma
a
n
m
=
+
=
+
,
ﻓﻲ ﺤﺎﻟﺔ
G
ﺘﺒﺩﻴﻠﻴﺔ
(
)
mb
ma
b
a
m
+
=
+
ﻟﻜل،
Z
m
∈
ﻭ
G
b
a
∈
,
ﻭﺒﻬﺫﺍ ﻓﺈﻥ ﺍﻟﺘﻁﺒﻴﻕ
ﺍﻵﺘﻲ
( )
Ζ
→
×
Ζ
G
ma
a
m
:
,
a
ﻴﺤﻭل ﺍﻟﻤﺠﻤﻭﻋﺔ
A
ﺇﻟﻰ
Z
-
ﻤﻭﺩﻭل
.
ﻓﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ
G
ﺘﺸﻜل ﻤﺠﻤﻭﻋﺔ ﺍﻟﻌﻨﺎﺼﺭ ﺫﺍﺕ،
ﺍﻟﺭﺘﺏ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
tors
G
ﻤﻥ
G
ﻭﺘﺴﻤﻰ ﺒﺯﻤﺭﺓ ﺍﻟﻔﺘل،
(torsion subgroup)
.
.1
7
ﻟﻴﻜﻥ
F
ﺤ
ﻘﻼﹰ ﻤﺎ
.
ﺇﻥ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﻨﻭﻉ
n
n
×
لـﻘﺤﻟﺍ ﻥﻤ ﺎﻬﻠﻤﺍﻭﻋ ﺫﺨﺄﺘ ﻲﺘﻟﺍ
F
ﻭﺍﻟﺘﻲ ﻤﺤﺩﺩ ﻜل ﻤﻨﻬﺎ ﻻ ﻴﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ ﺘﺸﻜل ﺯﻤﺭﺓ
( )
GL F
ﺘ
ﺔـﻤﺎﻌﻟﺍ ﺔﻴﻁﺨﻟﺍ ﺓﺭﻤﺯﻟﺎﺒ ﻰﻤﺴ
(general linear group)
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
.
ﺃﻤﺎ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ
V
ﻭﻫﻭ
F
-
ﻲ ﺫﻭـﻬﺠﺘﻤ ﺀﺎﻀﻓ
ﻓﻬﻭ ﻴﺸﻜل ﺍﻟﺯﻤﺭﺓ،ﻪﺘﻨﻤ ﺩﻌﺒ
( )
GL V
ـﺘﺩﻋﻰ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺨﻁﻴﺔ ﺍﻟﻌﺎﻤﺔ ﻟ
V
.
ﻪ ﺇﺫﺍـﻨﺃ ﻅﺤﻼﻨ
ـﻜﺎﻥ ﻟ
V
n
لـﺜﺎﻤﺘﻟﺍ ﻑﺭﻌﻨﻟ ﺩﻋﺍﻭﻗ ﺭﺎﺘﺨﻨ ﺫﺌﺩﻨﻋ ،ﺩﻌﺒ
( )
( )
GL
GL
→
n
V
F
ﺫﻱـﻟﺍ
ﻴﺭﺴ
ل
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﺇﻟﻰ ﻤﺼﻔﻭﻓﺎﺘﻪ ﻤﻊ ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺍﻟﻘﻭﺍﻋﺩ
.
2
ﺇﻥ
"
ﺍﻟﺯﻤﺭﺓ ﺍﻵﺒﻠﻴﺔ
"
ﺸﺎﺌﻌﺔ ﺃﻜﺜﺭ ﻤﻥ
"
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ
."
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١١
.1
8
ﻟﻴﻜﻥ
V
ﺍﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ ﺫﻭ ﺍﻟﺒﻌﺩ ﺍﻟﻤﻨﺘﻬﻲ ﻋﻠﻰ ﺍﻟﺤﻘل
F
.
لـﻴﻭﺤﺘﻟﺍ ﻥﺄـﺒ ﺭﻜﺫﺘﻨ
ﺜﻨﺎﺌﻲ ﺍﻟﺨ
ﻁﻴﺔ ﻋﻠﻰ
V
ﻫﻭ
ﺍﻟﺘﻁﺒﻴﻕ
:
φ
×
→
V
V
F
لـﻜ ﺩـﻨﻋ ﹰﺎﻴﻁﺨ ﻥﻭﻜﻴ ﻱﺫﻟﺍﻭ
ﻤﺘﻐﻴﺭ
.
ـﺇﻥ ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﻟ
φ
ﻫﻭ ﺘﻤﺎﺜل
V
V
→
:
α
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
6
(
)
(
)
,
,
φ α α
φ
=
v
w
v w
ﻟﻜل،
V
w
v
∈
,
ﺇﻥ ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﻴﺸﻜل ﺯﻤﺭﺓ
( )
Aut
φ
.
ﻟﺘﻜﻥ
n
e
e
e
,
...
,
,
2
1
ـ ﻗﺎﻋﺩﺓ ﻟ
V
ﻭﻟﺘﻜﻥ،
( )
(
)
n
j
i
j
i
e
e
P
≤
≤
=
,
1
,
φ
.
ﻤﺼﻔﻭﻓﺔ
φ
.
ﺎﺒﻕـﻁﺘ ﺓﺭﺎﺘﺨﻤﻟﺍ ﺓﺩﻋﺎﻘﻟﺍ ﻥﺇ
( )
Aut
φ
ـ ﻤ
ﺔـﻴﻤﺎﻅﻨﻟﺍ ﺕﺎﻓﻭﻔﺼـﻤﻟﺍ ﺓﺭـﻤﺯ ﻊ
(invertible)
A
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
3
( )
7
. .
=
tr
A P A
P
ﺇﺫﺍ ﻜﺎﻥ
φ
ﺃﻱ ﺃﻥ،ﹰﺍﺭﻅﺎﻨﺘﻤ
(
)
(
)
v
w
w
v
,
,
φ
φ
=
ﻟﻜل،
V
w
v
∈
,
ﺘ،ﺫﺎﺸ ﺭﻴﻏﻭ
ﺴﻤﻰ
( )
Aut
φ
ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻌﺎﻤﺩﺓ
(orthogonal)
ـ ﻟ
φ
.
ﻭﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
φ
ﺃﻱ ﺃﻥ،ﹰﺎﻴﻔﻟﺎﺨﺘ
(
)
(
)
v
w
w
v
,
,
φ
φ
−
=
ﻟﻜل،
V
w
v
∈
,
ﻭﻏﻴﺭ
ﺸﺎﺫ
ﺘﺴﻤﻰ،
( )
Aut
φ
ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻌﺎﻤﺩﺓ ﺍﻟﺯﻭﺠﻴﺔ
(symplectic)
ـ ﻟ
φ
.
ﺫﻩـﻫ ﻲـﻓ
ـ ﺘﻭﺠﺩ ﻗﺎﻋﺩﺓ ﻟ،ﺔﻟﺎﺤﻟﺍ
V
ﻷﻱ ﻤﺼﻔﻭﻓﺔ
φ
ﺍﻟﺘﻲ ﺘﻜﻭﻥ
n
m
I
I
J
m
m
m
=
−
=
2
,
0
0
2
ﻭﺯﻤﺭﺓ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻨﻅﺎﻤﻴﺔ
A
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
m
m
tr
J
A
J
A
2
2
=
ﺘﺩﻋﻰ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻌﺎﻤﺩﺓ ﺍﻟﺯﻭﺠﻴﺔ
2
SP
m
.
3
ﻋﻨﺩﻤﺎ ﻨﺴﺘﺨﺩﻡ ﻗﺎﻋﺩﺓ ﻟﺘﻌﺭﻴﻑ
V
ﻓﻲ
n
F
ﻴﺼﺒﺢ،
φ
ﻋﻠﻰ ﺍﻟﺸﻜل
(
)
1
1
1
1
,
. .
n
n
n
n
a
b
b
a
a
p
a
b
b
M
M
a
L
M
ﺇﺫﺍ ﻜﺎﻨﺕ
A
ـ ﻤﺼﻔﻭﻓﺔ ﻟ
α
ﻤﻊ ﺍﻟﺤﻔﺎﻅ ﻋﻠ
ﻋﻨﺩﺌﺫ،ﺓﺩﻋﺎﻘﻟﺍ ﻰ
α
ﻴﻘﺎﺒل ﺍﻟﺘﻁﺒﻴﻕ
1
1
n
n
a
a
A
a
a
M
a
M
ﺘﺼﺒﺢ،ﻙﻟﺫﻟ
(6)
ﻋﻠﻰ ﺍﻟﺸﻜل
(
)
(
)
1
1
1
1
1
1
. . .
.
.
.
. ,
.
.
,
,
=
∈
t r
n
n
n
n
n
n
n
b
b
a
b
a
a
A
p
A
a
a
p
F
b
b
a
b
M
L
M
M
M
.
ـﻻﺨﺘﺒﺎﺭ ﻫﺫﻩ ﺍﻟﺤﺎﻻﺕ ﻋﻠﻰ ﻗﻭﺍﻋﺩ ﻗﺎﻨﻭﻨﻴﺔ ﻟﻤﺘﺠﻬﺎﺕ ﻟ
n
F
ﻨﺭﻯ ﺒﺄﻨﻪ ﻴﻜﺎﻓﺊ،
(7)
.
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٢١
9.1
)
a
(
ﻴﻤﻜﻥ ﺃﻥ ﻴﻌﺎﺩ
ﺘﺭﺘﻴﺏ
ﺸﺭﻭﻁ ﺍﻟﺯﻤﺭﺓ
)
G
2
ﻭ
G
3
(
ﻌﻑـﻀﺃ ﻁﻭﺭﺸـﺒ
)
ﺩـﺠﻭﻴ
ﺴﺎﺭـﻴﻟﺍ ﻥـﻤ ﺱﻭـﻜﻌﻤﻭ ﺴﺎﺭﻴﻟﺍ ﻥﻤ ﺩﻴﺎﺤﻤ ﺭﺼﻨﻋ
) :(
G
`2
(
ﺩـﺠﻭﻴ
e
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ
a
a
e
=
∗
لــﻜﻟ
a
،
)
G
`3
(
لــﻜﻟ
G
a
∈
ﺩــﺠﻭﻴ ،
G
a
∈
′
ﻭﻥــﻜﻴ ﺙــﻴﺤﺒ
e
a
a
=
∗
′
.
ﻟﻜﻲ ﻨﺒﻴﻥ ﺒﺄﻥ ﻫﺫﻴﻥ ﺍﻟﺸﺭﻁﻴﻥ ﻴﻜﺎﻓﺌﺎﻥ
)
G
3
(
ﻭ
)
G
2
(
ﻟﻴﻜﻥ،
G
a
∈
ﻤﻥ،
G
3`
ﻨﺠﺩ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ
a
a
′′
′
,
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
e
a
a
=
∗
′
ﻭ
e
a
a
=
′
∗
′′
.
ﻋﻨﺩﺌﺫ
(
) (
) (
)
(
)
(
)
,
a
a
e
a
a
a
a
a
a
a
a
a
a
a
a
e
′
′
′′
′
′
∗
= ∗
∗
=
∗
∗
∗
′′
′
′
′′
′
=
∗
∗
∗
=
∗
=
ﻭﻫﺫﺍ ﻴﻜﺎﻓﺊ
)
G
3
(
ﻭ،
(
)
(
)
e
a
a
a
a
a
a
a
a
e
a
∗
=
∗
′
∗
=
∗
′
∗
=
∗
=
ﻭﻫﺫﺍ ﻴﻜﺎﻓﺊ
)
G
2
.(
)
b
(
ﻴﻤﻜﻥ ﺃﻥ ﺘﻌﺭﻑ ﺍﻟﺯﻤﺭﺓ ﻋﻠﻰ ﺃﻨﻬﺎ ﻤﺠﻤﻭﻋﺔ
G
ﺔـﻴﺌﺎﻨﺜ ﺔـﻴﻠﻤﻋ ﻊﻤ
( )
∗
ـ ﺒﺤﻴ
ﺙ
ﺘﺘﺤﻘﻕ ﺍﻟﺸﺭﻭﻁ
ﺍﻵﺘﻴ
ﺔ
) :
g
1
(
∗
، ﺘﺠﻤﻴﻌﻴﺔ
)
g
2
(
G
،ﺔـﻴﻟﺎﺨ ﺭﻴﻏ
)
g
3
(
لـﻜﻟ
G
a
∈
،
ﻴﻭﺠﺩ ﻋﻨﺼﺭ
G
a
∈
′
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
a
a
∗
′
ﻤﺤﺎﻴﺩ ﺍﻟﺯﻤﺭﺓ
.
ﺩـﻴﺎﺤﻤ ﺭﺼـﻨﻋ ﺩﺠﻭﻴ ﺎﻤﻜ
ﻭﺍﺤﺩ ﻋﻠﻰ ﺍﻷﻜﺜﺭ ﻓﻲ
ﻥـﻤ ،ﺎـﻬﻴﻠﻋ ﺔـﻓﺭﻌﻤﻟﺍ ﺔـﻴﻌﻴﻤﺠﺘﻟﺍ ﺔﻴﺌﺎﻨﺜﻟﺍ ﺔﻴﻠﻤﻌﻟﺍ ﻊﻤ ﺔﻋﻭﻤﺠﻤﻟﺍ
ﺍﻟﻭﺍﻀﺢ ﺃﻥ ﻫ
ﺫﻩ ﺍﻟﺸﺭﻭﻁ ﺘﻜﺎﻓﺊ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺘﻲ ﻓﻲ
)
a
.(
ﻲـﻓ ﺔﻴﺭﻐـﺼﺃ ﺔـﻋﻭﻤﺠﻤ ﺩﺠﻭﺘ
، ﻤﺜﻼﹰ، ﺔﺜﻼﺜﻟﺍ ﻁﻭﺭﺸﻟﺍ ﻥﻤ ﻥﻴﻁﺭﺸ ﺎﻬﻴﻓ ﻕﻘﺤﺘﻴ ﻲﺘﻟﺍ ﺕﻻﺎﺤﻟﺍ
(
)
+
,
ﺸﺭﻁﻴﻥـﻟﺍ ﻕﻘﺤﺘ
ﺒﻴﻨﻤﺎ ﻻ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ،ﻲﻨﺎﺜﻟﺍﻭ لﻭﻷﺍ
ﺸﺭﻁﻴﻥ ﺍﻷﻭلـﻟﺍ ﻕـﻘﺤﺘ ﺔﻴﻟﺎﺨﻟﺍ ﺔﻋﻭﻤﺠﻤﻟﺍﻭ ،ﺙﻟﺎﺜﻟﺍ
ﻭﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﻨﻭﻉ،ﻲﻨﺎﺜﻟﺍ ﻁﺭﺸﻟﺍ ﻕﻘﺤﺘ ﻻﻭ ،ﺙﻟﺎﺜﻟﺍﻭ
2
2
×
ﺫـﺨﺄﺘ ﻲﺘﻟﺍ
ﻋﻭﺍﻤﻠﻬﺎ ﻤﻥ ﺤﻘل ﻤﺎ ﻭﺘﺤﻘﻕ
BA
AB
B
A
−
=
∗
ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁﻴﻥ ﺍﻟﺜﺎﻨﻲ ﻭﺍﻟﺜﺎﻟﺙ ﻭﻻ
ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ ﺍﻷﻭل
.
ﺠﺩﻭل ﺍﻟﺠﺩﺍﺀﺍﺕ
ﺇﻥ
ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﺜﻨﺎﺌﻴﺔ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻴﻤﻜﻥ ﻭﺼﻔﻬﺎ ﺒﺎﻟﺠﺩﻭل
ﺍﻵﺘﻲ
ﻤﻥ ﺍﻟﺠﺩﺍﺀﺍﺕ
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٣١
e a b c ...
...
...
...
...
2
2
2
2
c
cb
ca
ce
bc
b
ba
be
ac
ab
a
ae
ec
eb
ea
e
e
a
b
c
M
ﻴﻜﻭﻥ
e
ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎﻴﺩ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﺃﻭل ﻋﻤﻭﺩ ﻭ ﺃﻭل ﺴﻁﺭ ﻤﻥ ﺍﻟﺠﺩﻭل ﻫﻭ
ﺭﺍﺭـﻜﺘ
ﻟﻌﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ
.
ﻊـﻤ ﻲﻟﺩﺎﺒﺘ ﺩﻭﻤﻋ لﻜﻭ ﺭﻁﺴ لﻜ ﻥﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﹰﺍﺩﻭﺠﻭﻤ ﺱﻭﻜﻌﻤﻟﺍ ﻥﻭﻜﻴﻭ
ﻋﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ
)
ﺍﻨﻅﺭ
1.2
)
e
.((
ﺇﺫﺍ ﻭﺠﺩ
n
ﻹﺜﺒﺎﺕ ﺍﻟﺨﺎﺼﺔ ﺍﻟﺘﺠﻤﻴﻌﻴﺔ ﻴﺠﺏﺫﺌﺩﻨﻋ ،ﹰﺍﺭﺼﻨﻋ
ﺃﻥ ﻨﺨﺘﺎﺭ
3
n
ﻤﻌﺎﺩﻟﺔ
.
ﻫﺫﺍ ﻴﻘﺘﺭﺡ ﻟﻨﺎ ﺨﻭﺍﺭﺯﻤﻴ
ﺔـﻴﻬﺘﻨﻤ ﺔﺒﺘﺭﺒ ﺓﺎﻁﻌﻤﻟﺍ ﺭﻤﺯﻟﺍ لﻜ ﺩﺎﺠﻴﻹ ﺔ
n
لـﻜ ،ﺩـﻴﺩﺤﺘﻟﺎﺒﻭ ،
ﺍﻟﺠﺩﺍﻭل ﺍﻟﻀﺭﺒﻴﺔ ﺍﻟﻤﺤﺘﻤﻠﺔ ﻭﺍﻟﺘﺄﻜﺩ ﻤﻥ ﺍﻟﻤﺴﻠﻤﺎﺕ
.
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
n
ﻐﻴﺭﺍﹰـﺼ
ﻭﻫﺫﺍ ﻟﻴﺱ ﻤﻭﻀﻭﻋﻴﺎﹰ،ﹰﺍﺩﺠ
!
ﺇﻥ ﺍﻟﺠﺩﻭل
ﻴﺤﺘﻭ
ﻱ ﻋﻠﻰ
2
n
ﻊـﻀﻭﻤ لﻜﻟ ﺎﻨﺤﻤﺴ ﺍﺫﺈﻓ ،ﻊﻀﻭﻤ
ﺃﻥ ﻴﺜﺒﺕ ﺃﻱ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
n
ﻨﺤﺼل ﻋﻠﻰﺫﺌﺩﻨﻋ ،
2
n
n
ﻲـﺘﻟﺍ ﹰﺍﺩﺠ لﻴﻠﻗ لﻤﺘﺤﻤ لﻭﺩﺠ
ﺘﻌﺭﻑ ﺯﻤﺭﺍﹰ
.
ﻴﻭﺠﺩ،ﹰﻼﺜﻤ
64
8
6277101735386680763835789423207666416102355444464034512896
=
ﻋﻤﻠﻴﺔ ﺜﻨﺎﺌﻴﺔ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻜﻭﻨﺔ ﻤﻥ
8
ﻟﻜﻥ ﻴﻭﺠﺩ ﺨﻤﺱ ﺼﻔﻭﻑ ﻤﺘ،ﺭﺼﺎﻨﻋ
ﻤﺎﺜﻠﺔ ﻤﻥ ﺍﻟﺯﻤﺭ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
8
(4.12)
.
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
(subgroups)
ﻗﻀﻴﺔ
1
.
10
ﻟﺘﻜﻥ
S
ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻭﻏﻴﺭ ﺨﺎﻟﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
.
ﺇﺫﺍ ﻜﺎﻥ
S
1
:
S
ab
S
b
a
∈
⇒
∈
,
ﻭ،
S
2
:
S
a
S
a
∈
⇒
∈
−
1
ﻓﺈﻥ ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﻤﺫﺌﺩﻨﻋ
ﻌﺭﻓﺔ ﻋﻠﻰ
G
ﺘﺤﻭل
S
ﺇﻟﻰ ﺯﻤﺭﺓ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
)
S
1
(
ﺇﻥ ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﺜﻨﺎﺌﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ
G
ﺔـﻴﺌﺎﻨﺜﻟﺍ ﺔـﻴﻠﻤﻌﻟﺍ ﻑﺭﻌﺘ
S
S
S
→
×
ﻋﻠﻰ
S
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻨﻬﺎ ﺘﺠﻤﻴﻌﻴﺔ،
.
ﺒﻔﺭﺽ ﺃﻥ
S
ﺼﺭـﻨﻌﻟﺍ ﻰﻠﻋ ﻱﻭﺤﺘ
a
،لـﻗﻷﺍ ﻰـﻠﻋ
ﻤﻌﻜﻭﺴﻪ
1
−
a
ﻭﺍﻟﺠﺩﺍﺀ،
1
−
=
aa
e
.
ﻭﺃﺨﻴﺭﺍﹰ ﻤﻥ
)
S
2
(
ﻥـﻤ ﺭﺼﻨﻋ ﻱﺃ ﺱﻭﻜﻌﻤ ﻥﺃ ﻥﻴﺒﻨ
S
ﻴﻨﺘﻤﻲ ﺇﻟﻰ
S
.
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ﺘﺴﻤﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻏﻴﺭ
ﺍﻟﺨﺎﻟﻴﺔ
S
ﺍﻟﺘﻲ ﺘﺤﻘﻕ
)
S
1
(
ﻭ
)
S
2
(
ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻋﻨﺩﻤﺎ
ﺘﻜﻭﻥ
S
ﻤﻨﺘﻬﻴﺔ ﻓﺈﻥ ﺍﻟﺸﺭﻁ
)
S
1
(
ﺸﺭﻁـﻟﺍ ﻰـﻟﺇ ﻱﺩﺅـﻴ
)
S
2
:(
ﻴﻜﻥـﻟ
S
a
∈
ﺫـﺌﺩﻨﻋ ،
{
}
...
,
,
2
a
a
ﺔــﺒﺘﺭ ﻥﻭــﻜﺘ ﻙﻟﺫــﻟﻭ ،
a
ﺄﻥــﺒ لﻭــﻘﻨﻭ ،ﺔــﻴﻬﺘﻨﻤ
e
a
n
=
ﺍﻵﻥ ﺇﻥ،
S
a
a
n
∈
=
−
−
1
1
.
ﺇﻥ ﺍﻟﻤﺜﺎل
(
) (
)
+ ⊂
+
,
,
ﻴﺒﻴﻥ ﺒﺄﻥ ﺍﻟﺸﺭﻁ
)
S
1
(
ﻻ ﻴﻘﺘﻀﻲ
)
S
2
(
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
S
ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.
ﻤﺜﺎل
11.1
ﺇﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
G
ﻫﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
( ) {
}
G
x
xg
gx
G
g
G
Z
∈
∀
=
∈
=
,
:
ﻭﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
G
.
ﻗﻀﻴﺔ
12.1
ﺇﻥ ﺘﻘﺎﻁﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻫﻭ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻏﻴﺭ ﺨﺎﻟﻴﺔ ﻷﻨﻬﺎ ﺘﺤﻭﻱ ﻋﻠﻰ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎ
ﻴﺩ
e
ﻭﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻥ،
ﺍﻟﺸﺭﻁﻴﻥ
)
S
1
(
ﻭ
)
S
2
(
ﻤﺤﻘﻘﺎﻥ
.
ﻤﻼﺤﻅﺔ
13.1
ﻭـﻫ ﻱﺭﺒﺠ ﻉﻭﻀﻭﻤ ﻥﻤ ﺔﻴﺌﺯﺠﻟﺍ ﺕﺎﻋﻭﻀﻭﻤﻟﺍ ﻊﻁﺎﻘﺘ ﻥﻭﻜﻴ ﺔﻤﺎﻌﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
ﻤﻭﻀﻭﻋﺔ ﺠﺯﺌﻴﺔ
.
ﺔـﻴﺌﺯﺠﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ ﻊﻁﺎﻘﺘﻭ ،ﺔﻴﺌﺯﺠ ﺔﻘﻠﺤ ﻭﻫ ﺔﻴﺌﺯﺠﻟﺍ ﺕﺎﻘﻠﺤﻟﺍ ﻊﻁﺎﻘﺘ ﹰﻼﺜﻤ
ﻭﻫﻜﺫﺍ،ﻲﺌﺯﺠ لﻭﺩﻭﻤ ﻭﻫ
.
ﻗﻀﻴﺔ
14.1
ﻟﻜل ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ
X
ﻤﻥ ﺯﻤﺭﺓ
G
ﻲـﻓ ﺔﻴﺭﻐﺼﺃ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﺩﺠﻭﺘ ،
G
ﺘﺤﻭﻱ
X
.
ﻭﺘﺘﻜﻭﻥ ﻤﻥ ﺠﻤﻴﻊ ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻟﻌﻨﺎﺼﺭ ﻤﻥ
X
ﻭﻤﻌﻜﻭﺴﺎﺘﻬﺎ
)
ﺴﻤﺎﺡـﻟﺍ ﻊﻤ
ﺒﺎﻟﺘﻜﺭﺍﺭ
.(
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
S
ﺘﻘﺎﻁﻊ ﺠﻤﻴﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ
X
ﺭﺓـﻤﺯ ﺎﻬﻨﺈﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﺘﺤﻭﻱ
X
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺒﺄﻨﻬﺎ ﺃﺼﻐﺭ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ،
G
ﺘﺤﻭﻱ
X
.
ﻭﻤﻥ
ﺍﻟﻭﺍﻀﺢ ﺃﻴﻀﺎﹰ ﺒﺄﻨﻬﺎ ﺘﺤﻭﻱ ﻋﻠﻰ ﺠﻤﻴﻊ ﺍﻟﺠﺩﺍﺀﺍﺕ ﻟﻌﻨﺎﺼﺭ ﻤﻥ
X
ﺭـﺼﺎﻨﻌﻟﺍ ﻩﺫﻫ ﺕﺎﺴﻭﻜﻌﻤﻭ
.
ﻟﻜﻥ ﺇﻥ ﻤﺠﻤﻭﻋﺔ ﻫﺫﻩ ﺍﻟﺠﺩﺍﺀﺍﺕ ﺘﺤﻘ
ﻕ
ﺍﻟﺸﺭﻁﻴﻥ
)
S
1
(
ﻭ
)
S
2
(
ﻲـﻓ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻲﻬﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ
G
ﺘﺤﻭﻱ
X
.
ﻭﻟﺫﻟﻙ ﻓﻬﻲ ﺘﺴﺎﻭﻱ
S
.
ﻨﺭﻤﺯ ﻟﻠﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
S
ﺎﻟﺭﻤﺯـﺒ ﺔﻘﺒﺎﺴﻟﺍ ﺔﻴﻀﻘﻟﺍ ﻲﻓ ﺓﺎﻁﻌﻤﻟﺍ
X
ﺎﻟﺯﻤﺭﺓـﺒ ﻰﻤﺴـﺘﻭ ،
ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔ
X
.
،ﻤﺜﻼﹰ
{ }
e
=
φ
.
ﻲـﻓ ﺭﺼـﻨﻋ لـﻜ ﺔﺒﺘﺭ ﺕﻨﺎﻜ ﺍﺫﺇ
X
ﻤﺜﻼﹰ ﺇﺫﺍ ﻜﺎﻨﺕ،ﺔﻴﻬﺘﻨﻤ
G
ﺭـﺼﺎﻨﻋ ﻥـﻤ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺕﺍﺀﺍﺩﺠﻟﺍ لﻜ ﺔﻋﻭﻤﺠﻤ ﺫﺌﺩﻨﻋ ،ﺔﻴﻬﺘﻨﻤ
X
ﺘﺸﻜل
ﺯﻤﺭﺓ ﻭﺘﺴﺎﻭﻱ
X
.
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ﻨﻘﻭل ﺒﺄﻥ
X
ﺘﻭﻟﺩ
G
ﺇﺫﺍ ﻜﺎﻥ
X
G
=
ﻜل ﻋﻨﺼﺭ ﻓﻲ،ﻥﺃ ﻱﺃ ،
G
ﻰـﻠﻋ ﺏـﺘﻜﻴ
ﺸﻜل ﺠﺩﺍﺀ ﻤﻨﺘﻪ ﻟﻌﻨﺎﺼﺭ ﻤﻥ
X
ﻭﻤﻌﻜﻭﺴﺎﺘﻬﺎ
.
ﺼﺭـﻨﻌﻟﺍ ﺔﺒﺘﺭ ﻥﺄﺒ ﻅﺤﻼﻨ
a
ﺭﺓـﻤﺯﻟﺍ ﻥـﻤ
ﻴﺴﺎﻭﻱ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴ
ﺔ
a
ﺍﻟﻤﻭﻟﺩﺓ ﺒﻪ
.
ﺃﻤﺜﻠﺔ
.1
15
ﺍﻟﺯﻤﺭ ﺍﻟﺩﺍﺌﺭﻴﺔ
.
،ﺭﻫﺎـﺼﺎﻨﻋ ﻥﻤ ﺩﺤﺍﻭ ﺭﺼﻨﻌﺒ ﺓﺩﻟﻭﻤ ﺕﻨﺎﻜ ﺍﺫﺇ ﺔﻴﺭﺌﺍﺩ ﺓﺭﻤﺯﻟﺍ ﻥﺄﺒ لﺎﻘﻴ
ﺇﺫﺍ ﻜﺎﻨﺕ،ﻪﻨﺃ ﻱﺃ
r
G
=
ﻟﺒﻌﺽ ﺍﻟﻌﻨﺎﺼﺭ
G
r
∈
.
ﺇﺫﺍ ﻜﺎﻨﺕ ﺭﺘﺒﺔ
r
ﺴﺎﻭـﺘﻭ ﺔﻴﻬﺘﻨﻤ
ﻱ
n
ﻋﻨﺩﺌﺫ،
{
}
2
mod
=
≈
↔
n
i
n
G
e r r
r
C
r
i
n
, ,
, ... ,
,
ﻭﻴﻤﻜﻥ ﺃﻥ ﻨﺘﺼﻭﺭ ﺒﺄﻥ
G
ﻋﺒﺎﺭﺓ ﻋﻥ ﺘﻨﺎﻅﺭﺍﺕ ﺩﻭﺭﺍﻨﻴﺔ ﺤﻭل ﻤﺭﻜﺯ ﻤﻀﻠﻊ ﻤﻨﺘﻅﻡ ﻤﺅﻟﻑ ﻤﻥ
n
-
ﻀﻠﻊ
.
ﺇﺫﺍ ﻜﺎﻨﺕ ﺭﺘﺒﺔ
r
ﻋﻨﺩﺌﺫ،ﺔﻴﻬﺘﻨﻤ ﺭﻴﻏ
{
}
i
r
C
r
r
e
r
r
G
i
i
i
↔
≈
=
∞
−
−
,
...
,
,
...
,
,
,
,
...
,
,
...
1
.
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺔﻴﺭﺌﺍﺩ ﺓﺭﻤﺯ ﹰﺎﻤﺎﻤﺘ ﺩﺠﻭﻴ ، لﺜﺎﻤﺘﻟﺍ ﻰﺘﺤ ،ﺍﺫﻬﺒﻭ
n
لـﻜﻟ
∞
≤
n
.
ﺴﺘﺨﺩﻡـﻨﺴ
ﻤﺴﺘﻘﺒﻼﹰ ﺍﻟﺭﻤﺯ،
n
C
ﻷﻱ ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
)
ﻟﻴﺱ ﺒﺎﻟﻀﺭﻭﺭﺓ
Ζ
Ζ
n
ﺃﻭ
Ζ
.(
1
16.
ﺯﻤﺭ ﺩﻴﻬﻴﺩﺭﺍل
4
n
D
.
ﻟﻜل
3
≥
n
،
n
D
ﻫﻲ ﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﺍﺕ ﻟﻤﻀﻠﻊ ﻤﻨﺘﻅﻡ ﻤﺅﻟﻑ
ﻤﻥ
n
-
ﻀﻠﻊ
5
.
ﻋﺩﺩ ﺭﺅﻭﺴﻪ
n
...
,
2
.
1
ﺎﺭﺏ ﺍـﻘﻋ ﻩﺎﺠﺘﺍ ﺱﻜﻌﺒ ﺔﻤﻗﺭﻤ
ﺴﺎﻋﺔـﻟ
.
ﻴﻜﻥـﻟ
r
ﺍﻟﺩﻭﺭﺍﻥ ﺒﺯﺍﻭﻴﺔ ﺘﺴﺎﻭﻱ
n
π
2
)
ﻟﺫﻟﻙ
n
i
i
mod
1
+
a
(
ﻭﻟﻴﻜﻥ،
s
ﻭﺭـﺤﻤﻠﻟ ﺱﺎﻜﻌﻨﺍ
=)
ﻭﺭــﺤﻤﻟﺍ لﻭــﺤ ﻥﺍﻭﺭﺩ
(
ﺭﺃﺱــﻟﺍ ﻥــﻤ
1
ﻀﻠﻊــﻤﻟﺍ ﺯــﻜﺭﻤ ﻰــﻟﺇ
)
ﻭﻥــﻜﻴ ﻙﻟﺫــﻟ
n
i
i
i
mod
2
−
+
a
.(
ﻋﻨﺩﺌﺫ
(
)
s
r
sr
r
srs
s
r
n
n
1
1
2
;
1
;
1
−
−
=
⇒
=
=
=
ﺘﻘﺘﻀﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﺎ
ﻴﻠﻲ
{
}
1
1
, , ...,
, ,
, ...,
−
−
=
n
n
n
D
e r
r
s rs
r
s
ﺃﻥ ﻋﻨﺎﺼﺭ ﻫﺫﻩ،ﺔﺴﺩﻨﻬﻟﺍ ﻥﻤ ،ﹰﺎﺤﻀﺍﻭ ﻥﻭﻜﻴ
ﻭ،ﺔﻔﻠﺘﺨﻤ ﻥﻭﻜﺘ ﺔﻋﻭﻤﺠﻤﻟﺍ
ﻟﺫﻟﻙ
2
=
n
D
n
.
4
ﻨﺭﻤﺯ ﻟﻬﺫﻩ ﺍﻟﺯﻤﺭﺓ ﺒﺎﻟﺭﻤﺯ
n
D
ﺃﻭ ﺒﺎﻟﺭﻤﺯ
n
D
2
ﻤﻌﺘﻤﺩﺍﹰ
ﻋﻠﻰ
ﺍﻟﻤﺅﻟﻑ
)
ـﻜﺎﻟﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﺒ
n
ﻀﻠﻊ
(
ﺃﻭ
ﻤﺠﺭﺩﺍﹰ
.
5
"
ﺒﺸﻜل ﺃﺴﺎﺴﻲ
"
n
D
ﻑ ﻋﻠﻰ ﺃﻨﻬﺎ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥﺭﻌﺘ ﻥﺃ ﻥﻜﻤﻴ
n
S
ـ ﻤﻭﻟﺩﺓ ﺒ
(
)
r : i
i 1 mod n
+
a
ﻭ
(
)
s : i
n
2 1 mod n
+ −
a
.
ـ ﻜل ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒﺫﺌﺩﻨﻋ
n
D
ﻴﻤﻜﻥ ﺃﻥ ﺘﺒﺭﻫﻥ ﺒﺩﻭﻥ ﺍﻟﺭﺠﻭﻉ ﺇﻟﻰ ﺍﻟﻬﻨﺩﺴﺔ
)
ﺃﻭ ﺘﻭﻀﻴﺢ ﻟﻠﻘﺎﺭﺉ
.(
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ﻟﻴﻜﻥ
t
ﺍﻨﻌﻜﺎﺴﺎﹰ ﻟﻠﻤﺤﻭﺭ ﺍﻟﻤﺎﺭ ﻤﻥ ﻤﻨﺘﺼﻑ ﺍﻟﻀﻠﻊ ﺍﻟﻭﺍﺼل ﺒﻴﻥ ﺍﻟﺭﺃﺴﻴﻥ
1
ﻭ
2
ﺯـﻜﺭﻤﻭ
ﺍﻟﻤﻀﻠﻊ
)
ﻟﺫﻟﻙ
n
i
i
i
mod
3
−
+
a
.(
ﻋﻨﺩﺌﺫ
ts
r
=
.
ﺈﻥـﻓ ﻲﻟﺎـﺘﻟﺎﺒﻭ
t
s
D
n
,
=
ﻭﻴﻜﻭﻥ
( )
( )
2
2
2
2
,
,
st
e
ts
e
t
e
s
=
=
=
=
ﻨﻌﺭﻑ
1
D
ﻟﻴﻜﻭﻥ
{ }
r
C
,
1
2
=
ﻑﺭـﻌﻨﻭ
2
D
ﻭﻥـﻜﻴﻟ
{
}
rs
s
r
C
C
,
,
,
1
2
2
=
×
.
ﺘﺴﻤﻰ ﺍﻟﺯﻤﺭﺓ
2
D
ﺃﻭ،ﻥﻴﻼﻜ ﺓﺭﻤﺯﺒ ﹰﺎﻀﻴﺃ
4
-
ﺯﻤﺭﺓ
.
ﻨﻼﺤﻅ ﺒﺄﻥ
3
D
ﻫﻲ ﺯﻤﺭﺓ ﻜل ﺘﺒﺩﻴﻼﺕ
ﺍﻟﻤﺠﻤﻭﻋﺔ
{ }
3
,
2
,
1
.
ﻭﻫﻲ ﺃﺼﻐﺭ ﺯﻤﺭﺓ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ
.
17.1
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
Q
.
ﻟﺘﻜﻥ
−
−
=
0
1
1
0
a
ﻭ
−
=
0
1
1
0
b
.
ﻋﻨﺩﺌﺫ
3
1
2
2
4
,
,
1
a
bab
b
a
a
=
=
=
−
ﻭﻤﻨﻪ
(
)
b
a
ba
3
=
.
ﺇﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ
GL
2
(C)
ﺍﻟﻤﻭﻟﺩﺓ ﺒﺎﻟﻌﻨﺼﺭﻴﻥ
a
ﻭ
b
ﻫﻲ
{
}
b
a
b
a
ab
b
a
a
a
e
Q
3
2
3
2
,
,
,
,
,
,
,
=
ﻭﻴﻤﻜﻥ ﺃﻥ ﺘﻭﺼﻑ ﺍﻟﺯﻤﺭﺓ
Q
ﻋﻠﻰ ﺃﻨﻬﺎ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
{
}
k
j
i
±
±
±
±
,
,
,
1
ﻥـﻤ
ﺎﻋﻲــــﺒﺭﻟﺍ ﺭــــﺒﺠﻟﺍ
H
.
ﺄﻥــــﺒ ﺭﻜﺫــــﺘﻨ
k
j
i
R
R
R
R
H
⊕
⊕
⊕
=
1
ﻤﻊ ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ
ﻴﻠﻲ
ji
k
ij
j
i
−
=
=
=
−
=
,
1
2
2
ﻴﺘﻤﺩﺩ ﺍﻟﺘﻁﺒﻴﻕ
b
j
a
i
a
a
,
ﺒﺸﻜل ﻭﺤﻴﺩ ﺇﻟﻰ ﺍﻟﺘﺸﺎﻜل
( )
2
M
→
H
C
ﻲـﻓ
R
–
ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﺘﻁﺒﻴﻘﺎﺕ ﺍﻟﺯﻤﺭﺓ،ﺭﻭﺒﺠ
j
i,
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ
b
a,
.
18.1
ﻨﺘﺫﻜﺭ
ﺒﺄﻥ
n
S
ﻫﻲ ﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻼﺕ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
n
...
,
2
,
1
.
ﺎﺭﺓـﺒﻋ ﺔﻠﻗﺎﻨﻤﻟﺍ ﻥﺇ
ﻋﻥ ﺘﺒﺩﻴﻠﺔ ﻴﻅﻬﺭ ﻓﻴﻬﺎ ﻋﻨﺼﺭﺍﻥ ﺃﻤﺎ ﺒﻘﻴﺔ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻬﻲ ﺜﺎﺒﺘﺔ
.
ﺩ ﺃﻥـﺠﻨ ﺔﻟﻭﻬﺴـﺒ
n
S
ﺩﺓـﻟﻭﻤ
ﺒﻤﻨﺎﻗﻼﺘﻬﺎ
)
ﺍﻨﻅﺭ
(4.25)
ﻓﻴﻤﺎ
ﻴﻠﻲ ﻴﺘﻀﺢ ﺍﻟﻤﻌﻨﻰ ﺒﺩﻗﺔ ﺃﻜﺜﺭ
(
ﺯﻤﺭ ﺒﺭﺘﺏ ﺼﻐﻴﺭﺓ
(Groups of small order)
ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺭﺘﺒﺔ ﺃﺼﻐﺭ ﻤﻥ
16
ﺘﻜﻭﻥ ﻤﺘ
ﺔـﻤﺌﺎﻘﻟﺍ ﻲـﻓ ﻲـﺘﻟﺍ ﺭﻤﺯﻟﺍ ﻯﺩﺤﺇ ﻊﻤ ﹰﺎﻤﺎﻤﺘ ﺔﻠﺜﺎﻤ
ﺍﻵﺘﻴ
ﺔ:
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٧١
ﺍﻟﺭﺘﺒﺔ
1
2
3
4
5
6
7
8
ﺍﻟﺯﻤﺭﺓ
1
C
2
C
3
C
4
2
,
C
D
5
C
6
3
,
C
S
7
C
5
ﺯ
ﻤﺭ
ﺓ
ﺍﻟﺭﺘﺒﺔ
9
10
11
12
13
14
15
16
ﺍﻟﺯﻤﺭ
ﺓ
9
3
3
,
C
C
C
×
10
5
,
C
D
11
C
5
ﺯﻤﺭ
ﺓ
13
C
14
7
,
C
D
15
C
14
ﺯﻤﺭﺓ
ﺍﻟﺭﺘﺒﺔ
:8
4
2
2
2
4
2
8
,
,
,
,
D
Q
C
C
C
C
C
C
×
×
×
ﺍﻟﺭ
ﺘﺒﺔ
12
:
12
2
6
2
3
4
3
4
,
,
,
,
ϕ
×
×
×
C
C
C
C
S
A
C
C
ﻟﻜل ﻋﺩﺩ ﺃﻭﻟﻲ
p
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﻭﺍﺤﺩﺓ ﻓﻘﻁ،
)
ﺘﺤﺕ ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل
(
، ﻭﻫﻲ ﺒﺎﻟﺘﺤﺩﻴﺩ،
p
C
)
ﺍﻨﻅﺭ
1.27
(
ﻭﺯﻤﺭﺘﺎﻥ ﻓﻘﻁ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
2
p
، ﻭﻫﻤﺎ ﺒﺎﻟﺘﺤﺩﻴﺩ،
p
p
C
C
×
ﻭ
2
p
C
)
ﺍﻨﻅﺭ
4.18
.(
ﻟﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
6
،
)
ﺍﻨﻅﺭ
4.23
(
ﻭﻤﻥ ﺃﺠل ﺍﻟﺭﺘﺒﺔ،
8
،
)
ﺍﻨﻅﺭ
4.21
(
ﻭﻤﻥ ﺃﺠل،
ﺍﻟﺭﺘﺒﺔ
12
،
)
ﺍﻨﻅﺭ
5.16
(
ﻭﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺭﺘﺏ،
10
،
14
ﻭ
15
،
)
ﺍﻨﻅﺭ
5.14
.(
ﺒﺄﻥ ﺍﻟﻘﻭﻯ ﺍﻷﻜﺜﺭ ﻜﺒﺭﺍﹰ ﻟﻸﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﻟﺘﻲ،ﺔﺒﻭﻌﺼﺒ ﺩﺠﻨ
ﺘﻘﺴﻡ
n
ﺭﺍﹰـﻤﺯ ﺎﻨﻴﺩﻟ لﻜﺸﺘﺴ ،
ﺃﻜﺜﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
.
ﺇﺫﺍ ﻜﺎﻥ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
( )
n
f
ﻋﺩﺩ ﺍﻟﺼﻔﻭﻑ ﺍﻟﻤﺘﻤﺎﺜﻠﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺫﻱ ﺍﻟﺭﺘﺒﺔ
n
ﻋﻨﺩﺌﺫ،
( )
( ) ( )
2
2
1
27
+
≤
o
e n
f n
n
ﺤﻴﺙ
( )
n
e
.
ـﺍﻷﺱ ﺍﻷﻜﺒﺭ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ ﺍﻟﻘﺎﺴﻡ ﻟ
n
ﻭ
( )
0
1
→
o
ﺎ ﺃﻥـﻤﻜ
( )
∞
→
n
e
)
ﺍﻨﻅﺭ
Pyber 1993
.(
ﻓﻲ
2001
ﺴﺎﻭﻱـﺘ ﻭﺃ ﺭﻐﺼﺃ ﺏﺘﺭﻟﺍ ﻥﻤ ﺭﻤﺯﻟﺍ ﻥﻤ ﺔﻀﺌﺎﻓﻭ ﺔﻠﻤﺎﻜ ﺔﻤﺌﺎﻗ ﺕﺩﺠﻭ ،
2000
–
ﻭﻴﻭﺠﺩ ﺘﻤﺎﻤﺎﹰ،لﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ
49,910,529,484
(Besche et al. 2001)
.
ﺍﻟﺘﺸﺎﻜﻼﺕ
(Homomorphisms)
ﺘﻌﺭﻴﻑ
19.1
ﺇﻥ ﺍﻟﺘﺸﺎﻜل ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺇﻟﻰ ﺯﻤﺭﺓ ﺜﺎﻨﻴﺔ
G
′
ﻕـﻴﺒﻁﺘ ﻭﻫ
G
G
′
→
:
α
،
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
( ) ( )
b
a
ab
α
α
α
=
ﻟﻜل،
G
b
a
∈
,
.
ﻭﺍﻟﺘﻤﺎﺜل ﻫ
ﻭ ﺘﺸﺎﻜل ﺘﻘﺎﺒل
.
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻤﺤﺩﺩ،ﹰﻼﺜﻤ
( )
det GL
→
X
n
F
F
:
ﻫﻭ ﺘﺸﺎﻜل
.
20.1
ﻟﻴﻜﻥ
α
ﺘﺸﺎﻜﻼﹰ ﻤﺎ ﻟﻜل ﻋﻨﺼﺭ
n
a
a
a
,
...
,
,
2
1
ﻤﻥ
G
،
(
)
(
)
(
)
( ) (
)
( )
( )
m
m
m
m
a
a
a
a
a
a
a
a
a
a
a
α
α
α
α
α
α
...
...
...
...
...
1
2
1
2
1
2
1
=
=
=
=
ﻟﻜل،ﺹﺎﺨ لﻜﺸﺒﻭ
1
≥
m
،
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٨١
( )
8
( )
( )
α
α
=
m
m
a
a
ﺒﺎﻹﻀﺎﻓﺔ ﺇﻟﻰ ﺫﻟﻙ
( )
( )
( ) ( )
e
e
ee
e
α
α
α
α
=
=
ﻭﻤﻨﻪ،
( )
e
e
=
α
)
ﺒﺘﻁﺒﻴﻕ
1.2 (a
.(
ﻭﺃﻴﻀﺎﹰ
( )
( )
( )
( )
a
a
e
a
a
a
a
e
aa
α
α
α
α
1
1
1
1
−
−
−
−
=
=
⇒
=
=
ﻭﺒﻬﺫﺍ ﻴﻜﻭﻥ
( )
( )
1
1
−
−
=
a
a
α
α
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ
)
(8
لـﻜﻟ ﺔﻘﻘﺤﻤ
Ζ
∈
m
.
ﺈﻥـﻓ ،ﻙﻟﺫـﻟ
ﺘﺸﺎﻜﻼﺕ ﺍﻟﺯﻤﺭ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺘﻜﻭﻥ ﺃﻴﻀﺎﹰ ﺘﺸﺎﻜﻼﺕ
Z
-
ﻤﻭﺩﻭل
.
ﻜل ﺴﻁﺭ ﻤﻥ ﺍﻟﺠﺩﺍﺀﺍﺕ ﻓﻲ ﺠﺩﻭل ﺍﻟﺯﻤﺭ ﻴﺸﻜل ﺘﺒﺩﻴﻠﺔ ﻟﻌﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ،ﹰﺎﻘﺒﺎﺴ ﺎﻨﻅﺤﻻ ﺎﻤﻜ
.
ﻭﻫﺫﻩ ﺘﺴﻤﺢ ﻟﺯﻤﺭﺓ ﻭﺍﺤﺩﺓ ﻟﺘﺘﺤﻭل ﺇﻟﻰ ﺯﻤﺭﺓ ﺘﺒﺩﻴﻼﺕ،ﻲﻠﻴﺎﻜ ﺓﺭﻤﺯ ﻙﻟﺫ ﻰﻠﻋ لﺎﺜﻤﻭ
.
ﻤﺒﺭﻫﻨﺔ
21.1
)
ﻜﺎﻴﻠﻲ
(
ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻤﺘﺒﺎﻴﻥ ﻗﺎﻨﻭﻨﻲ ﻤﻥ ﺍﻟﺸﻜل
( )
Sym
α
→
G
G
:
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻜل
G
a
∈
ﻑﺭﻌﻨ ،
G
G
a
L
→
:
ﺒﺎﻟﻌﻼﻗﺔ
ax
x a
)
ﺒﺎﻟﻀﺭﺏ ﻤﻥ ﺍﻟﻴﺴﺎﺭ
ـﺒ
a
.(
ﻟﻜل
G
x
∈
،
(
)( )
( )
(
)
( )
( ) ( ) ( )
x
ab
x
ab
bx
a
x
b
a
x
b
a
L
L
L
L
L
L
=
=
=
=
o
ﻪــــــﻨﻤﻭ
( )
=
L
L
L
ab
a
b
o
.
ﺎ ﺃﻥـــــــﻤﻜ
=
L
e
id
ﺎﻟﻲـــــــﺘﻟﺎﺒﻭ
( )
( )
′
′
=
=
L
L
L
L
a
a
id
a
a
o
o
ﻭﻥـﻜﻴ ﻲﻟﺎﺘﻟﺎﺒﻭ
L
a
، ﺃﻱ ﺃﻥ،ﹰﻼﺒﺎـﻘﺘ
( )
Sym
∈
L
a
G
.
ﺈﻥـﻓ ﺫـﺌﺩﻨﻋ
L
a
a a
ﺸﺎﻜلـﺘ
( )
Sym
→
G
G
ﻭﻫﻭ ﻤﺘﺒﺎﻴﻥ ﻭﺫﻟﻙ ﺤﺴﺏ ﻗﺎﻨﻭﻥ ﺍﻟﺤﺫﻑ،
.
ﻨﺘﻴﺠﺔ
22.1
ﻜل ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻤﻥ ﺭﺘﺒﺔ
n
ﻴﻤﻜﻥ ﺃﻥ ﺘﺼﺒﺢ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
n
S
.
ﺇﻥ،ﻅﺤﻟﺍ ﺀﻭﺴﻟ
n
S
ﺎـﻬﻴﻓ ﻥﻭـﻜﻴ ﻲـﺘﻟﺍ ﺔﻟﺎﺤﻟﺍ ﺀﺎﻨﺜﺘﺴﺎﺒ ،ﺎﻬﻌﻤ لﻤﺎﻌﺘﻨ ﻥﺄﺒ ﹰﺍﺩﺠ ﺓﺭﻴﺒﻜ
n
ﺼﻐﻴﺭﺍﹰ
.
ﺴﻨﺭﻯ ﻓﻴﻤﺎ ﺒﻌﺩ
(4.22)
ﺒﺄﻥ
G
ﺔـﺒﺘﺭﺒ ﺕﻼﻴﺩﺒﺘﻟﺍ ﺓﺭﻤﺯ ﻲﻓ ﺓﺭﻭﻤﻐﻤ ﹰﺎﺒﻟﺎﻏ ﻥﻭﻜﺘﺴ
ﺃﺼﻐﺭ ﺒﻜﺜﻴﺭ ﻤﻥ
!
n
.
ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺍﻟﻤﺭﺍﻓﻘﺔ
(Cosets)
ﻟﺘﻜﻥ
S
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻟﻴﻜﻥ،
{
}
{
}
S
s
sa
Sa
S
s
as
aS
∈
=
∈
=
,
,
، ﻴﻜﻭﻥ،ﺔﻘﻘﺤﻤ ﺔﻴﻌﻴﻤﺠﺘﻟﺍ ﺔﺼﺎﺨﻟﺍ ﻥﺃ ﺎﻤﺒ
( ) ( )
S
ab
bS
a
=
ﺎﻟﺭﻤﺯـﺒ ﺎﻬﻟ ﺯﻤﺭﻨ ﻥﺃ ﻥﻜﻤﻴﻭ ،
abS
.
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٩١
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻥـﻤ ﻲﺘﻟﺍ ﺕﺎﻋﻭﻤﺠﻤﻟﺍ ﻰﻤﺴﺘ ،
ﺸﻜلـﻟﺍ
aH
ـﺒﺎﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻟ
H
ﻓﻲ
G
ﺸﻜلـﻟﺍ ﻥـﻤ ﻥﻭـﻜﺘ ﻲﺘﻟﺍ ﺕﺎﻋﻭﻤﺠﻤﻟﺍ ﻰﻤﺴﺘ ﻭ ،
Ha
ـﺒﺎﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﻤﻴﻨﻴﺔ ﻟ
H
ﻓﻲ
G
.
ﻭﺒﻤﺎ ﺃﻥ
H
e
∈
، ﻓﺈﻥ،
H
aH
=
ﺎﻥـﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ
H
a
∈
.
ﻤﺜﺎل
23.1
ﻟﺘﻜﻥ
(
)
2
,
=
+
G
ﻭﻟﻴﻜﻥ،
H
ﺠﺯﺌﻴﺎﹰ ﺒﻌﺩﻩﺀﺎﻀﻓ
1
)
ﺩﺃـﺒﻤ ﻥﻤ ﺭﺎﻤ ﻡﻴﻘﺘﺴﻤﻟﺍ
ﺍﻹﺤﺩﺍﺜﻴﺎﺕ
.(
ﺍﻟﻤﺭﺍﻓﻘﺎﺕﺫﺌﺩﻨﻋ
)
ﺍﻟﻴﺴﺎﺭﻴﺔ ﻭ ﺍﻟﻴﻤﻴﻨﻴﺔ
(
ـﻟ
H
ﺔـﻴﺯﺍﻭﻤ ﺕﺎﻤﻴﻘﺘﺴﻤ ﻥﻋ ﺓﺭﺎﺒﻋ
ـﻟ
H
.
ﻗﻀﻴﺔ
24.1
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
.
a
(
ﺃﻱ ﻋﻨﺼﺭ
G
a
∈
ﺴﺎﺭﻴﺔـﻴﻟﺍ ﺔﻘﻓﺍﺭﻤﻟﺍ ﻲﻓ ﻊﻘﻴ
C
ــ ﻟ
H
ﺎﻥـﻜ ﺍﺫﺇ ﻁـﻘﻓﻭ ﺍﺫﺇ
aH
C
=
.
(b
ﺍﻟﻤﺭﺍﻓﻘﺘﺎﻥ ﺍﻟﻴﺴﺎﺭﻴﺘﺎﻥ ﺇﻤﺎ ﺃﻥ ﺘﻜﻭﻨﺎ ﻤﻨﻔﺼﻠﺘﻴﻥ ﺃﻭ ﻤﺘﺴﺎﻭﻴﺘﻴﻥ
.
(c
bH
aH
=
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
H
b
a
∈
−
1
.
(d
ﺃﻱ ﻤﺭﺍﻓﻘﺘﻴﻥ ﻴﺴﺎﺭﻴﺘﻴﻥ ﺘﺤﺘﻭﻴﺎﻥ ﻋﻠﻰ ﻨﻔﺱ ﺍﻟﻌﺩﺩ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
)
ﻤﻥ ﺍﻟﻤﻤﻜﻥ
ﺎـﻨﻭﻜﺘ ﻥﺃ
ﻏﻴﺭ ﻤﻨﺘﻬﻴﺘﻴ
ﻥ
.(
ﺍﻟﺒﺭﻫﺎﻥ
.
)
a
(
ﻤﻥ ﺍﻟﻤﺅﻜﺩ ﻤﻥ ﺃﻥ
aH
a
∈
.
ﺇﺫﺍ ﻜﺎﻥ،ﺱﻜﻌﻟﺍ
a
ﺴﺎﺭﻴﺔـﻴﻟﺍ ﺔﻘﻓﺍﺭﻤﻟﺍ ﻲﻓ ﻊﻘﻴ
bH
ﻴﻜﻭﻥﺫﺌﺩﻨﻋ ،
bh
a
=
ﻟﺒﻌﺽ ﺍﻟﻌﻨﺎﺼﺭ ﻤﺜل
h
ﻭﻟﻬﺫﺍ،
bH
bhH
aH
=
=
b)
(
ﺇﺫﺍ ﻜﺎﻥ
C
ﻭ
C
′
ﻏﻴﺭ ﻤﻨﻔﺼﻠﺘﻴ
ﺘﺫﺌﺩﻨﻋ ،ﻥ
ﺤﻭﻴﺎﻥ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﺸﺘﺭﻙ
a
ﻭﻴﻜﻭﻥ،
aH
C
=
ﻭ
aH
C
=
′
ﻭﺫﻟﻙ ﻤﻥ
(a)
.
c)
(
ﺎﻥــــﻜ ﺍﺫﺇ
H
b
a
∈
−
1
،
ﺫــــﺌﺩﻨﻋ
bH
a
H
1
−
=
ﻭﻥــــﻜﻴ ﻪــــﻨﻤﻭ ،
bH
bH
aa
aH
=
=
−
1
.
ﺎﻥـــﻜ ﺍﺫﺇ ،ﺱﻜﻌﻟﺎـــﺒﻭ
bH
aH
=
ﻭﻥـــﻜﻴ ﺫـــﺌﺩﻨﻋ ،
bH
a
H
1
−
=
ﻭﻤﻨﻪ،
H
b
a
∈
−
1
.
d)
(
ﺍﻟﺘﻁﺒﻴﻕ
( )
bh
ah
ba
L
a
:
1
−
ﺘﻘﺎﺒل
bH
aH
→
.
ﺇﻥ
ﺍﻟﺩﻟﻴ
ل
(index)
(
)
H
G :
ﻟﻠﺯﻤﺭﺓ
H
ﻓﻲ
G
ـ ﻫﻭ ﻋﺩﺩ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻟ
H
ﻓﻲ
G
6
.
،ﻤﺜﻼﹰ
(
)
1
:
G
ﻫﻲ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ
G
.
6
ﺇﻥ،ﹰﺎﺤﻴﻀﻭﺘ ﺭﺜﻜﺃ
(
)
H
G :
ﻫﻲ ﻗﺩﺭﺓ ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
G
a
aH
∈
,
.
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ﺇﻥ ﺍﻟﻤ
ﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻟﻠﺯﻤﺭﺓ
H
ﻓﻲ
G
.
ـﺘﺸﻜل ﺘﻐﻁﻴﺔ ﻟ
G
،
(24.1 b)
ﻪـﻨﺄﺒ ﻥﻴـﺒﺘ
ـﻴﺸﻜل ﺘﺠﺯﺌﺔ ﻟ
G
.
ﺍﻟﺸﺭﻁﺎﻥ،ﻯﺭﺨﺃ ﺕﺎﻤﻠﻜﺒﻭ
a
ﻭ
b
ﺴﺎﺭﻴﺔـﻴﻟﺍ ﺔـﻘﻓﺍﺭﻤﻟﺍ ﺱﻔﻨ ﻲﻓ ﻥﺎﻌﻘﻴ
ﺍﻟﺘﻲ ﺘﺸﻜل ﻋﻼﻗﺔ ﺘﻜﺎﻓﺅ ﻋﻠ
ﻰ
G
.
ﻤﺒﺭﻫﻨﺔ
(25.1)
)
ﻻﻏﺭﺍﻨﺞ
(
ﻟﺘﻜﻥ
G
ﻋﻨﺩﺌﺫ،ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ
(
) (
)(
)
1
:
:
1
:
H
H
G
G
=
ﺭﺘﺒﺔ ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻨﺘﻬﻴﺔ ﺘﻘﺴﻡ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ،ﺹﺎﺨ لﻜﺸﺒ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ـﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻟ
H
ﻓﻲ
G
ـ ﺘﺸﻜل ﺘﺠﺯﺌﺔ ﻟ
G
.
ﻴﻭﺠﺩ
(
)
H
G :
،ﻤﺭﺍﻓﻘﺔ
ﻭﻜل ﻤﻨﻬﺎ ﺘﺤﻭﻱ ﻋﻠﻰ
(
)
1
:
H
ﻋﻨﺼﺭ
.
ﻨﺘﻴﺠﺔ
(26.1)
ﺭﺘﺒﺔ ﻜل ﻋﻨﺼﺭ ﻤﻥ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﻘﺴﻡ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﺘﻁﺒﻴﻕ ﻤﺒﺭﻫﻨﺔ ﻻﻏﺭﺍﻨﺞ ﻋﻠﻰ
g
H
=
ﻭﻤﻥ،
(
)
( )
g
o
H
=
1
:
.
ﻤﺜﺎل
27.1
ﻟﺘﻜﻥ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
،
p
ﻜل ﻋﻨﺼﺭ ﻴﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔﺫﺌﺩﻨﻋ ،ﻲﻟﻭﺃ ﺩﺩﻋ
1
ﺃﻭ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﻟﻜﻥ
e
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻭﺤﻴﺩ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،
G
ﺼﺭـﻨﻋ ﻱﺄﺒ ﺓﺩﻟﻭﻤ
e
a
≠
.
ﺘﻜﻭﻥ،ﺹﺎﺨ لﻜﺸﺒﻭ
G
ﻭﺒﺎﻟﺘﺎﻟﻲ،ﺔﻴﺭﺌﺍﺩ
p
C
G
≈
.
ﺒﻴلـﺴ ﻰﻠﻋ ،ﻪﻨﺄﺒ ﻥﻴﺒﻴ ﺍﺫﻫ
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﻭﺍﺤﺩﺓ ﻓﻘﻁ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،لﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘﻭ ،لﺎﺜﻤﻟﺍ
1,000,000,007
ﻭ
)
ﻷ
ﻥ ﺍﻟﻌﺩﺩ
ﺃﻭﻟﻲ
.(
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﻥﺎـﺘﺭﻤﺯ ﺩﺠﻭﺘ ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
1,000,000,014,000,000,049
)
ﺭـﻅﻨﺍ
18.4
.(
28.1
،ﺔـﻴﻨﻴﻤﻴﻟﺍ ﺕﺎـﻘﻓﺍﺭﻤﻟﺍ ﺔﻋﻭﻤﺠﻤ ﻭ ﺔﻴﺭﺎﺴﻴﻟﺍ ﺕﺎﻘﻓﺍﺭﻤﻟﺍ ﺔﻋﻭﻤﺠﻤ ﻥﻴﺒ لﺒﺎﻘﺘ ﻕﻴﺒﻁﺘ ﺩﺠﻭﻴ
1
−
↔
Ha
aH
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
(
)
H
G :
ﻫﻭ
ﻋﺩﺩ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﻤﻴﻨﻴﺔ
ـﻟ
H
ﻓﻲ
G
.
،ﻟﻜﻥ
ﺇﻥ ﺍﻟﻤﺭﺍﻓﻘﺔ ﺍﻟﻴﻤﻴﻨﻴﺔ ﻟﻥ ﺘﺴﺎﻭﻱ ﺍﻟﻤﺭﺍﻓﻘﺔ،ﺔﻤﺎﻌﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
ﺍﻟﻴﺴﺎﺭﻴﺔ
).
ﺍﻨﻅﺭ
33.1
.(
29.1
ﻴﻭﺠﺩ ﻟﻤﺒﺭﻫﻨﺔ ﻻﻏﺭﺍﻨﺞ ﻤﻌﻜﻭﺱ ﺠﺯﺌﻲ ﻭﻫﻭ
:
ﻟﻴﻜﻥ
p
ﻋﺩﺩﺍﹰ ﺃﻭﻟﻴﺎﹰ ﻴﻘﺴﻡ ﺭﺘﺒﺔ
G
ﺤﻴﺙ،
(
)
1
:
G
m
=
ﻋﻨﺩﺌﺫ،
ﻓﺈﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
)
ﻲـﺸﻭﻜ ﺔﻨﻫﺭﺒﻤ
13.4
(
،
ﺇﺫﺍ ﻜﺎﻥ
n
p
ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ ﻭﻴﻘﺴﻡ
m
،
(
)
1
:
G
m
=
ﻭﻱـﺤﺘ ﺫﺌﺩﻨﻋ ،
G
ﺭﺓـﻤﺯ ﻰـﻠﻋ
ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
p
)
ﻤﺒﺭﻫﻨﺔ ﺴﻴﻠﻭ
2.5
.(
ﻨﻼﺤﻅ ﺃﻥ
4
-
ﺯﻤﺭﺓ
2
2
C
C
×
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
4
،
ﻟﻜﻥ ﻻ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
ﻭﺃﻥ ﺍﻟﺯﻤﺭﺓ،
4
A
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
ﻭﻱـﺤﺘ ﻻ ﻥـﻜﻟﻭ ،
ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘ
ﺒﺔ
6
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
13.4
.(
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ ﺍﻟﻨﺘﻴﺠﺔ،ﻡﻴﻤﻌﺘﻟﺎﺒﻭ
ﺍﻵﺘﻴ
.ﺔ
ﻗﻀﻴﺔ
30.1
ﻷﻱ ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ
K
H
⊃
ﻤﻥ
G
ﻴﻜﻭﻥ،
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١٢
(
) (
)(
)
K
H
H
G
K
G
:
:
:
=
)
،ﻨﻌﻨﻲ ﺒﺫﻟﻙ ﺃﻨﻪ
ﺇﻤﺎ ﻜﻠﺘﺎﻫﻤﺎ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺘﻴﻥ
ﺃﻭ ﺘﻜﻭﻨﺎﻥ ﻤﻨﺘﻬﻴﺘﺎﻥ ﻭﻤﺘﺴﺎﻭﻴﺘﺎﻥ
.(
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻜﺘﺏ
∈
=
C
i
i I
G
g H
)
ﺍﻻﺠﺘﻤﺎﻉ
ﻤﻨﻔﺼل
(
ﻭ،
∈
=
C
i
j J
H
h K
)
ﺎﻉـﻤﺘﺠﻻﺍ
ﻤﻨﻔﺼل
.(
ﺩﺩـﻌﻟﺎﺒ ﺔـﻴﻨﺎﺜﻟﺍ ﺓﺍﻭﺎﺴـﻤﻟﺍ ﺏﺭﻀﺒ
i
g
ﺄﻥـﺒ ﺩـﺠﻨ ،
∈
=
C
i
i
i
j J
g H
g h K
)
ﺍﻻﺠﺘﻤﺎﻉ
ﻤﻨﻔﺼل
(
ﻭ ﺒﻬﺫﺍ ﻴﻜﻭﻥ،
∈ ×
=
C
i
i
i j I J
G
g h K
,
)
ﺍﻻﺠﺘﻤﺎﻉ
ﻤﻨﻔﺼل
.(
ـﻭﻫﺫﺍ ﻴﺒ
ﻴﻥ
ﺒﺄﻥ
(
)
(
)(
)
K
H
H
G
J
I
K
G
:
:
:
=
=
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ
(Normal subgroups)
ﻟﺘﻜﻥ
S
ﻭ
T
ﻤﺠﻤﻭﻋﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻟﻴﻜﻥ،
{
}
T
t
S
s
st
ST
∈
∈
=
,
,
.
ﻴﻜﻭﻥ،ﺔﻘﻘﺤﻤ ﺔﻴﻌﻴﻤﺠﺘﻟﺍ ﺔﺼﺎﺨﻟﺍ ﻥﺃ ﺎﻤﺒ
( ) ( )
T
S
R
T
S
R
=
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻤﻜﻥ ﺃﻥ ﻨﺭﻤﺯ ﻟﻬﺫﻩ،
ﺍﻟﻤﺠﻤﻭﻋﺔ ﺒﺎﻟﺭﻤﺯ
T
S
R
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
N
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺎﻟﺭﻤﺯـﺒ ﺎـﻬﻟ ﺯـﻤﺭﻨﻭ ،
G
N <
ﺎﻥـﻜ ﺍﺫﺇ ،
N
gNg
=
−
1
ﻟﻜل
G
g
∈
.
ﻤﻼﺤﻅﺔ
1.31
ﻟ
ﻜﻲ ﻨﺒﻴﻥ
ﺒ
ﺄﻥ
N
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ،
N
gNg
⊂
−
1
ﻟﻜل
G
g
∈
ﺼﺭـﻨﻌﻟﺎﺒ ﻥﻴـﻤﻴﻟﺍ ﻥﻤﻭ ﺭﺎﺴﻴﻟﺍ ﻥﻤ ﺀﺍﻭﺘﺤﻻﺍ ﺍﺫﻫ ﺏﺭﻀﺒ ﻪﻨﻷ ،
1
−
g
ﻭ
g
ﻰـﻠﻋ
ﺍﻟﺘﺭﺘﻴﺏ ﻨﺤﺼل ﻋﻠﻰ ﺍﻻﺤﺘﻭﺍﺀ
1
−
⊂
gNg
N
ﻭﻨﻌﻴﺩ ﻫﺫﻩ ﺍﻟﻜﺘﺎﺒﺔ ﻤﻊ،
1
−
g
ﻟﻜل
g
ﻴﻌﻁﻲ
ﺒﺄﻥ
1
−
⊂
gNg
N
ﻟﻜل
g
.
ﺍﻟﻤﺜﺎل
ﺍﻵﺘﻲ
ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ ﺩﺠﻭﻴ ﻥﺃ ﻥﻜﻤﻴ ﻪﻨﺄﺒ ﻥﻴﺒﻴ
N
ﻥـﻤ
ﺍﻟﺯﻤﺭﺓ
G
ﻭ
G
ag
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
N
gNg
⊂
−
1
ﻟﻜﻥ
N
gNg
≠
−
1
.
ﺎلـﺜﻤ
32.1
ﻟﺘﻜﻥ
( )
Q
GL
G
2
=
ﺘﻜﻥـﻟﻭ ،
Ζ
∈
=
n
n
H
,
1
0
1
ﺈﻥـﻓ ﺫـﺌﺩﻨﻋ ،
H
ﺯﻤﺭﺓ
ﺠﺯﺌﻴﺔ ﻤﻥ
G
ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﺇﻥ،
Ζ
H
.
ﻟﺘﻜﻥ
=
1
0
0
5
g
.
ﻋﻨﺩﺌﺫ
=
=
−
−
0
0
5
1
1
0
0
5
1
0
5
5
1
0
1
1
1
n
n
g
n
g
ﻭﺒﺎﻟﺘﺎﻟﻲ
H
gHg
≠
−
⊂
1
.
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٢٢
ﻗﻀﻴﺔ
33.1
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
N
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ﻜل ﻤﺭﺍﻓﻘﺔ ﻴﺴﺎﺭﻴﺔ
، ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ،ﹰﺎﻀﻴﺃ ﺔﻴﻨﻴﻤﻴ ﺔﻘﻓﺍﺭﻤ ﻲﻫ
Ng
gN
=
ﻟﻜل
G
g
∈
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻭﺍﻀﺢ ﺃﻥ
Ng
gN
N
gNg
=
⇔
=
−
1
ﺇﺫﺍ ﻜﺎﻨﺕ،ﺍﺫﻬﻟ
N
ـﺔ ﻴﻤﻴﻨﻴـﻘﻓﺍﺭﻤ ﻲﻫ ﺔﻴﺭﺎﺴﻴ ﺔﻘﻓﺍﺭﻤ لﻜ ﺫﺌﺩﻨﻋ ،ﺔﻴﻤﻅﺎﻨ
ﺔ
)
،ﺔـﻘﻴﻘﺤﻟﺍ ﻲـﻓ
Ng
gN
=
.(
ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﻤﺭﺍﻓﻘﺔ ﺍﻟﻴﺴﺎﺭﻴﺔ،ﺱﻜﻌﻟﺎﺒﻭ
gN
ﺫـﺌﺩﻨﻋ ،ﹰﺎﻀﻴﺃ ﺔﻴﻨﻴﻤﻴ ﺔﻘﻓﺍﺭﻤ ﻲﻫ
ﺔـﻴﻨﻴﻤﻴﻟﺍ ﺔﻘﻓﺍﺭﻤﻟﺍ ﻥﻭﻜﺘ ﺕﻨﺃ ﺏﺠﻴ
Ng
ﻥـﻤ
(1.24 a)
.
ﺎﻟﻲـﺘﻟﺎﺒﻭ
Ng
gN
=
ﺫﻟﻙـﻟﻭ ،
N
gNg
=
−
1
.
34.1
ﺘﻨﺹ ﺍﻟﻘﻀﻴﺔ
ﻋﻠﻰ ﺃﻨﻪ
ﻟﻜﻲ ﺘﻜﻭﻥ،
N
لـﻜ لـﺠﺃ ﻥـﻤ ﺎﻨﻴﺩﻟ ﻥﻭﻜﻴ ﻥﺃ ﺏﺠﻴ ،ﺔﻴﻤﻅﺎﻨ
G
g
∈
ﻭ
N
n
∈
،
ﺍﻟﻌﻨﺼﺭ
n
′
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
g
n
gn
′
=
)
ﻟﻜل،ﺊﻓﺎﻜﻤ لﻜﺸﺒﻭ
G
g
∈
ﻭ
N
n
∈
ﻴﻭﺠﺩ،
n
′
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
n
g
ng
′
=
.(
ﺄﻥـﺒ لﻭـﻘﻨ ﻲﻜﻟ ،ﻯﺭﺨﺃ ﺓﺭﺎﺒﻌﺒﻭ
N
ﻨﺎﻅﻤﻴﺔ ﻫﺫﺍ ﻴﻜﺎﻓﺊ ﺍﻟﻘﻭل ﺒﺄﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ
G
ﻰـﻠﻋ ﺭﺜﺅﻴﻟ ﻪﻌﻀﻭﻤ ﺭﻴﻐﻴﻟ ﻙﺭﺤﺘﻴ ﻥﺃ ﻪﻨﻜﻤﻴ
ﻋﻨﺼﺭ ﻤﻥ
N
ﻟﻴﺴﺘﺒﺩل ﺒﻌﻨﺼﺭ ﺁﺨﺭ ﻤﻥ
N
.
ﻤﺜﺎل
35.1
(a)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ ﻴﺴﺎﻭﻱ
2
ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ
.
ﺘﻜﻥـﻟ ،لـﻌﻔﻟﺎﺒﻭ
G
g
∈
،
H
g
∉
.
ﻋﻨﺩﺌﺫ
C
gH
H
G
=
)
ﺍﻻﺠﺘﻤﺎﻉ
ﻤﻨﻔﺼل
.(
ﻭﺒﺎﻟﺘﺎﻟﻲ
ﻓﺈﻥ
gH
ﻤﺘﻤﻡ
H
ﻲـﻓ
G
.
ﻭﺒﻨﻔﺱ ﺍﻟﺘﺭﺘﻴﺏ ﻴﺒﻴﻥ ﻟﻨﺎ ﺒﺄﻥ
Hg
ﻤﺘﻤﻡ
H
ﻓﻲ
G
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻜﻭﻥ،
Hg
gH
=
.
(b)
ﻲــــﻫ لﺍﺭﺩــــﻴﻬﻴﺩ ﺓﺭــــﻤﺯ ﻥﺃ ﺭــــﺒﺘﻌﻨ
ﺸﻜلــــﻟﺍ ﻰــــﻠﻋ
{
}
s
r
s
r
r
e
D
n
n
n
1
1
,
...
,
,
...
,
,
−
−
=
.
لـﻴﻟﺩ ﻥﺈﻓ ﺫﺌﺩﻨﻋ
{
}
1
,
...
,
,
−
=
n
n
r
r
e
C
ﻴﺴﺎﻭﻱ
2
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﻨﺎﻅﻤﻴﺔ،
.
ﻟﻜل
3
≥
n
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
{ }
s
e,
ﺔ ﻷﻥـﻴﻤﻅﺎﻨ ﺕﺴـﻴﻟ
{ }
s
e
s
r
sr
r
n
,
2
1
∉
=
−
−
.
(c)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﺘﻜﻭﻥ ﻨ
ﺎﻅﻤﻴﺔ
)
ﻭﺍﻀﺢ
(
ﺤﻴﺤﺎﹰـﺼ ﺱﻴﻟ ﺱﻜﻌﻟﺍ ﻥﻜﻟ ،
:
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
Q
ﻟﻜﻥ ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻴﻬﺎ ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ،ﺔﻴﻠﻴﺩﺒﺘ ﺕﺴﻴﻟ
)
ﺭﻴﻥـﻤﺘﻟﺍ ﺭﻅﻨﺍ
1.1
.(
ﻨﻘﻭل ﺃﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺒﺴﻴﻁﺔ ﺇﺫﺍ ﻟﻡ ﺘﺤﻭ ﺯﻤﺭﺍﹰ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﺒﺎﺴﺘﺜﻨﺎﺀ
G
ﻭ
{ }
e
.
ﻡـﻅﻌﻤ
ﺍﻟﺯﻤﺭ ﻴﻤﻜﻨﻬﺎ ﺃﻥ ﺘﺤﻭﻱ ﻋﻠﻰ ﻜﺜﻴﺭ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ
–
ﻴﻠﻭـﺴ ﺕﺎﻨﻫﺭﺒﻤ ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
(§5)
ﺔـﻴﺭﺌﺍﺩﻟﺍ ﺭﻤﺯﻟﺍ ﺍﺩﻋﺎﻤ ﺔﻬﻓﺎﺘ ﺭﻴﻏ ﺔﻴﺌﺯﺠ ﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ لﻜ ﻰﻠﻋ ﻕﺒﻁﺘ
ﺒﺭﺘﺏ ﺃﻭﻟﻴﺔ
.
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ﻗﻀﻴﺔ
36.1
ﻟﺘﻜﻥ
H
ﻭ
N
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻤﻥ
G
ﻭ
N
ﻋﻨﺩﺌﺫ،ﺔﻴﻤﻅﺎﻨ
HN
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻭﺇﺫﺍ ﻜﺎﻨﺕ
H
ﻓﺈﻥﺫﺌﺩﻨﻋ ﺔﻴﻤﻅﺎﻨ
HN
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴ
ﺔ ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ
HN
ﻭ،ﺔﻴﻟﺎﺨ ﺕﺴﻴﻟ
( )( )
34.1
hn
h n
hh n n
HN
′ ′
′ ′′ ′∈
=
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ ﻤﻐﻠﻘﺔ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻌﻤﻠﻴﺔ ﺍﻟﻀﺭﺏ
.
ﺒﻤﺎ ﺃﻥ
( )
34.1
1
1
1
1
hn
n h
h n
HN
−
−
−
−
′
=
∈
=
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،ﺭﺌﺎﻅﻨﻟﺍ ﺏﻴﻜﺭﺘﻟ ﺔﺒﺴﻨﻟﺎﺒ ﺔﻘﻠﻐﻤ ﹰﺎﻀﻴﺃ ﻲﻫﻭ
HN
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
.
ﻋﻨﺩﺌﺫ
1
1
1
.
,
−
−
−
=
=
gHNg
gHg
gNg
HN
ﻟﻜل
G
g
∈
.
ﺇﻥ ﺘﻘﺎﻁﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﻫﻭ ﺃﻴﻀﺎﹰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
(cf 13.1)
.
ﻴﻤﻜﻨﻨﺎ،ﻙﻟﺫﻟ
ﻑ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔﺭﻌﻨ ﻥﺃ
X
ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﺭـﻤﺯﻟﺍ ﻊﻁﺎـﻘﺘ ﻥﻭﻜﺘﻟ
ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ
X
.
ﻭﺘﻜﻭﻥ ﻋﻠﻰ ﺸﻜل ﺤﺩﻭﺩ ﻤﻥ
X
ﻭﻫﻲ ﻤﻌﻘﺩﺓ ﻗﻠﻴﻼﹰ
.
ﻨﻘﻭل
ﺒﺄﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
X
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻨﺎﻅﻤﻴﺔ ﺇﺫﺍ ﻜﺎﻥ
X
gXg
⊂
−
1
ﻟﻜل
G
g
∈
.
ﺘﻤﻬﻴﺩﻴﺔ
37.1
ﺇﺫﺍ ﻜﺎﻨﺕ
X
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔﺫﺌﺩﻨﻋ ،ﺔﻴﻤﻅﺎﻨ
X
ﺍﻟﻤﻭﻟﺩﺓ ﺒﻬﺎ ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ
ﺘﻁﺒﻴﻕ
"
ﺍﻟﺘﺭﺍﻓﻕ ﻤﻊ
g
"
،
1
−
gag
a a
ﻋﺒﺎ،
ﺸﺎﻜلـﺘ ﻥـﻋ ﺓﺭ
G
G
→
.
ﺇﺫﺍ
ﻜﺎﻥ
X
a
∈
، ﻨﻘﻭل ﺒﺄﻥ،
m
x
x
a
...
1
=
ﻤﻊ ﻜل
i
x
ﺃﻭ ﻤﻌﻜﻭﺴﻬﺎ ﻓﻲ
X
ﻋﻨﺩﺌﺫ،
(
) (
)
1
1
1
1
...
−
−
−
=
∈
m
gag
gx g
gx g
X
ﺘﻤﻬﻴﺩﻴﺔ
38.1
ﻷﻱ
ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ
X
ﻲـﻓ
G
ﺔـﻴﺌﺯﺠﻟﺍ ﺔـﻋﻭﻤﺠﻤﻟﺍ ،
1
−
∈
U
g G
gXg
ﻭﻫﻲ ﺃﺼﻐﺭ ﻤﺠﻤﻭﻋﺔ ﻨﺎﻅﻤﻴﺔ ﺘﺤﻭﻱ ﻋﻠﻰ،ﺔﻴﻤﻅﺎﻨ
X
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻭﺍﻀﺢ
.
ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻘﻀﻴﺔ،ﺔﻘﺒﺎﺴﻟﺍ ﺕﺎﻴﺩﻴﻬﻤﺘﻟﺍ ﻊﻤﺠﺒ
ﺍﻵﺘﻴ
.ﺔ
ﻗﻀﻴﺔ
39.1
ﺔـﻋﻭﻤﺠﻤﻟﺎﺒ ﺓﺩـﻟﻭﻤﻟﺍ ﺔـﻴﻤﻅﺎﻨﻟﺍ ﺔﻴﺌﺯﺠﻟﺍ ﺓﺭﻤﺯﻟﺍ
ﺔـﻴﺌﺯﺠﻟﺍ
X
ﻲـﻓ
G
ﻲـﻫ
1
−
∈
U
g G
gXg
.
ﺍﻟﻨﻭﺍﺓ ﻭﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
(Kernels and quotients)
ﺇﻥ ﻨﻭﺍﺓ ﺍﻟﺘﺸﺎﻜل
G
G
′
→
:
α
ﻫﻲ
( )
( )
{
}
Ker
α
α
=
∈
=
g
G
g
e
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٤٢
ﺇﺫﺍ ﻜﺎﻥ
α
ﻋﻨﺩﺌﺫ،ﹰﺎﻨﻴﺎﺒﺘﻤ
( ) { }
Ker
α
=
e
.
ﺇﺫﺍ ﻜﺎﻥ،ﺱﻜﻌﻟﺍ
( ) { }
Ker
α
=
e
ﺈﻥـﻓ ،
α
ﻷﻥ،ﻥﻴﺎﺒﺘﻤ
( )
( )
( )
g
g
e
g
g
e
g
g
g
g
′
=
⇒
=
′
⇒
=
′
⇒
′
=
−
−
1
1
α
α
α
ﻗﻀﻴﺔ
40.1
ﻨﻭﺍﺓ ﺃﻱ ﺘﺸﺎﻜل ﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻭﺇﺫﺍ ﻜﺎﻥ،ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﺎﻬﻨﺃ ﺢﻀﺍﻭﻟﺍ ﻥﻤ
( )
Ker
α
∈
a
ﻟﺫﻟﻙ ﻓﺈﻥ،
( )
e
a
=
α
ﻭ،
G
g
∈
ﻋﻨﺩﺌﺫ،
(
)
( ) ( ) ( )
( ) ( )
e
g
g
g
a
g
gag
=
=
=
−
−
−
1
1
1
α
α
α
α
α
α
ﻭﺒﺎﻟﺘﺎﻟﻲ
( )
1
Ker
α
−
∈
gag
.
ﺇﻥ ﻨﻭﺍﺓ ﺍﻟﺘﺸﺎﻜل،ﹰﻼﺜﻤ
( )
det GL
→
X
n
F
F
:
ﻭﻉـﻨﻟﺍ ﻥﻤ ﺕﺎﻓﻭﻔﺼﻤﻟﺍ ﻥﻤ ﺓﺭﻤﺯ ﻲﻫ
n
n
×
ﻭﺍﻟﺘﻲ ﻤﺤﺩﺩﻫﺎ ﻴﺴﺎﻭﻱ
1
ـ
ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ
( )
GL
n
F
ﺔـﺼﺎﺨﻟﺍ ﺔﻴﻁﺨﻟﺍ ﺓﺭﻤﺯﻟﺎﺒ
ﻤﻥ ﺍﻟﺩﺭﺠﺔ
n
.
ﻗﻀﻴﺔ
41.1
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻫﻲ ﻨﻭﺍﺓ ﻟﺘﺸﺎﻜل ﺯﻤﺭﻱ
.
ﺇﺫﺍ ﻜﺎﻨﺕ،ﺭﺜﻜﺃ لﻴﺼﻔﺘﺒﻭ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﻭﺤﻴﺩﺓ ﺘﺒﻨﻰ ﻋﻠﻰ ﺍﺫﺌﺩﻨﻋ ،
ﻟﻤﺠﻤﻭﻋﺔ
N
G
ﻭﺍﻟﻤﺅﻟﻔﺔ
ﻤﻥ ﻤﺭﺍﻓﻘﺎﺕ
N
ﻓﻲ
G
ﺒﺤﻴﺙ
N
G
G
aN
a
→
:
a
ﻴﻜﻭﻥ ﺘﺸﺎﻜل ﺯﻤﺭﻱ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺭﻑـﻌﻨﻭ ،ﺔﻴﺭﺎﺴﻴ ﺕﺎﻘﻓﺍﺭﻤﻜ ﺕﺎﻘﻓﺍﺭﻤﻟﺍ ﺏﺘﻜﻨ
( )( ) ( )
N
ab
bN
aN
=
.
ﺎ ﺃﻥـﻨﻴﻠﻋ
ﻨﺒﺭﻫﻥ
(i)
، ﺇﻨﻪ ﻤﻌﺭﻑ ﺠﻴﺩﺍﹰ
ﻭ
(ii)
ﻴﻌﻁﻲ ﺯﻤﺭﺓ ﺘﺒﻨﻰ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ
.
ﻴﻜﻭﻥﺫﺌﺩﻨﻋﻭ
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻥ ﺍﻟﺘﻁﺒﻴﻕ
gN
g a
ﺘﺸﺎﻜل ﻨﻭﺍﺘﻪ
N
.
)i
(
ﻟﺘﻜﻥ
N
a
aN
′
=
ﻭ
N
b
bN
′
=
ﺄﻥـﺒ ﻥﻴﺒﻨ ﻥﺃ ﺎﻨﻴﻠﻋ ،
N
b
a
abN
′
′
=
.
ﻟﻜﻥ
( ) ( )
33.1
33.1
abN
a bN
a b N
aNb
a Nb
a b N
′
′
′ ′
′ ′
=
=
=
=
=
)
ii
(
ﺍﻟﻤﺭﺍﻓﻕ،ﺩﻴﻜﺄﺘﻟﺎﺒ ﻲﻌﻴﻤﺠﺘ ﺀﺍﺩﺠﻟﺍ ﻥﺇ
N
ﻭ،ﻱﺩﺎﻴﺤﻟﺍ ﺭﺼﻨﻌﻟﺍ ﻭﻫ
N
a
1
−
ﻭـﻫ
ﻤﻌﻜﻭﺱ
aN
.
42.1
ﺭﺓــﻤﺯﻟﺍ ﻥﺇ
N
G
ﺴﻤﺔــﻘﻟﺍ ﺓﺭــﻤﺯﺒ ﻰﻤﺴــﺘ
7
ـــﻟ
G
ﻰــﻠﻋ
N
.
ﻕــﻴﺒﻁﺘﻟﺍ
N
G
G
aN
a
→
:
a
ﻴﺘﻤﺘﻊ ﺒﺎﻟﺨﺎﺼﺔ ﺍﻟﻌﺎﻤﺔ
ﺍﻵﺘﻴ
ﺔ
:
ﻷﻱ ﺘﺸﺎﻜل
G
G
′
→
:
α
ﻥـﻤ
ﺍﻟﺯﻤﺭ ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( ) { }
e
N
=
α
ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻭﺤﻴﺩ،
G
N
G
′
→
ﺒﺤﻴﺙ ﻴﺠﻌل ﺍﻟﻤﺨﻁﻁ
ﺍﻵﺘﻲ
ﺘﺒﺎﺩﻟﻴﺎﹰ
:
7
ﺘﺴﻤﻰ ﻋﻨﺩ
ﺒﻌﺽ ﺍﻟﻤﺅ
ﻟﻔﻴﻥ
ـ ﺒ
”
ﺯﻤﺭﺓ ﺍﻟﺨﺎﺭﺝ
"
ﺒﺩﻻﹰ ﻤﻥ
"
ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ
."
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٥٢
N
G
G
aN
a
→
a
ﻨﻼﺤﻅ ﺒﺄﻨﻪ ﻟﻜل،ﻙﻟﺫ ﻥﺎﻴﺒﺘﻟ
( )
( ) ( )
( )
g
n
g
gn
N
n
α
α
α
α
=
=
∈
,
ﺈﻥـﻓ ﻙﻟﺫﻟﻭ ،
α
ﺜﺎﺒﺘﺔ ﻋﻠﻰ ﻜل ﻤﺭﺍﻓﻘﺔ ﻴﺴﺎﺭﻴﺔ
gN
ـ ﻟ
N
ﻓﻲ
G
.
ﻑ ﺍﻟﺘﻁﺒﻴﻕﺭﻌﻨ ﻙﻟﺫﻟﻭ
( )
( )
g
gN
G
N
G
α
α
α
=
′
→
,
:
ﻭ
α
ﺘﺸﺎﻜل ﻷﻥ
( )( )
(
)
(
)
( )
( ) ( )
( ) ( )
N
g
gN
g
g
g
g
N
g
g
N
g
gN
′
=
′
=
′
=
′
=
′
α
α
α
α
α
α
α
ﺇﻥ ﻭﺤﺩﺍﻨﻴﺔ
α
ﺘﺘﻀﺢ ﻤﻥ ﺃﻥ
N
G
G
→
ﻏﺎﻤﺭ
.
ﻤﺜﺎل
43.1
(a)
ﻨﻌﺘﺒﺭ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
Ζ
m
ﻓﻲ
Ζ
.
ﻭﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ
Ζ
Ζ
m
ﺔـﻴﺭﺌﺍﺩ ﺓﺭﻤﺯ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
m
.
(b)
ﻟﻴﻜﻥ
L
ﻤﺴﺘﻘﻴﻤﺎﹰ ﻤﺎﺭﺍﹰ ﻤﻥ ﻤﺒﺩﺃ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﻓﻲ
2
.
ﻓﺈﻥﺫﺌﺩﻨﻋ
2
L
ﻴﺘﻤﺎﺜل ﻤﻊ
)
ﻷﻨﻪ ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ ﻋﻠﻰ
ﺫﻭ ﺍﻟﺒﻌﺩ ﻭﺍﺤﺩ
(
(c)
ﻟﻜل
2
≥
n
ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ،
{ }
s
e
r
D
n
,
=
)
ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.(
ﺤﻭل ﻤﺒﺭﻫﻨﺎﺕ ﺍﻟﺘﻤﺎﺜل
(Theorems concerning homomorphisms)
ﺍﻟﻤﺒﺭﻫﻨﺎﺕ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﻘﻁﻊ ﺍﻟﺠﺯﺌﻲ ﺘﺴﻤﻰ ﺃﺤﻴﺎﻨﺎﹰ ﺒﻤﺒﺭﻫﻨﺎﺕ ﺍﻟﺘﻤﺎﺜل
)
، ﺍﻟﺜﺎﻨﻴﺔ، ﻰﻟﻭﻷﺍ
.( ...
ﺘﺤﻠﻴل ﺍﻟﺘﺸﺎﻜﻼﺕ ﺇﻟﻰ ﻋﻭﺍﻤل
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﺼﻭﺭﺓ ﺍﻟﺘﻁﺒﻴﻕ
T
S
→
:
α
ﻫﻲ
( )
( )
{
}
S
s
s
S
∈
=
,
α
α
.
ﻤﺒﺭﻫﻨﺔ
44.1
)
ﺍﻟﻤﺒﺭﻫﻨﺔ ﺍﻷﺴﺎﺴﻴﺔ ﻟﺘﺸﺎﻜﻼﺕ ﺍﻟﺯﻤﺭ
(
ﻷﻱ ﺘﺸﺎﻜل
G
G
′
→
:
α
، ﻤﻥ ﺍﻟﺯﻤﺭ
ﺘﻜﻭﻥ ﺍﻟﻨﻭﺍﺓ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓ
ﻲ
G
ﺍﻟﺼﻭﺭﺓ،
I
ـ ﻟ
α
ﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
′
،
ﺇﻥ
α
ﻴﺘﺤﻠل ﺒﺎﻟﺤﺎﻟﺔ ﺍﻟﻁﺒﻴﻌﻴﺔ ﺇﻟﻰ ﺘﻁﺒﻴﻕ
ﻭﺘﻁﺒﻴﻕ ﻤﺘﺒﺎﻴﻥ،لﺜﺎﻤﺘﻭ ،ﺭﻤﺎﻏ
:
G`
α
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٦٢
G
G
′
ﺎﻥـﻫﺭﺒﻟﺍ
.
ﻟﻘﺩ ﺭﺃﻴﻨﺎ ﻓﻲ
(40.1)
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ﻲﻫ ﺓﺍﻭﻨﻟﺍ ﻥﺃ
G
.
ﺎﻥـﻜ ﺍﺫﺇ
( )
a
b
α
=
ﻭ
( )
a
b
′
=
′
α
ﺫـــﺌﺩﻨﻋ ،
( )
a
a
b
b
′
=
′
α
ﻭ،
( )
1
1
−
−
=
a
b
α
ﺫﻟﻙـــﻟﻭ ،
( )
def
α
=
I
G
ﻴﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
ﻓﻲ
G
′
.
ﺘﺒﻴﻥ ﺍﻟﺨﺎﺼﺔ ﺍﻟﻌﺎﻤﺔ ﻟﺯﻤﺭﺓ
ﺍﻟﻘﺴﻤﺔ
(42.1)
ﺃﻥ ﺍﻟﺘﻁﺒﻴﻕ
( )
I
G
x
x
→
:
α
a
ﺸﺎﻜلـﺘﻟﺍ ﻑﺭـﻌﻴ
I
N
G
→
:
α
ﺒﺎﻟﻌﻼﻗﺔ
( )
( )
g
gN
α
α
=
.
ﻭﺇﺫﺍ ﻜﺎﻥ،ﺭﻤﺎﻏ لﻜﺎﺸﺘﻟﺍ ﺍﺫﻫ ﻥﺇ
( )
e
gN
=
α
ﻓﺈﻥﺫﺌﺩﻨﻋ ،
( )
Ker
α
∈
=
g
N
ﻭﻤﻨﻪ ﻓﺈﻥ،
ﻨﻭﺍﺓ
α
ﺘﺎﻓﻬﺔ
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻟﺘﺸﺎﻜل ﻤﺘﺒﺎﻴﻥ
.
ﻤﺒﺭﻫﻨﺔ ﺍﻟﺘﻤﺎﺜل
ﻤﺒﺭﻫﻨﺔ
45.1
)
ﻤﺒﺭﻫﻨﺔ ﺍﻟﺘﻤﺎﺜل
(
ﻟﺘﻜﻥ
H
ﻭ
N
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻤﻥ
G
ﻭ
N
ـ ﻨﺎﻅﻤﻴ
ﺔ
.
ﻋﻨﺩﺌﺫ
HN
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
،
N
H I
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
H
ﻭﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ،
(
)
N
HN
N
H
H
hN
N
H
h
→
I
a
I
:
ﺘﻤﺎﺜل
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻘﺩ ﺭﺃﻴﻨﺎ ﻓﻲ
(36.1)
ﺃﻥ
HN
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
.
ﻟ
ﻨﺄﺨﺫ ﺍﻟﺘﻁﺒﻴﻕ
hN
h
N
G
H
a
,
→
ﻨﻭﺍﺘﻪ،لﻜﺎﺸﺘ ﻥﻋ ﺓﺭﺎﺒﻋ ﻕﻴﺒﻁﺘﻟﺍ ﺍﺫﻫ ﻥﺇ
N
H I
ﻭﺍﻟﺘﻲ ﻫﻲ ﻨﺎﻅﻤﻴﺔ ﻓﻲ،
H
.
ﻤﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
44.1
ﻭﻤﻥ ﺍﻟﺘﻁﺒﻴﻕ ﻴﻨﺘﺞ ﻟﺩﻴﻨﺎ ﺍﻟﺘﻤﺎﺜل،
I
N
H
H
→
I
ﺤﻴﺙ
I
ﺼﻭﺭﺓ ﺍﻟﺘﻁﺒﻴ
ﻕ
.
ﻥـﻜﻟ
I
ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﺸﻜل
hN
ﻤﻊ
H
h
∈
، ﺃﻱ ﺃﻥ،
N
HN
I
=
.
ﻤﺒﺭﻫﻨﺔ ﺍﻟﺘﻘﺎﺒل
ﺘﺒﻴﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
ﺍﻵﺘﻴ
ﺔ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
G
ـ ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ ﻟ
G
ﺭـﻤﺯﻟﺍ ﻥـﻤ ﺔﻜﺒـﺸ ﻥﺈﻓ ﺫﺌﺩﻨﻋ ،
ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ـ ﺘﺘﺤﻜﻡ ﺒﺒﻨﻴﺔ ﺸﺒﻜﺔ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻟ
G
ﻕـﻴﺒﻁﺘﻟﺍ ﺓﺍﻭـﻨ ﻰـﻠﻋ ﻊـﻘﺘ ﻲﺘﻟﺍ
G
G
→
.
G/N
ﻤﺘﺒﺎﻴﻥ
ﻏﺎﻤﺭ
gN
g a
( )
g
gN
α
a
I
ﺘﻤﺎﺜل
α
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٧٢
ﻤﺒﺭﻫﻨﺔ
46.1
)
ﻤﺒﺭﻫﻨﺔ ﺍﻟﺘﻘﺎﺒل
(
ﺘﻜﻥـﻟ
G
G
→
:
α
ﻴﻜﻥـﻟﻭ ،ﺭﻤﺎـﻏ ﻱﺭـﻤﺯ لﻜﺎﺸـﺘ
( )
Ker
α
=
N
.
ﻴﻭﺠﺩ ﺘﻁﺒﻴﻕ ﺘﻘﺎﺒلﺫﺌﺩﻨﻋ
}
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ
G
{
↔
}
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ
G
ﻭﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ
N
{
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ
N
ﻓﻬﻲ ﺘﻘﺎﺒل
( )
H
H
α
=
ﺔـﻴﺌﺯﺠﻟﺍ ﺓﺭـﻤﺯﻟﺍﻭ
H
ﻓﻲ
G
لـﺒﺎﻘﺘ ﻲﻬﻓ
( )
H
H
1
−
=
α
.
ﺎﻥـﻜ ﺍﺫﺇ ،ﻙﻟﺫـﻟ ﺔﻓﺎـﻀﻹﺎﺒﻭ
H
H
↔
ﻭ
H
H
′
↔
′
ﻋﻨﺩﺌﺫ،
(a)
H
H
H
H
′
⊂
⇔
′
⊂
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ،
(
)
(
)
H
H
H
H
:
:
′
=
′
(b)
ﺘﻜﻭﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻭﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻨﺘﺞ،
ﻟﺩﻴﻨﺎ ﺍﻟﺘﻤﺎﺜل
→
G H
G H
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻤﻥ ﺍﻟﺴﻬﻭﻟﺔ ﺭﺅﻴﺔ ﺒﺄﻥﺫﺌﺩﻨﻋ ،
( )
H
1
−
α
ﺯﻤﺭﺓ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﺘﺤﻭﻱ
N
ﻭﺇﺫﺍ ﻜﺎﻨﺕ،
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻓﺈﻥﺫﺌﺩﻨﻋ ،
( )
H
α
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
)
ﺍﻨﻅﺭ
44.1
.(
،ﻤﻥ ﺍﻟﻭﺍﻀﺢ
( )
HN
H
=
−
1
αα
ﺴﺎﻭﻱـﺘ ﻲﺘﻟﺍﻭ ،
H
ﺇﺫﺍ
ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
N
H
⊃
ﻭ،
( )
H
H
=
−
1
αα
.
ﻓﺈﻥ ﺍﻟﻌﻤﻠﻴﺘﻴﻥ ﺘﻌﻁﻴﺎﻥ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺘﻘﺎﺒل،ﻙﻟﺫﻟ
ﺍﻟﻤﻁﻠﻭﺏ
.
ﺃﻤﺎ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺒﺎﻗﻴﺔ ﻓﻤﻥ ﺍﻟﺴﻬل ﺒﺭﻫﺎﻨﻬﺎ
.
لـﻴﻠﺤﺘﻟﺍ ،ﹰﻼﺜﻤ
∈
′ =
C
i
i I
H
a H
ــ ﻟ
H
′
ــﺴﺎﺭﻴﺔ ﻟـﻴﻟﺍ ﺕﺎـﻘﻓﺍﺭﻤﻟﺍ ﻥـﻤ لﺼـﻔﻨﻤ ﻉﺎﻤﺘﺠﺍ ﻰﻟﺇ
H
ﻴﻌﻁﻴ
ﺸﺎﺒﻬﺎﹰـﻤ ﹰﻼﻴـﻠﺤﺘ ﺎـﻨ
∈
′ =
C
i
i I
H
a H
ـ ﻟ
H
′
.
ﻨﺘﻴﺠﺔ
47.1
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺔـﻋﻭﻤﺠﻤ ﻥﻴﺒ لﺒﺎﻘﺘ ﻕﻴﺒﻁﺘ ﺩﺠﻭﻴ ﺫﺌﺩﻨﻋ ،
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻰـﻠﻋ ﺔـﻴﻭﺎﺤﻟﺍﻭ
N
ﻲـﻓ ﺔـﻴﺌﺯﺠﻟﺍ ﺭـﻤﺯﻟﺍ ﺔـﻋﻭﻤﺠﻤﻭ
,
G N
،
↔
H
H N
ﺘﻜﻭﻥ،ﻙﻟﺫ ﻰﻠﻋ ﹰﺓﻭﻼﻋ ،
H
ﻲـﻓ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
ﻁ ﺇﺫﺍـﻘﻓﻭ ﺍﺫﺇ
ﻜﺎﻨﺕ
N
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
N
G
،
G
ﻭﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻨﺘﺞ ﻟﺩﻴﻨﺎ ﺍﻟﺘﻤﺎﺜل،
(
) (
)
G H
G N
H N
→
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺎـﻬﻴﻓ ﻥﻭـﻜﻴ ﻲـﺘﻟﺍ ﺔـﻨﻫﺭﺒﻤﻟﺍ ﻥـﻤ ﺔـﺼﺎﺨﻟﺍ ﺔـﻟﺎﺤﻟﺍ ﻩﺫـﻫ
α
ﻕـﻴﺒﻁﺘﻟﺍ ﻭـﻫ
N
G
G
gN
g
→
:
a
.
ﻤﺜﺎل
48.1
ﻟﺘﻜﻥ
4
D
G
=
ﻭﻟﺘﻜ
ﻥ
N
ﺯﻤﺭﺘﻬﺎ ﺍﻟﺠﺯﺌﻴﺔ
2
r
.
ﺒﺎﻟﺭﺠﻭﻉ ﺇﻟﻰ
(16.1)
ﺩـﺠﻨ
ﺃﻥ
3
1
r
srs
=
−
ﻭﻟﺫﻟﻙ،
( )
2
2
1
3
2
−
=
=
sr s
r
r
.
ﻟﺫﻟﻙ ﻓﺈﻥ
N
ﻨﺎﻅﻤﻴﺔ
.
ﺭﺘﻴﻥـﻤﺯﻟﺍ ﻥﺇ
G
ﻭ
N
G
ﻟﻬﻤﺎ
ﺍﻟﺸﺒﻜﺔ
ﺍﻵﺘﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
PDF created with pdfFactory Pro trial version
٨٢
ﺍﻟﺠﺩﺍﺀ
ﺍﺕ
ﺍﻟﻤﺒﺎﺸﺭ
ﺓ
(Direct product)
ﻟﺘﻜﻥ
G
ﻭﻟﺘﻜﻥ،ﺓﺭﻤﺯ
k
H
H
H
,
,...
,
2
1
ﺯﻤﺭﺍﹰ ﺠﺯﺌﻴﺔ ﻤﻥ
G
.
ﻨﻘﻭل ﺒﺄﻥ
G
ﺩﺍﺀـﺠ ﻲﻫ
ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
i
H
ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﻁﺒﻴﻕ
(
)
G
H
H
H
h
h
h
h
h
h
k
k
k
→
×
×
×
...
:
...
,
...
,
,
2
1
2
1
2
1
a
ﺘﻤﺎﺜﻼﹰ ﻤﻥ ﺍﻟﺯﻤﺭ
.
ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻜل ﻋﻨﺼﺭ
g
ﻤﻥ
G
ﺸﻜلـﻟﺎﺒ ﺩﻴﺤﻭ ﺸﻜلﺒﻭ ﺏﺘﻜﻴ ﻥﺃ ﻥﻜﻤﻴ
i
i
k
H
h
h
h
h
g
∈
=
,
...
2
1
ﺎﻥـــﻜ ﺍﺫﺇﻭ ،
k
h
h
h
g
...
2
1
=
ﻭ
k
h
h
h
g
′
′
′
=
′
...
2
1
،
ﻋﻨﺩﺌﺫ
( ) (
) (
)
1 1
2
2
′
′
′
′
=
k
k
gg
h h
h h
h h
...
.
ﺇﻥ ﺍﻟﻘﻀﺎﻴﺎ
ﺍﻵﺘﻴ
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺯﻤﺭﻫﺎ ﺍﻟﺠﺯﺌﻴﺔﺀﺍﺩﺠ ﺓﺭﻤﺯﻟﺍ ﻥﻭﻜﺘ ﻲﻜﻟ ﺓﺩﻋﺎﻗ ﺎﻨﻴﻁﻌﺘ ﺔ
.
ﻗﻀﻴﺔ
49.1
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺯﻤﺀﺍﺩﺠ
ﺭﻫﺎ ﺍﻟﺠﺯﺌﻴﺔ
2
1
, H
H
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
(a)
2
1
H
H
G
=
(b)
{ }
e
H
H
=
2
1
I
(c)
ﻜل ﻋﻨﺼﺭ ﻤﻥ
1
H
ﻴﺘﺒﺎﺩل ﻤﻊ ﻜل ﻋﻨﺼﺭ ﻤﻥ
2
H
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺀﺍﺩﺠ
ﺯﻤﺭﻫﺎ ﺍﻟﺠﺯﺌﻴﺔ
2
1
, H
H
ﻓﺈﻥﺫﺌﺩﻨﻋ ،
(a)
ﻭ
(c)
ﻭ
(b)
ﻤﻥ ﺃﺠل ﺃﻱ،ﻪﻨﻷ ﻕﻘﺤﻤ
2
1
H
H
g
I
∈
ﻴﺘﻁﺎﺒﻕ ﺍﻟﻌﻨﺼﺭ،
(
)
1
,
−
g
g
ﻊـﻤ
e
ﺴﺒﺔـﻨﻟﺎﺒ
ﻟﻠﺘﻁﺒﻴﻕ
(
)
2
1
2
1
,
h
h
h
h
a
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﺴﺎﻭﻱ،
( )
e
e,
.
،ﺒﺎﻟﻌﻜﺱ
(c)
ﻴﻘﺘﻀﻲ ﺒﺄﻥ
(
)
2
1
2
1
,
h
h
h
h
a
ﻭ،لﻜﺎﺸﺘ
(b)
ﻴﺒﻴﻥ ﺒﺄﻥ ﺍﻟﺘﻁﺒﻴﻕ ﻤﺘﺒﺎﻴﻥ
:
{ }
e
H
H
h
h
e
h
h
=
∈
=
⇒
=
−
2
1
1
2
1
2
1
I
.
s
r
s
r
1
r
s
r ,
2
rs
r ,
2
2
r
1
rs
s
r
2
s
r
3
s
4
D
2
4
r
D
PDF created with pdfFactory Pro trial version
٩٢
ﻤﻥ،ﹰﺍﺭﻴﺨﺃ
(a)
ﻨﺠﺩ ﺒﺄﻨﻪ ﻏﺎﻤﺭ
.
ﻗﻀﻴﺔ
50.1
ﺘﺸﻜل ﺍﻟﺯﻤﺭﺓ
G
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺯﻤﺭﻫﺎ ﺍﻟﺠﺯﺌﻴﺔﺀﺍﺩﺠ
2
1
, H
H
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫ
ﺍ ﻜﺎﻥ
(a)
2
1
H
H
G
=
(b)
{ }
e
H
H
=
2
1
I
(c)
ﻜل ﻤﻥ
2
1
, H
H
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻘﺩ
ﺒﺭﻫﻨﺕ ﻫﺫﻩ ﺍﻟﺸﺭﻭﻁ
ﻓﻲ ﺍﻟﻘﻀﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ
،
ﺒﻘﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ ﻜل ﻋﻨﺼﺭ
1
h
ﻤﻥ
1
H
ﻴﺘﺒﺎﺩل ﻤﻊ ﻜل ﻋﻨﺼﺭ
2
h
ﻤﻥ
2
H
.
ﺇﻥ ﺍﻟﻤﺒﺎﺩل
[
]
(
)
(
)
def
1
1
1
1
1
1
1
2
1
2 1
2
1 2 1
2
1
2 1
2
−
−
−
−
−
−
=
=
=
h h
h h h h
h h h
h
h h h h
,
.
ﻴﻨﺘﻤﻲ ﺇﻟﻰ
2
H
ﻷﻥ
2
H
ﻭﻴﻨﺘﻤﻲ ﺇﻟﻰ،ﺔﻴﻤﻅﺎﻨ
1
H
ﻭ ﻤﻥ،ﺔﻴﻤﻅﺎﻨ ﺎﻬﻨﻷ
(b)
ﻨﺠﺩ ﺒﺄﻨﻪ ﻴﺴﺎﻭﻱ
e
،
ﻭﺒﺎﻟﺘﺎﻟﻲ
1
2
2
1
h
h
h
h
=
.
ﻗﻀﻴﺔ
51.1
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺯﻤﺭﻫﺎ ﺍﻟﺠﺯﺌﻴﺔﺀﺍﺩﺠ
k
H
H
H
,
,...
,
2
1
ﻁـﻘﻓﻭ ﺍﺫﺇ
ﺇﺫﺍ ﻜﺎﻥ
(a)
k
H
H
H
G
...
2
1
=
(b)
(
)
{ }
e
H
H
H
H
H
k
j
j
j
=
+
−
...
...
1
1
1
I
،
ﻟﻜل
j
(c)
ﻜل ﻤﻥ
k
H
H
H
,
,...
,
2
1
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
،ﻟﺯﻭﻡ ﺍﻟﺸﺭﻁ ﻭﺍﻀﺢ
ﺴﻭﻑ ﻨﺒﺭﻫﻥ ﻋﻠﻰ ﻜﻔﺎﻴﺔ ﺍﻟﺸﺭﻁ
.
ﻤﻥ ﺃﺠل
2
=
k
،
ﻓﻘﺩ
ﺎـﻨﻤﻗ
ﺒﺒﺭﻫ
ﺎ
ﻨﻪ
ﺴﺎﺒﻘﺎﹰ
ﻭﻟﺫﻟﻙ ﺴﻨﻨﺎﻗﺵ ﺒﺎﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ،
k
.
ﺇﻥ
ﺍﻟﻤﻨﺎﻗﺸﺔ ﺒﺎﻻﺴﺘﻘﺭﺍﺀ
ﻲـﻓ ﺔﻤﺩﺨﺘﺴـﻤﻟﺍ
(36.1)
ﺘﺒﻴﻥ ﺒﺄﻥ
1
2
1
...
−
k
H
H
H
ﻲـﻓ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
.
ﺸﺭﻭﻁـﻟﺍ ﻥﺇ
(a,b,c)
ﻤﺤﻘﻘﺔ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
1
2
1
,
...
,
,
−
k
H
H
H
ﻓﻲ
1
2
1
...
−
k
H
H
H
ﻥـﻤﻭ ﻙﻟﺫـﻟﻭ ،
ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻲ ﻴﻜﻭﻥ
(
)
1
2
1
1 2
1
1
2
1
1
2
1
,
,... ,
...
:
...
...
k
k
k
k
h h
h
h h
h
H
H
H
H H
H
−
−
−
−
×
× ×
→
a
ﺘ
ﻤﺎﺜﻼﹰ
.
ﺇﻥ ﺍﻟﺯﻤﺭﺘﻴﻥ
1
2
1
,
,...
,
−
k
H
H
H
ﻭ
k
H
ﺘ
ﺤﻘﻘﺎﻥ ﻓﺭﻀﻴﺎﺕ
(50.1)
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،
(
)
(
)
G
H
H
H
hh
h
h
k
k
k
k
→
×
−
1
1
...
:
,
a
ﺘﻤﺎﺜل ﺃﻴﻀﺎﹰ
.
ﺇﻥ ﺘﺭﻜﻴﺏ ﻫﺫﻩ ﺍﻟﺘﻤﺎﺜﻼﺕ
(
) (
)
(
)
1
1
1
1
2
1
1
2
1
k
k
k
k
k
k
k
k
k
h
h
h
h
h
h h
hh
H
H
H
H
H H
H
H
G
−
−
−
×
× ×
×
→
×
→
a
a
,...,
...
,
...
...
ﻴﺭﺴل
(
)
k
h
h
h
,
...
,
,
2
1
ﺇﻟﻰ
k
h
h
h
...
2
1
.
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٠٣
ﺍﻟﺯﻤﺭ ﺍﻟﺘﺒ
ﺩﻴﻠﻴﺔ
(Commutative groups)
ﻭﺩﻭﻻﺕـﻤﻟﺍ ﺔﻨﻫﺭﺒﻤ ﻲﻓ ﺀﺯﺠﻜ ﺭﺒﻜﺃ لﻜﺸﺒ ﺕﺴﺭﺩ ﺩﻴﻟﻭﺘﻟﺍ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺔﻴﻠﻴﺩﺒﺘﻟﺍ ﺭﻤﺯﻟﺍ ﻑﻴﻨﺼﺘ ﻥﺇ
ﻓﻴﻤﺎ ﻴﻠ،ﻙﻟﺫ لﻤﻜﻨ ﻥﺃ لﺠﺃ ﻥﻤ ،ﻥﻜﻟ ،ﺔﻴﺴﻴﺌﺭﻟﺍ ﺕﻻﺎﻴﺩﻴﻹﺍ ﺔﻘﻠﺤ ﻰﻠﻋ
ﻲ
ﺍﻟﺸﺭﺡ ﺍﻷﻭﻟﻲ
.
ﻟﺘﻜﻥ
M
ﺇﻥ ﺍﻟ،ﻊﻤﺠﻟﺍ ﺔﻴﻠﻤﻋ ﻊﻤ ،ﺔﻴﻠﻴﺩﺒﺘ ﺓﺭﻤﺯ
ﺯﻤﺭﺓ
k
x
x ...
,
1
ﻤﻥ
M
ﺩﺓـﻟﻭﻤﻟﺍ ﻭ
ﺒﺎﻟﻌﻨﺎﺼﺭ
k
x
x
,
...
,
1
ﺘﺘﻜﻭﻥ ﻤﻥ ﺍﻟﻤﺠﻤﻭﻉ
Ζ
∈
∑
i
i
i
m
x
m
,
.
ﺘﺴﻤﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
{
}
k
x
x
,
...
,
1
ـ ﻗﺎﻋﺩﺓ ﻟ
M
ﺇﺫﺍ ﻜﺎﻨﺕ ﺘﻭﻟﺩ
M
ﻭ
1 1
...
0
0,
+
+
= ⇒
=
∈
Z
k
k
i
i
i
m x
m x
m x
m
ﻟﻜل
i
ـ ﺃﻥ ﻟ،ﺢﻀﺍﻭﻟﺍ ﻥﻤ
M
ﺔـﻴﺭﺌﺍﺩ ﺭـﻤﺯﻟ ﹰﺍﺭـﺸﺎﺒﻤ ﹰﺎﻋﻭﻤﺠﻤ ﻥﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﺔﻴﻬﺘﻨﻤ ﺓﺩﻋﺎﻗ
ﻤﻨﺘﻬﻴﺔ
.
ﺘﻤﻬﻴﺩﻴﺔ
52.1
ﺒﻔﺭﺽ ﺃﻥ
M
ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔ
{
}
k
x
x
,
...
,
1
ﺘﻜﻥـﻟﻭ
k
c
c
,
,...
1
ﺩﺍﺩـﻋﺃ
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،ﺔﺤﻴﺤﺼ
(
)
1
,
,...
gcd
1
=
k
c
c
.
ﺘﻭﺠﺩ ﻤﻭﻟﺩﺍﺕﺫﺌﺩﻨﻋ
k
y
y
,
,...
1
ﻥـﻤ
M
،
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
k
k
x
c
x
c
y
+
+
=
...
1
1
1
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
0
<
i
c
ﻨﻐﻴﺭ ﺇﺸﺎﺭﺍﺕ ﻜل ﻤﻥ،
i
c
ﻭ
i
x
.
ﺭﺽ ﺃﻥـﻔﻨ ﻥﺄﺒ ﺎﻨﻟ ﺢﻤﺴﻴ ﺍﺫﻫ
ﻜل ﻤﻥ
Ν
∈
i
c
.
ﺒﺎﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ
k
c
c
s
+
+
=
...
1
.
ﺎﻥـﻜ ﺍﺫﺇ ﺔـﻘﻘﺤﻤ ﺔﻴﺩﻴﻬﻤﺘﻟﺍ ﻥﻭﻜﺘ
1
=
s
ﻨﻔﺭﺽ ﺃﻥ،
1
>
s
.
ﻥـﻤ ﻥﺍﺭﺼﻨﻋ لﻗﻷﺍ ﻰﻠﻋ ﺩﺠﻭﻴ ،ﺫﺌﺩﻨﻋ
i
c
ﺩﻭﻤﻴـﻌﻤ ﺭـﻴﻏ
،ﻥ
،ﻴﻜﻥــــﻟﻭ
0
2
1
>
≥
c
c
.
ﺍﻵﻥ
ﺇﻥ
k
x
x
x
x
x
M
,
...
,
,
,
3
2
1
1
+
=
ﻭ
ﺎــــﻤﺒ
ﺃﻥ
(
)
1
,
,...
,
,
gcd
3
2
2
1
=
−
k
c
c
c
c
c
، ﺒﺎﻻﺴﺘﻘﺭﺍﺀ،ﻪﻨﻤﻭ ،
k
y
y
M
,
,...
1
=
ﺤﻴﺙ،
(
)
(
)
k
k
k
k
x
c
x
c
x
c
x
c
x
x
c
x
c
c
y
+
+
=
+
+
+
+
+
−
=
...
...
1
1
3
3
2
1
2
1
2
1
ﻤﺒﺭﻫﻨﺔ
53.1
ﻜل ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻟﻬﺎ ﻗﺎﻋﺩﺓ
.
ﺍﻟﺒﺭﻫ
ﺎﻥ
.
ﻨﻨﺎﻗﺵ ﺒﺎﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ ﻋﺩﺩ ﻤﻭﻟﺩﺍﺕ
M
.
ﺇﺫﺍ ﻜﺎﻨﺕ
M
ﺘﻜﻭﻥ،ﺩﺤﺍﻭ ﺭﺼﻨﻌﺒ ﺓﺩﻟﻭﻤ
ﻭﻟﺫﻟﻙ ﻨﻔﺭﺽ ﺒﺄﻨﻪ ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ،ﺔﻬﻓﺎﺘﻟﺍ ﺔﻟﺎﺤﻟﺍ
1
>
k
لـﻗﻷﺍ ﻰـﻠﻋ ﺩﻟﻭﻤ
.
ﺔـﻋﻭﻤﺠﻤﻟﺍ ﻥـﻤ
{
}
k
x
x
,
...
,
1
ـ ﺍﻟﻤﻭﻟﺩﺓ ﻟ
M
ﻤﻊ
k
ﻋﻨﺼﺭ ﻴﻭﺠﺩ ﻋﻨﺼﺭ ﻭﺍﺤﺩ
1
x
ﻲـﻫ ﻪﺘﺒﺘﺭ ﻥﻭﻜﺘ ﺙﻴﺤﺒ
ﺍﻟﺭﺘﺒﺔ ﺍﻷﺼﻐﺭ
.
ﻨﺒﻴﻥ ﻻﺤﻘﺎﹰ ﺒﺄﻥ
M
ـ ﻫﻲ ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ ﻟ
1
x
ﻭ
k
x
x
,
,...
2
.
ﻭﻫﺫﺍ
ﻭﺫﻟ،ﻥﺎﻫﺭﺒﻟﺍ لﻤﻜﻴ
ﺎﻨﻲ ﻭـﺜﻟﺍ لﻭﺩﻭﻤﻟﺍ ﺩﻋﺍﻭﻗ ﻥﺃ ﻥﻤﻭ ﻲﺌﺍﺭﻘﺘﺴﻻﺍ ﺽﺭﻔﻟﺍ ﻥﻤ ﻙ
1
x
ﺸﻜﻼﻥـﻴ
ـﻗﺎﻋﺩﺓ ﻟ
M
.
ﺇﺫﺍ ﻟﻡ ﺘﻜﻥ
M
ـ ﻤﺠﻤﻭﻋﺎﹰ ﻤﺒﺎﺸﺭﺍﹰ ﻟ
1
x
ﻭ
k
x
x
,
,...
2
ﺘﻭﺠﺩ ﺍﻟﻌﻼﻗﺔﺫﺌﺩﻨﻋ ،
( )
9
1 1
2
2
...
0,
+
+
+
=
k
k
m x
m x
m x
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١٣
ﺤﻴﺙ
0
1
1
≠
x
m
.
ﻴﻜﻥـﻟ
(
)
0
,
,...
gcd
1
>
=
k
m
m
d
ﻴﻜﻥـﻟﻭ ،
=
i
i
c
m
d
.
ﻥـﻤ
ﻭ،ﺔﻴﺩﻴﻬﻤﺘﻟﺍ
1
, ...,
=
k
M
y
y
ﻭ
k
k
x
c
x
c
y
+
+
=
...
1
1
.
ﻟﻜﻥ
0
...
2
2
1
1
1
=
+
+
+
=
k
k
x
m
x
m
x
m
dy
ﻭ
( )
1
1
x
o
m
d
<
≤
ﻭﻫﺫﺍ ﺘﻨﺎﻗﺽ،
.
ﻨﺘﻴﺠﺔ
54.1
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
ﻭ ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ
n
ﻋﻨﺼﺭ ﻋﻠﻰ ﺍﻷﻜﺜﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
،ﻤﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻟﺩﻴﻨﺎ
r
n
n
C
C
G
×
×
≈
...
1
ﻤﻥ ﺃﺠل
N
n
i
∈
.
ﻭﻤﻥ ﺍﻟﻔﺭﺽ ﻴﻨﺘﺞ
ﺃﻥ
i
n
ﺃﻭﻟﻴﺔ ﻨﺴﺒﻴﺎﹰ
.
ﻟﻴﻜﻥ
i
a
لـﻤﺎﻌﻟﺍ ﺩﻟﻭﻴ ﹰﺍﺭﺼﻨﻋ
i
.
ﺫـﺌﺩﻨﻋ
(
)
r
a
a
,
,...
1
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
r
n
n ...
1
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺘﻭﻟﺩ،
G
.
ﻤﻼﺤﻅﺔ
55.1
ﻟﻴﻜﻥ
F
ﺤﻘل ﻤﺎ
.
ﺇﻥ ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻓﻲ
X
F
ﺭﺓـﻴﺜﻜﻟ ﺭﻭﺫـﺠ ﻲﻫ
ﺍﻟﺤﺩﻭﺩ
1
−
n
X
.
ﻷﻥ ﺍﻟﺘﺤﻠﻴل ﻭﺤﻴﺩ ﻓﻲ
X
F
ﻴﻭﺠﺩ ﻤﻨﻬﺎ،
n
ﺠﺫﺭ
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،ﺭﺜﻜﻷﺍ ﻰﻠﻋ
ﺍﻟﻨﺘﻴﺠﺔ ﺘﺒﻴﻥ ﺒﺄﻥ ﺃﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻨﺘﻬﻴﺔ ﻤﻥ
X
F
ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ
.
ﻤﺒﺭﻫﻨﺔ
56.1
ﻴﻤﻜﻥ ﺃﻥ
ﺘﻤﺜل ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻏﻴﺭ ﺍﻟﺘﺎﻓﻬﺔ
M
ﺒﺎﻟﺸﻜل
( )
10
1
...
∞
≈
× ×
×
s
r
n
n
M
C
C
C
ﻤﻥ ﺃﺠل ﺃﻋﺩﺍﺩ ﺼﺤﻴﺤﺔ ﻤﻌﻴﻨﺔ
2
,
,...
1
≥
s
n
n
ﻭ
0
≥
r
.
ﻭﺒﺎﻹﻀﺎﻓﺔ ﻟﺫﻟﻙ ﻓﺈﻥ
(a)
r
ﻴﻌﺭﻑ
M
ﺒﺸﻜل ﻭﺤﻴﺩ
(b)
ﻴﻤﻜﻥ ﺍﺨﺘﻴﺎﺭ
i
n
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
2
1
≥
n
ﻭ
s
s
n
n
n
n
1
2
1
,
,...
−
ﻓﻬﻲ ﺘﻌﺭﻑﺫﺌﺩﻨﻋﻭ ،
M
ﺒﺸﻜل ﻭﺤﻴﺩ
.
(c)
ﻴﻤﻜﻥ ﺍﺨﺘﻴﺎﺭ
i
n
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺘﺤﺩﺩ،ﺔﻴﻟﻭﺃ ﺩﺍﺩﻋﻷ ﻯﻭﻗ ﻥﻭﻜﺘ ﺙﻴﺤﺒ
M
ﺒﺸﻜل ﻭﺤﻴﺩ
.
ﻴﺩﻋﻰ ﺍﻟﻌﺩﺩ
r
ﻤﺭﺘﺒ
ﺔ
(rank)
M
.
ﺘﺘﻌﺭﻑ
M
ﺒﻁﺭﻴﻘﺔ ﻭﺤﻴﺩﺓ ﺒﻭﺍﺴﻁﺔ
r
ﻨﻌﻨﻲ ﺒﺫﻟﻙ،
ـﺃﻥ ﻜل ﺘﺤﻠﻴﻠﻴﻥ ﻟ
M
ﻤﻥ ﺍﻟﺸﻜل
(10)
ﺴﻴﻜﻭﻥ ﻋﺩﺩ ﺍﻟﺯﻤﺭ ﻤﻥ،
∞
C
ﻨﻔﺴﻪ
)
ﻰـﻟﺇ ﺔﺒﺴﻨﻟﺎﺒﻭ
i
n
ﻓﻲ
(b)
ﻭ
(c)
ﻨﺒﺭﻫﻥ ﺒﺸﻜل ﻤﺸﺎﺒﻪ
.(
ﺼﺤﻴﺤﺔـﻟﺍ ﺩﺍﺩـﻋﻷﺍ ﻰﻤﺴﺘ
s
n
n
,
,...
1
ﻲـﻓ
(b)
ـﺍﻟﻘﻭﺍﺴﻡ ﺍﻟﻼﻤﺘﻐﻴﺭﺓ ﻟ
M
.
ﺘﻌﻨﻲ
(c)
ﺃﻥ
M
ﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺜل
( )
11
1
1
...
,
1
∞
≈
× ×
×
≥
e
et
t
r
i
p
p
M
C
C
C
e
ﻤﻥ ﺃﺠل ﻗﻭﻯ ﻷﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ ﻤﻌﻴﻨﺔ
i
e
i
p
)
ﻴﻤﻜﻥ ﺘﻜﺭﺍﺭ ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ
(
ﺇﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ،
t
e
t
e
p
p
,
,...
1
1
ﺘﻌﺭﻑ
M
ـ ﻭﺘﺴﻤﻰ ﺒﺎﻟﻘﻭﺍﺴﻡ ﺍﻻﺒﺘﺩﺍﺌﻴﺔ ﻟ،ﺩﻴﺤﻭ لﻜﺸﺒ
M
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺍﻹﺜﺒﺎﺕ ﺍﻷﻭل ﻫﻭ ﻤﻌ
ﻨﻰ ﺁﺨﺭ ﻟﻠﻤﺒﺭﻫﻨﺔ
53.1
.
)
a
(
ﻟﻜل ﻋﺩﺩ ﺃﻭﻟﻲ
p
ﻻ ﻴﻘﺴﻡ ﺃﻱ ﻤﻥ
i
d
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٢٣
(
)
(
)
r
r
p
pC
C
pM
M
Ζ
Ζ
≈
≈
∞
∞
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
r
ﻫﻭ ﺍﻟﺒﻌﺩ ﻟﻠﺯﻤﺭﺓ
pM
M
ﺒﺎﻋﺘﺒﺎﺭﻩ
p
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
.
(b,c)
ﺇ
ﺫﺍ ﻜﺎﻥ
( )
1
,
gcd
=
n
m
ﻋﻨﺩﺌﺫ،
n
m
C
C
×
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺭﺼﻨﻋ ﻰﻠﻋ ﻱﻭﺘﺤﺘ
mn
ﻭﻟﺫﻟﻙ،
( )
12
×
≈
m
n
mn
C
C
C
ﺒﺎﺴﺘﺨﺩﺍﻡ
(12)
ﻟﺘﺤﻠﻴل
i
n
C
ﺇﻟﻰ ﺠﺩﺍﺀ ﺯ
ﻤﺭ ﺩﺍﺌﺭﻴﺔ ﺭﺘﺒﺔ ﻜل ﻤﻨﻬﺎ ﻗﻭﻯ ﻷﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ
.
ﻭﻫﺫﺍ ﻗﺩ
،ﺘﺤﻘﻕ
(12)
ﻲـﻓ لـﻴﻠﺤﺘﻟﺎﻜ لـﻴﻠﺤﺘ لﻴﻜﺸـﺘﻟ لﻤﺍﻭﻋ ﻊﻴﻤﺠﺘﻟ ﻡﺩﺨﺘﺴﺘ ﻥﺃ ﻥﻜﻤﻴ
(b)
،ﺜﻼﹰـﻤ ،
ei
s
i
n
p
C
C
=
∏
ﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﺠﺩﺍﺀ ﻤﻥ ﺃﺠل ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ ﻤﺨﺘﻠﻔﺔ ﻤﻥ،
i
p
ﻭ
i
e
ﻭﺓـﻗ ﺭﺒﻜﺃ
ﻟﻠ
ﻌﺩﺩ ﺍﻷﻭﻟﻲ
i
p
.
ﻟﺒﺭﻫﺎﻥ ﺍﻟﻭﺤﺩﺍﻨﻴﺔ ﻓﻲ
(b)
ﻭ
(c)
ﻴﻤﻜﻥ ﺃﻥ ﻨﺴﺘﺒﺩل،
M
ﺔـﻴﺌﺯﺠﻟﺍ لـﺘﻔﻟﺍ ﺓﺭﻤﺯﺒ
)
ﺫﻟﻙـﻟﻭ
ﻨﻔﺭﺽ ﺃﻥ
0
=
r
.(
ﺇﻥ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ
p
ﺴﻴﻜﻭﻥ ﺃﻱ ﻋﺩﺩ ﺃﻭﻟﻲ
i
p
ﻓﻲ
(11)
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ
ﻜﺎﻨﺕ
M
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ،
p
ﻤﻜﺭﺭ
a
ﺭﺓـﻤ
ﺤﻴﺙ
a
p
ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ ﺍﻟﺘﻲ ﺘﻘﺴﻡ
p
.
،ﻭﺒﺸﻜل ﻤﺸﺎﺒﻪ
2
p
ﻥـﻤ ﹰﺎﻀـﻌﺒ ﻡﺴﻘﻴﺴ
ﺍﻟﻌﻨﺎﺼﺭ
i
e
i
p
ﻓﻲ
(11)
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﺤﻭﺕ
M
ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
ﻓﻲ ﺘﻠﻙ ﺍﻟﺤﺎﻟﺔ،
ﺴﻭﻑ ﻴﻘﺴﻡ ﺘﻤﺎﻤﺎﹰ
b
ﺍﻷﻋﺩﺍﺩ
i
e
i
p
ﺤﻴﺙ،
b
b
a
p
p
2
−
ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ
M
ﻤﻥ ﺍﻟﺭﺘﺒﺔ ﺍﻟﺘﻲ
ﺘﻘﺴﻡ
2
p
.
ـ ﻨﺠﺩ ﺒﺄﻥ ﺍﻟﻘﻭﺍﺴﻡ ﺍﻷﻭﻟﻴﺔ ﻟ،ﺔﻘﻴﺭﻁﻟﺍ ﻩﺫﻬﺒ ﺔﻌﺒﺎﺘﻤﻟﺎﺒ
M
ﺩﺩـﻋ ﻥﻤ ﻑﺭﻌﺘ ﻥﺃ ﻥﻜﻤﻴ
ﻋﻨﺎﺼﺭ
M
ﺍﻟﺘﻲ
ﺭﺘﺒﺔ ﻜل ﻤﻨﻬﺎ ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
.
ﺃﻭ ﻴﻤﻜﻥ ﺃﻥ ﺘﺒﺭﻫﻥ،ﺔﻴﻟﻭﻷﺍ ﻡﺴﺍﻭﻘﻟﺍ ﻥﻤ ﺓﺭﻴﻐﺘﻤﻼﻟﺍ لﻤﺍﻭﻌﻟﺍ ﺔﻴﻨﺍﺩﺤﻭ ﻰﻟﺇ لﺼﻭﺘﻨ ﻥﺃ ﻥﻜﻤﻴ
ﻤﺒﺎﺸﺭﺓﹰ
:
ﻟﻴﻜﻥ
s
n
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ ،ﺭﻔﺼﻟﺍ ﻥﻤ ﺭﺒﻜﺃ ﺢﻴﺤﺼ ﺩﺩﻋ ﺭﻐﺼﺃ
0
=
M
n
s
،
1
−
s
n
ﺃﺼﻐﺭ ﻋﺩﺩ ﺼﺤﻴﺢ ﺃ
ﻜﺒﺭ ﻤﻥ ﺍﻟﺼﻔﺭ ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺘﻜﻭﻥ
0
1
=
−
M
n
s
، ﺩﺍﺌﺭﻴﺔ
2
−
s
n
ﺃﺼﻐﺭ
ﻋﺩﺩ ﺼﺤﻴﺢ ﺒ
ﺤﻴﺙ ﺇﻥ
2
−
s
n
ﻭﻫﻜﺫﺍ،ﻥﻴﺘﻴﺭﺌﺍﺩ ﻥﻴﺘﺭﻤﺯﻟ ﺀﺍﺩﺠ لﻜﺸ ﻰﻠﻋ لﺜﻤﻴ ﻥﺃ ﻥﻜﻤﻴ
.
ﺘﺒﺭﻫﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺍﻟﺘﻲ ﺘﻜﻤل ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
:
ﻜل ﺯﻤﺭﺓ ﻜﻬﺫﻩ
ﺘﻜﻭﻥ ﻤﺘﻤﺎﺜﻠﺔ
ﺘﻤﺎﻤﺎﹰ ﻤﻊ ﺇﺤﺩﻯ ﺍﻟﺯﻤﺭ
ﺍﻵﺘﻴ
ﺔ
r
r
n
n
n
n
n
n
C
C
r
1
2
1
,
,...
,
...
1
−
×
×
ﺭﺘﺒﺔ ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ ﺘﺴﺎﻭﻱ
r
n
n
.
...
1
.
ﻜل ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،ﹰﻼﺜﻤ
90
ﺘﻤﺎﺜل ﺘﻤﺎﻤﺎﹰ ﻭﺍﺤﺩﺓ
ﻤﻥ ﺍﻟﺯﻤﺭ
ﺍﻵﺘﻴ
ﺔ
90
C
ﺃﻭ
30
3
C
C
×
-
ﻨﻼﺤﻅ ﺃ،ﻙﻟﺫ ﺢﻴﻀﻭﺘﻟ
ﺏـﺠﻴ ﺭﻴﻐﺘﻤﻻ لﻤﺎﻋ ﺭﺒﻜﺃ ﻥ
ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﻌﺎﻤل ﻤﻥ ﺍﻟﺭﺘﺒﺔ
90
ﻭ ﺍﻟﻘﺎﺒل ﻟﻠﻘﺴﻤﺔ ﻋﻠﻰ ﻜل ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﻷﺼﻐﺭ ﻤﻥ
90
.
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ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺨﻁﻴﺔ ﻟﻠﺯﻤﺭ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ
ﻟﻴﻜﻥ
( )
{
}
1
,
=
∈
=
z
C
z
C
µ
.
ﺔـﻴﻬﺘﻨﻤ ﺭـﻴﻏ ﺓﺭﻤﺯ ﻩﺫﻫ
.
ﺤﻴﺢـﺼ ﺩﺩـﻋ ﻱﻷ
n
،
ﺍﻟﻤﺠﻤﻭﻋﺔ
( )
C
n
µ
ـ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ ﺍﻟﻘﺎﺴﻤﺔ ﻟ
n
ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻲـﻓ ،
ﺍﻟﺤﻘﻴﻘﺔ
( )
{
} {
}
1
/
2
,
...
,
,
1
1
0
,
−
=
−
≤
≤
=
n
n
m
i
n
n
m
e
C
ξ
ξ
µ
π
ﺤﻴﺙ
n
i
n
e
/
2
π
ξ
=
ﺍﻟﺠﺫﺭ ﺍﻟﻨﻭﻨﻲ ﺍﻻﺒﺘﺩﺍﺌﻲ ﻟﻠﻭﺍﺤﺩ
.
ﺇﻥ ﺍﻟﻤﻴﺯﺓ ﺍﻟﺨﻁﻴﺔ
)
ﻁـﻘﻓ ﺓﺯﻴﻤﻟﺍ ﻭﺃ
(
ﺭﺓـﻤﺯﻠﻟ
G
ﺸﺎﻜلـﺘﻟﺍ ﻭـﻫ
( )
C
G
µ
→
.
ﺇﻥ
ﺍﻟﻤﺠﻤﻭﻋﺔ
∨
G
ﻤﻊ ﻫﺫﻩ ﺍﻟﻤﻴﺯﺍﺕ ﺘﺼﺒﺢ ﺯﻤﺭﺓ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻌﻤﻠﻴﺔ ﺍﻟﺠﻤﻊ
(
)( )
( ) ( )
χ χ
χ
χ
′
′
+
=
g
g
g
ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺜﻨﻭﻴﺔ
.
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ،ﹰﻼﺜﻤ
( )
1
χ
χ
a
ﺘﻤﺎﺜل ﻤﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺜﻨﻭﻴ
ﺔ
∨
Z
ﺇﻟﻰ
( )
C
µ
.
ﻨﺴﻤﻲ ﺍﻟﺘﺸﺎﻜل
1
a
a
ﺒﺎﻟﻤﻴﺯﺓ ﺍﻟﺘﺎﻓﻬﺔ
)
ﺃﻭ ﺍﻷﺴﺎﺴﻴﺔ
.(
ﺍﻟﺒﺎﻗﻲ ﺍﻟﺘﺭﺒﻴﻌﻲ ﺒﺎﻟﻤﻭﺩﻭل
p
ﻟﻸﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ
a
ﺍﻟﺘﻲ ﻻ ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
ﺔـﻓﺭﻌﻤ
ﺒﺎﻟﺸﻜل
1
1
Ζ Ζ
=
−
p
a
p
ﺇﻥ ﻫﺫﺍ ﻴﻌﺘﻤﺩ ﻓﻘﻁ ﻋﻠﻰ،ﺢﻀﺍﻭﻟﺍ ﻥﻤ
a
ﺒﺎﻟﻤﻭﺩﻭل
p
ﻥـﻤ لـﻜ ﻥﺎﻜ ﺍﺫﺇﻭ ،
b
a,
ﺒﻼﻥـﻘﻴ ﻻ
ﻰــــﻠﻋ ﺔﻤﺴــــﻘﻟﺍ
p
ﺫــــﺌﺩﻨﻋ ،
=
P
b
p
a
p
ab
ﺈﻥــــﻓ ﻙﻟﺫــــﻟ ﻭ ،
[ ]
(
)
{ }
( )
2
:
1
χ
µ
Ζ Ζ → ± =
a
a
a
p
C
b
ﻫﻲ ﻤﻴﺯﺓ
(
)
χ
Ζ Ζ
p
.
ﻟﻨﺭﻤﺯ ﺍﻵﻥ ﻷﻱ ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﺒﺎﻟﺭﻤﺯ
( )
n
µ
)
ﻟﻴﺱ ﺒﺎﻟﻀﺭﻭﺭﺓ
( )
C
n
µ
(
ﻭ،
ﻟﻠﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺫﺍﺕ ﺍﻟﻘﻭﺓ
(exponent)
n
ﺒﺎﻟﺭﻤﺯ،
(
)
Hom
,
µ
∨
=
n
G
G
.
ﻤﺒﺭﻫﻨﺔ
57.1
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺒﺩﺒﻠﻴﺔ ﻤﻨﺘﻬﻴﺔ ﺒﻘﻭﺓ
n
.
(a)
ﺇﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺜﻨﻭﻴﺔ
∨
G
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ
G
.
(b)
ﺍﻟﺘﻁﺒﻴﻕ
∨
∨
→
G
G
ﺍﻟﺫﻱ ﺘﺭﺴل ﺃﻱ ﻋﻨﺼﺭ
a
ﻤﻥ
G
ﺯﺓـﻴﻤﻟﺍ ﻰﻟﺇ
( )
χ
χ
a
a
ﻓﻲ
∨
G
ﻴﻜﻭﻥ ﺘﻤﺎﺜﻼﹰ
.
،ﻭﺒﻌﺒﺎﺭﺓ ﺃﺨﺭﻯ
∨
≈
G
G
ﻭ
∨∨
G
G
.
a
ﻤﺭﺒﻊ ﻓﻲ
ﻏﻴﺭ ﺫﻟﻙ
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ﺍﻟﺒﺭﻫﺎﻥ
.
ﺍﻟﺒﺭﻫﺎﻥ ﻭﺍﻀﺢ ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭ ﺍﻟﺩ
ﻭ،ﺔﻴﺭﺌﺍ
(
)
∨
∨
∨
×
≅
×
H
G
H
G
.
58.1
ﻨﻌﻨﻲ ﺒﺄﻥ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻁﺒﻴﻌﻲ
∨
∨
→
G
G
ﺔـﻨﻫﺭﺒﻤ ﻥـﻤ ﺔـﺼﺎﺨ ﺔﻟﺎﺤ ﻥﻭﻜﻴﻭ لﺜﺎﻤﺘ
)
ﺒﻭﻨﺘﺭﻴﺎﻏﻴﻥ
.(
ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﺘﻌﺘﺒﺭ ﺍﻟﺯﻤﺭ ﻓﻲ ﺍﻟﻁﺒﻭﻟﻭﺠﻴﺎ،ﺔﻟﺎﺤﻟﺍ ﻡﻴﻤﻌﺘﺒﻭ
.
ﺎـﻤﻜ ،ﹰﻼﺜـﻤ
،ﺎﺒﻘﺎﹰــــﺴ ﺎــــﻨﻅﺤﻻ
( )
µ
∨
Ζ
C
لــــﻜ ،
Ζ
∈
m
ﺯﺓــــﻴﻤ ﻑﺭﻌﺘــــﺴ
( )
( )
C
C
m
µ
µ
ξ
ξ
→
:
a
ﻴﻭﺠﺩ ﺍﻟﻌﺩﻴﺩ ﻤﻥ ﺍﻟﺘﺸﺎﻜﻼﺕ،
( )
( )
C
C
µ
µ
→
ﻟﻴﺴﺕ ﻤﻥ
ﻟﻜﻥ ﻫﺫﻩ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﺴﺘﻤﺭﺓ ﻓﻘﻁ،لﻜﺸﻟﺍ ﺍﺫﻫ
.
ﻟﺘﻜﻥ
G
ﺔـﻴﺠﻭﻟﻭﺒﻁﻟﺍ ﻲﻓ ﺓﺎﻁﻌﻤ ﺔﻴﻠﻴﺩﺒﺘ ﺓﺭﻤﺯ
، ﺍﻟﻤﺘﺭﺍﺼﺔ ﺍﻟﻤﺤﻠﻴﺔ ﻷﺠل ﻋﻤﻠﻴﺎﺕ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﺴﺘﻤﺭﺓ
ﺭﺓـﻤﺯﻟﺍ ﺫـﺌﺩﻨﻋ
∨
G
ﻭﺍﺹـﺨﻟﺍ ﻊـﻤ
ﺍﻟﻤﺴﺘﻤﺭﺓ
( )
C
G
µ
→
ﺎﻟﻲـﺘﻟﺎﺒﻭ ،ﹰﺎـﻴﻠﻜ ﺔـﺼﺍﺭﺘﻤ ﻥﻭﻜﺘﻟ ﺔﻴﻌﻴﺒﻁﻟﺍ ﺔﻴﺠﻭﻟﻭﺒﻁﻟﺍ ﺓﺯﻴﻤﻟﺍ ﺎﻬﻟ
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻁﺒﻴﻌﻲ
∨
∨
→
G
G
ﻴﻜﻭﻥ ﺘﻤﺎﺜﻼﹰ ﺤﺴﺏ ﻤﺒﺭﻫﻨﺎﺕ
)
ﺒﻭﻨﺘﺭﻴﺎﻏﻴﻥ
(
ﺍﻟﺜﻨﻭﻴﺔ
.
ﻤﺒﺭﻫﻨﺔ
59.1
)
ﻋﻼﻗﺎﺕ ﺍﻟﺘﻌﺎﻤ
ﺩ
(
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻨﺘﻬﻴﺔ
.
ﻷﻱ ﻤﻴﺯﺘﻴﻥ
χ
ﻭ
ψ
ﻲـﻓ
G
،
( )
( )
1
;
0
;
χ ψ
χ
ψ
χ ψ
−
∈
=
=
≠
∑
a G
G
a
a
ﻭﺒﺤﺎﻟﺔ ﺨﺎﺼﺔ
( )
0
χ
∈
=
∑
a G
G
a
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
ψ
χ
=
ﻋﻨﺩﺌﺫ،
( )
( )
1
1
χ
ψ
−
=
a
a
ﺴﺎﻭﻱـﻴ ﻉﻭـﻤﺠﻤﻟﺍ ﻥﺈﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ ،
G
ﻓﻴﻤﺎ ﻋﺩﺍ ﺫﻟﻙ ﻴﻭﺠﺩ،
G
b
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
( )
( )
b
b
ψ
χ
≠
.
ﺤﻴﺙ
a
لـﻜ ﺢﺴـﻤﺘ
G
ﻭﻜﺫﻟﻙ،
ab
ﺃﻴﻀﺎﹰ
،
ﻭﻤﻨﻪ
( )
( )
( ) ( )
( )
( )
( )
( )
( )
1
1
1
1
a G
a G
a G
a
a
ab
ab
b
b
a
a
χ
ψ
χ
ψ
χ
ψ
χ
ψ
−
−
∈
∈
−
−
∈
=
=
∑
∑
∑
ﻷﻥ
( )
( )
1
1
≠
−
b
b
ψ
χ
ﻭﺒﺎﻟﺘﺎﻟﻲ،
( )
( )
1
0
χ
ψ
−
∈
=
∑
a G
a
a
.
ﻨﺘﻴﺠﺔ
60.1
ﻷﻱ
G
a
∈
ﻴﻜﻭﻥ
ﺇﺫﺍ ﻜﺎﻨﺕχ ﺔﻬﻓﺎﺘ
ﻓﻲ ﺍﻟﺤﺎﻻﺕ ﺍﻷﺨﺭﻯ
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٥٣
( )
0
χ
∨
∈
=
∑
a G
G
a
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﺘﻁﺒﻴﻕ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻋﻠﻰ
∨
G
ﻨﻼﺤﻅ ﺃﻥ،
( )
∨
∨
G
G
.
ﺘﻤﺎﺭﻴﻥ
ﺇﻥ
ﻫﺫﻩ ﺍﻟﺘﻤﺎﺭﻴﻥ ﻤﻭﻀﺤﺔ ﺒﺎﻟﺸﺭﺡ ﻜﻤﺎ ﻫﻭ ﻤﻁﻠﻭﺏ ﻟﻴﺅﺨﺫ ﺒﻬﺎ
.
1-1
ﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﺭﺘﺒﺔﻴﺒ
2
ﻊـﻴﻤﺠ ﻊـﻤ لﺩﺎـﺒﺘﻴ ،
ﻋﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ
Q
ﺍﺴﺘﻨﺘﺞ ﺃﻥ،
Q
ﺘﻤﺎﺜل
4
D
ﻲـﻓ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ لﻜ ﻥﺃ ﻭ ،
Q
ﻭﻥـﻜﺘ
ﻨﺎﻅﻤﻴﺔ
.
2-1
ﻨﺄﺨﺫ ﺍﻟﻌﻨﺎﺼﺭ
−
−
=
−
=
1
1
1
0
0
1
1
0
b
a
ﻓﻲ
( )
2
GL
Z
.
ﻥ ﺃﻥﻴﺒ
1
1
3
=
=
4
a
و
b
ﻟﻜﻥ ﺭﺘﺒﺔ ﺍﻟﻌﻨﺼﺭ،
ab
ﻭﺒﺎﻟﺘﺎﻟﻲ،ﺔﻴﻬﺘﻨﻤ ﺭﻴﻏ
ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ
b
a,
ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.
3-1
ﻥ ﺃﻥ ﻜل ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺭﺘﺒﺘﻬﺎ ﻋﺩﺩ ﺯﻭﺠﻲ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﺭﺘﺒﺘﻪ ﺘﺴﺎﻭﻱﻴﺒ
2
.
4-1
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺩﻟﻴﻠﻬﺎ ﻴﺴﺎﻭﻱ
n
.
ﺎﻥـﻜ ﺍﺫﺇ ﻪﻨﺃ ﻥﻴﺒ
G
g
∈
،
ﻓﺈﻥﺫﺌﺩﻨﻋ
N
g
n
∈
.
ﻥـﻜﺘ ﻡﻟ ﺍﺫﺇ ﹰﺎﺌﻁﺎﺨ ﺍﺫﻫ ﻥﻭﻜﻴ ﻥﺃ ﻥﻜﻤﻤﻟﺍ ﻥﻤ ﻥﺃ ﻪﻴﻓ ﻥﻴﺒﺘ ﹰﻻﺎﺜﻤ ﻲﻁﻋﺃ
N
ﻨﺎﻅﻤﻴﺔ
.
5-1
ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﺫﺍﺕ ﻗﻭﺓ ﻤﻨﺘﻬﻴﺔ ﺇﺫﺍ ﻭﺠﺩ
0
>
m
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ
e
a
m
=
لـﻜﻟ
G
a
∈
ﻭﻫﻭ ﺃﺼﻐﺭ ﻋﺩﺩ ﻴﺤﻘﻕ،
ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺴﻨﺩﻋﻭﻩ ﻓﻴﻤﺎ ﺒﻌﺩ ﻗﻭﺓ
G
.
ﻥ ﺃﻥ ﻜل ﺯﻤﺭﺓﻴﺒ
ﻗﻭﺘﻬﺎ ﻴﺴﺎﻭﻱ
2
ﺘﻜﻭﻥ ﺘﺒﺩﻴﻠﻴﺔ
.
6-1
ﻴﻘﺎل ﻋﻥ ﺍﻟﺯﻤﺭﺘﻴﻥ
H
ﻭ
H
′
ﺃﻨﻬﻤﺎ ﻗﺎﺒﻠﺘﺎﻥ ﻟﻠﻘﻴﺎﺱ ﺇﺫﺍ ﻜﺎﻥ ﺩﻟﻴل
H
H
′
I
ﻤﻨﺘﻪ ﻓﻲ
ﻜل ﻤﻥ ﺍﻟﺯﻤﺭ
ﺘﻴﻥ
H
ﻭ
H
′
.
ﺔـﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻟﺍ ﻰﻠﻋ ﺅﻓﺎﻜﺘ ﺔﻗﻼﻋ لﻜﺸﺘ ﺱﺎﻴﻘﻟﺍ ﺔﻴﻠﺒﺎﻗ ﻥﺃ ﻥﻴﺒ
ﻓﻲ
G
.
ﺇﺫﺍ ﻜﺎﻥ
=
a
e
ﻓﻲ ﺍﻟﺤﺎﻻﺕ ﺍﻷﺨﺭﻯ
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٦٣
ﺍﻟﻔﺼل ﺍﻟﺜﺎﻨﻲ
ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ،ﺕﺎﻤﻴﺩﻘﺘﻟﺍﻭ ﺓﺭﺤﻟﺍ ﺭﻤﺯﻟﺍ
Free Groups and Presentations Coxeter groups
ﻤﻥ ﺍﻟﻤﻔﻴﺩ
ﺩﺍﺌﻤﺎﹰ ﻟ
ﻭ
ﺼﻑ ﺯﻤﺭﺓ
ﺇﻋﻁﺎﺀ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﻭﻟﺩﺍﺕ ﻭﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﺭﻭﺍﺒﻁ ﻟﻠﻤﻭﻟﺩﺍﺕ
ﺍﻟﺘﻲ ﺘﺴﺘﻨﺘﺞ ﻤﻨﻬﺎ
ﻋﻼﻗﺎﺕ
ﺃﺨﺭﻯ
.
ﺭﺓـﻤﺯﻟﺍ ﻑـﺼﻭ ﻥـﻜﻤﻴ ،ﹰﻼﺜﻤ
n
D
ﺩﺍﺕـﻟﻭﻤﻟﺎﺒ
s
,
r
ﻭ
ﺍﻟﻌﻼﻗﺎﺕ
e
srsr
,
e
s
,
e
r
2
n
=
=
=
ﻨﻌﻁﻲ ﺍﻟﻘﻴﻤﺔ ﻟﻬﺫﺍ ﺍﻟﻤﻌﻨﻰ،لﺼﻔﻟﺍ ﺍﺫﻫ ﻲﻓ
.
ﺃﻭﻻﹰ ﻨﺤ
ﺔـﻋﻭﻤﺠﻤ ﻰﻠﻋ ﺓﺭﺤﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻑﻴﺭﻌﺘﻟ ﺝﺎﺘ
X
ﻤﻥ ﺍﻟﻤﻭﻟﺩﺍﺕ
ـ
ـﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒ
X
ﻭﻥـﻜﺘ ﻲـﺘﻟﺍ ﻙـﻠﺘ ﺀﺎﻨﺜﺘـﺴﺎﺒ ﺕﺎﻗﻼﻋ ﻭﻥﺩﺒ
ﻤﺴﺘﻨﺘﺠﺔ ﻤﻥ ﻤﺴﻠﻤﺎﺕ ﺍﻟﺯﻤﺭﺓ
.
ﻷﻥ
ﺍﻟﻨﻅﺎﺌﺭ
ﻗﺩ
ﺘﺴﺒﺏ
ﺍﻟﺘﺒﺎﺴﺎﺕ
ﻨﻘﻭﻡ ﺃﻭﻻﹰ ﺒـﺴ ،
ﺎﻟ
ﻰـﻠﻋ ﺔـﺴﺍﺭﺩ
ﺃﻨﺼﺎﻑ ﺍﻟﺯﻤﺭ
٨
،
ﻭ
ﺍﻟﺘﻲ ﻨﻌ
ﻨﻲ ﺒﻬﺎ ﺍﻟﻤﺠﻤﻭﻋﺔ
S
ﺼﺭﻫﺎـﻨﻋﻭ ﺎـﻬﻴﻠﻋ ﺔﻓﺭﻌﻤ ﺔﻴﺌﺎﻨﺜ ﺔﻴﻠﻤﻋ ﻊﻤ
ﺩــﻴﺎﺤﻤﻟﺍ
e
.
ﺸﺎﻜلــﺘﻟﺍ ﻥﺇ
S
S
′
→
:
α
ﻕــﻘﺤﻴ ﻕــﻴﺒﻁﺘ ﻭــﻫ ﺭــﻤﺯﻟﺍ ﻑﺎﺼــﻨﺃ ﻥــﻤ
( )
( ) ( )
b
a
ab
α
α
α
=
ﻟﻜل
S
b
a
∈
,
ﻭ
( )
e
e
=
α
.
ـﻋﻨﺩﺌ
ﺫ
α
لـﻜ ﻰـﻠﻋ ﻅﻓﺎـﺤﻴ
ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
ﺃﻨﺼﺎﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ
)
(Free semigroups
ﻟﺘﻜﻥ
{
}
...
,
,
,
c
b
a
X
=
)
ﻤﻥ ﺍﻟﻤﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﻏﻴﺭ
ﻤﻨﺘﻬﻴﺔ
(
ﺔـﻤﻠﻜﻟﺍ ،ﺯﻭﻤﺭﻟﺍ ﻥﻤ ﺔﻋﻭﻤﺠﻤ
(word)
ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﻤﻨﺘﻬﻴﺔ ﻤﻥ ﺭﻤﻭﺯ
X
ﻭﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺍﻟﺘﻜﺭﺍﺭ
ﻓﻴﻬﺎ
.
،ﻤﺜﻼﹰ
ﺇﻥ
b
aabac
aa
,
,
ﻜﻠﻤﺎﺕ ﻤﺨﺘﻠﻔﺔ
.
، ﻤﺜﻼﹰ،ﺎﻤﻬﻨﻴﺒ ﺞﻤﺩﻟﺎﺒ ﻙﻟﺫﻭ ﻥﻴﺘﻤﻠﻜ ﺏﺭﻀ ﻥﻜﻤﻴ
aaaaaabac
aabac
aaaa
=
∗
ﻫﺫﺍ ﻴﻌﺭﻑ ﻋﻠﻰ ﻜل ﻤﺠﻤﻭﻋﺔ ﺍﻟﻜﻠﻤﺎﺕ ﻋﻤﻠﻴﺔ ﺜﻨﺎﺌﻴﺔ ﺘﺠﻤﻴﻌﻴﺔ
.
ﻨﺭﻤﺯ ﻟﻬﺎ،ﺔﻴﻟﺎﺨ ﺔﻴﻟﺎﺘﺘﻤ ﺩﺠﻭﺘ ﺎﻤﻜ
ﺒﺎﻟﺭﻤﺯ
1
) .
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴ
ﻜﻭﻥ ﻓﻴﻬﺎ ﺍﻟﺭﻤﺯ
1
ﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻑـﻠﺘﺨﻤ ﺯﻤﺭﺒ ﻪﻟ ﺯﻤﺭﻨ ،
.(
8
ﻭﻴﺴﻤﻰ ﺃﻴﻀﺎﹰ ﻤﻭﻨﻭﺌﻴﺩ
.
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٧٣
ﻓﺈﻥ ﺍﻟﺭﻤﺯﺫﺌﺩﻨﻋ
1
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎﻴﺩ
.
ﻨﻜﺘﺏ
SX
ﺔـﻴﻠﻤﻌﻟﺍ ﻩﺫﻫ ﻊﻤ ﺕﺎﻤﻠﻜﻟﺍ ﻥﻤ ﺔﻋﻭﻤﺠﻤﻟ
ﺍﻟﺜﻨﺎﺌﻴﺔ
.
ﻋﻨﺩﺌﺫ
SX
ﺭﺓـﺤﻟﺍ ﺭﺓـﻤﺯﻟﺍ ﻑﺼـﻨ ﻰﻤﺴﺘﻭ ،ﺭﺓﻤﺯ ﻑﺼﻨ
(
free semigroups
)
ﻋﻠﻰ
X
.
ﺇﺫﺍ ﻁﺎﺒﻘﻨﺎ ﺍﻟﻌﻨﺼﺭ
a
ﻤﻥ
X
ﻤﻊ ﺍﻟﻜﻠﻤﺔ
a
ﺘﺼﺒﺢ،
X
ﻥـﻤ ﺔـﻴﺌﺯﺠ ﺔـﻋﻭﻤﺠﻤ
SX
ﻭ
ﺘﻭﻟﺩﻫﺎ
)
ﺃﻱ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ ﻨﺼﻑ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤ
ﻥ
SX
ﺘﺤﻭﻱ
X
.(
ﻕـﻴﺒﻁﺘﻠﻟ ،لﺎﺤ ﺔﻴﺃ ﻰﻠﻋ
SX
X
→
ﺍﻟﺨﺎﺼﺔ ﺍﻟﻌﺎﻤﺔ
ﺍﻵﺘﻴ
ﺔ
:
ﻤﻥ ﺃﺠل ﺃﻱ ﺘﻁﺒﻴﻕ ﻟﻠﻤﺠﻤﻭﻋﺎﺕ
S
X
→
:
α
ﻥـﻤ
X
ﺇﻟﻰ ﻨﺼﻑ ﺍﻟﺯﻤﺭﺓ
S
ﻓﺈﻨﻪ ﻴﻭﺠﺩ ﺘﺸﺎﻜ،
ل ﻭﺤﻴﺩ
S
SX
→
ﻴﺠﻌل ﺍﻟﻤﺨﻁﻁ
ﺍﻵﺘﻲ
ﺘﺒ
ﺎﺩ
ﻟﻴﺎﹰ
:
a
a
X
SX
→
a
ـ ﻴﺄﺨﺫ ﺍﻟﺘﻤﺩﻴﺩ ﺍﻟﻭﺤﻴﺩ ﻟ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
α
ﺍﻟﻘﻴﻡ
ﺍﻵﺘﻴ
ﺔ:
( )
(
)
( ) ( ) ( )
...
...
,
1
1
a
b
d
dba
S
α
α
α
α
α
=
=
ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ
( Free groups )
ﻨﺭﻴﺩ ﺃﻥ ﻨﺒﻨﻲ ﺯﻤﺭﺓ
FX
ﺘﺤﻭﻱ
X
ـ ﻭﻟﻬﺎ ﻨﻔﺱ ﺍﻟﺨﻭﺍﺹ ﺍﻟﻌﺎﻤﺔ ﻟ
SX
ﻤﻊ
"
ﺭﺓـﻤﺯ ﻑﺼﻨ
"
ﺒﺩﻻﹰ ﻤﻥ
"
ﺯﻤﺭﺓ
."
ﻨﻌﺭﻑ
X
′
ﻋﻠﻰ ﺃﻨﻬﺎ ﻤﺠﻤﻭﻋﺔ ﺤﺎﻭﻴﺔ ﻋﻠﻰ
ﺭﻤﻭﺯ ﻤﻥ
X
ﺎﻓﻲـﻀﺇ ﺯﻤﺭ ﻊﻤ
ـ ﻨﺭﻤﺯ ﺒ،ﹰﺎﻀﻴﺃ
1
−
a
ﻟﻜ،
ل
X
a
∈
ﻭﺒﻬﺫﺍ ﻴﻜﻭﻥ،
{
}
,...
,
,
,
1
1
−
−
=
′
b
b
a
a
X
ﻟﻴﻜﻥ
W
′
ﻤﺠﻤﻭﻋﺔ ﺍﻟﻜﻠﻤﺎﺕ ﺍﻟﺘﻲ ﺭﻤﻭﺯﻫﺎ ﻤﻥ
X
′
.
ﺔـﻴﻠﻤﻌﻟ ﺔﺒﺴـﻨﻟﺎﺒ ﺓﺭـﻤﺯ ﻑﺼﻨ ﺢﺒﺼﺘ
ﻭﻟﻜﻨﻬﺎ ﻟﻴﺴﺕ ﺯﻤﺭﺓ ﻷﻥ،ﺞﻤﺩﻟﺍ
1
−
a
ـ ﻟﻴﺱ ﻨﻅﻴﺭﺍﹰ ﻟ
a
ﺩﻭﺩـﺤﻟﺍ ﻑﺫﺤ ﻥﻜﻤﻴ ﻻﻭ ،ﻥﻵﺍ ﻰﺘﺤ
ﺍﻟﻭﺍﻀﺤﺔ ﻓﻲ ﺍﻟﻜﻠﻤﺎﺕ ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﺸﻜل
ﺍﻵﺘﻲ
:
...
...
...
...
1
a
a
أو
aa
-1
−
ﺘﺴﻤﻰ ﺍﻟﻜﻠﻤﺔ ﻤﺨﺘﺯﻟﺔ
(reduced)
ﺇﺫ ﻜﺎﻨﺕ ﻻ ﺘﺤﻭﻱ ﻋﻠﻰ ﺃﻱ ﻤﻥ ﺍﻷﺯﻭﺍﺝ
a
a
أو
aa
-1
1
−
.
ﺒﺎﻟﺒﺩﺀ ﺒﺎﻟﻜﻠﻤﺔ
w
ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻤ،
ﻨﺘﻬﻴﺔ ﻤﻥ ﺍﻻﺨﺘﺼﺎﺭﺍﺕ ﻟﻨﺼل ﺇﻟﻰ ﺍﻟﻜﻠﻤﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ
)
ﻤﻥ ﺍﻟﻤﺤﺘﻤل ﺃﻥ ﺘﻜﻭﻥ ﺍﻟﺨﺎﻟﻴﺔ
(
ﻭ،
ﺍﻟﺘﻲ ﺴﻨﺴﻤﻲ ﻜل ﻤﻨﻬﺎ ﺒﺎﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل
(reduced form)
0
w
ـ ﻟ
w
.
، ﻤﺜﻼﹰ،ﺕﺍﺭﺎﺼﺘﺨﻻﺍ ﻩﺫﻫ لﻴﻜﺸﺘﻟ ﺔﻔﻠﺘﺨﻤ ﻕﺭﻁ ﺩﺎﺠﻴﺇ ﻥﻜﻤﻴ
S
α
PDF created with pdfFactory Pro trial version
٨٣
ca
bb
ca
a
a
cbb
a
c
c
a
cabb
ca
ca
cc
ca
c
aa
c
ca
c
a
bb
ca
→
→
→
→
→
→
−
−
−
−
−
−
−
−
−
−
−
−
1
1
1
1
1
1
1
1
1
1
1
1
ﻭﻀ
ﻌﻨﺎ ﺨﻁﻭﻁﺎﹰ ﺘﺤﺕ
ﺍﻷﺯﻭﺍﺝ ﺍﻟﻤﺨﺘﺼﺭﺓ
.
ﻨﻼﺤﻅ ﺒﺄﻥ ﺍﻟﺭﻤﺯ ﺍﻟﻤﺘﻭﺴﻁ
a
ﺎﻟﺭﻤﺯـﺒ لﺯـﺘﺨﻤ
ﺍﻟﻤﺨﺘﻠﻑ ﻋﻨﻪ
1
−
a
ﻭﺍﻟﺤ،
ﺩﻭﺩ ﺍﻟﻤﺨﺘﻠﻔﺔ ﺒﻘﻴﺕ ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ
)
ﺍﻟﻜﻠﻤﺔ
ca
ﻰـﻨﻤﻴﻟﺍ ﺔﻬﺠﻟﺍ ﻲﻓ ﻲﺘﻟﺍ
ﻤﻥ ﺍﻻﺨﺘﺼﺎﺭ
ﻭ،لﻭﻷﺍ
ﺍﻟﻜﻠﻤﺔ
ca
ﻓﻲ ﺍﻟﺠﻬﺔ ﺍﻟﻴﺴﺭﻯ ﻤﻥ ﺍﻻﺨﺘﺼﺎﺭ
ﺎﻨﻲـﺜﻟﺍ
.(
ﻰـﻟﺇ ﺎﻨﻠـﺼﻭ
ﻭﺍﻟﻨﺘﻴﺠﺔ،ﺎﻬﺴﻔﻨ ﺔﺠﻴﺘﻨﻟﺍ
ﺍﻵﺘﻴ
ﺔ ﺴﺘﺒﻴﻥ ﻟﻨﺎ ﺒﺄﻥ ﻫﺫﺍ ﺴﻴﺤﺼل ﺩﺍﺌﻤﺎﹰ
.
ﻗﻀﻴﺔ
1.2
ﻴﻭﺠﺩ ﺸﻜل ﻤﺨﺘﺯل ﻭﺍﺤﺩ ﻓﻘﻁ ﻟﻜل ﻜﻠﻤﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺴﺘﺨﺩﻡ ﻤﺒﺩﺃ ﺍﻻﺴﺘﻘﺭﺍﺀ ﺍﻟﺭﻴﺎﻀﻲ ﻋﻠﻰ ﻁﻭل ﺍﻟﻜﻠﻤﺔ
w
.
ﺇﺫﺍ ﻜﺎﻨﺕ
w
ﻋﻨﺩﺌﺫ،ﺔﻟﺯﺘﺨﻤ
ﻓﺈﻥ
ﺍﻟﺒﺭﻫﺎﻥ ﻤﺤﻘﻕ
.
ﺎـﻬﻟ ﻲـﺘﻟﺍ ﺝﺍﻭﺯﻷﺍ ﺩـﺤﺃ ﺎﻨﻴﺩﻟ ﻪﻨﺃ ﹰﻻﻭﺃ ﺽﺭﻔﻨ ﻯﺭﺨﻷﺍ ﺕﻻﺎﺤﻟﺍ ﻲﻓ ﺎﻤﺃ
ﺍﻟﺸﻜل
1
0
0
a a
−
،
،ﺃﻭ
1
0
0
a a
−
ﻷﻥ ﺍﻟﺘﺭﺘﻴﺏ ﻨﻔﺴﻪ ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ
.
ـ ﻨﻔﺭﺽ ﺃﻥ ﺃﻱ ﺸﻜﻠﻴﻥ ﻤﺨﺘﺯﻟﻴﻥ ﻟ
w
ﺍﻟ
ﻠ
ﺼﺎﺭﺍﺕـﺘﺨﻻﺍ ﻥﻤ ﺔﻴﻟﺎﺘﺘﻤﺒ ﺎﻤﻬﻴﻠﻋ لﺼﺤﻨ ﻥﻴﺫ
،
ﻭﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
1
0
0
−
a
a
ﻤﺨﺘﺼﺭﺍﹰ ﻓﻲ ﺍﻟﺒﺩﺍﻴﺔ
،
ﺭﺽـﻔﻟﺍ ﻕﻴﺒﻁﺘ ﻊﻴﻁﺘﺴﻨ ﺎﻨﻨﻷ ﻙﻟﺫﻭ ،ﻥﺎﻴﻭﺎﺴﺘﻤ
ﺍﻻﺴﺘﻘﺭﺍﺌﻲ ﻋﻠﻰ ﺍﻟﻜﻠ
ﻤﺔ
)
ﺍﻷﻗﺼﺭ
(
ﺍﻟﺘﻲ ﺤﺼﻠﻨﺎ ﻋﻠﻴﻬﺎ ﺒﺎﻻﺨﺘﺼﺎﺭ
1
0
0
−
a
a
.
ﺒﻌﺩ ﺫﻟﻙ ﻨﻔﺭﺽ
ـ ﺃﻥ ﺃﻱ ﺸﻜﻠﻴﻥ ﻤﺨﺘﺯﻟﻴﻥ ﻟ
w
ﻥـﻤ ﺔـﻴﻟﺎﺘﺘﻤﺒ ﺎـﻤﻬﻴﻠﻋ ﺎﻨﻠﺼـﺤ ﻥﻴﺫـﻟﺍ
ﺍﻻﺨﺘﺼﺎﺭﺍﺕ
،
ﻭﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
1
0
0
−
a
a
ﻤﺨﺘﺼﺭﺍﹰ ﻓﻲ ﺒﻌﺽ ﺍﻟﻨﻘﺎﻁ
،
ﻷﻥ ﻨﺘﻴﺠﺔ ﻫﺫﻩ،ﻥﺎﻴﻭﺎﺴﺘﻤ
ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﻤﻥ ﺍﻻﺨﺘﺼﺎﺭﺍﺕ ﻟﻥ ﺘﺘﺄﺜﺭ ﺇﺫﺍ ﻜﺎﻥ
1
0
0
−
a
a
ﻤﺨﺘﺼﺭﺍﹰ ﻓﻲ ﺍ
ﻟﺒﺩﺍﻴﺔ
.
ﻨﻌﺘﺒﺭ ﺃﻥ ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل،ﹰﺍﺭﻴﺨﺃ
0
w
ﻲـﺘﻟﺍﻭ ﺕﺍﺭﺎﺼﺘﺨﻻﺍ ﻥﻤ ﺔﻴﻟﺎﺘﺘﻤﺒ ﻪﻴﻠﻋ ﺎﻨﻠﺼﺤ ﻱﺫﻟﺍ
ﻻ ﻴﻜﻭﻥ ﻓﻴﻬﺎ ﺃﻱ ﺍﺨﺘﺼﺎﺭ ﻴﻨﺘﺞ
1
0
0
−
a
a
ﻤﺒﺎﺸﺭﺓﹰ
.
ﺒﻤﺎ ﺃﻥ
1
0
0
−
a
a
ﻟﻥ ﻴﺒﻘﻰ ﻓﻲ
0
w
ﺩﻫﺎـﻨﻋ ،
ﺴﻴﺨﺘﺼﺭ ﻭﺍﺤﺩﺓ ﻋﻠﻰ ﺍﻷﻗل ﻤﻥ
1
0
−
a
، ﺃﻭ،
0
a
ﻓﻲ ﺒﻌﺽ ﺍﻟﻨﻘﺎﻁ
.
،ﺴﻪـﻔﻨ ﺝﻭﺯﻟﺍ ﺭﺼﺘﺨﻴ ﻡﻟ ﺍﺫﺇ
ﺍﻻﺨﺘﺼﺎﺭ ﺍﻷﻭل ﺍﻟﺫﻱ ﻴﺘﻀﻤﻥ ﺍﻟﺯﻭﺝ ﻴﻜﻭﻥ ﻤﻥ ﺍﻟﺸﻜلﺫﺌﺩﻨﻋ
1
1
1
0
0
0
0
0
0
...
...
...
a a a
أو
a a a
−
−
−
/ /
/ /
ﺤﻴﺙ ﻭﻀﻊ ﺨﻁﺎﹰ ﺘﺤﺕ ﺍﻟﺯﻭﺝ ﺍﻷﺴﺎﺴﻲ
.
ﻟﻜﻥ ﺍﻟﻜﻠﻤﺔ ﺍﻟﺘ
ﺼﺎﺭﺍﺕـﺘﺨﻻﺍ ﻩﺫﻫ ﺩﻌﺒ ﺎﻬﻴﻠﻋ ﺎﻨﻠﺼﺤ ﻲ
ﻲﺴﺎﺴﻷﺍ ﺝﻭﺯﻟﺍ ﺭﺼﺘﺨﻨ ﻥﺃ ﻥﻜﻤﻴ ﺍﺫﻬﻟﻭ ،ﹰﺍﺭﺼﺘﺨﻤ ﻲﺴﺎﺴﻷﺍ ﺝﻭﺯﻟﺍ ﻥﺎﻜ ﺍﺫﺇ ﺎﻤﻴﻓ ﺎﻬﺴﻔﻨ ﻥﻭﻜﺘﺴ
ﺒﺩﻻﹰ ﻤﻨﻬﺎ
.
ﻭﺒﺫﻟﻙ ﺴﻨﻌﻭﺩ ﺇﻟﻰ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﺒﺭﻫﻨﺕ
.
ﻨﻘﻭل ﺃﻥ ﺍﻟﻜﻠﻤﺘﻴﻥ
w
ﻭ
w
′
ﻤﺘﻜﺎﻓﺌﺘﺎﻥ
(equivalent)
ﻭﻨﺭﻤﺯ ﻟﻬﻤ،
ﺎﻟﺭﻤﺯـﺒ ﺎ
w
w
′
،
ﺇﺫﺍ ﻜﺎﻥ ﻟﻬﻤﺎ ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘ
ﺯل ﻨﻔﺴﻪ
.
ﻭﻫﻲ ﺘﺸﻜل ﻋﻼﻗﺔ ﺘﻜﺎﻓﺅ
)
ﻭﺍﻀﺤﺔ
.(
ﻗﻀﻴﺔ
2.2
ﺃﻱ ﺃﻥ،ﺎﻤﻬﻟ ﺔﺌﻓﺎﻜﻤ ﺔﻤﻠﻜ ﻭﻫ ﻥﻴﺘﺌﻓﺎﻜﺘﻤ ﻥﻴﺘﻤﻠﻜ ﺀﺍﺩﺠ ﻥﺇ
,
w
w
v
v
wv
w v
′
′
′ ′
⇒
PDF created with pdfFactory Pro trial version
٩٣
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
0
w
ﻭ
0
v
ﺸﻜﻼﻥ ﻤﺨﺘﺯﻻﻥ ﻟﻠﻜﻠﻤﺘﻴﻥ
w
ﻭ
v
ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ
.
ﺼلـﺤﻨ ﻲﻜﻟ
ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل ﻟﻠﻜﻠﻤﺔ
wv
ﻴﻥـﺘﻤﻠﻜﻟﺍ ﻲﻓ ﻥﻜﻤﻤ ﻭﻫ ﺎﻤ لﻜ ﹰﻻﻭﺃ ﻑﺫﺤﻨ ﻥﺃ ﺏﺠﻴ ،
w
ﻭ
v
ﻋﻠﻰ ﺤﺩﺍٍلﻜ
.
ﺤﺘﻰ ﻨﺤﺼل ﻋﻠﻰ ﺸﻜل ﺍﻟﻜﻠﻤﺔ
0
0
w v
ﻭﻤﻥ ﺜﻡ ﻨﺘﺎﺒﻊ ﺍﻟﺤﺫﻑ
.
ﺘﺞـﻨﻴ ﺍﺫـﻬﺒﻭ
ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل ﻟﻠﻜﻠﻤﺔ
0
0
w v
ﻤﻥ ﺍﻟﻜﻠﻤﺔ
wv
.
ﺔـﻤﻠﻜﻟﺍ لﺠﻷ ﺔﻘﻘﺤﻤ ﺔﻬﺒﺎﺸﻤﻟﺍ ﺔﻟﺎﺤﻟﺍﻭ
w v
′ ′
،
ﻟﻜﻥ
)
ﻤﻥ ﺍﻟﻔﺭﺽ
(
ـﺍﻟﺸﻜﻼﻥ ﺍﻟﻤﺨﺘﺯﻻﻥ ﻟ
w
ﻭ
v
ــ ﻴﻜﺎﻓﺌﺎﻥ ﺍﻟﺸﻜﻠﻴﻥ ﺍﻟﻤﺨﺘﺯﻟﻴﻥ ﻟ
w
′
ﻭ
v
′
ﻭﺒﻬﺫﺍ ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻨﺘﻴﺠﺔ ﻨﻔﺴﻬﺎ ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ،
.
ﻟﺘﻜﻥ
FX
ﻤﺠﻤﻭﻋﺔ ﺠﻤﻴﻊ ﺼﻔﻭﻑ ﺍﻟﺘﻜﺎﻓﺅ ﻟﻠﻜﻠﻤﺎﺕ
.
ﺇﻥ ﺍﻟﻘﻀﻴﺔ
2.2
ﺔـﻴﻠﻤﻌﻟﺍ ﻥﺄـﺒ ﻥﻴـﺒﺘ
ﺍﻟﺜﻨﺎﺌﻴﺔ ﻋﻠﻰ
W
′
ﺘﻌﺭﻑ ﻋﻤﻠﻴﺔ ﺜﻨﺎﺌﻴﺔ ﻋﻠﻰ
FX
ﺼﻑـﻨ ﻰـﻟﺇ ﺢـﻀﺍﻭ لﻜﺸﺒ لﻭﺤﺘﺘ ﻲﺘﻟﺍ ،
ﺯﻤﺭﺓ
.
ﻷﻥ،ﺭﺌﺎﻅﻨﻟﺍ ﻰﻠﻋ ﹰﺎﻀﻴﺃ ﻱﻭﺤﺘ ﻲﻫﻭ
(
)
(
)
1
1
1
1
...
...
1
ab
gh
h g
b a
−
−
−
−
ﻭﺒﻬﺫﺍ ﺘﺼﺒﺢ
FX
ﺘﺴﻤﻰ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺤﺭﺓ،ﺓﺭﻤﺯ
(free group)
ﻋﻠﻰ
X
.
ﺔـﺼﻼﺨﻟﺍﻭ
:
ﺇﻥ
ﻋﻨﺎﺼﺭ
FX
ﻤﻤﺜﻠﺔ ﺒﻜﻠﻤﺎﺕ ﻤﻥ
X
′
ﺘﻤﺜل ﺍﻟﻜﻠﻤﺘﺎﻥ ﺍﻟﻌﻨﺼﺭ ﻨﻔﺴﻪ ﻤﻥ؛
FX
ﻁ ﺇﺫﺍـﻘﻓﻭ ﺍﺫﺇ
ــ ﻭ ﺘﻤﺜل ﺍﻟﻜﻠﻤﺔ ﺍﻟﺨﺎﻟﻴﺔ ﺒ؛ﺞﻤﺩﻟﺍ ﺔﻘﻴﺭﻁﺒ ﺀﺍﺩﺠﻟﺍ ﻑﺭﻌﻴﻭ ؛ﻪﺴﻔﻨ لﺯﺘﺨﻤﻟﺍ لﻜﺸﻟﺍ ﺎﻤﻬﻟ ﻥﺎﻜ
1
؛
ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻨﻅﺎﺌﺭ ﺒﻁﺭﻴﻘﺔ ﺒﺴﻴﻁﺔ
.
ﻜل ﻜﻠﻤﺔ ﻤﻥ،ﻲﻟﺎﺘﻟﺎﺒﻭ
FX
، ﺘﻤﺜل ﺒﻜﻠﻤﺔ ﻤﺨﺘﺯﻟﺔ ﻭﺤﻴﺩﺓ
ﻤﻊ ﻋﻤﻠﻴﺔ ﺍﻟﺠﺩﺍﺀ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻴﻬﺎ ﻭﻫﻲ ﻋﻤﻠﻴﺔ ﺍﻟ
ﺩﻤﺞ ﻟﻠﻭﺼﻭل ﺇﻟﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل
.
ﺇﺫﺍ ﻁﺎﺒﻘﻨﺎ
a
X
∈
ﺒﺼﻑ ﺍﻟﺘﻜﺎﻓﺅ ﻟﻠﻜﻠﻤﺔ
a
)
ﺍﻟﻤﺨﺘﺯﻟﺔ
(
ﻭلـﺤﺘﺘ ﺫﺌﺩﻨﻋ ،
X
ﺎـﻬﻨﺃ ﻰـﻠﻋ
ﺘﺘﻁﺎﺒﻕ ﻤﻊ ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻓﻲ
FX
-
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻨﻬﺎ ﺘﻭﻟﺩ
FX
.
ﺇﻥ ﺍﻟﻘﻀﻴﺔ
ﺍﻵﺘﻴ
ﻲـﻫ ﺔ
ﻤﻠﺨﺹ ﻟﻤﻌﻨﻰ ﺍﻟﺤﻘﻴﻘﺔ ﺍﻟﺫﻱ ﺘﻨﺹ ﻋﻠﻰ ﺃﻨﻪ ﻻ ﺘﻭﺠﺩ ﻋﻼﻗﺎﺕ ﺒﻴﻥ ﻋﻨﺎﺼﺭ
X
ﺭـﺒﺘﻌﺘ ﺎﻤﺩـﻨﻋ
ﻜﻌﻨﺎﺼﺭ ﻤﻥ
FX
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻌﻨﺎﺼﺭ ﺍﻟﻤﻔﺭﻭﻀﺔ ﻓﻲ ﻤﺴﻠﻤﺎﺕ ﺍﻟﺯﻤﺭﺓ
.
ﻗﻀﻴﺔ
3.2
ﻷﻱ ﺘﻁﺒﻴﻕ
: X
G
α
→
ﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﺇﻟﻰ ﺍﻟﺯﻤﺭﺓ
G
ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻭﺤﻴﺩ،
FX
G
→
ﻴﺠﻌل ﺍﻟﻤﺨﻁﻁ
ﺍﻵﺘﻲ
ﺘﺒﺎﺩﻟﻴﺎﹰ
:
a
a
X
FX
→
a
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺄﺨﺫ ﺍﻟﺘﻁﺒﻴﻕ
: X
G
α
→
.
ﻭﻨﻤﺩﺩﻩ ﺇﻟﻰ ﺍﻟﺘﻁﺒﻴﻕ
X
G
′ →
ﺸﻜلـﻟﺎﺒ ﻰﻁﻌﻤﻟﺍﻭ
( )
( )
1
1
a
a
α
α
−
−
=
.
ﺒﻤﺎ ﺃﻥ
G
ﺩﺩـﻤﻨ ،ﺓﺭـﻤﺯ ﻑﺼﻨ ،ﺔﺼﺎﺨ ﺔﻟﺎﺤﻜ ،
α
ﺸﺎﻜلـﺘﻟﺍ ﻰـﻟﺇ
ﻷﻨﺼﺎﻑ ﺍﻟﺯﻤﺭ
SX
G
′ →
.
ﻫﺫﺍ ﺍﻟﺘﻁﺒﻴﻕ ﺴﻴﺭ
ﻲـﻓ ﻪﺴﻔﻨ ﺭﺼﻨﻌﻟﺍ ﻰﻟﺇ ﺔﺌﻓﺎﻜﺘﻤﻟﺍ ﺕﺎﻤﻠﻜﻟﺍ لﺴ
G
،
ﻼلـﺨ ﻥـﻤ لﻤﺍﻭﻋ ﻰﻟﺇ لﻠﺤﺘﻴ ﻑﻭﺴ ﻙﻟﺫﻟﻭ
FX
SX
′
=
.
ﺇﻥ
ﺎﺘﺞـﻨﻟﺍ ﻕـﻴﺒﻁﺘﻟﺍ
G
α
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٠٤
FX
G
→
ﻫﻭ ﺘﺸﺎﻜل ﺯﻤﺭﻱ
.
ﻭﻫﻭ ﻭﺤﻴﺩ ﻷﻨﻨﺎ ﻨﻌﻠﻡ ﺒﺄﻨﻪ ﻤﻌﺭﻑ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﻤﻭﻟﺩﺍﺕ
ـﻟ
FX
.
ﻤﻼﺤﻅﺔ
4.2
ﻴﺘﺼﻑ ﺍﻟﺘﻁﺒﻴﻕ
:
,
X
FX x
x
ι
→
a
ﺒﺎﻟﺨﺎﺼﺔ ﺍﻟﻌﺎﻤﺔ
ـﺍﻵﺘﻴ
ﺔ
:
ﺎﻥـﻜ ﺍﺫﺇ
: X
F
ι
′
′
→
ﺩـﻴﺤﻭ لـﺜﺎﻤﺘ ﺩـﺠﻭﻴ ﺫـﺌﺩﻨﻋ ،ﺎﻬﺴـﻔﻨ ﺔـﻤﺎﻌﻟﺍ ﺔـﺼﺎﺨﻟﺍ ﻪﻟﻭ ﺭﺨﺁ ﻕﻴﺒﻁﺘ
: FX
F
α
′
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
α ι ι
′
=
o
،
FX
ﻨﺘﺫﻜﺭ ﺍﻟﺒﺭﻫﺎﻥ
:
ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ
ι
ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻭﺤﻴﺩ،ﻡﺎﻋ لﻜﺸﺒ
: FX
F
α
′
→
ﻭﻥـﻜﻴ ﺙﻴﺤﺒ
α ι ι
′
=
o
ﻭﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ؛
ι
′
ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻭﺤﻴﺩ،ﻡﺎﻋ لﻜﺸﺒ
: F
FX
β
′ →
ﻭﻥـﻜﻴ ﺙﻴﺤﺒ
β ι ι
′ =
o
ﺇﻥ،ﻥﻵﺍ ؛
(
)
β α ι ι
=
o
o
ﻟﻜﻥ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ،
ι
ﻴﻜﻭﻥ،
F X
id
ﺸﺎﻜلـﺘﻟﺍ ﻭـﻫ ،
ﺍﻟﻭﺤﻴﺩ
FX
FX
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
FX
id
ι ι
=
o
ﻭﺒﺎﻟﺘﺎﻟﻲ،
FX
id
β α
=
o
، ﻭﺒﺸﻜل ﻤﺸﺎﺒﻪ،
F
id
α β
′
=
o
ﻭﺒﻬﺫﺍ ﻓﺈﻥ؛
α
ﻭ
β
ﺘﻤﺎﺜﻼﻥ ﻋﻜﻭﺴﺎﻥ
.
ﻨﺘﻴﺠﺔ
5.2
ﻜل ﺯﻤﺭﺓ ﻫﻲ ﺯﻤﺭﺓ ﻗﺴﻤﺔ ﻟﺯﻤﺭﺓ ﺤﺭﺓ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺨﺘﺎﺭ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻤﻥ ﻤﻭﻟﺩﺍﺕ ﺍﻟﺯﻤﺭﺓ
G
)
،ﻤﺜﻼﹰ
X
G
=
(
ﻭﻟﺘﻜﻥ،
F
ﺯﻤﺭﺓ
ﺤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔ
X
.
ﻤﻥ
(3.2)
ﻕـﻴﺒﻁﺘﻟﺍ ﺩﺩﻤﻴ ،
:
a
a X
G
→
a
ﺸﺎﻜلـﺘﻟﺍ ﻰـﻟﺇ
F
G
→
ﻋﺒﺎﺭﺓ ﻋﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺘﺤﻭﻱ،ﺓﺭﻭﺼﻟﺍ ﻥﻭﻜﻭ ،
X
ﻓﻬﻲ ﺘﺴﺎﻭﻱ،
G
.
ﺇﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺤﺭﺓ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
{ }
X
a
=
ﺘﻜﻭﻥ ﺒﺒﺴﺎﻁﺔ ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
C
∞
ﻤﻭﻟﺩﺓ
ﺒﺎﻟﻌﻨﺼﺭ
a
ﻟﻜﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺤﺭﺓ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻜﻭﻨﺔ ﻤﻥ ﻋﻨﺼﺭﻴﻥ ﺘﻜﻭﻥ ﻤﻌﻘﺩﺓ ﺠﺩﺍﹰ،
.
ﺒﻌﺽ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﻬﺎﻤﺔ ﻋﻠﻰ ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ،ﻥﺎﻫﺭﺒ ﻥﻭﺩﺒﻭ ،ﻥﻵﺍ ﺵﻗﺎﻨﺄﺴ
.
ﻤﺒﺭﻫﻨﺔ
6.2
(Nielsen- Schreier)
9
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﺤﺭﺓ ﻫﻲ ﺯﻤﺭﺓ ﺤﺭﺓ
.
ﻴﺴﺘﺨﺩﻡ ﺍﻟﺒﺭﻫﺎﻥ
ﻭﺒﺸﻜل ﺨﺎﺹ ﻓﻲ ﺍﻟﻔﻀﺎﺀﺍﺕ ﺍﻟﻤﺘﺭﺍﺼﺔ،ﺔﻴﺠﻭﻟﻭﺒﻁﻟﺍ ﻲﻓ لﻀﻓﻷﺍ
-
ﺭـﻅﻨﺍ
Serre 1980
ﺃﻭ
Rotman 1995
ﺍﻟﻤﺒﺭﻫﻨﺔ،
11.44
.
9
Nielsen
(1921)
ﺕـﺒﻠﻟ ﺔﻴﻤﺯﺭﺍﻭﺨ ﻰﻁﻋﺃ ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓﻭ ،ﺩﻴﻟﻭﺘﻟﺍ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺔﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻟﺍ لﺠﺃ ﻥﻤ ﺍﺫﻫ ﻥﻫﺭﺒ
،ﻓﻴﻤﺎ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﻜﻠﻤﺔ ﺘﻘﻊ ﻓﻲ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
Schreier (1927)
ﺒﺭﻫﻥ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ
.
X
F
′
α
ι
ι
′
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١٤
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺘﺎﻥ
FX
ﻭ
FY
ﻥـﻤ لﻜﻟ ﻥﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﻥﻴﺘﻠﺜﺎﻤﺘﻤ
X
ﻭ
Y
ﺩﺭﺓـﻘﻟﺍ
ﻨﻔﺴﻬﺎ
.
ﻭﻟﺫﻟﻙ ﻴﻤﻜﻥ ﺃﻥ ﻨﻌﺭﻑ ﻤﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺤﺭﺓ
G
ﺒ
ﺭﺓـﺤ ﺓﺩـﻟﻭﻤ ﺔﻋﻭﻤﺠﻤ ﻱﺃ ﺭﺓﺩﻗ ﺎﻬﻨﺄ
)
ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻟﻠﺯﻤﺭﺓ
G
ﻭﺍﻟﺘﻲ ﻤﻥ ﺃﺠﻠﻬﺎ ﻴﻜﻭﻥ ﺍﻟﺘﺸﺎﻜل
FX
G
→
ﺍﻟ
ﻲـﻓ ﻰﻁﻌﻤ
(3.2)
ﺘﻤﺎﺜﻼﹰ
.(
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻤﻥ ﺯﻤﺭﺓ ﺤﺭﺓ
G
.
ﺔـﻴﻤﺯﺭﺍﻭﺨ ﺩـﺠﻭﺘ ﺫﺌﺩﻨﻋ
ﻟﺘﺸﻜﻴل ﻤﻥ ﺃﻱ ﻤﺠﻤﻭﻋﺔ ﻤﻨﺘﻬﻴﺔ ﻤﻥ ﻤﻭﻟﺩﺍﺕ
H
ﻤﺠﻤﻭﻋﺔ ﻤﻨﺘﻬﻴﺔ ﺤﺭﺓ ﻤﻥ ﺍﻟﻤﻭﻟﺩﺍﺕ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻭﻜﺎﻥ
(
)
:
G H
i
= < ∞
ﺘﻜﻭﻥﺫﺌﺩﻨﻋ ،
H
ﻥـﻤ ﺓﺭـﺤ ﺓﺭﻤﺯ
ﺍﻟﺭﺘﺒﺔ
1
ni
i
− +
ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﺭﺘﺒﺔ،ﺹﺎﺨ لﻜﺸﺒﻭ
H
ﺔـﺒﺘﺭ ﻥﻤ ﺭﺒﻜﺃ
F
.
ﺭـﻅﻨﺍ ،ﻥﻴﻫﺍﺭـﺒﻠﻟ
Rotman
1995
ﺍﻟﻔﺼل،
11
ﻭ،
Hall 1959
ﺍﻟﻔﺼل،
7
.
ﺍﻟﻤﻭﻟﺩﺍﺕ ﻭﺍﻟﻌﻼﻗﺎﺕ
(Generators and relation)
ﻨﻌﺘﺒﺭ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻭ ﺍﻟﻤﺠﻤﻭﻋﺔ
R
ﻭﺯـﻤﺭ ﻥﻤ ﺔﻠﻜﺸﺘﻤﻟﺍ ﺕﺎﻤﻠﻜﻟﺍ ﻥﻤ ﺔﻔﻟﺅﻤﻟﺍ
X
′
.
لـﻜ
ﻋﻨﺼﺭ ﻤﻥ
R
ﻴﻤﺜل ﻋﻨﺼﺭ ﻟﻠﺯﻤﺭﺓ ﺍﻟﺤﺭﺓ
FX
،
ﺇﻥ
ـ ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ ﻟ
G
ﻤﻥ
FX
ﻰـﻠﻋ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﻤﻭﻟﺩﺓ ﺒﻬﺫﻩ ﺍﻟﻌﻨﺎﺼﺭ ﺘﻤﻠﻙ
X
ﻜﻤﻭﻟﺩﺍﺕ
(generators)
ﻟﻬﺎ ﻭ
R
ﻜ
ﻌﻼﻗﺎﺕ
(relation)
)
ﺃﻭ ﻜﻤﺠﻤﻭﻋﺔ ﻤﻥ
ﺍﻟ
ﻌﻼﻗﺎﺕ ﺍﻟﻤﻌﺭﻓﺔ
.(
ﻭﻴﻘﺎل ﺃﻴﻀﺎﹰ
ﺒ
ﺄﻥ
(
)
,
X R
ﺘﻘﺩﻴﻡ
ﻟﻠﺯﻤﺭﺓ
G
ـ ﻭﻨﺭﻤﺯ ﻟ،
G
ﺒﺎﻟﺭﻤﺯ
X R
.
ﻤﺜﺎل
7.2
(a)
ﺇﻥ ﻟﺯﻤﺭﺓ ﺩﻴﻬﻴﺩﺭﺍل
n
D
ﺍﻟﻤﻭﻟﺩﺍﺕ
,
r s
ﻭﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﻤﻌﺭﻓﺔ
2
,
,
n
r
s
srsr
)
ﺍﻨﻅﺭ
9.2
ﻟﻠﺒﺭﻫﺎﻥ
(
(b)
ﺔـﻤﻤﻌﻤﻟﺍ ﺔـﻴﻋﺎﺒﺭﻟﺍ ﺓﺭـﻤﺯﻟﺍ
(generalized quaternion group)
,
3
n
Q
n
≥
،
ﻤﻭﻟﺩﺍﺘﻬﺎ
a
ﻭ
b
ﻭﺍﻟﻌﻼﻗﺎﺕ
10
ﻫﻲ
1
2
2
2
2
1
1
1,
,
n
n
a
a
b
bab
a
−
−
−
−
=
=
=
ﻤﻥ ﺃﺠل
3
n
=
ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ ﻓﻲ
(17.1)
.
ﺎـﻬﺘﺒﺘﺭ ،ﺔﻤﺎﻌﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
2
n
)
ﺔـﻓﺭﻌﻤﻟ
ﺍﻟ
ﺩـﻴﺯﻤ
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ،ﺎﻬﻨﻋ
4-2
.(
10
ﺒ
ﻜﻼﻡ
ﺃ
ﺩﻕ
ﺴﺄﻗﻭل ﺒﺄﻥ ﺍﻟﻌﻼﻗﺎﺕ،
1
2
2
2
2
1
,
,
n
n
a
a
b
bab a
−
−
−
−
.
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٢٤
(c)
ﺼﺭﺍﻥــﻨﻌﻟﺍ لﺩﺎــﺒﺘﻴ
a
ﻭ
b
ﺎﻥــﻜ ﺍﺫﺇ ﻁــﻘﻓﻭ ﺍﺫﺇ ﺓﺭــﻤﺯﻟﺍ ﻲــﻓ
ﺎــﻬﻟﺩﺎﺒﻤ
(commutator)
[ ]
1
1
,
a b
aba b
−
−
=
ﻴﺴﺎﻭﻱ
1
.
ﺔـﻴﻠﻴﺩﺒﺘﻟﺍ ﺓﺭﺤﻟﺍ ﺓﺭﻤﺯﻟﺍ
(free abelian
group)
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻭﻟﺩﺍ
ﺕ
1
,...,
n
a
a
ﻟﻬﺎ ﺍﻟﻤﻭﻟﺩﺍﺕ
1
,...,
n
a
a
ﻭﺍﻟﻌﻼﻗﺎﺕ
,
,
i
j
a a
i
j
≠
ﺍﻨﻅﺭ،ﺔﻠﺜﻤﻷﺍ ﺔﻴﻘﺒ لﺠﺃ ﻥﻤ
1967
Massey
ﻥـﻤ ﺓﺩـﻴﺠ ﺔـﻴﻤﻜ ﻰـﻠﻋ ﻱﻭـﺘﺤﻴ ﻱﺫﻟﺍ ،
ﺍﻟﺘﺩﺍﺨﻼﺕ ﺒﻴﻥ ﻨﻅﺭﻴﺔ ﺍﻟﺯﻤﺭ
ﻭ
ﺍﻟﻁﺒﻭﻟﻭﺠﻴﺔ
.
ﻀﺎﺀﺍﺕـﻔﻟﺍ ﻥﻤ لﺎﻜﺸﺃ ﺓﺩﻋ لﺠﺃ ﻥﻤ ،ﹰﻼﺜﻤ
ﺔـﻴﻤﺯﺭﺍﻭﺨ ﺩـﺠﻭﺘ ،ﺔـﻴﺠﻭﻟﻭﺒﻁﻟﺍ
ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺘﻘﺩﻴﻡ ﺍﻟﺯﻤﺭ ﺍﻷﺴﺎﺴﻴﺔ
.
(d)
ﻰـﻠﻋ ﺓﺭـﺤ ﺓﺭﻤﺯ ﻲﻫ ﺓﺩﺤﺍﻭ ﺔﻜﺭﺤﺘﻤ ﺔﻁﻘﻨ ﻊﻤ ﺡﻭﺘﻔﻤ ﺹﺭﻘﻟ ﺔﻴﺴﺎﺴﻷﺍ ﺓﺭﻤﺯﻟﺍ
σ
ﺤﻴﺙ
σ
ﺩﻭﺭﺍﻥ ﺤﻭل ﺍﻟﻨﻘﻁﺔ
(ibid II 5.1)
.
(e)
ﺍﻟﺯﻤﺭﺓ ﺍﻷﺴﺎﺴﻴﺔ ﻟﻜﺭﺓ ﻤﻊ
r
ﻨﻘﻁﺔ ﻤﺘﺤﺭﻜﺔ ﻟﻬﺎ ﺍﻟﻤﻭﻟﺩﺍﺕ
1
,...,
r
σ
σ
)
ﺤﻴﺙ
i
σ
ﻫﻭ
ﺍﻟﺩﻭﺭﺍﻥ ﺤﻭل
i
ﻨﻘﻁﺔ
(
ﻭ
ﺍﻟﻌﻼﻗﺔ ﺍﻟﻭﺤﻴﺩﺓ
1
...
1
r
σ σ
=
(f)
ﻥ ﺍﻟـﻤ ﺹﺍﺭـﺘﻤﻥ ﺍﻟﺎـﻤﻴﺭ ﺢﻁﺴـﻟ ﺔﻴـﺴﺎﺴﻷﺍ ﺓﺭﻤﺯﻟﺍ
ﻭﻉـﻨ
g
ﻪـﻟ
2g
ﺩـﻟﻭﻤ
1
1
, ,...,
,
g
g
u v
u
v
ﻭ ﺍﻟﻌﻼﻗﺔ ﺍﻟﻭﺤﻴﺩﺓ
1
1
1
1
1
1
1
1
...
1
g
g
g
g
u v u v
u v u v
−
−
−
−
=
)
ibid IV
ﺍﻟﺘﻤﺭﻴﻥ
5.7
.(
ﻀﻴﺔــﻗ
8.2
ﺘﻜﻥــﻟ
G
ﺩﻴﻡــﻘﺘﻟﺎﺒ ﺔــﻓﺭﻌﻤ ﺓﺭــﻤﺯ
(
)
,
X R
.
ﺭﺓــﻤﺯ ﻱﻷ
H
ﻭ
ﻕــﻴﺒﻁﺘﻟﺍ
: X
H
α
→
ﺍﻟﺫﻱ ﻴﺭﺴل ﻜل ﻋﻨﺼﺭ ﻤﻥ
R
ﺇﻟﻰ
1
)
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻭﺍﻀﺤﺔ
11
(
ﻴﻭﺠﺩ ﺘﺸﺎﻜل،
ﻭﺤﻴﺩ
G
H
→
ﻴﺠﻌل ﺍﻟﻤﺨﻁﻁ
ﺍﻵﺘﻲ
ﺘﺒﺎﺩﻟﻴﺎﹰ
:
a
a
X
G
→
a
ﺎﻥـﻫﺭﺒﻟﺍ
.
ﻨﻌﻠﻡ
ﺭﺓـﺤﻟﺍ ﺭـﻤﺯﻠﻟ ﺔﻤﺎﻌﻟﺍ ﺔﺼﺎﺨﻟﺍ ﻥﻤ
(3.2)
ﺄﻥـﺒ
α
ﺸﺎﻜلـﺘﻟﺍ ﻰـﻟﺇ ﺩﺩـﻤﺘﻴ
FX
H
→
ﺍﻟﺫﻱ ﻨﺭﻤﺯ ﻟﻪ ﺜﺎﻨﻴﺔﹰ ﺒﺎﻟﺭﻤﺯ،
α
.
ﻟﺘﻜﻥ
R
ι
ﺼﻭﺭﺓ
R
ﻓ
ﻲ
FX
.
ﻤﻥ ﺍﻟﻔﺭﺽ
( )
Ker
ι
α
⊂
R
ﻭﻟﺫﻟﻙ ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ،
N
ـ ﻤﻭﻟﺩﺓ ﺒ
R
ι
ﻲـﻓ ﺓﺍﻭﺘﺤﻤ ﻲﻫﻭ
11
ﻜل ﻋﻨﺼﺭ ﻤﻥ
R
ﻴﻤﺜل ﺒﻌﻨﺼﺭ ﻭﺍﺤﺩ ﻤﻥ
X
ﻭ ﺒﺸﺭﻁ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺘﻤﺩﻴﺩ ﺍﻟﻭﺤﻴﺩ،
α
ـ ﻟ
FX
ﻴﺭﺴل ﻜل ﻫﺫﻩ
ﺍﻟﻌﻨﺎﺼﺭ ﺇﻟﻰ
1
.
H
α
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٣٤
( )
Ker
α
.
ﻤﻥ ﺍﻟﺨﺎﺼﺔ ﺍﻟﻌﺎﻤﺔ ﻟﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
(42.1)
ﻴﺘﺤﻠل،
α
ﻼلـﺨ ﻥـﻤ لـﻤﺍﻭﻋ ﻰﻟﺇ
FX N
G
=
.
ﻭ،ﺩﻭﺠﻭﻟﺍ ﻥﻫﺭﺒﻴ ﺍﺫﻫ
ﺃﻤﺎ
ﺍﻟﻭﺤﺩﺍﻨﻴﺔ
ﻓ
ﺘﺄﺘﻲ ﻤﻥ ﻜﻭﻨ
ﻨﺎ
ﻨﻌﻠﻡ ﺒﺄﻥ ﺍﻟﺘﻁﺒﻴﻕ ﻤﻌﺭﻑ
ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﻤﻭﻟﺩﺍﺕ
X
.
ﻤﺜﺎل
9.2
ﻟﺘﻜﻥ
2
, ;
,
,
n
G
a b a b baba
=
.
ﻨﺒﺭﻫﻥ ﺒﺄﻥ
G
ﺩﺭﺍلـﻴﻬﻴﺩ ﺓﺭـﻤﺯ لﺜﺎﻤﺘ
n
D
)
ﺍﻨﻅﺭ
1.16
.(
ﻷﻥ ﺍﻟﻌﻨﺼﺭﻴﻥ
,
n
r s
D
∈
ﺘﺤﻘﻕ
ﻴﻤﺩﺩ ﺍﻟﺘﻁﺒﻴﻕ، ﻁﻭﺭﺸﻟﺍ ﻩﺫﻫ
{ }
,
,
,
n
a b
D
a
r b
s
→
a
a
ﺒﺸﻜل ﻭﺤﻴﺩ ﺇﻟﻰ ﺍﻟﺘﺸﺎﻜل
n
G
D
→
.
ﺇﻥ ﻫﺫﺍ ﺍﻟﺘﺸﺎﻜل ﻏﺎﻤﺭ ﻷﻥ
r
ﻭ
s
ﺩﺍﻥـﻟﻭﻴ
n
D
.
ﺇﻥ
ﺍﻟﻤﻌﺎﺩﻻﺕ
2
1
1,
1,
n
n
a
b
ba
a
b
−
=
=
=
ﺘﺒﻴﻥ ﺃﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ
G
ﻴﻤﺜل ﺒﺈﺤﺩﻯ ﺍﻟﻌﻨﺎﺼﺭ
ﺍﻵﺘﻴ
،ﺔ
1
1
1,...,
, ,
,...,
n
n
a
b ab
a
b
−
−
ﻭﻤﻨﻪ
2
≤
=
n
G
n
D
.
ﻟﺫﻟﻙ ﻴﻜﻭﻥ ﺍﻟﺘﺸﺎﻜل ﻏﺎﻤﺭﺍﹰ
)
ﻭﻫﺫﻩ ﺍﻟﺭﻤﻭﺯ ﺘﻤﺜل ﻋﻨﺎﺼﺭ ﻤﺨﺘﻠﻔﺔ ﻤﻥ
G
.(
، ﺒﺸﻜل ﻤﺸﺎﺒﻪ
( )
2
2
, ;
,
,
n
n
G
a b a b
ba
D
=
ﺒﺎﻟﻌﻼﻗﺔ
,
a
s b
t
a
a
.
ﺍﻟﺘﻘﺩﻴﻡ ﺍﻟﻤﻨﺘﻬﻲ ﻟﻠﺯﻤﺭ
(Finitely presented groups)
ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﻤﻨﺘﻬﻴ
ﺔ ﺍﻟﺘﻘﺩﻴﻡ ﺇﺫﺍ ﻜﺎﻥ ﻟﻬﺎ ﺍﻟﺘﻘﺩﻴﻡ
(
)
,
X R
ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﻜل ﻤﻥ،
X
ﻭ
R
ﻤﻨﺘﻬﻴﺔ
.
ﻤﺜﺎل
10.2
ﻨﻌﺘﺒﺭ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.
ﻟﺘﻜﻥ
X
G
=
ﻭﻟﺘﻜﻥ،
R
ﻤﺠﻤﻭﻋﺔ ﺍﻟﻜﻠﻤﺎﺕ
{
}
1
−
= ∈
abc
ab
c
G
ﺃﺩﻋﻲ ﺒﺄﻥ
(
)
,
X R
ﺘﻘﺩﻴﻡ ﻟﻠﺯﻤﺭﺓ
G
ﻭﻟﻬﺫﺍ ﻓﺈﻥ،
G
ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻘﺩﻴﻡ
.
ﻟﺘﻜﻥ
G
X R
′ =
.
ﺇﻥ ﺍﻟﺘﻤﺩﻴﺩ
:
a
a X
G
→
a
ـ ﻟ
FX
ﻴﺭﺴل ﻜل ﻋﻨﺼﺭ ﻤﻥ
R
ﺇﻟﻰ
1
ﺫﻟﻙـﻟﻭ ،
ﻨﻌﺭﻑ ﺍﻟﺘﺸﺎﻜل
G
G
′ →
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻏﺎﻤﺭﺍﹰ ﺒﺸﻜل ﻭﺍﻀﺢ،
.
ﻟﻜﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ
G
′
لـﺜﻤﻴ
ﺒﻌﻨﺼﺭ ﻤﻥ
X
ﻭ،
ﻤﻨﻪ
G
G
′ ≤
.
ﻟﺫﻟﻙ ﻓﺈﻥ ﺍﻟﺘﺸﺎﻜل ﻏﺎﻤﺭ
.
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٤٤
ﺯﻤﺭﺓ ﺒﺘﻘﺩﻴﻡ،ﺔﻟﻭﻬﺴﺒﻭ ،ﻑﻴﺭﻌﺘ ﻥﻜﻤﻴ ﻪﻨﺃ ﺎﻤﺒ
ﺒﻴﻨﻤﺎ ﺘ،ﻪﺘﻨﻤ
ﻌﺭ
ﻴ
ﻑ ﺨﻭﺍﺹ ﺍﻟﺯﻤﺭﺓ
ﻫﻭ
ﺭـﻤﺃ
ﺒﻐﺎﻴﺔ ﺍﻟﺼﻌﻭﺒﺔ
-
ﻜﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ،ﹰﺎﻤﺎﻤﺘ ﺓﺭﻴﻐﺼ ﻥﻭﻜﺘ ﻥﺃ ﻥﻜﻤﻴ ﻲﺘﻟﺍ ،ﺓﺭﻤﺯﻟﺍ ﻑﺭﻌﻨ ﺎﻨﻨﺄﺒ ﻅﺤﻼﻨ
ﻟﺯﻤﺭﺓ ﺤﺭﺓ ﻜﺒﻴﺭﺓ ﺒﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻜﺒﻴﺭﺓ
.
ﻨﺴﺘﻌﺭﺽ ﺒﻌﺽ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺴﻠﺒﻴﺔ
.
ﻤﺴﺄﻟﺔ ﺍﻟﻜﻠﻤﺔ
(The word problem)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻌﺭﻓﺔ ﺒﺎﻟﺘﻘﺩﻴﻡ ﺍﻟﻤﻨﺘﻬﻲ
(
)
,
X R
.
ﻤﺴﺄﻟﺔ ﺍﻟﻜﻠﻤﺔ ﻟﻠﺯﻤﺭﺓ
G
ﺎ ﺇﺫﺍـﻤﻴﻓ لﺄﺴـﺘ
ﺘﻭﺠﺩ ﺨﻭﺍﺭﺯﻤﻴﺔ
)
ﻁﺭﻴﻘﺔ ﻤﺤﻜﻤﺔ
(
ﻟﻠﺠﺯﻡ ﻓﻴﻤﺎ ﺇﺫﺍ
ﻜﺎﻨﺕ ﺍﻟﻜﻠﻤﺔ ﻋﻠﻰ
X
′
ﺘﻤﺜل
1
ﻲـﻓ
G
.
ﺇﻥ
ﺍﻟﺠﻭﺍﺏ ﺴﻠﺒﻲ
.
ﻥ ﻜل ﻤﻥﻴﺒ
Novicov
ﻭ
Boone
ﺒ
ﺄﻨﻪ ﺘﻭﺠﺩ ﺯﻤﺭ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻘﺩﻴﻡ
G
ﻲـﺘﻟﺍﻭ
ﻟﻴﺴﺕ ﻟﻬﺎ ﺨﻭﺍﺭﺯﻤﻴﺔ ﻜﻬﺫﻩ
.
ﻭﺒﺎﻟﻁﺒﻊ ﺘﻭﺠﺩ ﺯﻤﺭ ﺃﺨﺭﻯ ﻴﻭﺠﺩ ﻟﻬﺎ ﺨﻭﺍﺭﺯﻤﻴﺔ
.
ﺇﻥ ﺍﻷﻓﻜﺎﺭ ﻨﻔﺴﻬﺎ ﻗﺎﺩﺕ ﺇﻟﻰ ﺍﻟﻨﺘﻴﺠﺔ
ﺍﻵﺘﻴ
ﺔ
:
ﻻ ﺘﻭﺠﺩ ﺨﻭﺍﺭﺯﻤﻴﺔ ﺍﻟﺘﻲ ﺴﺘﺤﺩﺩ ﻷﻱ ﺘﻘﺩﻴﻡ ﻤﻨﺘﻪ
،ﻯﻭـﻘﻟﺍ ﺔـﻤﻴﺩﻋ ،لـﺤﻠﻟ ﺔﻠﺒﺎﻗ ،ﺔﻴﻠﻴﺩﺒﺘ ،ﺔﻴﻬﺘﻨﻤ ،ﺔﻬﻓﺎﺘ ﺔﻠﺒﺎﻘﻤﻟﺍ ﺓﺭﻤﺯﻟﺍ ﺕﻨﺎﻜ ﺍﺫﺇ ﺎﻤﻴﻓ ﻱﺭﺎﻴﺘﺨﺍ
ﺃﻡ ﺘﺤﻭﻱ ﻋﻠﻰ ﻤﺴﺄﻟﺔ ﺍﻟﻜﻠﻤﺔ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل،ﻻ ﻡﺃ ،ﺓﺭﺤ ،لﺘﻔﻟﺍ ﺓﺭﺤ ،لﺘﻔﻠﻟ ﺔﻠﺒﺎﻗ ،ﺔﻁﻴﺴﺒ
.
ﺍﻨﻅﺭ
Rotman 1995
ﺍﻟﻔﺼل،
12
ﻟﻠﺒﺭﻫﺎﻥ ﻋﻠﻰ ﻫﺫﻩ ﺍﻟﺤﺎﻻﺕ،
.
ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ
( The Burnside problem )
ﻨﺘﺫﻜﺭ ﺒﺄﻨﻪ ﻴﻘﺎل ﻋﻥ ﺍﻟﺯﻤﺭﺓ ﺒﺄﻨﻬﺎ ﻤﻥ ﺍﻟﻘﻭﺓ
e
ﺇﺫﺍ ﻜﺎﻥ
1
e
g
=
لـﻜﻟ
g
G
∈
ﻭ
e
ﻭـﻫ
ﺃﺼﻐﺭ ﻋﺩﺩ ﻁﺒﻴﻌﻲ ﻴﺤﻘﻕ ﻫﺫﻩ ﺍﻟﺨﺎﺼﺔ
.
ﻤﻥ ﺍﻟﺴﻬل ﺃﻥ ﻨﻜﺘﺏ ﺒﻌﺩ
ﺫﻟﻙ
ﺭـﻴﻏ ﺭـﻤﺯﻟﺍ ﻥﻋ ﺔﻠﺜﻤﺃ
ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﻤﻭﻟﺩﺓ ﺒﻌﺩﺩ ﻤ
ﻨﺘﻪ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺭﺘﺒﺔ ﻤﻨﺘﻬﻴﺔ
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
1-2
(
ﻟﻜﻥ ﻫل ﺘ،
ﺩـﺠﻭ
؟ﺯﻤﺭ ﻜﻬﺫﻩ ﺒﻘﻭﺓ ﻤﻨﺘﻬﻴﺔ
)
ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ
(
.
ﻓﻲ ﻋﺎﻡ
1968
ﻜﺎﻨﺕ ﺇﺠﺎﺒﺔ،
Adjan
ﻭ
Novikov
ﻨﻌﻡ
:
ﺘﻭﺠﺩ ﺯﻤﺭ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻭﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ ﺒﻘﻭﺓ ﻤﻨﺘﻬﻴﺔ
.
ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ ﺍﻟﻤﺤﺩﺩﺓ
(The Restricted Burnside problem)
ﺇﻥ ﺯﻤﺭﺓ ﺒﺭﻨﺴﺎﻴﺩ
(Burnside
group)
ﺍﻟﺘﻲ ﻗﻭﺘﻬﺎ
e
ﻋﻠﻰ
r
ﻤﻭﻟﺩ
( )
B
,
r e
ﺭﺓـﻤﺯ ﻲﻫ
ﺍﻟﻘﺴ
ﻤﺔ ﻟﺯﻤﺭﺓ ﺤﺭﺓ ﻋﻠﻰ
r
ﻤﻭﻟﺩ ﺒﺎﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻭﻟﺩﺓ ﺒﻜل ﻗﻭﻯ
e
.
ﺘﺘﺴﺎﺀل
ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ
ﻓﻴﻤﺎ ﺇﺫﺍ
ﻜﺎﻥ
( )
B
,
r e
ﻴﻡـﻘﻟﺍ ﺽـﻌﺒ ﺀﺎﻨﺜﺘـﺴﺎﺒ ﺔـﻴﻬﺘﻨﻤ ﺭﻴﻏ ﻥﻭﻜﺘ ﻲﻜﻟ ﻡﻭﻠﻌﻤﻟﺍ ﻥﻤﻭ ،ﻪﺘﻨﻤ
ـﺍﻟﺼﻐﻴﺭﺓ ﻟ
r
ﻭ
e
.
ﺇﻥ ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ
ﺍﻟﻤﺤﺩﺩﺓ ﺘ
ﺘﺴﺎﺀ
ل ﻓﻴﻤﺎ ﺇﺫﺍ
( )
B
,
r e
ﻴﺤ
ـﺘ
ﻁـﻘﻓ ﻱﻭ
،ﻋﻠﻰ ﻋﺩﺩ ﻤﻨﺘﻪ ﻤﻥ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
ﺘ،ﺊﻓﺎﻜﺘﻤ لﻜﺸﺒﻭ
ﺴﺄل ﻓﻴﻤﺎ ﺇﺫﺍ
ﻜﺎﻨﺕ
ﺭﺓـﻤﺯ ﺩـﺠﻭﺘ
ﻗﺴﻤﺔ ﻭﺍﺤﺩﺓ ﻤﻨﺘﻬﻴﺔ ﻤﻥ
( )
B
,
r e
ﺤﺎﻭﻴﺔ
ﻋﻠﻰ ﻜل ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻷﺨﺭﻯ ﻜﺯﻤﺭ ﻗﺴﻤﺔ
.
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٥٤
ﺇﻥ ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
)
ﺍﻨﻅﺭ
p48
(
ﻴﺒﻴﻥ
ﻰ ﺃﻥـﻠﻋ ﻥﻫﺭﺒﻨ ﻲﻜﻟ ﻪﻨﺃ
( )
B
,
r e
ﻴﺤﺘﻭ
ﻱ ﺩﺍﺌﻤﺎﹰ ﻋﻠﻰ ﻋﺩﺩ ﻤﻨﺘﻪ
ﻤﻥ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ،
e
ﺩﺩـﻌﻟ ﺓﻭﻗ ﻭﻫ
ﺃﻭﻟ
ﻲ
.
ﻫﺫﺍ ﻤﺎ ﻗﺎﻡ ﺒﻪ
Zelmanov
Efim
ﻓﻲ ﻋﺎﻡ
1989
ﻥـﻤ ﺭـﻜﺒﻤ لﻤﻋ ﺩﻌﺒ
Kostrikin
.
ﺍﻨﻅﺭ
Feit 1995
.
ﺨﻭﺍﺭﺯﻤﻴﺔ ﺘﻭﺩ
-
ﻜﻭﻜﺴﺘﻴﺭ
(Todd-Coxeter algorithm)
ﻴﻭﺠﺩ ﺒﻌﺽ ﺍﻟﺘﻘﺩﻴﻤﺎﺕ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﻤﻨﻅﻭﺭﺓ
ﺍﻟﺴﻠﻴﻤﺔ ﺘﻤﺎﻤﺎﹰ ﻭ
ﺭـﻤﺯ ﻑﺭـﻌﺘﻟ ﺔـﻤﻭﻠﻌﻤ ﻥﻭﻜﺘ ﻲﺘﻟﺍ
ﻟﻜﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟﺘﻠﻙ ﺍﻟﺘﻤﺜﻴﻼﺕ،ﹰﺎﻤﺎﻤﺘ ﺓﺭﻴﻐﺼ
ﺴﻴﻜﻭﻥ ﺒﺭﻫﺎﻥ ﻫﺫﺍ ﺼﻌﺒﺎﹰ ﺠﺩﺍﹰ
.
ﻲـﺴﺎﻴﻘﻟﺍ ﻡﺩـﻘﺘﻟﺍ ﻥﺇ
ﻟﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻫﺫﻩ ﺴﻴﻜﻭﻥ ﺒﺩﺭﺍﺴﺔ ﺨﻭﺍﺭﺯﻤﻴﺔ ﺘﻭﺩ
-
ﻜﻭﻜﺴﺘﻴﺭ
)
ﺍﻨﻅﺭ ﺍﻟﻔﺼل
ﺍﻟﺭﺍﺒﻊ
.(
ﺴﻭﻑ ﻨﻁﻭﺭ ﻁﺭﻕ ﻤﺨﺘﻠﻔﺔ
ﻟﺘﻤﻴﻴﺯ ﺍﻟﺯﻤﺭ ﻋﻥ ﺘﻘﺩﻴﻤﺎﺘﻬﺎ
)
ﺍﻨﻅﺭ ﺇﻟﻰ ﺍﻟﺘﻤﺎﺭﻴﻥ ﺃﻴﻀﺎﹰ
.(
Maple
ﺇﻥ ﻤﺎ
ﻴﻠﻲ ﻨﺴﺨﺔ ﻟﺠﻠﺴﺔ ﻤﺩﻭﻨﺔ ﻤﻥ
Maple
:
maple [This starts maple on a sun, pc, … ]
with(group) ; [This loads the group package, and lists
some of the available commands .]
G: =grelgroup ({a,b},{[a,a,a,a] , [b,b] , [b,a,b,a]}) ;
[This defines G to be the group with generators a , b and
relations aaaa , bb , and baba ; use 1/a for the inverse of a .]
grouporder (G) ;
[This attempts to find the order of the group G .]
H: =subgrel ({x =[a,a] ,y =[b]} , G) ;
[This defines H to be the subgroup of G with
generators x =aa and y = b]
pres (H) ; [This computes a presentation of H]
quit [This exits maple]
To go help on a command , type ? command
ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ
(Coxeter groups)
ﺇﻥ ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ
Coxeter system)
(
ﺯﻭﺝـﻟﺍ ﻭﻫ
(
)
,
G S
ﺭﺓـﻤﺯﻟﺍ ﻥـﻤ ﻑـﻟﺅﻤﻟﺍ
G
ـﺔ ﻤـﻋﻭﻤﺠﻤﻭ
ﺩﺍﺕـﻟﻭﻤﻟﺍ ﻥ
S
ﺭﺓـﻤﺯﻠﻟ
G
ﺸﻜلــﻟﺍ ﻥـﻤ ﺕﺎـﻗﻼﻌﻟ ﻁـﻘﻓ ﺔﻋﻭــﻀﻭﻤﻟﺍ
( )
( )
,
1
=
m s t
st
ﺤﻴﺙ،
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٦٤
(13)
( )
( )
( )
( )
,
1
,
,
2
,
,
=
≥
=
m s s
all s
m s t
m s t
m t s
ﻋﻨﺩﻤﺎ ﻻ ﺘﻭﺠﺩ ﻋﻼﻗﺔ ﺒﻴﻥ
s
ﻭ
t
ﻨﻀﻊ،
( )
,
m s t
= ∞
.
ﻴﻜﻭﻥ ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ ﻤﻌﺭﻓﺎﹰﺫﺌﺩﻨﻋ
ﺒﺎﻟﻤﺠﻤﻭﻋﺔ
S
ﻭﺍﻟﺘﻁﺒﻴﻕ
{ }
:
m S
S
× →
∞
U
ﻤﺤﻘﻘﺎﹰ
(13)
ﻋﻨﺩﺌﺫ،
G
S R
=
ﺤﻴﺙ
( )
( )
( )
{
}
,
,
=
< ∞
m s t
R
st
m s t
ﺇﻥ ﺯﻤﺭ ﻜ
ﻭﻜﺴﺘﻴﺭ ﻫﻲ ﺍﻟﺯﻤﺭ ﺍﻟﺘﻲ ﺘﻅﻬﺭ ﻜﺠﺯﺀ ﻤﻥ ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ
.
ﺔـﻋﻭﻤﺠﻤﻟﺍ ﺓﺭﺩﻗ ﻥﺇ
S
ﺘﺴﻤﻰ ﻤﺭﺘﺒﺔ
(rank)
ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ
.
ﺃﻤﺜﻠﺔ
11.2
ﺇﻥ ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ،لﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ
1
ﻫﻭ
{ }
(
)
2
,
C
s
ﻓﻘﻁ
.
12.2
ﺃﻨﻅﻤﺔ ﻜﻭﻜﺴﺘﻴﺭ ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ
2
ﻴﻜﻭﻥ
ﺩﻟﻴﻠﻬﺎ
( )
,
2
m s t
≥
.
(a)
ﺇﺫﺍ ﻜﺎﻥ
( )
,
m s t
ﻋﺩﺩﺍﹰ ﺼﺤﻴﺤﺎﹰ
n
ﻨﻅﺎﻡ ﻜﻭﻜﺴﺘﻴﺭ ﻫﻭﺫﺌﺩﻨﻋ ،
{ }
(
)
,
,
G s t
ﺤﻴﺙ
( )
2
2
, ;
,
,
n
n
G
s t s t
st
D
=
)
ﺍﻨﻅﺭ
2.9
.(
،ﺒﺸﻜل ﺨﺎﺹ
s
t
≠
ﻭ
st
ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ
n
.
(b)
ﺎﻥــﻜ ﺍﺫﺇ
( )
,
m s t
= ∞
ﻭــﻫ ﺭﻴﺘﺴــﻜﻭﻜ ﻡﺎــﻅﻨ ﺫــﺌﺩﻨﻋ ،
{ }
(
)
,
,
G s t
ﺙــﻴﺤ
2
2
, ;
,
G
s t s t
=
.
ﻨﻌﺘﺒﺭ ﺍﻟﺘﻁﺒﻴﻕ
{ }
( )
2
,
GL
→
s t
،
( )
( )
def
def
1 2
1
0
,
0
1
2
1
σ
σ
−
=
=
−
a
a
s
t
s
t
ﻜﻤﺎ ﺃﻥ
2
2
1
s
t
σ
σ
= =
ﺸﺎﻜلـﺘﻟﺍ ﻰـﻟﺇ ﻕـﻴﺒﻁﺘﻟﺍ ﺍﺫﻫ ﺩﺩﻤﻴ ،
( )
2
GL
→
G
.
ﺍﻵﻥ
( )
3
2
2
1
s
t
σ σ
−
=
−
ﻭ
ﻤﻨﻪ
( ) ( ) ( ) ( ) ( )
1
1
1
1
1
,
2
1
1
0
0
1
s
t
s
t
σ σ
σ σ
=
=
=
+
ﻭﺒﺎﻟﺘﺎﻟﻲ
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٧٤
(
)
( ) ( ) ( )
1
1
1
2
0
0
1
m
s
t
m
σ σ
=
+
ﺍﻟﺫﻱ ﻴﺒﻴﻥ ﺒﺄﻥ
s
t
σ σ
ﻤﻥ ﺭﺘﺒﺔ
12
ﻭﻤﻨﻪ ﻓﺈﻥ،ﺔﻴﻬﺘﻨﻤ ﺭﻴﻏ
st
ﻤﻥ ﺭﺘﺒﺔ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ ﺃﻴﻀﺎﹰ
.
13.2
ﻟﺘﻜﻥ
n
V
=
ﻤﺯﻭﺩﺍﹰ ﺒﺎﻟﺸﻜل ﺍﻟﺘﻨﺎﻅﺭﻱ ﺍﻟﺜﻨﺎﺌﻲ ﺍﻟﻤﻭﺠﺏ ﺍﻟﻘﺎﻨﻭﻨﻲ
( )
( )
(
)
1
1
,
i
i
i
i
i n
i n
x
y
x y
≤ ≤
≤ ≤
=
∑
ﺍﻻﻨﻌﻜﺎﺱ
(reflection)
ﻫﻭ ﺍﻟﺘﻁﺒ
ﻴﻕ ﺍﻟﺨﻁﻲ
:
s V
V
→
ﺭـﻴﻏ ﺕﺎـﻬﺠﺘﻤﻟﺍ لـﺴﺭﻴ ﻱﺫﻟﺍ
ﺍﻟﺼﻔﺭﻴﺔ
α
ﺇﻟﻰ
α
−
ﻭﻴﺜﺒﺕ ﻨﻘﺎﻁ ﺍﻟﺴﻁﻭﺡ ﺍﻟﺯﺍﺌﺩﺓ
H
α
ﻰـﻠﻋ ﺓﺩﻤﺎﻌﺘﻤﻟﺍ
α
.
ﺏـﺘﻜﻨ
s
α
ـﻟﻼﻨﻌﻜﺎﺱ ﺍﻟﻤﻌﺭﻑ ﺒ
α
ﻭﺍﻟﺫﻱ ﻴﻌﻁﻰ ﺒﺎﻟﺼﻴﻐﺔ،
( )
(
)
2
,
,
v
s v
v
α
α
α
α α
= −
ﻷﻥ ﻫﺫﺍ ﻴﻭﻀﺢ
α
ﻭ
H
α
ﻭﺒﺎﻟﺘﺎﻟﻲ،ﺩﺩﺤﻤ لﻜﺸﺒ
)
ﺒﺎﻟﺨﻁﻴﺔ
(
ﻋﻠﻰ ﻜل
V
H
α
α
=
⊕
.
ﺇﻥ
ﺍﻟﺯﻤﺭﺓ ﺍﻻﻨﻌﻜﺎﺴﻴﺔ ﺍ
ﻟﻤﻨﺘﻬﻴﺔ ﻫﻲ ﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﺎﻻﻨﻌﻜﺎﺴﺎﺕ
.
ﻟﺯﻤﺭ ﻤﺜل
G
ﻥ ﺃﻥـﻜﻤﻤﻟﺍ ﻥـﻤ ،
ﻨﺨﺘﺎﺭ ﻤﺠﻤﻭﻋﺔ
S
ﺘﻠﻙ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎـﻟ ﺓﺩﻟﻭﻤﻟﺍ ﺕﺎﺴﺎﻜﻌﻨﻻﺍ ﻥﻤ
(
)
,
G S
ﻨﻅﺎﻡ
Coxeter
.(Humphreys 1990, 1.9)
ﻓﺈﻥ ﺍﻟﺯﻤﺭ ﺍﻻﻨﻌﻜ،ﺍﺫﻬﻟ
ﺎﺴﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﺠﻤﻴﻊ ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ
)
، ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻨﻬﺎ ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
ibid.,6.4
.(
14.2
ﻟﺘﻜﻥ
n
S
ﺘﺅﺜﺭ ﻋﻠﻰ
n
، ﺒﺘﺒﺩﻴل ﺍﻹﺤﺩﺍﺜﻴﺎﺕ
(
)
( )
( )
(
)
1
1
,...,
,...,
n
n
a
a
a
a
σ
σ
σ
=
ﺍﻟﻤﻨﺎﻗﻠﺔ
( )
ij
ﺘﺒﺎﺩل ﺒﻴﻥ
i
ﻭ
j
ﻭﺘﺭﺴل ﺍﻟﻤﺘﺠﻪ،
0,..., 0,1, 0,..., 0, 1, 0,...
j
i
α
−
ﻭﻴﺘﺭﻙ ﻨﻘﺎﻁ ﺍﻟﺴﻁﺢ ﺍﻟﺯﺍﺌﺩ،ﻩﺭﻴﻅﻨ ﻰﻟﺇ
1
,...,
,...,
,...,
j
i
i
i
n
H
a
a
a
a
α
=
ﻤﺜﺒﺘﺎﹰ
.
ﻴﻜﻭﻥ،ﻙﻟﺫﻟ
( )
ij
ﺍﻨﻌﻜﺎﺴﺎﹰ
.
ﻜﺫﻟﻙ
n
S
ﻫﺫﺍ ﻴﺒ،ﺕﻼﻗﺎﻨﻤﻟﺎﺒ ﺓﺩﻟﻭﻤ
ﻴﻥ ﺒﺄﻨﻬﺎ ﺯﻤﺭﺓ ﺍﻨﻌﻜﺎﺴﻴﺔ
ﻤﻨﺘﻬﻴﺔ
)
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺯﻤﺭﺓ ﻜﻭﻜﺴﺘﻴﺭ
ﺃﻴﻀﺎﹰ
.(
ﺒﻨﻴﺔ ﺯﻤﺭ ﻜﻭﻜﺴﺘﻴﺭ
( The structure of Coxeter groups)
12
ـﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ ﺸﻜل ﺠﻭﺭﺩﺍﻥ ﺍﻟﻘﺎﻨﻭﻨﻲ ﻟ
s
t
σ σ
ﻫﻭ
( )
1
2
0
1
.
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٨٤
ﻤﺒﺭﻫﻨﺔ
15.2
ﻟﻴﻜﻥ
(
)
,
G S
ﻨﻅﺎﻡ
Coxeter
ﺍﻟﻤﻌﺭﻑ ﺒﺎﻟﺘﻁﺒﻴﻕ
{ }
:
m S
S
× →
∞
U
ﺍﻟﻤﺤﻘﻕ
(13)
.
(a)
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻘﺎﻨﻭﻨﻲ
S
G
→
ﻏﺎﻤﺭ
.
(b)
ﻜل
s
S
∈
ﻟﻪ ﺍﻟﺭﺘﺒﺔ
2
ﻓﻲ
G
.
(c)
ﻟﻜل
s
t
≠
ﻓﻲ
S
،
st
ﻤﻥ ﺍﻟﺭﺘﺒ
ﺔ
( )
,
m s t
ﻓﻲ
G
.
ﺴﻴﺸﻐل ﺍﻟﺒﺭﻫﺎﻥ ﺒﻘﻴﺔ ﻫﺫﻩ ﺍﻟﻔﻘﺭﺓ
.
ﻨﻼﺤﻅ ﺃﻥ ﺭﺘﺒﺔ
s
ﺘﺴﺎﻭﻱ
1
ﺃﻭ
2
ﻭﺇﻥ ﺭﺘﺒﺔ،
st
ﺘﻘﺴﻡ
( )
,
m s t
ﻭﻟﺫﻟﻙ ﺘﻨﺹ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻋﻠﻰ ﺃﻥ ﻋﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ،
S
ﺘﺒﻘﻰ ﻤﺨﺘﻠﻔﺔ ﻓﻲ
G
ﻭ ﺃﻥ ﻜل
s
ﻭﻜل
st
ﻟﻬﻤﺎ ﺍﻜﺒﺭ ﺭﺘﺒﺔ ﻤﻤﻜﻨﺔ
.
ﻟﻴﻜﻥ
ε
ﺍﻟﺘﻁﺒﻴﻕ
{ }
1
S
→ ±
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
1
s
ε
= −
ﻟﻜل
s
.
ﻋﻨﺩﺌﺫ
ε
لـﺴﺭﻴ
st
ﺇﻟﻰ
1
ﻷﻱ
,
s t
S
∈
ﻭﻟﺫﻟﻙ ﻓﻬﻭ ﻴﻤﺩﺩ ﺇﻟﻰ ﺘﺸﺎﻜل ﺍﻟﺯﻤﺭ،
{ }
1
G
→ ±
)
ﺍﻨﻅﺭ
8.2
.(
ﻜل
s
ﻴﻁﺎﺒﻕ
-1
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
2
.
ﻫﺫﺍ ﻴﺒﺭﻫﻥ
(b)
ﻭﻟﻜﻲ ﻨﺒﺭﻫﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺒﺎﻗﻴﺔ ﺴﻨﻌﺘﺒﺭ،
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
V
ﻤﻊ ﺍﻟﻘﺎﻋﺩﺓ
( )
s
s S
e
∈
ﺒﺩﻟﻴل
s
.
ﻰـﻠﻋ ﻑﺭﻌﻨ
V
"
ﻴﺎﹰـﺴﺩﻨﻫ
"
ﺎـﻬﻟ ﺩـﺠﻭﻴ ﻲـﺘﻟﺍ
"
ﺎﺕـﺴﺎﻜﻌﻨﺍ
"
ﺔـﻔﻠﺘﺨﻤ
,
s
s
S
σ
∈
ﺒ،
ﺤﻴﺙ ﺇﻥ
s
t
σ σ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
( )
,
m s t
.
ﻤﻥ
(8.2)
ﻕـﻴﺒﻁﺘﻟﺍ ﺩﺩﻤﻴ ،
s
s
σ
→
ﺇﻟﻰ ﺍﻟﺘﺸﺎﻜل ﺍﻟﺯﻤﺭﻱ
( )
GL
→
G
V
.
ﻜﻤﺎ ﺃﻥ
s
σ
ﻤﺨﺘﻠﻔﺔ
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ،
s
ﻲـﻓ ﹰﺎﻔﻠﺘﺨﻤ
G
ﻭ،
s
t
σ σ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
( )
,
m s t
ﻭﻴ،
ﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﻌﻨﺼﺭ
st
ﻤﻥ ﺍﻟﺭﺘﺒﺔ ﻨﻔﺴﻬﺎ
.
ﻨﻌﺭﻑ ﺍﻟﺸﻜل
B
ﺍﻟﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ ﺍﻟﺘﻨﺎﻅﺭﻱ ﻋﻠﻰ
V
ﺒﺎﻟﻘﺎﻋﺩﺓ ﺍﻵ
ﺘﻴﺔ
(
)
( )
(
)
( )
cos
,
,
B
,
1
s
t
m s t
if m s t
e e
π
−
≠ ∞
=
−
ﻜﻤﺎ ﺃﻥ
(
)
B
,
1
0
= ≠
s
t
e e
ﺇﻥ ﺍﻟﻤﺘﻤﻡ ﺍﻟﻌ،
ـﻤﻭﺩﻱ ﻟ
s
e
ﻤﻊ ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ
B
ﺴﻁﺢـﻟﺍ ﻭـﻫ
ﺍﻟﺯﺍﺌﺩ ﺍﻟﺫﻱ ﻻ ﻴﺤﻭﻱ
s
e
ﻭ ﻟﺫﻟﻙ،
s
s
V
e
H
=
⊕
.
ﺭﻑـﻌﻨ ﻥﺄﺒ ﺢﻤﺴﻴ ﺍﺫﻫﻭ
”
ﺎﺱـﻜﻌﻨﻻﺍ
"
ﺒﺎﻟﻘﺎﻋﺩﺓ
(
)
2
,
,
s
s
s
v
v
B v e e v
V
σ
= −
∈
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻥ
s
σ
ﻫﻭ
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ ﺍﻟﺫﻱ ﻴﺭﺴل
s
e
ﺎﻁـﻘﻨ ﺕﺒﺜﻤﻭ ﻩﺭﻴﻅﻨ ﻰﻟﺇ
s
H
،
ﺇﺫﺍﹰ
2
1
s
σ
=
ﻓﻲ
( )
GL V
.
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺒﺄﻥ
s
σ
ﻭﻤﻨﻪ ﺒﻘﻲ، ﻑﻠﺘﺨﻤ
ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ
s
t
σ σ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
( )
,
m s t
.
ﻟﻜل
,
s t
S
∈
ﻟﻴﻜﻥ،
,
s t
V
ﺍﻟﻔﻀﺎﺀ ﺫﻭ
2
-
ﺒﻌﺩ
s
t
e
e
⊕
.
ﻏﻴﺭ ﺫﻟﻙ
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٩٤
ﺘﻤﻬﻴﺩﻴﺔ
16.2
ﺇﻥ ﻤﻘﺼﻭﺭ
B
ـ ﻟ
,
s t
V
ﻤﻭﺠﺏ ﺘ
ﺎﻥـﻜ ﺍﺫﺇ ﹰﺎـﻤﺎﻤﺘ ﺏـﺠﻭﻤ ﻥﻭـﻜﻴﻭ ،ﹰﺎﺒﻴﺭﻘ
( )
,
m s t
≠ ∞
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
,
s
t
s t
v
ae
be
V
=
+
∈
.
ﺇﺫﺍ ﻜﺎﻥ
( )
,
m s t
≠ ∞
ﻋﻨﺩﺌﺫ،
( )
(
)
(
)
(
)
(
)
2
2
2
2
2
,
2
cos
cos
sin
0
B v v
a
ab
m
b
a b
m
b
m
π
π
π
=
−
+
= −
+
>
ﻷﻥ
(
)
sin
0
m
π
≠
.
ﺇﺫﺍ ﻜﺎﻥ
( )
,
= ∞
m s t
ﻋﻨﺩﺌﺫ،
( )
(
)
2
2
2
B
,
2
0
=
+
+
= +
≥
v v
a
ab
b
a
b
ﺘﻤﻬﻴﺩﻴﺔ
17.2
ﺍﻟﻤﻘﺼﻭﺭ ﻤﻥ
st
ﺇﻟﻰ
,
s t
V
ﻟﻪ ﺍﻟﺭﺘﺒﺔ
( )
,
m s t
.
ﺎﻥــﻫﺭﺒﻟﺍ
.
ﺎﻥــﻜ ﺍﺫﺇ
( )
,
m s t
≠ ∞
ﺸﻜلــﻟﺍ ﻥﺈــﻓ ،
,
s t
B V
ﻭ،ﹰﺎــﻤﺎﻤﺘ ﺏــﺠﻭﻤ
ﺫﻟﻙــﻟ
(
)
,
,
,
s t
s t
V
B V
ﻫﻭ ﻓﻀﺎﺀ ﺇﻗﻠﻴﺩﻱ
.
،ﻋﻼﻭﺓﹰ ﻋﻠﻰ ﺫﻟﻙ
s
σ
ﻭ
t
σ
ﺍﻨﻌﻜﺎﺴﺎﺕ ﻋﻠﻰ
,
s t
V
ﻲـﻓ
ﺎﻻﺕــــــﺤ
(13.2)
ﺎﻟﻤﺘﺠﻬﻴﻥــــــﺒ ﻑﺭــــــﻌﻤ
s
e
ﻭ
t
e
.
ﺎ ﺃﻥــــــﻤﻜ
(
)
( )
(
)
( )
(
)
,
cos
,
cos
,
s
t
B e e
m s t
m s t
π
π π
= −
=
−
ﺍﻟﺯﺍﻭﻴﺔ ﺒﻴﻥ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ،
ـﺍﻟﻤﺜﺒﺘﺔ ﺒ
s
e
ﻭ
t
e
ﺘﺴﺎﻭﻱ
( )
,
m s t
π
.
ﻤﻥ
(16.1)
ﻨﺭﻯ ﺒﺄﻥ،
s
σ
ﻭ
t
σ
ﺘﻭﻟﺩﺍﻥ ﺯﻤﺭﺓ
ﺩﻴﻬﻴﺩﺭﺍل
( )
,
m s t
D
ﻭ ﺃﻥ
s
t
σ σ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
( )
,
m s t
.
ﺇﺫﺍ ﻜﺎﻥ
( )
,
m s t
= ∞
ــ ﺍﻟﻤﺼﻔﻭﻓﺘﺎﻥ ﻟﺫﺌﺩﻨﻋ ،
s
σ
ﻭ
t
σ
ﻰـﻟﺇ ﺔﺒﺴـﻨﻟﺎﺒ
ﺩﺓـﻋﺎﻘﻟﺍ
{
}
,
s
t
e e
ﺍﻟﻤﻭﺠﻭﺩﺓ ﻓﻲ
(12.2)
ﻤﻥ ﺫﻟﻙ ﻴﻨﺘﺞ ﺒﺄﻥ،
s
t
σ σ
ﻤﻥ ﺭﺘﺒﺔ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.
ﺒﻤﺎ ﺃﻥ ﺭﺘﺒﺔ
st
ﺘﻘﺴﻡ ﺭﺘﺒﺔ
( )
,
m s t
ﻓﺎﻟﺘﻤﻬﻴﺩﻴﺔ ﺘﺒﻴﻥ ﺒﺄﻨﻪ ﻴﺴﺎﻭﻱ،
( )
,
m s t
.
ﺔــﻅﺤﻼﻤ
18.2
ﺇﻥ
ﺸﺎﻜلــﺘﻟﺍ
( )
GL
G
V
→
ﺔــﻨﻫﺭﺒﻤﻟﺍ ﻥﺎــﻫﺭﺒ ﻲــﻓ
2.12
ﺎﻤﺭــﻏ
(Humphreys 1990, 5.4)
ﻭﻟﻜﻥ ﻤﻥ ﺍﻟﺼﻌﺏ ﺒﺭﻫﺎﻨﻪ،
.
ﺘﻤﺎﺭﻴﻥ
2
-
1
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﻤﻭﻟﺩﺍﺕ
1
,...,
n
a
a
ﻭﺍﻟﻌﻼﻗﺎﺕ
,
1,
i
j
a a
i
j
=
≠
ﻲـﻫ ،
ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ
ﺤﺭﺓ ﻋﻠﻰ
1
,...,
n
a
a
] .
ﺍﺴﺘﺨﺩﻡ ﺍﻟﺨﻭﺍﺹ ﺍﻟﻌﺎﻤﺔ
Hinit
.[
2
-
2
ﻟﻴﻜﻥ
a
ﻭ
b
ﻋﻨﺼﺭﻴﻥ ﻤﻥ ﺯﻤﺭﺓ ﺍﺨﺘﻴﺎﺭﻴﺔ ﺤﺭﺓ
F
.
ﺒﺭﻫﻥ ﺃﻥ
:
(a)
ﺇﺫﺍ ﻜﺎﻥ
n
n
a
b
=
ﻤﻊ
1
n
>
ﻋﻨﺩﺌﺫ،
a
b
=
.
(b)
ﺇﺫﺍ ﻜﺎﻥ
m
n
n
m
a b
b a
=
ﻤﻊ
0
mn
≠
ﻋﻨﺩﺌﺫ،
ab
ba
=
.
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٠٥
(c)
ﺇﺫﺍ ﻜﺎﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ
n
x
a
=
ﺍﻟﺤل
x
ﻟﻜل
n
ﻋﻨﺩﺌﺫ،
1
a
=
.
2
-
3
ﻟﻴﻜﻥ
n
F
ﻴﺭﻤﺯ ﻟﺯﻤﺭﺓ ﺤﺭﺓ ﻋﻠﻰ
n
ﻤﻭﻟﺩ
.
ﺒﺭﻫﻥ
:
(a)
ﺇﺫﺍ ﻜﺎﻥ
n
m
<
،
ﻋﻨﺩﺌﺫ
n
F
ﺘﺘﻤﺎﺜل ﻤﻊ ﻜل ﻤﻥ ﺍﻟﺯﻤﺭﺓ
ــﺴﻤﺔ ﻟـﻘﻟﺍ ﺓﺭﻤﺯﻭ ﺔﻴﺌﺯﺠﻟﺍ
m
F
.
(b)
ﺒﺭﻫﻥ ﺃﻥ
1
1
F
F
×
ﻟﻴﺴﺕ ﺯﻤﺭﺓ ﺤﺭﺓ
.
(c)
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺭﻜﺯ
( )
1
n
Z F
=
ﺼﺤﻴﺢ ﻤﻥ ﺃﺠل
1
n
>
.
2
-
4
ﺒﺭﻫﻥ ﺃﻥ
n
Q
)
ﺍﻨﻅﺭ
2.7b
(
ﺘﺤﻭ
ﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺤﻴﺩﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭﺍﻟﺘﻲ ﻫﻲ،
( )
n
Z Q
.
ﺒﺭﻫﻥ ﺒﺄﻥ
( )
n
n
Q
Z Q
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ
1
2
n
D
−
.
2
-
5
(a)
ﻟﺘﻜﻥ
( )
4
2
2
, ;
,
,
G
a b a b
ab
=
.
ﺒﺭﻫﻥ ﺃﻥ
G
ﺘﻤﺎﺜل ﺯﻤﺭﺓ ﺩﻴﻬﻴﺩﺭﺍل
4
D
.
(b)
ﺒﺭﻫﻥ ﺃﻥ
2
, ;
,
G
a b a abab
=
ﺯﻤﺭﺓ ﻏﻴﺭ
ﻤﻨﺘﻬﻴﺔ
) .
ﺭﺓـﻤﺯﺒ ﹰﺓﺩﺎـﻋ ﻑﺭﻌﺘ ﻩﺫﻫ
ﺩﻴﻬﻴﺩﺭﺍل ﻏﻴﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
.(
2
-
6
ﻟﺘﻜﻥ
3
3
4
1
1
1
1
, , ;
,
,
,
,
G
a b c a b c acac
aba bc b
−
−
−
−
=
.
ﺒﺭﻫﻥ ﺃﻥ
G
ﻫﻲ ﺍﻟﺯﻤﺭﺓ
ﺍﻟﺘﺎﻓﻬﺔ
{ }
1
.
]
(
) (
)
3
3
1
1
aba
bcb
−
−
=
.[Hinit Expand
2
-
7
ﻟﺘﻜﻥ
F
ﺯﻤﺭﺓ ﺤﺭﺓ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
,
x y
ﻭﻟﺘﻜﻥ
2
G
C
=
ﺩـﻟﻭﻤﻟﺍ ﻊﻤ ،
1
a
≠
.
ﻭﻟﻴﻜﻥ
α
ﺍﻟﺘﺸﺎﻜل
F
G
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
( )
x
a
y
α
α
= =
.
ﺩﺓـﻟﻭﻤﻟﺍ ﺔﻋﻭﻤﺠﻤﻟﺍ ﺩﺠﻭﺃ
ﺍﻷﺼﻐﺭﻴﺔ ﻟﻨﻭﺍﺓ
α
.
؟ﻭﻫل ﺍﻟﻨﻭﺍﺓ ﺯﻤﺭﺓ ﺤﺭﺓ
2
-
8
ﻟﺘﻜﻥ
1 3
5
, ;
G
s t t s t
s
−
=
=
.
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻌﻨﺼﺭ
1
1
1
1
g
s t s tst st
− −
−
−
=
ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﻨﻭﺍﺓ ﻜل ﺘﻁﺒﻴﻕ ﻤﻥ ﺍﻟﺯﻤﺭ
ﺓ
G
ﺇﻟﻰ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.
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١٥
ﺍﻟﻔﺼل ﺍﻟﺜﺎﻟﺙ
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻭﺍﻟﺘﻤﺩﻴﺩﺍﺕ
Automorphisms and extensions
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭ
(Automorphisms of groups)
ﺇﻥ ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ
(Automorphism)
ﻟﺯﻤﺭﺓ
G
ﻫﻭ ﺘﻤﺎﺜل ﻤﻥ ﻫﺫﻩ
ﺴﻬﺎـﻔﻨ ﻰـﻟﺇ ﺓﺭـﻤﺯﻟﺍ
.
ﺍﻟﻤﺠﻤﻭﻋﺔ
( )
ut
A
G
ﻤﻥ ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭﺓ
G
ﺏـﻴﻜﺭﺘ ﺔـﻴﻠﻤﻌﻟ ﺔﺒﺴﻨﻟﺎﺒ ﺓﺭﻤﺯ ﺢﺒﺼﺘ
ﺍﻟﺘﻁﺒﻴﻘﺎﺕ
:
ﺔـﻴﻠﻤﻋ ﻭـﻫ ﺕﺎـﻘﻴﺒﻁﺘﻟﺍ ﺏﻴﻜﺭﺘ ،ﹰﺎﻀﻴﺃ ﻲﺘﺍﺫ لﺜﺎﻤﺘ ﻭﻫ ﻥﻴﻴﺘﺍﺫ ﻥﻴﻠﺜﺎﻤﺘ ﺏﻴﻜﺭﺘ ﻥﺇ
ﺘﺠﻤﻴﻌﻴﺔ ﺩﻭﻤﺎﹰ
)
ﺍﻨﻅﺭ
(5)
(
ﺍﻟﺘﻁﺒﻴ،
ﻕ ﺍﻟﻤﻁﺎﺒﻕ
g
g
a
ﺫﺍﺘﻲـﻟﺍ لﺜﺎﻤﺘﻟﺍ ،ﺩﻴﺎﺤﻤﻟﺍ ﺭﺼﻨﻌﻟﺍ ﻭﻫ
ﻭﺍﻟﺫﻱ ﻫﻭ ﺃﻴﻀﺎﹰ ﺘﻤﺎﺜل ﺫﺍﺘﻲ،ﺱﻭﻜﻌﻤ ﻪﻟ ﻥﻭﻜﻴﺴ ﻲﻟﺎﺘﻟﺎﺒﻭ ،ﹰﻼﺒﺎﻘﺘ ﻥﻭﻜﻴ
.
ﻟﻜل
g
G
∈
ﺍﻟﺘﻁﺒﻴﻕ،
g
i
"
ـﺍﻟﺘﺭﺍﻓﻕ ﺒﺎﻟﻨﺴﺒﺔ ﻟ
g
"،
1
:
x
gxg
G
G
−
→
a
ـﻫﻭ ﺘﻤﺎﺜل ﺫﺍﺘﻲ ﻟ
G
.
ﺩﺍﺨﻠﻲـﻟﺍ ﻲﺘﺍﺫﻟﺍ لﺜﺎﻤﺘﻟﺎﺒ لﻜﺸﻟﺍ ﺍﺫﻫ ﻥﻤ ﻲﺘﺍﺫﻟﺍ لﺜﺎﻤﺘﻟﺍ ﻰﻤﺴﻴ
(inner
Automorphisms)
ﺎﺭﺠﻲــﺨﻟﺍ لــﺜﺎﻤﺘﻟﺎﺒ ﻰﻤﺴــﻴﻓ ﺭــﺨﻵﺍ لــﺜﺎﻤﺘﻟﺍ ﺎــﻤﺃ ،
(outer
Automorphisms)
.
ﻨﻼﺤﻅ ﺃﻥ
( ) ( )
(
)
1
1
1
gh x gh
g hxh
g
−
−
−
=
ﺃﻱ ﺃﻥ،
( )
(
)
( )
gh
g
h
i
x
i
i
x
=
o
ﻭﻟﺫﻟﻙ ﻓﺈﻥ ﺍﻟﺘﻁﺒﻴﻕ
( )
:
ut
g
g
i
G
A
G
→
a
ﺘﺸﺎﻜل
.
ﻴﺭﻤﺯ ﻟﺼﻭﺭﺘﻪ ﺒﺎﻟﺸﻜل
( )
Inn G
.
ﻨﻭﺍﺘﻪ ﻫﻲ ﻤﺭﻜﺯ
G
،
( ) {
}
;
,
Z G
g
G gx
xg
x
G
=
∈
=
∀ ∈
ﻭﺒﻬﺫﺍ ﻨﺤﺼل ﻤﻥ
(44 .1)
ﻋﻠﻰ ﺍﻟﺘﻤﺎﺜل
( )
( )
Inn
→
G Z G
G
ﺇﻥ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
( )
Inn G
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻭﻫ
( )
Aut G
:
لـﻜﻟ
g
G
∈
ﻭ
( )
Aut G
α
∈
،
(
)
( )
( )
(
)
( )
( )
( )( )
1
1
1
1
.
.
. .
g
g
x
i
x
g
x g
g x
g
i
α
α
α
α
α
α
α
−
−
−
−
=
=
=
o
o
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٢٥
ﻤﺜﺎل
1.3
(a)
ﻟﺘﻜﻥ
p
n
G
=
F
.
ﺇﻥ ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭﺓ
G
ﺎﹰـﻤﺎﻤﺘ ﻲﻫ ﺔﻴﻠﻴﺩﺒﺘﻟﺍ ﺭﻤﺯﻟﺎﻜ
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭﺓ
G
ﻜﺎﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ ﻋﻠﻰ
p
F
ﺫﻟﻙـﻟ ،
( )
( )
Aut
GL
n
p
G
=
F
.
ﺒﻤﺎ ﺃﻥ
G
،ﺘﺒﺩﻴﻠﻴﺔ
ﻋﻨﺩﺌﺫ
ﻜل ﺍﻟﺘﻤﺎﺜﻼﺕ ﻏﻴﺭ ﺍﻟﺘﺎﻓﻬﺔ ﻓﻲ
G
ﺴﺘﻜﻭﻥ ﺨﺎﺭﺠﻴﺔ
.
(b)
ـ ﻜﺫﻟﻙ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺨﺎﺼﺔ ﻟ
(a)
ﻨﺭﻯ ﺒﺄﻥ،
(
)
( )
2
2
2
2
Aut
GL
C
C
×
=
F
.
(c)
ﺒﻤﺎ ﺃﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
Q
ﻫﻲ
2
a
ﻋ،
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎﺫﺌﺩﻨ
( )
2
2
2
Inn Q
Q
a
C
C
≈
×
،ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ
( )
4
Aut Q
S
≈
.
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
5-3
.
ﺍﻟﺯﻤﺭ ﺍﻟﺘﺎﻤﺔ
(Complete groups)
ﻑـــﻴﺭﻌﺘ
2.3
ﺔـــﻤﺎﺘ ﺓﺭـــﻤﺯﻟﺍ ﻥﻭـــﻜﺘ
(Complete)
ﻕـــﻴﺒﻁﺘﻟﺍ ﻥﺎـــﻜ ﺍﺫﺇ
( )
:
Aut
g
g
i
G
G
→
a
ﺘﻤﺎﺜﻼﹰ
.
ﻟ
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺘﺎﻤﺔ ﺇﺫﺍ،ﻙﻟﺫ
ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
(a)
ﺍﻟﻤﺭﻜﺯ
( )
Z G
ﻟﻠﺯﻤﺭﺓ
G
ﻭ،ﹰﺎﻬﻓﺎﺘ
(b)
ﻜل ﺘﻤﺎﺜل ﺫﺍﺘﻲ ﻟﻠﺯﻤﺭﺓ
G
ﻴﻜﻭﻥ ﺩﺍﺨﻠﻴﺎﹰ
.
ﻤﺜﺎل
3.3
(a)
ﻟﻜل
2, 6
n
S
،n
≠
ﺯﻤﺭﺓ ﺘﺎﻤﺔ
.
ﺒﻤﺎ ﺃﻥ ﺍﻟﺯﻤﺭﺓ
2
S
ﺎﻟﻲـﺘﻟﺎﺒﻭ ﺔـﻴﻠﻴﺩﺒﺘ
(a)
ﻻ
،ﻴﺘﺤﻘﻕ
( )
( )
6
6
2
Aut
Inn
S
S
C
≈
ﻭﺒﺎﻟﺘﺎﻟﻲ
6
S
ﻻ ﻴﺤﻘﻕ
(b)
.
ﺭـﻅﻨﺍ
Rotman 1995
،
ﺍﻟﻤﺒﺭﻫﻨﺔ
7.5,7.10
.
(b)
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻋﻨﺩﺌﺫ،ﺔﻁﻴﺴﺒ ﺭﻴﻏ ﺔﻴﻠﻴﺩﺒﺘ ﺓﺭﻤﺯ
( )
Aut G
ﺘ
ﺔـﻤﺎ
.
ﺭـﻅﻨﺍ
Rotman
1995
ﺍﻟﻤﺒﺭﻫﻨﺔ،
.7.14
ﻤﻥ ﺍﻟﺘﻤﺭﻴﻥ
3-4
،
( )
2
2
3
GL
S
≈
F
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻟﺯﻤﺭ ﻏﻴﺭ ﺍﻟﻤﺘﻤﺎﺜﻠﺔ،
2
2
C
C
×
ﻭ
3
S
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻥ ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ
.
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﺍﻟﺯﻤﺭ ﺍﻟﺩﺍﺌﺭﻴﺔ
(Automorphisms of cyclic groups)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻭﻟﺘﻜﻥ،
G
a
=
.
ﻟﻴﻜﻥ
m
ﻋﺩﺩﺍﹰ ﺼﺤﻴﺤﺎﹰ
≤
1
.
ﺇﻥ
ﺍﻟﻤﻀﺎﻋﻑ ﺍﻷﺼﻐﺭ ﻟﻠﻌﺩﺩ
m
ﺍﻟﺫﻱ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
n
ﻫﻭ
(
)
.
gcd
,
n
m
m n
.
،ﻟﺫﻟﻙ
m
a
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
(
)
gcd
,
n
m n
ﺩﺍﺕـﻟﻭﻤ ﻥﺈـﻓ ﻲﻟﺎـﺘﻟﺎﺒﻭ ،
G
ﺭـﺼﺎﻨﻌﻟﺍ ﹰﺎـﻤﺎﻤﺘ ﻲـﻫ
m
a
ﻭ
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٣٥
(
)
gcd
,
1
m n
=
.
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ
α
ﻟﻠﺯﻤﺭﺓ
G
ﻴﺠﺏ ﺃﻥ ﻴﺭﺴل
a
ﺭﺓـﻤﺯﻠﻟ ﺭﺨﺁ ﺩﻟﻭﻤ ﻰﻟﺇ
G
ﻭﺒﺎﻟﺘﺎﻟﻲ،
( )
m
a
a
α
=
ﻟﻤﺠﻤﻭﻋ
ﺔ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
m
ﻊـﻤ ﹰﺎﻴﺒﺴـﻨ ﺔﻴﻟﻭﻷﺍ
n
.
ﻕـﻴﺒﻁﺘﻟﺍ
m
α a
ﻴﻌﺭﻑ ﺘﻤﺎﺜﻼﹰ
( ) (
)
Aut
n
C
n
×
→
Z n Z
ﺤﻴﺙ
ﻴﻌﺘﻤﺩ ﻫﺫﺍ ﺍﻟﺘﻤﺎﺜل ﻋﻠﻰ ﺍﺨﺘﻴﺎﺭ ﺍﻟﻤﻭﻟﺩﺍﺕ
a
ﻟﻠﺯﻤﺭﺓ
G
:
ﺇﺫﺍ ﻜﺎﻥ
( )
m
a
a
α
=
ﻤﻥﺫﺌﺩﻨﻋ ،
ﺃﺠل ﺃﻱ ﻋﻨﺼﺭ ﺁﺨﺭ
i
b
a
=
ﻟﻠﺯﻤﺭﺓ
G
،
( )
( )
( )
( )
( )
m
i
m
i
mi
i
b
a
a
a
a
b
α
α
α
=
=
=
=
=
ﺒﻘﻲ ﺃﻥ ﻨﺤﺩﺩ
(
)
n
×
Z n Z
.
ﺇﺫﺍ ﻜﺎﻥ
1
1
...
s
r
r
s
n
p
p
=
ﺘﺤﻠﻴل
n
ﺩﺍﺩـﻋﻷ ﻯﻭـﻗ ﺀﺍﺩﺠ ﻰﻟﺇ
ﻋﻨﺩﺌﺫ،ﺔﻴﻟﻭﺃ
(
)
1
1
1
...
,
mod
mod
,...
s
r
r
r
s
n
p
p
m
n
m
p
× ×
↔
Z
Z
Z
Z
Z
Z
ﺤﺴﺏ ﻤﺒﺭﻫﻨﺔ ﺍﻟﺒﻭﺍﻗﻲ ﺍﻟﺼﻴﻨﻴﺔ
.
ﻭﻟﺫﻟﻙ،ﺕﺎﻘﻠﺤ لﺜﺎﻤﺘ ﻥﻋ ﺓﺭﺎﺒﻋ ﺍﺫﻫ ﻥﺈﻓ
(
)
(
)
(
)
1
1
...
s
r
r
s
n
p
p
×
×
×
× ×
Z n Z
Z n Z
Z n Z
ﺒﻘﻲ ﺃﻥ ﻨﺄﺨﺫ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
r
n
p
=
،
p
ﺃﻭﻟﻲ
.
ﻨﻔﺭﺽ ﺃﻭﻻﹰ ﺃﻥ
p
ﻋﺩﺩ ﻓﺭﺩﻱ
.
ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
0,1,...,
1
r
p
−
ﻤﺠﻤﻭﻋﺔ ﺘﺎﻤﺔ ﻤﻥ ﻤﻤﺜﻼﺕ
r
p
Z
Z
ﻭ،
1
p
ﻰـﻠﻋ ﺔﻤﺴﻘﻠﻟ ﺔﻠﺒﺎﻘﻟﺍ ﺭﺼﺎﻨﻌﻟﺍ ﻩﺫﻫ ﻥﻤ
p
.
ﺎﻟﻲـﺘﻟﺎﺒﻭ
(
)
r
p
×
Z n Z
ﻥـﻤ
ﺍﻟﺭﺘﺒﺔ
(
)
1
1
r
r
r
p
p
p
p
p
−
−
=
−
.
ﻷﻥ
1
p
−
ﻭ
r
p
ﻥـﻤ ﻡﻠﻌﻨ ﻥﺤﻨ ،ﹰﺎﻴﺒﺴﻨ ﻥﺎﻴﻟﻭﺃ
(12)
ﺒﺄﻥ
(
)
r
p
×
Z n Z
ﻤﺘﻤﺎﺜل ﻤﻊ ﺍﻟﺠﺩﺍﺀ ﺍﻟﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺓ
A
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
p
−
ﻭﺍﻟﺯﻤﺭﺓ
B
ﻤﻥ
ﺍﻟﺭﺘﺒﺔ
1
r
p
−
.
ﻤﻥ ﺍﻟﺘﻁﺒﻴﻕ
(
)
(
)
r
p
p
p
×
×
×
→
=
Z n Z
Z n Z
F
ﻴﻨﺘﺞ ﺍﻟﺘﻤﺎﺜل
p
A
×
→
F
ﻭ ﻤﻥ ﻜﻭﻥ،
p
×
F
ﺯﻤﺭﺓ ﺠﺯ،
،لـﻘﺤﻠﻟ ﺔﻴﺒﺭﻀﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻥﻤ ﺔﻴﻬﺘﻨﻤ ﺔﻴﺌ
ﻭﻫﻲ ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ
(1.55)
.
ﻟﺫﻟﻙ
(
)
r
p
A
ξ
×
⊃ =
Z n Z
ﻤﻥ ﺃﺠل ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
ξ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
p
−
.
ﻨﺠﺩ ﺃﻭﻻﹰ ﺒﺄﻥ،ﻥﻴﺩﺤﻟﺍ ﻱﺫ ﺔﻴﺭﻅﻨ ﻡﺍﺩﺨﺘﺴﺎﺒ
1 p
+
ﻤﻥ
ﺔـﺒﺘﺭﻟﺍ
1
r
p
−
ﻓﻲ
(
)
r
p
×
Z n Z
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻭ ﻴﻭﻟﺩ،
B
.
ﻭﺒﻬﺫﺍ ﺘﻜﻭﻥ
(
)
r
p
×
Z n Z
ﺩﻫﺎـﻟﻭﻤ ﻊـﻤ ،ﺔﻴﺭﺌﺍﺩ
(
)
. 1
p
ξ
+
ﻭﻴﻤﻜﻥ ﺃﻥ ﻴﻜﺘﺏ ﻜل ﻋﻨﺼﺭ ﻭﺒﺸﻜل ﻭﺤﻴﺩ ﺒﺎﻟﺸﻜل،
(
) {
}
(
)
{
}
, gcd
,
1
n
m
n
m n
×
=
=
+
=
Z
Z
Z
ﻋﻨﺎﺼﺭ ﺍﻟﻭﺤﺩﺓ ﻓﻲ ﺍﻟﺤﻠﻘﺔ
n
Z
Zh
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٤٥
(
)
1
. 1
, 0
1, 0
j
i
r
p
i
p
j
p
ξ
−
+
≤ < −
≤ <
ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ
(
)
{
}
2
2
8
1, 3, 5, 7
3, 5
C
C
×
=
=
≈
×
Z Z
ﻟﻴﺴﺕ ﺩﺍﺌﺭﻴﺔ
.
ﺍﻟﻤﻠﺨﺹ
4.3
(a)
ﻟﻜل ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻓﻲ
G
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
n
،
( ) (
)
Aut G
n
×
→
Z n Z
.
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﻓﻲ
G
ﻴﻘﺎﺒل
[ ]
(
)
m
n
×
∈
Z n Z
ﻴﺭﺴل ﺍﻟﻌﻨﺼﺭ
a
ﻓﻲ
G
ﺇﻟﻰ
m
a
.
(b)
ﺇﺫﺍ ﻜﺎﻥ
1
1
...
s
r
r
s
n
p
p
=
؛
ﺘﺤﻠﻴل
n
ﺇﻟﻰ ﺠﺩﺍﺀ ﻗﻭﻯ ﻷﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ
i
p
ﻋﻨﺩﺌﺫ،
(
)
1
1
1
...
,
mod
mod
,...
s
r
r
r
s
n
p
p
m
n
m
p
× ×
↔
Z
Z
Z
Z
Z
Z
(c)
(
)
(
)
1
2
2
1
2
2
2
2
2,
2
r
r
r
p
p
r
p
p
r
C
p
p
C
C
C
−
−
−
×
=
=
>
≈
×
Z n Z
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ
(Characteristic subgroups)
ﺘﻌﺭﻴﻑ
5.3
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ
(Characteristic subgroup)
ﺭﺓـﻤﺯﻟﺍ ﻲﻓ
G
ﻲـﻫ
ﺍﻟﺯﻤﺭﺓ
H
ﺍﻟﺘﻲ ﺘﺤﻘﻕ
( )
H
H
α
=
ﻟﻜل ﺘﻤﺎﺜل ﺫﺍﺘﻲ
α
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
.
ﻭﺒﻨﻔﺱ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ
(31.1)
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ
( )
H
H
α
⊂
ﻟﻜل
( )
Aut G
α
∈
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ،ﺍﺫﻬﻟ
H
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻲـﻓ ﻲﻠﺨﺍﺩ لﺜﺎﻤﺘ لﻜﻟ ﺔﺒﺴﻨﻟﺎﺒ ﺔﺘﺒﺎﺜ ﺕﻨﺎﻜ ﺍﺫﺇ
G
ﻭﺘﻜﻭﻥ ﻤﻤﻴﺯﺓ ﺇﺫﺍ ﻜﺎﻨﺕ ﺜﺎﺒﺘﺔ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻜل ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ،
.
ﺭﺓـﻤﺯﻟﺍ ﻥﺇ ،ﺹﺎـﺨ لﻜﺸﺒ
ﺍﻟﺠﺯ
ﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻫﻲ ﺯﻤﺭﺓ ﻨﺎﻅﻤﻴﺔ
.
ﻤﻼﺤﻅﺔ
6.3
(a)
ﻨﺄﺨﺫ ﺍﻟﺯﻤﺭﺓ
G
ﻭﺍﻟﺯﻤﺭﺓ ﺍﻟﻨﺎﻅﻤﻴﺔ
N
.
ﺭﺓـﻤﺯﻠﻟ ﻲﻠﺨﺍﺩﻟﺍ ﻲﺘﺍﺫﻟﺍ لﺜﺎﻤﺘﻟﺍ ﻥﺇ
G
ﻴﺤﺩﺩ ﺘﻤﺎﺜﻼﹰ
ﺫﺍﺘﻴ
ﺎﹰ
ﻟﻠﺯﻤﺭﺓ
N
ﻭﺍﻟﺫﻱ ﻤﻥ ﺍ،
ﻟﻤﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﺨﺎﺭﺠﻴﺎﹰ
)
،ﻙـﻟﺫ ﻰـﻠﻋ لﺎﺜﻤ
ﺍﻨﻅﺭ
15.3
.(
ﻟﺫﻟﻙ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﻓﻲ
N
ﻟﻴﺴ
ﺕ
ﺒﺎﻟﻀﺭﻭﺭﺓ
ﻲـﻓ ﺔﻴﻤﻅﺎﻨ ﻥﻭﻜﺘ ﻥﺃ
G
.
ﺃﻴﻀﺎﹰ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻤﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ ﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ
.
(b)
ﺍﻟﻤﺭﻜﺯ
( )
Z G
ﻟﻠﺯﻤﺭﺓ
G
ﻷﻥ،ﺓﺯﻴﻤﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻭﻫ
ﻓﺮدي
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٥٥
( ) ( )
( ) ( )
,
,
zg
gz
g
G
z
g
g
z
g
G
α
α
α
α
=
∀ ∈ ⇒
=
∀ ∈
ﻭﻜﻤﺎ ﺃﻥ
g
ﻴﻤﺴﺢ ﻜل
G
،
( )
g
α
ﺃﻴﻀﺎﹰ ﻴﻤﺴﺢ ﻜل
G
.
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻊ ﺘﻌﺭﻴﻑ
ﻨﻅﺭﻴﺔ ﺍﻟﺯﻤﺭ ﺍﻟﻌﺎﻤﺔ ﻟﻜﻲ ﺘﻜﻭﻥ ﻤﻤﻴﺯﺓ
.
(c)
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
m
،ﺯﺓـﻴﻤﻤ ﻥﻭﻜﺘ ﻥﺃ ﺏﺠﻴ ﺫﺌﺩﻨﻋ ،
ﻷﻥ
( )
H
α
ﻫﻭ ﺃﻴﻀﺎﹰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
ﻓﻲ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
m
.
(d)
ﻭﻥـﻜﺘ ﻥﺃ ﻱﺭﻭﺭﻀﻟﺍ ﻥﻤ ﺱﻴﻟ ﻥﻜﻟ ﺔﻴﻤﻅﺎﻨ ﻭﻥﻜﺘ ﺔﻴﻠﻴﺩﺒﺘ ﺓﺭﻤﺯ ﻥﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ لﻜ
ﻤﻤﻴﺯﺓ
.
ﺍﻟﻔﻀﺎﺀ ﺍﻟﺠﺯﺌﻲ ﻤﻥ ﺍﻟﺒﻌﺩ،ﹰﻼﺜﻤ
1
ﻓﻲ
2
p
G
=
F
ﺴﺒﺔـﻨﻟﺎﺒ ﹰﺎﺘﺒﺎﺜ ﻥﻭﻜﻴ ﻥﻟ
( )
2
GL
p
F
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻻ ﻴﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ
.
ﺍﻟﺠﺩﺍﺀﺍﺕ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭﺓ
(Semidirect products)
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﻋﻨﺩﺌﺫ
ﻜل ﻋﻨﺼﺭ
g
G
∈
ﻴﻌﺭﻑ ﺘﺸﺎﻜﻼﹰ
ﺫﺍﺘﻴ
ﺎﹰ
ــ ﻟ
N
،
1
n
gng
−
a
ﻭﻫﺫﺍ ﻴﻌﺭﻑ ﺍﻟﺘﺸﺎﻜل،
( )
:
,
θ
→
a
g
G
Aut N
g
i
N
ﺇﺫﺍ ﻭﺠﺩﺕ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
Q
ﻓﻲ
G
ﺒ
ﺤﻴﺙ ﺇﻥ
G
G N
→
ﻴﻐﻤﺭ
Q
لـﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ
ﻓﻲ
G N
ﺃﺩﻋﻲ ﺒﺄﻨﻪ ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﻌﻴﺩ ﺒﻨﺎﺀﺫﺌﺩﻨﻋ ،
G
ﻤﻥ
Q
ﻭ
N
ــ ﻭﺍﻟﻤﻘﺼﻭﺭ ﻟ،
θ
ﻋﻠﻰ
Q
.
ﻭﺒﺎﻹﻀﺎﻓﺔ ﻟﺫﻟﻙ ﻴﻤﻜﻥ ﺃﻥ ﻨﻜﺘﺏ
ﺍﻟﻌﻨﺼﺭ
g
G
∈
ﻭﺒﺸﻜل ﻭﺤﻴﺩ ﺒﺎﻟﺸﻜل
,
,
g
nq n
N
q
Q
=
∈
∈
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ
q
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻭﺤﻴﺩ ﻓﻲ
Q
ﻴﻁﺎﺒﻕ
gN
G N
∈
ﻭ،
n
ﻴﺠﺏ ﺃﻥ ﻴﺴﺎﻭﻱ
1
gq
−
.
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺘﻘﺎﺒل،ﺍﺫﻬﻟ
1 1
G
N
Q
−
←→ ×
ﺇﺫﺍ ﻜﺎﻥ
g
nq
=
ﻭ
g
n q
′
′ ′
=
ﻋﻨﺩﺌﺫ،
( )( )
(
)
( )( )
1
.
.
gg
nq
n q
n qn q
n
q
n
θ
−
′
′ ′
′
′
′
′
=
=
=
ﺘﻌﺭﻴﻑ
7.3
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ
(Semidirect products)
ﻟﻠﺯﻤﺭﺘﻴﻥ
N
ﻭ
Q
ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﻨﺎﻅﻤﻴﺔ ﻭ
ﻤﻥ
G
G N
→
ﻴﻨﺘﺞ ﺍﻟﺘﻤﺎﺜل
Q
G N
→
.
ﺘﻜﻭﻥ،ﺊﻓﺎﻜﻤ لﻜﺸﺒ
G
ﺠﺩﺍﺀ ﺸﺒ
ﻪ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ
N
ﻭ
Q
ﺇﺫﺍ ﻜﺎﻥ
;
;
1
N
G
NQ
G
N
Q
=
=
<
I
ﻨﻼﺤﻅ ﺒﺄﻨﻪ ﻟﻴﺱ ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﺘﻜﻭﻥ
Q
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
G
ﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ
N
ﻭ
Q
ﻨﻜﺘﺏ،
θ
= ×
G
N
Q
)
ﺤﻴﺙ
( )
:
θ
→
Q
Aut N
(
ﺘﻌﻁﻲ ﺘﺄﺜﻴﺭ
Q
ﻋﻠﻰ
N
ﺒﺘﻤﺎﺜل ﺫﺍﺘﻲ ﺩﺍﺨﻠﻲ
.
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٦٥
ﻤﺜﺎل
8.3
(a)
ﻓﻲ
n
D
،
2
n
≥
ﻟﻴﻜﻥ،
n
C
r
=
ﻭ
2
C
s
=
ﻋﻨﺩﺌﺫ،
2
n
n
D
r
s
C
C
θ
θ
=
×
=
×
ﺤﻴﺙ
( )
( )
i
i
s
r
r
θ
−
=
)
ﺍﻨﻅﺭ
1.16
.(
(b)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻨﺎﻭﺒﺔ
n
A
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
n
S
)
ﺴﺎﻭﻱـﻴ ﺎـﻬﻠﻴﻟﺩ ﻥﻷ
2
(
ﻭ،
( )
{ }
12
Q
=
ﺘﻐﻤﺭ ﻓﻲ
n
n
S
A
)
ﺘﺤﺕ ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل
(
.
ﺒﻬﺫﺍ ﺘﻜﻭﻥ
2
n
n
S
A
C
θ
=
×
.
(c)
ﺇﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﺠﺩﺍﺀ
ﺸ
ﺭـﻴﻏ ﺔﻘﻴﺭﻁ ﻱﺄﺒ ﺭﺸﺎﺒﻤ ﻪﺒ
ﺘﺎﻓﻬﺔ
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
2-3
.(
(d)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
،
p
ﻻ ﺘﺸﻜل ﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ،ﻲﻟﻭﺃ
)
ﻷﻨ
ﻬﺎ
ﻭﻱـﺤﺘ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺤﻴﺩﺓ ﻓﻘﻁ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.(
(e)
ﻟﺘﻜﻥ
( )
GL
n
G
F
=
.
ﻟﺘﻜﻥ
B
ﺎـﻴﻠﻌﻟﺍ ﺔﻴﺜﻠﺜﻤﻟﺍ ﺕﺎﻓﻭﻔﺼﻤﻟﺍ ﺓﺭﻤﺯ ﻥﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
ﻓﻲ
G
،
T
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻘﻁﺭﻴﺔ ﻓﻲ
G
،
U
ﻥـﻤ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ
ﺯﻤﺭﺓ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺍﻟﻌﻠﻴﺎ ﺤﻴﺙ ﻜل ﻤﻌﺎﻤﻼﺘﻬﺎ ﺍﻟﻘﻁﺭﻴﺔ ﺘﺴﺎﻭﻱ ﺍﻟﻌﺩﺩ
1
.
ﻋﻨﺩﺌﺫ
،
ﺭﺽـﻔﺒ
2
n
=
،
( )
{ }
( )
{ }
( )
{ }
0
1
,
,
0
0
0 1
B
T
U
∗ ∗
∗
∗
=
=
=
∗
∗
ﻓﺈﻥ
U
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
B
،
UT
B
=
ﻭ،
{ }
1
U
T
=
I
.
،ﻟﺫﻟﻙ
θ
= ×
B
U
T
ﻨﻼﺤﻅ
ﻋﻨﺩﻤﺎ،
2
n
≥
ﻓﺈﻥ ﺘﺄﺜﻴﺭ،
T
ﻋﻠﻰ
U
، ﻤﺜﻼﹰ،ﹰﺎﻬﻓﺎﺘ ﺱﻴﻟ
( )( )
(
)
1
1
0
1
0
1
/
0
0
1
0
1
0
a
c
a
ac b
b
b
−
−
=
ﻭﻟﺫﻟﻙ ﻓﺈﻥ
B
ـ ﺸﺒﻪ ﻤﺒﺎﺸﺭ ﻟﺀﺍﺩﺠ ﺱﻴﻟ
T
ﻭ
U
.
ﻤﻥ ﺍﻟﺠﺩﺍﺀ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ،ﻪﻨﺄﺒ ﺎﻨﻴﺃﺭ ﺩﻘﻟ
θ
= ×
G
N
Q
ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﺜﻼﺜﻴﺔ،
( )
(
)
, , :
N Q
Q
Aut N
θ
→
ﻭﺃﻥ ﺍﻟﺜﻼﺜﻴﺔ ﺘﺤﺩﺩ
G
.
ﻨﺒﺭﻫﻥ ﺍﻵﻥ ﺒﺄﻥ ﻜل ﺜﻼﺜﻴﺔ
(
)
, ,
N Q
θ
ﺭﺘﻴﻥـﻤﺯﻟﺍ ﻥﻤ ﻑﻟﺄﺘﺘ
N
ﻭ
Q
ﻭ
ﻋﻠﻰ ﺍﻟﺘﺸﺎﻜل
( )
:
θ
→
Q
Aut N
ﺭـﺸﺎﺒﻤﻟﺍ ﻪﺒﺸ ﺀﺍﺩﺠﻟﺍ ﻥﻤ ﻲﺘﺄﻴ ﻱﺫﻟﺍ
.
،ﺔـﻋﻭﻤﺠﻤﻜ
ﻟﺘﻜﻥ
G
N
Q
= ×
ﻭﻟﻨﻌﺭﻑ،
( )(
)
( )( )
(
)
,
,
.
,
n q
n q
n
q
n
θ
′ ′
′
′
=
ﻗﻀﻴﺔ
9.3
ﺇﻥ ﻗﺎﻨﻭﻥ ﺍﻟﺘﺭﻜﻴﺏ ﺍﻟﺴﺎﺒﻕ ﻴﺠﻌل ﻤﻥ
G
ﺒﻪـﺸ ﺀﺍﺩـﺠﻟﺍ ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ ،ﻲﻫﻭ ﺓﺭﻤﺯ
ﺍﻟﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ
N
ﻭ
Q
.
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٧٥
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻜﺘﺏ
n
q
ﺒﺩﻻﹰ ﻤﻥ
( )( )
q
n
θ
ﻟﺫﻟﻙ ﻴﺼﺒﺢ ﻗﺎﻨﻭﻥ ﺍﻟﺘﺭﻜﻴﺏ ﺒﺎﻟﺸﻜل،
( )(
) (
)
,
,
.
,
n
n q
n q
n q
′
′ ′
′
=
ﻋﻨﺩﺌﺫ
( )(
)(
) (
) ( ) (
)(
)
(
)
,
,
,
.
.
,
,
,
,
n
n
n q
n q
n q
n q qq
qq q
n q
n q
n q
′
′′
′ ′
′′ ′′
′
′ ′′
′ ′
′′ ′′
=
=
ﻭﻟﻬﺫﺍ ﻓﺈﻥ ﻗﺎﻨﻭﻥ ﺍﻟﺘﺠﻤﻴﻊ ﻤﺤﻘﻕ
.
ﻷﻥ
( )
1
1
θ
=
ﻭ
( )( )
1
1
q
θ
=
( )( ) ( ) ( )( )
1,1
,
,
,
1,1
n q
n q
n q
=
=
ﻭﻟﺫﻟﻙ ﻓﺈﻥ
( )
1,1
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﺤﺎﻴﺩ
.
ﺒﺎﻟﺘﺎﻟﻲ
( )
(
)
( )
(
)
( )
1
1
1
1
1
1
,
,
1,1
,
,
n
n
n q
q
q
q
q
n q
−
−
−
−
−
−
=
=
ﻟﺫﻟﻙ ﻓﺈﻥ
(
)
1
1
1
,
n
q
q
−
−
−
ﻫﻭ ﻤﻌﻜﻭﺱ ﺍﻟﻌﻨﺼﺭ
( )
,
n q
،
ﺇﺫﺍﹰ
G
ﺢ ﺃﻥـﻀﺍﻭﻟﺍ ﻥـﻤﻭ ،ﺓﺭـﻤﺯ
,
N
G
NQ
G
=
<
ﻭ
1
N
Q
=
I
ﻭﺒﺎﻟﺘﺎﻟﻲ،
θ
= ×
G
N
Q
ﺩﻤﺎـﻨﻋ ،ﻙﻟﺫ ﻰﻠﻋ ﹰﺓﻭﻼﻋ
ﺘﻌﺘﺒﺭ ﻜل ﻤﻥ
N
ﻭ
Q
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻤﻥ
G
ﻓﺈﻥ ﺘﺄﺜﻴﺭ،
Q
ﻋﻠﻰ
N
ـ ﻤﻌﻁﻰ ﺒ
θ
.
ﺃﻤﺜﻠﺔ
10.3
ﺍﻟﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
.
ﻟﻴﻜﻥ
θ
ﺍﻟﺘﺸﺎﻜل ﻏﻴﺭ ﺍﻟﺘﺎﻓﻪ
)
ﺍﻟﻭﺤﻴﺩ
(
( )
4
3
2
C
Aut C
C
→
ﺫﻟﻙ ﺍﻟﺫﻱ ﻴﺭﺴل ﻤﻭﻟﺩ،ﺩﻴﺩﺤﺘﻟﺎﺒ
4
C
ﺇﻟﻰ ﺍﻟﺘﻁﺒﻴﻕ
2
a
a
a
.
ﻋﻨﺩﺌﺫ
def
3
4
G
C
C
θ
=
×
ﺭﺓـﻤﺯ
ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
ﻟﻴﺴﺕ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ،
4
A
.
ﺇﺫﺍ ﺭﻤﺯﻨﺎ ﻟﻤﻭﻟﺩﺍﺕ
3
C
ﻭ
4
C
ﺎﻟﺭﻤﻭﺯـﺒ
a
ﻭ
b
ﻋﻨﺩﺌﺫ،
a
ﻭ
b
ﻴﻭﻟﺩﺍﻥ
G
ﻭﺘ،
ﻜﻭﻥ ﻟﺩﻴﻨﺎ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﻤﻌﺭﻓﺔ
3
4
1
2
1,
1,
a
b
bab
a
−
=
=
=
11.3
ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﻤﺒﺎﺸﺭﺓ
(Direct product)
.
ﺍﻟﺘﻘﺎﺒل ﻟﻠﻤﺠﻤﻭﻋﺎﺕ
( ) ( )
,
,
:
n q
n q
N
Q
N
Q
θ
× → ×
a
ﺎﻥـﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﺭﻤﺯﻟﺍ ﻥﻤ ﹰﻼﺜﺎﻤﺘ ﻥﻭﻜﻴ
θ
ﻪـﻓﺎﺘﻟﺍ لﻜﺎﺸـﺘﻟﺍ
( )
Q
Aut N
→
، ﺃﻱ ﺃﻥ،
( )( )
q
n
n
θ
=
ﻟﻜل
,
q
Q n
N
∈
∈
.
12.3
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
6
.
ﺇﻥ ﻜﻼﹰ ﻤﻥ
3
S
ﻭ
6
C
ﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺓ
3
C
ﻋﻠﻰ
2
C
-
ﺘﻘﺎﺒﻼﻥ ﺍﻟﺘﺸﺎﻜﻠﻴﻥ
( )
2
2
3
C
C
Aut C
→
.
13.3
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
)
ﺍﻟﻌﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
(
.
ﻟﺘﻜﻥ
N
a
=
ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ
ﺍﻟﺭﺘﺒﺔ
2
p
ﻭﻟﺘﻜﻥ،
Q
b
=
ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺤﻴﺙ،
p
ﺭﺩﻱـﻓ ﻲـﻟﻭﺃ ﺩﺩﻋ
.
ﻋﻨﺩﺌﺫ
1
Aut
p
p
N
C
C
−
≈
×
)
ﺍ
ﻨﻅﺭ
4.3
(
ﻭ،
p
C
ــ ﻤﻭﻟﺩ ﺒ
1
:
p
a
a
α
+
a
)
ﻅ ﺃﻥـﺤﻼﻨ
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٨٥
( )
2
1 2
,...
p
a
a
α
+
=
.(
ﺭﻑـــﻌﻨ
( )
Aut
Q
N
→
ﺸﻜلـــﻟﺎﺒ
b
a
a
.
ﺭﺓـــﻤﺯﻟﺍ
def
G
N
Q
θ
=
×
ﻤﻭﻟﺩﺍﺘﻬﺎ
,
a b
ﻭ ﻋﻼﻗﺎﺘﻬﺎ ﺍﻟﻤﻌﺭﻓﺔ
2
1
1
1,
1,
p
p
p
a
b
bab
a
−
+
=
=
=
ﻫﻲ ﺯﻤﺭﺓ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
ﻭﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
2
p
.
14.3
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
)
ﻻ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
.(
ﺘﻜﻥـﻟ
,
N
a b
=
ﺍﻟﺠﺩﺍﺀ ﺍﻟﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ ﺍﻟﺩﺍﺌﺭﻴﺘﻴﻥ
a
ﻭ
b
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺘـﻟﻭ ،
ﻜﻥ
Q
c
=
ﺭﺓـﻤﺯ
ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﻨﻌﺭﻑ
( )
:
Aut
Q
N
θ
→
ﺒﺤﻴﺙ،ﹰﻼﻜﺎﺸﺘ ﻥﻭﻜﻴ ﻲﻜ
( )
( )
( )
( )
,
i
i
i
c
a
ab
c
b
b
θ
θ
=
=
)
ﺇﺫﺍ ﻋﻴﻨﺎ
N
ﻜﺯﻤﺭﺓ ﺠﻤﻌﻴﺔ
2
p
N
=
F
ﻤﻊ
,
a b
ﺫـﺌﺩﻨﻋ ،ﺔﻴﻨﻭﻨﺎﻘﻟﺍ ﺓﺩﻋﺎﻘﻟﺍ ﺭﺼﺎﻨﻋ
( )
i
c
θ
ﻫﻭ
ﺘﻤﺎﺜل ﺫﺍﺘﻲ ﻋﻠﻰ
N
ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻤﺼﻔﻭﻓﺔ
1
0
1
i
.(
ﺍﻟﺯﻤﺭﺓ
def
G
N
Q
θ
=
×
ﻥـﻤ ﺓﺭﻤﺯ
ﺍﻟﺭﺘﺒﺔ
3
p
ﻤﻭﻟﺩﺍﺘﻬﺎ،
, ,
a b c
ﻭﻋﻼﻗﺎﺘﻬﺎ ﻤﻌﺭﻓﺔ ﺒﺎﻟﺸﻜل
[ ]
[ ]
1
1,
,
,
1
,
p
p
p
a
b
c
ab
cac
b a
b c
−
=
=
=
=
= =
ﻷﻥ
1
b
≠
ﺇﻥ ﺍﻟﻤﺴﺎﻭﺍﺓ ﺍﻟﻭﺴﻁﻰ ﺘﺒﻴﻥ ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﻟﻴﺴﺕ ﺘﺒﺩﻴﻠﻴﺔ،
.
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
p
ﻓﺈﻥ،ﹰﺎﻴﺩﺭﻓ
ﻜل ﺍﻟﻌﻨﺎﺼﺭ ﺴﺘﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺼﺭـﻨﻌﻟﺍ ﺀﺎﻨﺜﺘﺴﺎﺒ
1
.
ﺩﻤﺎـﻨﻋ
4
2,
p
G
D
=
≈
ﺫـﻟﺍ ،
ﻱ
ﻴﺤﺘﻭ
ﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
2
.
ﺔـﻔﻠﺘﺨﻤ ﺕﻼﻴﺜﻤﺘ ﺓﺭﻤﺯﻠﻟ ﻥﻭﻜﻴ ﻥﺃ ﻥﻜﻤﻴ ﻪﻨﺄﺒ ﺎﻨﻟ ﻥﻴﺒﻴ ﺍﺫﻫ
ﺘﻤﺎﻤﺎﹰ ﻜﺯﻤﺭﺓ ﺍﻟﺠﺩﺍﺀ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ
:
(
)
3.8a
4
4
2
2
2
2
D
C
C
C
C
C
θ
θ
≈
×
≈
×
×
ﻟﻜل ﻋﺩﺩ ﺃﻭﻟﻲ ﻓﺭﺩﻱ
p
،
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﻏﻴﺭ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
ﺭﺓـﻤﺯﻟﺍ ﻊﻤ ﺔﻠﺜﺎﻤﺘﻤ
ﻓﻲ
(13.3)
ﺇﺫﺍ ﺤﻭﺕ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
ﻭ
ﺘﻜﻭﻥ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ ﺍﻟﺯﻤﺭﺓ ﻓﻲ
(14.3)
ﺇﺫﺍ
ﻟﻡ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻜﻬﺫﺍ
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
3-4
.(
ﺩـﺠﻭﺘ ،لـﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ ،ﺹﺎﺨ لﻜﺸﺒ
ﺯﻤﺭﺘﺎ
ﻥ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺘﻴﻥ ﻓﻘﻁ ﻤﻥ ﺍﻟﺭﺘﺒ
ﺔ
3
p
.
15.3
ﺠﻌل ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﺍﻟﺨﺎﺭﺠﻲ ﺩﺍﺨﻠﻴﺎﹰ
.
ﻟﻴﻜﻥ
α
ﻭﻥـﻜﻴ ﻥﺃ ﻥـﻜﻤﻤﻟﺍ ﻥـﻤ ،ﹰﺎﻴﺘﺍﺫ ﹰﻼﺜﺎﻤﺘ
ﻟﻠﺯﻤﺭﺓ،ﹰﺎﻴﺠﺭﺎﺨ
N
.
ﺇﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﺼﺒﺢـﻴ ﺙﻴﺤﺒ ﺎﻤ ﺔﻘﻴﺭﻁﺒ
α
ﻤﻘﺼﻭﺭﺍﹰ ﻟﻠﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﺍﻟﺩﺍﺨﻠﻲ ﻟﻠﺯﻤﺭﺓ
G
ﺭﺓـﻤﺯﻟﺍ ﻰﻠﻋ
N
.
ﻟﺒ
ـﺭﻫ
ﻴﻜﻥـﻟ ،ﻙـﻟﺫ ﻥﺎ
( )
:
Aut
C
N
θ
∞
→
.
ﺘﺸﺎﻜﻼﹰ ﻴﺭﺴل ﺍﻟﻤﻭﻟﺩ
a
ﻤﻥ
C
∞
ﻰـﻟﺇ
( )
Aut N
α
∈
ﺘﻜﻥـﻟ ﻭ
G
N
C
θ
∞
= ×
.
ﺼﺭـــﻨﻌﻟﺍ
( )
1,
g
a
=
ﺭﺓـــﻤﺯﻟﺍ ﻥـــﻤ
G
ﺔـــﺼﺎﺨﻟﺍ ﻕـــﻘﺤﻴ
( )
( )
(
)
1
,1
,1
g n
g
n
α
−
=
ﻟﻜل،
n
N
∈
.
PDF created with pdfFactory Pro trial version
٩٥
ﻗﺎ
لـﺜﺎﻤﺘ ﻰـﻟﺇ ﺓﺭـﺸﺎﺒﻤﻟﺍ ﻪﺒـﺸ ﺕﺍﺀﺍﺩـﺠﻟﺍ لﻴﻭﺤﺘﻟ ﺓﺩﻋ
(Making outer
automorphisms inner)
ﺴ
ﺎﻥـﺘﻴﺜﻼﺜﻟﺍ ﺩﺩـﺤﺘ ﺎﻤﺩـﻨﻋ ﻙـﻟﺫﻭ ﺓﺩـﻋﺎﻗ ﺎﻨﻴﺩـﻟ ﻥﻭﻜﻴ ﻥﺄﺒ ﺎﻨﻟ ﹰﺍﺩﻴﻔﻤ ﻥﻭﻜﻴ
(
)
, ,
N Q
θ
ﻭ
(
)
, ,
N Q
θ
′
ﺯﻤﺭﺍﹰ ﻤﺘﻤﺎﺜﻠﺔ
.
ﺘﻤﻬﻴﺩﻴﺔ
16.3
ﺇﺫﺍ ﻜﺎﻥ
θ
ﻭ
θ
′
ﻤﺘﺭﺍﻓﻘ
ﻴ
ﺃﻱ ﺃﻨﻪ ﻴﻭﺠﺩ،ﻥ
( )
Aut N
α
∈
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ ،
( )
( )
1
q
q
θ
α θ
α
−
′
=
o
o
ﻟﻜل
q
Q
∈
ﻋﻨﺩﺌﺫ،
N
Q
N
Q
θ
θ
′
×
≈ ×
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺄﺨﺫ ﺍﻟﺘﻁﺒﻴﻕ
( )
( )
(
)
:
,
,
,
N
Q
N
Q
n q
n
q
θ
θ
γ
α
′
×
→ ×
a
ﻋﻨﺩﺌﺫ
( ) (
)
( )
(
)
( )
(
)
( ) ( ) ( )
(
)
(
)
( )
( )
(
)
( )
(
)
(
)
( )
( )( )
(
)
(
)
1
,
.
,
,
.
,
.
,
.
,
.
,
,
n q
n q
n
q
n
q
n
q
n
n
q
n
n
q
n
γ
γ
α
α
α
θ
α
α
α θ
α
α
α
α θ
−
′ ′
′
′
=
′
′
′
=
′
′
=
′
′
=
o
o
ﻭ
( ) (
)
(
)
( )( )
(
)
( )
( )( )
(
)
(
)
,
.
,
.
,
.
,
γ
γ
θ
α
α θ
′ ′
′
′
=
′
′
=
n q
n q
n
q
n
n
q
n
ﻭﻤﻨﻪ ﻓﺈﻥ
γ
ﻤﻌﻜﻭﺴﻪ،ﻱﺭﻤﺯ لﻜﺎﺸﺘ
( )
( )
(
)
1
,
,
n q
n
q
α
−
a
لـﺜﺎﻤﺘ ﻭـﻬﻓ ﻲﻟﺎـﺘﻟﺎﺒﻭ ،
ﺯﻤﺭﻱ
.
ﺘﻤﻬﻴﺩﻴﺔ
17.3
ﺇﺫﺍ ﻜﺎﻥ
θ θ α
′
=
o
ﺤﻴﺙ،
( )
Aut Q
α
∈
ﻋﻨﺩﺌﺫ،
.
N
Q
N
Q
θ
θ
′
×
≈ ×
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ
( )
( )
(
)
,
,
n q
n
q
α
a
ﻫﻭ ﺘﻤﺎﺜل
N
Q
N
Q
θ
θ
′
×
→ ×
.
ﺘﻤﻬﻴﺩﻴﺔ
18.3
ﺇﺫﺍ ﻜﺎﻨﺕ
Q
ﺩﺍﺌﺭﻴﺔ ﻭ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
( )
Q
θ
ﻤﻥ
( )
Aut N
ﻤﺘﺭﺍﻓ
ﻘﺔ ﻤﻊ
( )
Q
θ
′
ﻋﻨﺩﺌﺫ،
.
N
Q
N
Q
θ
θ
′
×
≈ ×
PDF created with pdfFactory Pro trial version
٠٦
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
a
ﻤﻭﻟﺩ ﻟﻠﺯﻤﺭﺓ
Q
ﻴﻭﺠﺩ ﺍﻟﻌﺩﺩﺫﺌﺩﻨﻋ ،
i
ﻭ
( )
Aut N
α
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
.
( )
( )
1
.
.
i
a
a
θ
α θ
α
−
′
=
ﺍﻟﺘﻁﺒﻴﻕ
( )
( )
(
)
,
,
i
n q
n
q
α
a
ﻫﻭ ﺍﻟﺘﻤﺎﺜل
N
Q
N
Q
θ
θ
′
×
→ ×
.
ﺘﻤﺩﻴﺩﺍﺕ ﺍﻟﺯﻤﺭ
(Extensions of groups)
ﺇﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﻭ ﺍﻟﺘﺸﺎﻜﻼﺕ
1
1
N
G
ι
π
→
→ →
ﺘﻜﻭﻥ ﺘﺎﻤﺔ ﺇﺫﺍ ﻜﺎﻥ
ι
ﻭ،ﹰﺎﻨﻴﺎﺒﺘﻤ
π
ﻭ،ﹰﺍﺭﻤﺎـﻏ
( )
( )
Ker
Im
π
ι
=
.
ﺫﻟﻙـﻟ
( )
N
ι
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
)
ـﺘﻤﺎﺜل ﺒﺎﻟﻨﺴﺒﺔ ﻟ
ι
ﻟﻠﺯﻤﺭﺓ
N
(
ﻭ
( )
G
N
Q
ι
→
.
ﻏﺎﻟﺒﺎﹰ ﻤﺎ ﻨﻁﺎﺒﻕ
N
ﻤﻊ ﺍﻟﺯﻤﺭﺓ
ﺍﻟﺠﺯﺌﻴﺔ
( )
N
ι
ﻓﻲ
G
ﻭ
Q
ﻤﻊ ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ
G N
.
ﺩﺍﹰـﻴﺩﻤﺘ لـﺜﻤﺘ ﺎـﻬﻨﺄﺒ ﹰﺎﻀﻴﺃ ﺭﻴﺸﺘ ﺔﻘﺒﺎﺴﻟﺍ ﺔﻤﺎﺘﻟﺍ ﺔﻴﻟﺎﺘﺘﻤﻟﺍ ﻥﺇ
(Extension)
ﺭﺓـﻤﺯﻠﻟ
Q
ﺒﺎﻟﺯﻤﺭﺓ
N
)
ﻭﺃﻴﻀﺎﹰ ﺘﺸﻜل ﺘﻤﺩﻴﺩﺍﹰ ﻟﻠﺯﻤﺭﺓ
N
ﺒﺎﻟﺯﻤﺭﺓ
Q
ﺅﻟﻔﻴﻥـﻤﻟﺍ ﺽـﻌﺒ ﺩﻨﻋ
.(
ﻭﻥـﻜﻴ
ﺍﻟﺘﻤﺩﻴﺩ ﻤﺭﻜﺯﻴﺎﹰ
(central)
ﺇﺫﺍ ﻜﺎﻥ
( )
( )
N
Z G
ι
⊂
.
ﺍﻟﺠﺩﺍﺀ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ،ﹰﻼﺜﻤ
N
Q
θ
×
ﻴﺘﺤﻭل ﺇﻟﻰ ﺘﻤﺩﻴﺩ ﻟﻠﺯﻤﺭﺓ
Q
ﺒﺎﻟﺯﻤﺭﺓ
N
،
1
1
N
N
Q
Q
θ
→
→ ×
→ →
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻤﺭﻜﺯﻴﺎﹰ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
θ
ﺘﺸﺎﻜﻼﹰ ﺘﺎﻓﻬﺎﹰ
.
ﻴﻘﺎل
ﻋﻥ ﺘﻤﺩﻴﺩﺍ ﺍﻟ
ﺯﻤﺭﺓ
Q
ﺒﺎﻟﺯﻤﺭﺓ
N
ﺃﻨﻬﻤﺎ
ﻤﺘﻤﺎﺜﻼﻥ ﺇﺫﺍ ﻭﺠﺩ ﻤﺨﻁﻁﺎﹰ ﺘﺒﺎﺩﻟﻴﺎﹰ
1
1
N
G
Q
→
→ → →
1
1
N
G
Q
′
→
→
→ →
ﻴﻘﺎل ﺒﺄﻥ ﺘﻤﺩﻴﺩ
Q
ـ ﺒ
N
،
1
1
N
G
ι
π
→
→ →
ﻤﻨﻔﺼﻼﹰ
(split)
ﺭـﺸﺎﺒﻤﻟﺍ ﻪﺒـﺸ ﺀﺍﺩـﺠﻟﺎﺒ ﻑﺭـﻌﻤﻟﺍ ﺩﻴﺩﻤﺘﻟﺍ ﻊﻤ ﹶﻼﺜﺎﻤﺘﻤ ﻥﺎﻜ ﺍﺫﺇ
N
Q
θ
×
.
ﺍﻟﺸﺭﻁﺎﻥ
ﺍﻵﺘﻴ
ﺎﻥ ﻤﺘﻜﺎﻓﺌﺎﻥ
:
(a)
ﻴﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
Q
G
′ ⊂
ﺒ
ﺤﻴﺙ ﺇﻥ
π
ﻴﻌﻁﻲ
ﺍﻟﺘﻤﺎﺜل
Q
Q
′ →
ﺃﻭ،
(b)
ﻴﻭﺠﺩ ﺘﺸﺎﻜل
:
s Q
G
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
id
π
=
o s
.
ﻟﻥ ﻴﻜﻭﻥ ﺍﻟﺘﻤﺩﻴﺩ ﻤﻨﻔﺼﻼﹰ،ﺔﻤﺎﻌﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
.
ﺍﻟﺘﻤﺩﻴﺩ،ﹰﻼﺜﻤ
≈
1
1
N
Q
Q N
→
→ →
→
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١٦
ﺤﻴﺙ
N
ﺃﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
ﻓﻲ ﺍﻟﺯﻤﺭ
ﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
Q
ﻭ
2
1
1
p
p
p
C
C
C
→
→
→
→
ﻟﻴﺱ ﻤﻨﻔﺼﻼﹰ
.
ﻨﺴﺘﻌﺭﺽ ﻗﺎﻋﺩﺘﻴﻥ ﻟﻜﻲ ﻴﻜﻭﻥ ﺍﻟﺘﻤﺩﻴﺩ ﻤﻨﻔﺼﻼﹰ
.
ﻤﺒﺭﻫﻨﺔ
19.3
(Schur – Zassenhaus)
ﺔـﻴﻟﻭﺃ ﺩﺍﺩﻋﺃ ﺎﻬﺒﺘﺭ ﻲﺘﻟﺍ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺭﻤﺯﻟﺍ ﺩﻴﺩﻤﺘ ﻥﺇ
ﻨﺴﺒﻴﺎﹰ ﻴﻜﻭﻥ ﻤﻨﻔﺼﻼﹰ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
Rotman 1995, 7.41
.
ﻗﻀﻴ
ﺔ
20.3
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﺇﺫﺍ ﻜﺎﻨﺕ،
N
ﺈﻥـﻓ ﺫﺌﺩﻨﻋ ،ﺔﻤﺎﺘ
N
ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺓ
N
ﻤﻊ ﻤﻤﺭﻜﺯ
N
ﻓﻲ
G
،
( )
{
}
def
,
G
C
N
g
G gn
ng
n
N
=
∈
=
∀ ∈
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
( )
G
Q
C
N
=
.
ﺴﻨﺒﻴﻥ ﺒﺄﻥ
N
ﻭ
Q
ﺘﺤﻘﻘﺎﻥ ﺸﺭﻭﻁ ﺍﻟﻘﻀﻴﺔ
50.1
.
، ﻨﻼﺤﻅ ﺃﻭﻻﹰ ﺒﺄﻨﻪ
1
:
−
→
a
n
gng
N
N
ﻟﻜل
∈
g
G
ﺭﺓـﻤﺯﻠﻟ ﻲـﺘﺍﺫ لﺜﺎﻤﺘ
N
ﻭ،
)
ﻷﻥ
N
ﺘﺎﻤﺔ
(
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﺍﻟﺩﺍﺨﻠﻲ ﻤﻌﺭﻓﺎﹰ ﺒﻌﻨﺼﺭ،
γ
ﻓﻲ
N
ﻟﻬﺫﺍ،
1
1
gng
n
γ γ
−
−
=
ﻟﻜل،
n
N
∈
ﺄﻥـﺒ ﺔﻘﺒﺎﺴـﻟﺍ ﺔﻟﺩﺎﻌﻤﻟﺍ ﻥﻴﺒﺘ
1
g
Q
γ
−
∈
ﺎﻟﻲـﺘﻟﺎﺒﻭ ،
( )
1
g
g
NQ
γ γ
−
=
∈
.
ﺎ ﺃﻥـﻤﺒ
g
ﻨﻜﻭﻥ ﻗﺩ ﺒﻴﻨﺎ ﺒﺄﻥ،ﻱﺭﺎﻴﺘﺨﺍ
G
NQ
=
.
ﻨﻼﺤﻅ ﻓﻴﻤﺎ ﻴﻠ
ﻲ
ﺒﺄﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ
N
Q
I
ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﻤﺭﻜﺯ
N
ﺎﹰـﻬﻓﺎﺘ ﻥﻭـﻜﻴ ﻱﺫﻟﺍ ،
)
ﻷﻥ
N
ﺘﺎﻤﺔ
(
ﻭﺒﺎﻟﺘﺎﻟﻲ،
1
N
Q
=
I
.
ﻷﻱ ﻋﻨﺼﺭ،ﹰﺍﺭﻴﺨﺃ
g
nq
G
=
∈
،
(
)
1
1
1
gQg
n qQq
n
Q
−
−
−
=
=
)
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﻜل
ﻋﻨﺼﺭ ﻤﻥ
N
ﻴﺘﺒﺎﺩل ﻤﻊ ﻜل ﻋﻨﺼﺭ ﻤﻥ
Q
.(
ﻟﺫﻟﻙ ﻓﺈﻥ
Q
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ
G
.
ﺇﻥ ﺍﻟﺘﻤﺩﻴﺩ
ﻴﻌﻁﻲ ﺍﻟﺘﺸﺎﻜل
( )
:
Aut
G
N
θ
′
→
، ﻭﺒﺎﻟﺘﺤﺩﻴﺩ،
( )( )
1
g
n
gng
θ
−
′
=
ﻟﻴﻜﻥ
q
G
∈
%
ﻴﺭﺴل ﺇﻟﻰ
q
ﻓﻲ
Q
ﻕـﻴﺒﻁﺘﻟﺍ ﻕـﻓﻭ
θ
ﺼﻭﺭﺓـﻟﺍ ﺫـﺌﺩﻨﻋ ،
( )
q
θ
′
%
ﻲـﻓ
( )
( )
Aut
Inn
N
N
ﺘﻌﺘﻤﺩ ﻋﻠﻰ
q
ﻟﺫﻟﻙ ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﺘﺸﺎﻜل،ﻁﻘﻓ
1
1
N
G
Q
→
→ → →
PDF created with pdfFactory Pro trial version
٢٦
( )
( )
( )
def
:
Out
Aut
Inn
Q
N
N
N
θ
→
=
ﻴﻌﺘﻤﺩ ﺍﻟﺘﻁﺒﻴﻕ
θ
ﻭﻨﻜﺘﺏ،ﺕﺍﺩﻴﺩﻤﺘﻠﻟ ﺔﻠﺜﺎﻤﺘﻤﻟﺍ ﻑﻭﻔﺼﻟﺍ ﻰﻠﻋ ﻁﻘﻓ
(
)
1
Ext
,
Q N
θ
ﻟﻤﺠﻤﻭﻋﺔ
ﺍﻟﺼﻔﻭﻑ ﺍﻟﻤﺘﻤﺎﺜﻠﺔ ﻟﻠﺘﻤﺩﻴﺩﺍﺕ ﻤﻊ
θ
.
ﻟﻘﺩ ﺩﺭﺴﺕ ﻫﺫﻩ ﺍﻟﻤﺠ
ﻤﻭﻋﺎﺕ ﻋﻠﻰ ﻨﻁﺎﻕ ﻭﺍﺴﻊ
.
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
N
ﻭ
Q
ﺘﺒﺩﻴﻠﻴﺘﻴﻥ ﻭ
θ
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ،ﹰﺎﻬﻓﺎﺘ
G
ﺩـﺠﻭﺘﻭ ،ﹰﺎﻀـﻴﺃ ﺔﻴﻠﻴﺩﺒﺘ
ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﺘﺒﻨﻰ ﻓﻭﻕ
(
)
1
Ext
,
Q N
.
ﻋﻼ
ﺭﺘﻴﻥـﻤﺯﻠﻟ ﺔﻴﺘﺍﺫﻟﺍ ﺕﻼﻜﺎﺸﺘﻟﺍ ﻥﺈﻓ ،ﻙﻟﺫ ﻰﻠﻋ ﹰﺓﻭ
N
ﻭ
Q
ﺘﺅﺜﺭ ﻜﺘﺄﺜﻴﺭ ﺍﻟﺘﺸﺎﻜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻋﻠﻰ
(
)
1
Ext
,
Q N
.
ـ ﺒﺎﻟﻀﺭﺏ ﺒ،ﺔﺼﺎﺨ ﺔﻟﺎﺤﻜ
m
ﻓﻭﻕ
N
ﺃﻭ
Q
ـ ﻴﻨﺘﺞ ﺍﻟﻀﺭﺏ ﺒ
m
ﻓﻭﻕ
(
)
1
Ext
,
Q N
.
ﺕـﻨﺎﻜ ﺍﺫﺇ ،ﺍﺫﻬﻟ
N
ﻭ
Q
ﻤﺨﺘﺯﻟﺘﻴ
ﻥ ﻋﻠﻰ
m
ﻭ
n
ﻋﻨﺩ،ﺏﻴﺘﺭﺘﻟﺍ ﻰﻠﻋ
ﺌﺫ
(
)
1
Ext
,
Q N
ﻰـﻠﻋ ﹰﻻﺯـﺘﺨﻤ ﻥﻭﻜﻴ
m
ﻭ
n
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻜﻭﻥ ﻤﺨﺘﺯﻻﹰ ﻋﻠﻰ،
(
)
gcd
,
m n
.
ﺔـﻨﻫﺭﺒﻤ ﺕﺒﺜﻴ ﺍﺫﻫ
Schur – Zassenhaus
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
.
ﻤﺨﻁﻁ ﻫﻭﻟﺩﺭ
(The Holder program)
.
ﺴﺄﻭﻟﻲ ﺍﻻﻫ
ﺘﻤﺎﻡ ﺍﻷﻜﺒﺭ ﺇﺫﺍ ﻜﺎﻥ ﻤﻥ ﺍﻟﻤﻤﻜﻥ
ﺇﻋﻁﺎﺀ ﻨﻅﺭﺓ ﻋﺎﻤﺔ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻜﺎﻤﻠﺔ ﻤﻥ
ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
Otto Holder, Math. Ann., 1892
ﻨﺘﺫﻜﺭ ﺒﺄﻥ
G
ﺭـﻴﻏ ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺭـﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ ﻻ ﺎﻤﺩﻨﻋ ﺔﻁﻴﺴﺒ ﺓﺭﻤﺯ ﻥﻭﻜﺘ
ﺍﻟﺯﻤﺭﺘﻴﻥ
1
ﻭ
G
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ،ﻯﺭﺨﺃ ﺕﺎﻤﻠﻜﺒ
G
ﻐﺭـﺼﺃ ﺭﻤﺯﻟ ﹰﺍﺩﻴﺩﻤﺘ ﻥﻜﺘ ﻡﻟ ﺍﺫﺇ ﺔﻁﻴﺴﺒ
ﻤﻨﻬﺎ
.
ﻜل ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻴﻤﻜﻥ ﺃﻥ ﻴﺤﺼل ﻋﻠﻴﻬﺎ ﺒﺄﺨﺫ ﺘﻤﺩﻴﺩﺍﺕ ﻤﻜﺭﺭﺓ ﻟﺯﻤﺭ ﺒﺴﻴﻁﺔ
.
ﻥـﻜﻤﻴ ﺍﺫﻬﻟ
ﺃﻥ ﺘﻌﺘﺒﺭ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺒﺴﻴﻁﺔ ﻜﻘﺎﻋﺩﺓ ﻟﺒﻨﺎﺀ ﺨﻼﻴﺎ ﻟﺠﻤﻴﻊ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
ﺇﻥ ﻤﺸﻜ
ﻠﺔ ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺘﻘﻊ ﻓﻲ ﺠﺯﺃﻴﻥ
:
A
.
،ﺘﺼﻨﻴﻑ ﺠﻤﻴﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
B
.
ﺘﺼﻨﻴﻑ ﺠﻤﻴﻊ ﺘﻤﺩﻴﺩﺍﺕ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
ﺴﻴﻁﺔـﺒﻟﺍ ﺔـﻴﻬﺘﻨﻤﻟﺍ ﺭـﻤﺯﻟﺍ ﻑﻴﻨﺼﺘ
(Classification of finite simple
groups)
ﺘﻭﺠﺩ ﻗﺎﺌﻤﺔ ﻜﺎﻤﻠﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺒﺴﻴﻁﺔ
.
ﻭﻫﻲ
(a)
ﺍﻟﺯﻤﺭ ﺍﻟﺩﺍﺌﺭﻴﺔ ﺍ
،ﻟﺘﻲ ﺭﺘﺒﻬﺎ ﻋﺩﺩﺍﹰ ﺃﻭﻟﻴﺎﹰ
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(b)
ﺍﻟﺯﻤﺭ ﺍﻟﻤﺘﻨﺎﻭﺒﺔ
n
A
ﻟﻜل
5
n
≥
)
ﺍﻨﻅﺭ ﺍﻟﻔﺼل
ﺍﻵﺘﻲ
(،
(c)
ﻭ،ﺕﺎﻓﻭﻔﺼﻤﻟﺍ ﺭﻤﺯ ﻥﻤ ﺔﻴﻬﺘﻨﻤ ﺭﻴﻏ ﺓﺩﺩﺤﻤ ﺕﺎﻋﻭﻤﺠﻤ
(d)
"
ﺍﻟﺯﻤﺭ ﺍﻟﻤﺘﻔﺭﻗﺔ
"
26
ﺇﻥ ﺍﻟﺼﻑ ﺍﻷﻜﺒﺭ ﻫﻭ
(c)
ﻟﻜﻥ،
"
ﺍﻟﺯﻤﺭ ﺍﻟﻤﺘﻔﺭﻗﺔ
"
26
ﺘﺄ
ﺼﻐﻴﺭـﻟﺍ ﺎﻫﺩﺩﻋ ﻥﻤ ﺭﺜﻜﺃ ﹰﺎﻤﺎﻤﺘﻫﺍ ﺫﺨ
ﺍﻟﺫﻱ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﻤﻘﺘﺭﺤﺎﹰ
.
ﺴﻴﺩـﻟﺍ ،ﻡﻬﻨـﻤ ﺭـﺒﻜﺃ ﺎـﻬﻨﺄﺒ ﺕﺍﺭﻭﺼـﺘ ﻻﺇ ﻪﻴﺩﻟ ﺱﻴﻟ ﺽﻌﺒﻟﺍ
Fischer-Griess
ﻴﻜﻭﻥ ﻗﺩ ﺍﻨﺘﻘل ﺇﻟﻰ ﺘﺭﻜﻴﺏ ﺃﻜﺜﺭ ﻤﻥ ﻤﺠﻤﻭﻋﺔ،
.
ﻨﺄﺨﺫ،ﺕﺎﻓﻭﻔﺼﻤﻟﺍ ﺓﺭﻤﺯ ﻰﻠﻋ لﺎﺜﻤﻜﻭ
( ) {
}
def
SL
,
, det
1
m
q
q
m m A
A
=
×
∈
=
F
F
ﻫﻨﺎ
n
q
p
=
،
p
ﻭ،ﻲﻟﻭﺃ
q
F
ﺍﻟﺤﻘل ﻤﻊ
q
ﻋﻨﺼﺭ
.
ﺎﻥـﻜ ﺍﺫﺇ ﺔﻁﻴﺴﺒ ﻥﻭﻜﺘ ﻥﻟ ﺓﺭﻤﺯﻟﺍ ﻩﺫﻫ
2
q
≠
ﻷﻥ ﺍﻟﻤﺼﻔﻭﻓﺔ ﺍﻟﺴﻠﻤﻴﺔ،
0
0
0
0
,
1
0
0
m
ξ
ξ
ξ
ξ
=
L
O
L
لـﺠﺃ ﻥﻤ ﺯﻜﺭﻤﻟﺍ ﻲﻓ ﻥﻭﻜﺘ ،
ﺃﻱ
m
ﺍﻟﺫﻱ ﻴﻘﺴﻡ
1
q
−
ﻫﻲ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﺘﻲ ﻓﻲ ﺍﻟﻤﺭﻜﺯ،ﻩﺫﻫ ﻥﻜﻟ ،
ﻭﺘﻜﻭﻥ،ﻁﻘﻓ
ﺍﻟﺯﻤﺭ
( )
( )
{
}
def
PSL
SL
m
q
n
q
=
F
F
ﺒﺴﻴﻁﺔ ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
3
m
≥
(Rotman 1995, 8.23)
ﻭﻋﻨﺩﻤﺎ
2
m
=
ﻭ
3
q
>
(ibid.
8.13)
.
ﻤﻥ ﺃﺠل ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
3
m
=
ﻭ
2
q
=
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ،
4-6
)
ﺄﻥـﺒ ﻅﺤﻼﻨ
( )
( )
3
2
2
PSL
GL
F
F
.(
ﻴﻤﻜﻥ ﺃﻥ ﻨﺤﺼل ﻋﻠﻰ ﺯﻤﺭ ﺒﺴﻴﻁﺔ ﻤﻨﺘﻬﻴﺔ ﺃﺨﺭﻯ ﻤﻥ ﺍﻟﺯﻤﺭ ﻓﻲ
(8.1)
.
B
ﺘﺼﻨﻴﻑ ﻜل ﺘﻤﺩﻴﺩﺍﺕ ﺍ
ﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
،ﺃﺼﺒﺢ ﻤﻌﺭﻭﻓﺎﹰ ﻟﻨﺎ ﺍﻟﻜﺜﻴﺭ ﻋﻥ ﺘﻤﺩﻴﺩﺍﺕ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
ﻓ
ﺘﻤﺩﻴﺩ ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻭﺍﺤﺩﺓ ﺒﺯﻤﺭﺓ،ﹰﻼﺜﻤ
ﺃﺨﺭﻯ
.
ﻋﻼﻭﺓ ﻋﻠﻰ ﺫﻟﻙ
ﻜﺘﺏ،
Solomon
(2001,p347)
:
...
ﺇﻥ ﺘﺼﻨﻴﻑ ﻜل ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬ
ﻴﺔ ﻏﻴﺭ ﻗﺎﺒل ﻟﻠﺘﻁﺒﻴﻕ ﺒﺸﻜل ﺘﺎﻡ
.
ﻭ
ﻻ ﺘﻭﺠﺩ ﺨﺒﺭﺍﺕ
ﺴﺎﺒﻘﺔ ﺘﺒﻴﻥ ﺒﺄﻥ ﻤﻌﻅﻡ ﺍﻟﺯ
ﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻲ ﺘﻭﺠﺩ ﻓﻲ
"
ﺍﻟﻁﺒﻴﻌﺔ
... "
ﺘﻜﻭﻥ
"
ﻤﻐﻠﻘﺔ
"
، ﻭﺯﻤﺭ ﻫﻴﺴﻴﻨﺒﻴﺭﻍ،لﺍﺭﺩﻴﻬﻴﺩ ﺭﻤﺯﻜ ﺎﻤ ﺭﻤﺯﻟ ﺔﺒﺴﻨﻟﺎﺒ ﻭﺃ ﺔﻁﻴﺴﺒﻟﺍ ﺭﻤﺯﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ ﺎﻤﺇ
ﺍﻟﺦ
...
ﻭﺍﻟﺘﻲ ﺘﺄﺘﻲ ﺒﺸﻜل ﻁﺒﻴﻌﻲ ﻓﻲ ﺩﺭﺍﺴﺔ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ،
.
ﻓﻲ ﻋﺎﻡ،ﻙﻟﺫ لﺒﻗ ﺎﻨﻅﺤﻻ ﺎﻤﻜ
2001
ﻗﺎﺌﻤﺔ ﻜﺎﻤﻠﺔ ﺼﺤﻴﺤﺔ ﻤﻥ ﺍﻟﺯﻤﺭ،
ﻭﻓﺭﺓـﺘﻤ ﺕﻨﺎﻜ ﺔﻴﻬﺘﻨﻤﻟﺍ
ﻓﻘﻁ ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭ ﺍﻟﺘﻲ ﺭﺘﺒﺘﻬﺎ ﺘﺼل ﺇﻟﻰ
2000
ﺘﻘﺭﻴﺒ
ﻭﻋﺩﺩ ﺍﻟﺯﻤﺭ ﻓﻲ ﺍﻟﻘﺎﺌﻤﺔ ﻏﺎﻤﺭ،ﹰﺎ
.
ﺍﻟﻤﺭﻜﺯ
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ﻨﻼﺤﻅ ﺒﺄﻥ ﺤﻠﻡ
ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻴﺭﺠﻊ ﻋﻠﻰ ﺍﻷﻗل
ﺇﻟﻰ
.Holder 1892
ﻥـﻜﻟ
ﺍﻹﺴﺘﺭﺍﺘﻴﺠﻴﺔ ﺍﻟﻭﺍﻀﺤﺔ ﻟ
ﺘﺤﻘﻴﻕ
ﻫﺫﻩ ﻟﻡ ﺘﺒﺩﺃ ﺒﺎﻟﻅﻬﻭﺭ ﺤﺘﻰ ﻋﺎﻡ
1950s
ﻋﻨﺩﻤﺎ ﻋﻤل ﺒﺭﻭ،
ﺭ ﻭـﻴﻴ
ﻤﻘﺘﺭﺤﻭﻥ ﺁﺨﺭﻭﻥ ﻋﻠﻰ ﺃﻥ ﺍﻟﻤﻔﺘﺎﺡ
ﻴﻜﻤﻥ
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺭـﺼﺎﻨﻌﻟﺍ ﺕﺍﺯـﻜﺭﻤﻤ ﺔﺴﺍﺭﺩ ﻲﻓ
2
)
ﺍﻟﻤﻤﺭﻜﺯﺍﺕ ﺍﻻﻟﺘﻔﺎﻓﻴﺔ
.(
ﺒﺭﻭﻴﻴﺭ ﻭ ﻓﻭﻟﻴﻴﺭ،ﹰﻼﺜﻤ
(1995)
ﺔـﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ ﻱﺃ لﺠﺃ ﻥﻤ ،ﻪﻨﺄﺒ ﺎﻨﻴﺒ
H
ﺘﻜﻭﻥ ﻤﺤﺩﺩﺍﺕ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻤﻊ ﻤﻤﺭﻜﺯ ﺍﻟﺘﻔﺎﻓﻲ ﻤﺘﻤﺎﺜل ﻤﻊ،
H
ﺴﺄﻟﺔـﻤ ﻭـﻫ
ﻤﻨﺘﻬﻴﺔ
.
ﺘﺭﺍﺘﻴﺠﻴﺔـﺴﻹﺍ ﺕﺤﺒﺼﺃ ﻙﻟﺫﻟﻭ ،ﺔﻴﺎﻐﻠﻟ ﺔﺴﻠﺴ ﺔﻟﺄﺴﻤﻟﺍ ﻥﺄﺒ ﹰﺎﻨﻴﺒﻤ لﻤﻋ ﺩﻌﺒ ﺎﻤﻴﻓ
:
(a)
ﺇﻥ
ﻗﺎﺌﻤﺔ ﺍﻟﺯﻤﺭ
H
،
ﻭ،ﺔـﻴﻬﺘﻨﻤﻟﺍ ﺔﻁﻴﺴﺒﻟﺍ ﺭﻤﺯﻟﺍ ﺽﻌﺒﻟ ﺔﻴﻓﺎﻔﺘﻟﺍ ﺕﺍﺯﻜﺭﻤﻤﻜ ﺭﺒﺘﻌﺘ ﻥﺃ ﻥﻜﻤﻴ ﻲﺘﻟﺍ
(b)
ﻟﻜل ﺯﻤﺭﺓ
H
ﻓﻲ
(a)
ﻗﺎﺌﻤﺔ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻟﻠﺯﻤﺭﺓ
H
ﺯـﻜﺭﻤﻤﻜ ﺭـﺒﺘﻌﺘ ﻲﺘﻟﺍ
ﺍﻟﺘﻔﺎﻓﻲ
.
ﻫﺫﺍ ﻴﻘﺘﺭﺏ ﺒﺘﻁﺒﻴﻘﻪ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ،ﻊﺒﻁﻟﺎﺒ
ﺍﻟﺭﺘﺒﺔ
2
ﻟﻜﻥ،
ﻫﻨﺎﻙ
ﻜل ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻤﻨﺘﻬﻴﺔ ﺭﺘﺒ،ﻥﺄﺒ لﻭﻘﻴ ﻡﻴﺩﻗ ﻥﻴﻤﺨﺘ
ﻲـﺠﻭﺯ ﺩﺩـﻋ ﺎـﻬﺘ
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭﺫﻟﻙ ﺤﺴﺏ ﻤﺒﺭﻫﻨﺔ ﻜﻭﺸﻲ
(13.4)
ﺭـﻤﺯﻟﺍ ﺍﺩﻋﺎﻤ
ﺍﻟﺩﺍﺌﺭﻴﺔ ﺍﻟﺘﻲ ﺭﺘﺒﻬﺎ ﻋﺩﺩﺍﹰ ﺃﻭﻟﻴﺎﹰ
.
ﺴﻭﻥـﺒﻤﻭﺜﻭ ﺕﻴﻓ لﺒﻗ ﻥﻤ ﻥﻴﻤﺨﺘﻟﺍ ﺍﺫﻫ ﻥﺎﻫﺭﺒ ﻊﻤ
(1963)
،
ﺇﻥ
ﺍﻟﺠﻬﺩ ﺍﻟﺫﻱ ﺒﺫل ﻹﻜﻤﺎل ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺒﺩﺃ ﺒﺠﺩﻴﺔ
.
ﻟﻘﺩ
ﺃﻋ
ﻠ
ـﺼﻨﺒﻑ ﺍﻟﺘـﺘﻟﺍ ﻥ
ﺎﻡ
ﻋﺎﻡ
1982
،
ﻭﻟﻜﻥ
ﻤﺎﺯﺍل ﻏﻴﺭ ﻤﻭﺜﻭﻕ ﺒﻪ
ﻷﻥ ﺍﻟﺒﺭﻫﺎﻥ،
ﻋﻠﻴﻪ
ﺼﻔﺤﺎﺕـﻟﺍ ﻥﻤ ﻑﻻﺁ ﻰﻠﻋ ﺩﻤﺘﻋﺍ
ﻓﻲ ﺃﺠﺯﺍﺀ
ﻭ،ﺃﺭﻘﺘ ﺎﻤ ﹰﺍﺭﺩﺎﻨ ﻲﺘﻟﺍ ﻑﺤﺼﻟﺍ ﻥﻤ
،ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ
ﻭ
ﺃﺜﻨﺎﺀ
ﺼﻴﺎﻏﺔ
ﺎﻥـﻫﺭﺒﻟﺍ
،
، ﺎﺩﺓـﻌﻟﺎﻜ
ﺍﻜﺘﺸﻔﺕ
ﺜﻐﺭﺓ
.
ﻭ،ﺩـﺤﻟﺍ ﺍﺫﻫ ﺩﻨﻋ ﺍﻭﻔﻗﻭﺘﻓ
ﺸﻭﺭﺍﺕـﻨﻤ ﻊـﻤ
Aschbacher
ﻭ
Smith
2004
ﺃﺼﺒﺢ ﺒﺸﻜل ﻋﺎﻡ ﻤﻘﺒﻭﻻﹰ
ﻟﺩﻴﻨﺎ
ﺒﺄﻥ ﺒﺭﻫﺎﻥ ﺘﺼﻨﻴﻑ ﺍﻟﺯﻤﺭ
ﺃﻤﺭ
ﻤﻌﻘﺩ ﺤﻘﺎﹰ
.
ﺍﻨﻅﺭ ﻜﺘﺎﺏ ﺭﻭﻨﺎﻥ،ﻑﻴﻨﺼﺘﻟﺍ ﺦﻴﺭﺎﺘﻟ ﻲﺒﻌﺸﻟﺍ ﻥﺎﻴﺒﻟﺍ لﺠﺃ ﻥﻤ
2006
ﻥـﻋ ﺭﺜﻜﺃ ﺔﻓﺭﻌﻤﻟ ﻭ ،
ﺍﻨﻅﺭ ﺍﻟﺠﺯﺀ ﺍﻟﻤﻌﺭﻭﺽ،ﻲﻨﻘﺘﻟﺍ ﻥﺎﻴﺒﻟﺍ
Solomon 2001
.
ﺘﻤﺎﺭﻴﻥ
1-3
ﻟﺘﻜﻥ
2
4
,
,
,
n
D
a b a
b abab
=
th
n
ﺯﻤﺭﺓ ﺩﻴﻬﻴﺩ
ﺭﺍل
.
ﺇﺫﺍ ﻜﺎﻥ
n
ﺭﻫﻥـﺒ ،ﻱﺩﺭﻓ
ﺃﻥ
2
2
,
n
n
D
a
a b
≈
×
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،
2
2
n
n
D
C
D
≈
×
.
2-3
ﻟﺘﻜﻥ
G
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
(1.17)
.
ﺒﺭﻫﻥ ﺃﻨﻪ ﻻ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ
G
ﺒﻪـﺸ ﺀﺍﺩﺠ لﻜﺸ ﻰﻠﻋ
ﻤﺒﺎﺸﺭ ﺒﺄﻱ ﻁﺭﻴﻘﺔ ﻏﻴﺭ ﺘﺎﻓﻬﺔ
.
3-3
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
mn
ﺒ
ﺤﻴﺙ
ـ ﻻ ﻴﻜﻭﻥ ﻟ
m
ﻭ
n
ﺃﻱ
ﺸﺘﺭﻙـﻤ لـﻤﺎﻋ
.
ﺇﺫﺍ
ﻜﺎﻨﺕ
G
ﺘﺤﻭ
ﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺍﺤﺩﺓ ﻓﻘﻁ
M
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
m
ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ﻰـﻠﻋ ﻭ
ﻭﺍﺤﺩﺓ ﻓﻘﻁ
N
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﺒﺭﻫﻥ ﺒﺄﻥ،
G
ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ
M
ﻭ
N
.
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٥٦
4-3
ﺒﺭﻫﻥ ﺃﻥ
( )
2
2
3
GL
S
≈
F
.
5-3
ﻟﺘﻜﻥ
G
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
(1.17)
.
ﺒﺭﻫﻥ ﺃﻥ
( )
4
Aut G
S
≈
.
6-3
ﺘﻜﻥــﻟ
G
ﻊــﻴﻤﺠ ﺔــﻋﻭﻤﺠﻤ
ﻲــﻓ ﺕﺎﻓﻭﻔﺼــﻤﻟﺍ
( )
3
GL
ﺸﻜلــﻟﺍ ﻥــﻤ
0
0
,
0
0
0
a
b
a
c
ad
d
≠
.
ﺃﺜﺒﺕ ﺃﻥ
G
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
( )
3
GL
ﺸﻜلـﺘ ﺎـﻬﻨﺃ ﻥﻫﺭﺒﻭ ،
ـﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ ﻟ
2
)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﻤﻌﻴﺔ
(
ﻋﻠﻰ
×
×
×
.
ﻭﻫل ﻫﻲ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﻬﺎﺘﻴﻥ
؟ﺍﻟﺯﻤﺭﺘﻴﻥ
7-3
ﺃﻭﺠﺩ ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟﻠﺯﻤﺭﺘﻴﻥ
C
∞
ﻭ
3
S
.
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٦٦
ﺍﻟﻔﺼل ﺍﻟﺭﺍﺒﻊ
ﺘﺄﺜﻴﺭ ﺯﻤﺭ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺎﺕ
Groups Acting on Sets
ﺘﻌﺭﻴﻑ ﻭ ﺃﻤﺜﻠﺔ
ﺘﻌﺭﻴﻑ
1.4
ﻟﺘﻜﻥ
X
ﻤﺠﻤﻭﻋﺔ ﻭ ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
.
ﺇﻥ ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻴﺴﺎﺭﻱ
(left action)
ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻫﻭ ﺘﻁﺒﻴﻕ
(
)
,
:
g x
gx G
X
X
×
→
a
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
(a)
1x
x
=
ﻟﻜل،
x
X
∈
:
(b)
(
)
(
)
1
2
1
2
g g
x
g
g x
=
ﻟﻜل،
1
2
,
,
g g
G x
X
∈
∈
.
ﺇﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻤﻊ ﺍﻟﺘﺄﺜﻴﺭ
)
ﺍﻟﻴﺴﺎﺭﻱ
(
ﻟﻠﺯﻤﺭﺓ
G
ﺘﺩﻋﻰ
G
-
ﻤﺠﻤﻭﻋﺔ ﻤﻥ
)
ﺍﻟﻴﺴﺎﺭ
.(
ﻟﻜل،ﻪﻨﺃ ﻁﻭﺭﺸﻟﺍ ﻲﻀﺘﻘﺘ
g
G
∈
ﻓﺈﻥ ﻟﻠﻤﻨﺎﻗﻠﺔ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻭﻓﻕ،
g
،
:
,
,
L
g
X
X
x
gx
→
a
ﻤﻌﻜﻭﺱ
( )
1
L
g
−
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،
L
g
، ﺃﻱ ﺃﻥ،لﺒﺎﻘﺘ
( )
Sym
L
g
X
∈
.
ﺴﻠﻤﺔـﻤﻟﺍ ﻥـﻤ
(b)
ﻨﺠﺩ ﺃﻥ
(14)
( )
:
Sym
L
g
g
G
X
→
a
ﺘﺸﺎﻜل
.
ــﺴﺎﺭﻱ ﻟـﻴﻟﺍ ﺭﻴﺜﺄـﺘﻟﺍ ﻥـﻤ ،ﻙﻟﺫـﻟ
G
ﻰـﻠﻋ
X
ﺸﺎﻜلـﺘﻟﺍ ﻰـﻠﻋ لﺼـﺤﻨ ،
( )
Sym
G
X
→
ـ ﻜل ﺘﺸﺎﻜل ﻜﻬﺫﺍ ﻴﻌﺭﻑ ﺘﺄﺜﻴﺭﺍﹰ ﻟ،ﺱﻜﻌﻟﺎﺒﻭ ،
G
ﻋﻠﻰ
X
.
ﻪـﻨﺄﺒ لﺎـﻘﻴ
ﺃﻤﻴﻥ
(faithful)
)
ﺃﻭ ﺩﻗﻴﻕ
(effective)
(
ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﺸﺎﻜل
(14)
ﺇﺫﺍ ﻜﺎﻥ،ﻪﻨﺃ ﻱﺃ ،ﹰﺎﻨﻴﺎﺒﺘﻤ
1
gx
x
g
= ⇒ =
ﻟﻜل
x
X
∈
ﻤﺜﺎل
2.4
(a)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﺘﻨﺎﻅﺭﻴﺔ
n
S
ﺘﺅﺜﺭ ﺒﺸﻜل ﺃﻤﻴﻥ ﻋﻠﻰ
{
}
1, 2,..., n
.
(b)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
H
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺘﺅﺜﺭ ﺒﺸﻜل ﺃﻤﻴﻥ ﻋﻠﻰ
G
ﺒﺎﻟﺘﺤﻭﻴل ﺍﻟﻴﺴﺎﺭﻱ
(
)
,
,
H
G
G
h x
hx
× →
a
(c)
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
G
.
ﺇﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺘﺅﺜﺭ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
ـﻟ
H
،
(
)
,
,
G G H
G H
g C
gC
×
→
a
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، ﻤﺜﻼﹰ،ﻥﺎﻜ ﺍﺫﺇ ،ﹰﺎﻨﻴﻤﺃ ﻥﻭﻜﻴ ﺭﻴﺜﺄﺘﻟﺍﻭ
H
G
≠
ﻭ
G
ﺒﺴﻴﻁﺔ
.
(d)
ﻜل ﺯﻤﺭﺓ
G
، ﺘﺅﺜﺭ ﻋﻠﻰ ﻨﻔﺴﻬﺎ ﺒﺎﻟﺘﺭﺍﻓﻕ
(
)
def
1
,
,
x
G G
G
g x
g
gxg
−
× →
=
a
ﻷﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
N
،
G
ﺘﺅﺜﺭ ﻋﻠﻰ
N
ﻭ
G N
ﺒﺎﻟﺘﺭﺍﻓﻕ
.
(e)
ﻤﻥ ﺃﺠل ﺃﻱ ﺯﻤﺭﺓ
G
،
( )
Aut G
ﺘﺅﺜﺭ ﻋﻠﻰ
G
.
(f)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺤﺭﻜﻴﺔ ﻏﻴﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺘﺄﺜﻴﺭ
(group of rigid motion)
ﻓﻲ
n
ﻫﻲ ﺍ
ﺭﺓـﻤﺯﻟ
G
ﻤﻥ ﺍﻟﺘﻘﺎﺒﻼﺕ
n
n
→
ﺍﻟﺘﻲ ﺘﺤﺎﻓﻅ ﻋﻠﻰ ﺍﻷﻁﻭﺍل
.
ﻋﻨﺩﺌﺫ
G
ﻰـﻠﻋ ﺭﺜﺅﺘ
n
ﻥـﻤ
ﺍﻟﻴﺴﺎﺭ
.
ﻴﻌﺭﻑ ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻴﻤﻴﻨﻲ
(right action)
X
G
G
× →
ﺒﺸﻜل ﻤ
ﺸﺎﺒﻪ
.
ﺄﺜﻴﺭـﺘﻟﺍ لـﻴﻭﺤﺘﻟ
ﻨﻀﻊ،ﻱﺭﺎﺴﻴ ﺭﻴﺜﺄﺘ ﻰﻟﺇ ﻲﻨﻴﻤﻴﻟﺍ
1
g
x
xg
−
∗ =
.
ــﻲ ﻟـﻨﻴﻤﻴ ﻲﻌﻴﺒﻁ ﺭﻴﺜﺄﺘ ﺩﺠﻭﻴ ،ﹰﻼﺜﻤ
G
ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﻤﻴﻨﻴﺔ ﻟﻠﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
H
ﻓﻲ
G
،ﺩـﻴﺩﺤﺘﻟﺎﺒ ،
(
)
,
C g
Cg
a
،
ﺍﻟﺫﻱ ﻤﻥ ﺍﻟﻤﻤﻜﻥ ﺃﻥ ﻴﺘﺤﻭل ﺇﻟﻰ ﺘﺄﺜﻴﺭ ﻴﺴﺎﺭﻱ
(
)
,
g C
gC
a
.
ﺍﻟﺘﻁﺒﻴﻕ ﻤﻥ
G
-
ﻤﺠﻤﻭﻋﺔ
)
،ﺃﻭ
G
-
ﺘﻁﺒﻴﻕ ﺃﻭ
G
-
ﺎﻓﺊـﻜﻤ ﻕـﻴﺒﻁﺘ
(
ﻕـﻴﺒﻁﺘﻟﺍ ﻭـﻫ
:X
Y
ϕ
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
.
( )
( )
,
,
gx
g
x
g
G x
X
ϕ
ϕ
=
∀ ∈
∈
ـﺍﻟﺘﻤﺎﺜل ﻟ
G
-
ﻤﺠﻤﻭﻋﺔ ﻫﻭ
G
-
ﻴﻜﻭﻥ ﻨﻅﻴﺭﻩ ﻫﻭ ﺃﻴﻀﺎﹰﺫﺌﺩﻨﻋ ،لﺒﺎﻘﺘ ﻕﻴﺒﻁﺘ
G
-
ﺘﻁﺒﻴﻕ
.
ﺍﻟﻤﺩﺍﺭﺍﺕ
(Orbits)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
.
ﻴﻘﺎل ﻋﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
S
X
⊂
ﺔـﺘﺒﺎﺜ ﺎﻬﻨﺃ
(stable)
ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺘﺄﺜﻴﺭ
G
ﺇﺫﺍ ﻜﺎﻥ
.
,
g
G x
S
gx
S
∈
∈ ⇒
∈
ﻴﻨﺘﺞ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
S
ﻤﻥ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
X
.
ﻨﻜﺘﺏ
~
G
x
y
ﺎﻥـﻜ ﺍﺫﺇ
y
gx
=
ﺒﻌﺽـﻟ ،
g
G
∈
.
ﻴﺔ ﻷﻥـﺴﺎﻜﻌﻨﺍ ﺔـﻗﻼﻌﻟﺍ ﻥﺇ
1
x
x
=
ﻭﺘﻨﺎﻅﺭ،
ﻴﺔ ﻷﻥ
1
y
gx
x
g y
−
=
⇒ =
)
ـﺒﺎﻟﻀﺭﺏ ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﺒ
1
g
−
ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻤﺴﻠﻤﺎﺕ
(
ﻭ،
ﻫﻲ
ﻤﺘﻌﺩﻴﺔ ﻷﻥ
.
( ) ( )
,
y
gx
z
g y
z
g
gx
g g x
′
′
′
=
=
⇒ =
=
ﻟﺫﻟﻙ ﻓﻬﻲ ﺘﺸﻜل ﻋﻼﻗﺔ ﺘﻜﺎﻓﺅ
.
ﻨﺩﻋﻭ ﺼﻔﻭﻑ ﺍﻟﺘﻜﺎﻓﺅ
G
-
ﻤﺩﺍﺭ
.
ﻟﻬﺫﺍ ﻓﺈﻥ
G
-
ﻤﺩﺍﺭ ﻴﺸﻜل
ﺘﺠﺯﺌﺔ ﻟﻠﻤﺠﻤﻭﻋﺔ
X
.
ﻨﻜﺘﺏ
\
G X
ﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺩﺍﺭﺍﺕ
.
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٨٦
ﻟﺩﻴﻨﺎ
ﺇﻥ،ﻑﻴﺭﻌﺘﻟﺍ ﻥﻤ
G
-
ﻤﺩﺍﺭ
ﻴﺤﻭ
ﻱ
0
x
ﺇﺫﺍ ﻜﺎﻥ
{
}
0
0
Gx
gx
g
G
=
∈
ﻭﻫﻲ ﺃﺼﻐﺭ ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ
G
-
ﺜﺎﺒﺘﺔ
X
ﺘﺤﻭﻱ
0
x
.
ﻤﺜﺎل
3.4
(a)
ﺒﻔﺭﺽ ﺃﻥ
G
ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﻭﻟﻴﻜﻥ،
G
α
∈
ﻋﻨﺼﺭﺍﹰ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
.
ﻋﻨﺩﺌﺫ
ﻤ
ﺩﺍﺭﺍﺕ
α
ﻫﻲ ﻤﺠﻤﻭﻋﺎﺕ ﻤﻥ ﺍﻟﺸﻜل
{
}
1
0
0
0
,
,...,
n
x
x
x
α
α
−
)
ﻰـﻠﻋ ﺔﻋﻭﻤﺠﻤﻟﺍ ﻱﻭﺤﺘ ﻥﺃ ﻥﻜﻤﻴ ﻙﻟﺫﻟﻭ ،ﺔﻔﻠﺘﺨﻤ ﺭﺼﺎﻨﻌﻟﺍ ﻩﺫﻫ ﻥﻭﻜﺘ ﻥﺃ ﻱﺭﻭﺭﻀﻟﺍ ﻥﻤ ﺱﻴﻟ
ﻋﺩﺩ ﺃﻗل ﻤﻥ
n
ﻋﻨﺼﺭ
.(
(b)
ﺇﻥ ﻤﺩﺍﺭﺍﺕ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﻤﻥ
G
ﺍﻟﺘﻲ ﺘﺅﺜﺭ ﻋﻠﻰ
G
ﻲـﻫ ﺭﺎﺴﻴﻟﺍ ﻥﻤ ﺀﺍﺩﺠﻟﺎﺒ
ـﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﻤﻴﻨﻴﺔ ﻟ
H
ﻓﻲ
G
.
ﻨﻜﺘﺏ
\
H G
ﺔـﻴﻨﻴﻤﻴﻟﺍ ﺕﺎـﻘﻓﺍﺭﻤﻟﺍ ﺔﻋﻭﻤﺠﻤﻟ
.
ﺸﻜلـﺒ
ﺇﻥ ﻤﺩﺍﺭﺍﺕ،ﻪﺒﺎﺸﻤ
H
ﺍﻟﺘﻲ ﺘﺅﺜﺭ ﻋﻠﻰ
G
ــﺴﺎﺭﻴﺔ ﻟـﻴﻟﺍ ﺕﺎﻘﻓﺍﺭﻤﻟﺍ ﻲﻫ ﻥﻴﻤﻴﻟﺍ ﻥﻤ ﺀﺍﺩﺠﻟﺎﺒ
H
ﻓﻲ
G
ﻭﻨﻜﺘﺏ،
G H
ﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
.
ﻰـﻠﻋ ﺓﺭﻤﺯﻟﺍ ﻥﻭﻨﺎﻗ ﻥﺄﺒ ﻅﺤﻼﻨ
G
ﻟﻥ ﻴﻨﺘﺞ ﻋﻨﻪ ﻗﺎﻨﻭﻥ ﺍﻟﺯﻤﺭﺓ ﻋﻠﻰ
G H
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻓﻴﻬﺎ
H
ﻨﺎﻅﻤﻴﺔ
.
(c)
ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭﺓ
G
ﺘﺩﻋﻰ ﺍﻟﻤﺩﺍﺭﺍﺕ ﺒﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ،ﻕﻓﺍﺭﺘﻟﺎﺒ ﺎﻬﺴﻔﻨ ﻰﻠﻋ ﺭﺜﺅﺘ ﻲﺘﻟﺍ
:
ﻟﻜل
x
G
∈
ﺇﻥ ﺼﻑ ﺘﺭﺍﻓﻕ،
x
ﻫﻭ ﺍﻟﻤﺠﻤﻭﻋﺔ
{
}
1
gxg
g
G
−
∈
ﻭﺍﻟﺘﻲ ﻫﻲ ﻤﺭﺍﻓﻘﺎﺕ
0
x
.
ﺇﻥ ﺼﻑ ﺘﺭﺍﻓﻕ
x
ﻴﺤﻭ
ﻱ ﺩﺍﺌﻤﺎﹰ
0
x
ﻥـﻤ ﻁـﻘﻓ ﻑﻟﺄﺘﻴ ﻭ ،
0
x
ﺇﺫﺍ
ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
0
x
ﻓﻲ ﻤﺭﻜﺯ
G
.
ﻓﻲ ﺍﻟﺠﺒﺭ ﺍﻟﺨﻁﻲ ﺘﻜﻭﻥ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
G
ﺴﺎﻭﻴﺔـﺘﻤ
ﻤﻊ
( )
GL
n
k
ﻭﺍﻟﺘﻲ ﺘﺴﻤﻰ ﺒ
ﻭ،ﻪﺒﺎﺸﺘﻟﺍ ﻑﻭﻔﺼ
ﺇﻥ ﻨﻅﺭﻴﺔ ﺍﻷﺸﻜﺎل ﺍﻟﻨﺴﺒﻴﺔ ﺍﻟﻘﺎﻨﻭ
ﺯﻭﺩـﺘ ﺔـﻴﻨ
ﺒﻤﺠﻤﻭﻋﺔ ﺘﻤﺜﻴﻼﺕ ﻟﺼﻔﻭﻑ ﺍﻟﺘﺭﺍ
ﻓﻕ
:
ﺘﻜﻭﻥ ﺍﻟﻤﺼﻔﻭﻓﺘﺎﻥ ﻤﺘﺸﺎﺒﻬﺘﻴﻥ
)
ﻤﺘﺭﺍﻓﻘﺘﻴﻥ
(
ﺇﺫﺍ ﻭ
ﻁ ﺇﺫﺍـﻘﻓ
ﻜﺎﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﺸﻜل ﺍﻟﻨﺴﺒﻲ ﺍﻟﻘﺎﻨﻭﻨﻲ
.
ﻨﻼﺤﻅ ﺒﺄﻨﻪ ﺘﻜﻭﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ
X
ﺜﺎﺒﺘﺔ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﺠﺘﻤﺎﻋﺎﹰ ﻟﻤﺩﺍﺭﺍﺕ
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠ،ﹰﻼﺜﻤ
ﺯﺌﻴﺔ
H
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﺠﺘﻤﺎﻋﺎﹰ ﻟﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
.
ﻴﻘﺎل ﺒﺄﻥ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
X
ﻤﺘﻌﺩﻴﺎﹰ
(transitive)
ﻭﻴﻘﺎل ﺒﺄﻥ،
G
ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﺸﻜلـﺒ
ﻤﺘﻌﺩ
(transitivity)
ﻷﺠل ﺃﻱ ﻋﻨﺼﺭﻴﻥ،ﻪﻨﺃ ﻱﺃ ،ﻁﻘﻓ ﺩﺤﺍﻭ ﺭﺍﺩﻤ ﺩﺠﻭ ﺍﺫﺇ ،
x
ﻭ
y
ﻥـﻤ
X
ﻴﻭﺠﺩ،
g
G
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
gx
y
=
.
ﺘﺩﻋﻰ ﺍﻟﺫﺌﺩﻨﻋ
ﺔـﻋﻭﻤﺠﻤ
X
G
-
ﺴﺔـﻨﺎﺠﺘﻤ
(homogeneous)
.
ﺇﻥ،ﹰﻼﺜﻤ
n
S
ﺘﺅﺜﺭ ﻋﻠﻰ
{
}
1, 2,..., n
ﺒﺸﻜل ﻤﺘﻌﺩ
.
ﺭﺓـﻤﺯ ﻱﺃ لـﺠﻷ
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٩٦
ﺠﺯﺌﻴﺔ
H
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻓﺈﻥ،
G
ﺘﺅﺜﺭ ﻋﻠﻰ
G H
ﺒﺸﻜل ﻤﺘﻌﺩ
ﺄﺜﻴﺭـﺘ ﻥﻜﻟ ،
G
ﻰـﻠﻋ
ﻨﻔﺴﻬﺎ ﻴﻜﻭﻥ ﻏﻴﺭ ﻤﺘﻌﺩ
ﺇﺫﺍ ﻜﺎﻥ
1
G
≠
ﻷﻥ
{ }
1
ﻫﻭ ﺼﻑ ﺘﺭﺍﻓﻕ ﺩﺍﺌﻤﺎﹰ
.
ﻴﻜﻭﻥ ﺘﺄﺜﻴﺭ ﺍﻟﺯﻤﺭﺓ
G
ﻋﻠﻰ
X
ﻤﺘﻌﺩﻴﺎﹰ ﺒﺸﻜل ﻤﻀﺎﻋﻑ
(doubly transitive)
ﺇﺫﺍ ﻜﺎﻥ
ﻷﻱ ﺯﻭﺠﻴﻥ
(
) (
)
1
2
1
2
,
,
,
x x
y y
ﻤﻥ ﻋﻨﺎﺼﺭ
X
ﺒﺤﻴﺙ
1
2
x
x
≠
ﻭ
1
2
y
y
≠
ﺩـﺠﻭﻴ ،
ﻋﻨﺼﺭ
)
ﻤﻔﺭﺩ
(
g
G
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
1
1
gx
y
=
ﻭ
2
2
gx
y
=
.
ﻨﻌﺭﻑ
k
-
ﻤﻀﺎﻋﻔ
ﺎﹰ
ﻤﺘﻌﺩﻴ
ﺎﹰ
(fold transitivity)
ﻟﻜل
3
k
≥
ﺒﺸﻜل ﻤﺸﺎﺒﻪ
.
ﺍﻟﻤﺜﺒﺘﺎﺕ
(Stabilizers)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ
X
.
ﺇﻥ ﺍﻟﻤ
ﺜﺒﺕ
(Stabilizer)
)
ﻭﺍﺹـﺨﻟﺍ ﺩﺤﻭﻤ ﻭﺃ
(isotropy
group)
(
ﺍﻟﻌﻨﺼﺭ
x
X
∈
ﺇﺫﺍ ﻜﺎﻥ
.
( )
{
}
Stab x
g
G gx
x
=
∈
=
ﻟﻜﻥ ﻟﻴﺱ ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﻴﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ،ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻭﻫﻭ
.
ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ
:
ﺘﻤﻬﻴﺩﻴﺔ
4.4
ﻷﻱ
g
G
∈
ﻭ
x
X
∈
،
( )
( )
1
Stab
.Stab
.
gx
g
x g
−
=
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
g x
x
′ =
ﻋﻨﺩﺌﺫ،ﹰﺍﺩﻴﺩﺤﺘ ،
(
)
1
,
gg g
x
gg x
gx
y
−
′
′
=
=
=
ﻭﺒﺎﻟﺘﺎﻟﻲ
( )
( )
1
.Stab
.
Stab
g
x g
gx
−
⊂
.
ﺇﺫﺍ ﻜﺎﻥ،ﺱﻜﻌﻟﺍ
( )
g
gx
gx
′
=
ﻋﻨﺩﺌﺫ،
(
)
( )
1
1
1
,
g g g x
g g
gx
g y
x
−
−
−
′
′
=
=
=
ﻭﻤﻨﻪ
( )
1
Stab
g g g
x
−
′ ∈
ﺃﻱ ﺃﻥ،
( )
1
.Stab
.
g
g
x g
−
′∈
.
ﻤﻥ ﺍﻟﻭﺍﻀﺢ
( )
( )
(
)
Stab
Ker
Sym
,
x X
x
G
x
∈
=
→
I
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ﻭﻫ ﻱﺫﻟﺍﻭ
G
.
ﺎﻥـﻜ ﺍﺫﺇ ﻁـﻘﻓﻭ ﺍﺫﺇ ﹰﺎـﻨﻴﻤﺃ ﺭﻴﺜﺄـﺘﻟﺍ ﻥﻭـﻜﻴ
( ) { }
Stab
1
x X
x
∈
=
I
.
ﻤﺜﺎل
5.4
(a)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ ﻨﻔﺴﻬﺎ ﺒﺎﻟﺘﺭﺍﻓﻕ
.
ﻋﻨﺩﺌﺫ
( )
{
}
Stab x
g
G gx
xg
=
∈
=
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٠٧
ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﺯﻤﺭﺓ ﺒﻤﻤﺭﻜﺯ
(centralizer)
( )
G
C
x
ﺍﻟﻌﻨﺼﺭ
x
ﻓﻲ
G
.
ﻥـﻤ ﻑﻟﺄﺘﺘ ﻲﻫﻭ
ﺠﻤﻴﻊ ﻋﻨﺎﺼﺭ
G
، ﻤﺭﻜﺯ،ﻪﻨﺃ ﻱﺃ ،ﻪﻌﻤ لﺩﺎﺒﺘﺘ ﻲﺘﻟﺍ
x
.
ﺇﻥ ﺍﻟﺘﻘﺎﻁﻊ
( )
{
}
,
G
x G
C
x
g
G gx
xg
x
G
∈
=
∈
=
∀ ∈
I
ﻫﻭ ﻤﺭﻜﺯ
G
.
(b)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ
G H
ﺴﺎﺭـﻴﻟﺍ ﻥﻤ ﺀﺍﺩﺠﻟﺎﺒ
.
ﺫـﺌﺩﻨﻋ
( )
Stab H
H
=
،
ﻭﻤﺜﺒﺕ
gH
ﻫﻭ
1
gHg
−
.
(c)
ﻟﺘﻜﻥ
G
ﻲـﻓ ﺓﺭﺜﺄﺘﻤﻟﺍ ﺭﻴﻏ ﺕﺎﻨﺍﺭﻭﺩﻟﺍ ﻥﻤ ﺓﺭﻤﺯ
n
)
(4.2f
.
ﺩﺃـﺒﻤ ﺯـﻜﺭﻤﻤ ﻥﺇ
ﺩﺓـﻤﺎﻌﺘﻤﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻭﻫ ﺕﺎﻴﺜﺍﺩﺤﻹﺍ
n
O
ـﻟﻠ
ﻰـﻠﻋ ﻲـﺴﺎﻴﻘﻟﺍ ﺏـﺠﻭﻤﻟﺍ ﻑﺭـﻌﻤﻟﺍ لﻜﺸ
n
)
Artin1991
ﺍﻟﻔﺼل،
4
،
5.16
.(
ﻟﻴﻜﻥ
(
)
,
n
T
+
ﻥـﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
ﻭﻴﻼﺕـﺤﺘﻟ
n
ﺘﻁﺒﻴﻕ ﻤﻥ ﺍﻟﺸﻜل،ﻪﻨﺃ ﻱﺃ ،
0
v
v
v
+
a
ﻟﺒﻌﺽ
0
n
v
∈
.
ﺈﻥـﻓ ﺫﺌﺩﻨﻋ
T
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻭ ﺃﻥ
G
T
O
θ
×
)
cf. Artin 1991
ﺍﻟﻔﺼل،
.
5,§2
.(
ﻷﺠل ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
S
ﻤﻥ
X
،
ﻨﻌﺭﻑ ﻤﺜﺒﺕ
(Stabilizer)
S
ﺒﺎﻟﺸﻜل
( )
{
}
Stab S
g
G gS
S
=
∈
=
ﻋﻨﺩﺌﺫ
( )
Stab S
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭ،
ﺒ
ﻨﻔﺱ ﺍﻟﺘﺭﺘﻴﺏ ﻜﻤﺎ ﻓﻲ ﺒﺭﻫﺎﻥ
(4.4)
ﺘﺒﻴﻥ ﺒﺄﻥ
( )
( )
1
Stab
.Stab
.
gS
g
S g
−
=
ﻤﺜﺎل
4.6
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ
G
ﻭ،ﻕﻓﺍﺭﺘﻟﺎﺒ
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺴﻤﻰـﻴ
ﻤﺜﺒﺕ
H
ﺒﺎﻟﻤﻨﻅﻡ
(normalizer)
( )
G
N
H
ـ ﻟ
H
ﻓﻲ
G
:
( )
{
}
1
G
N
H
g
G gHg
H
−
=
∈
=
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺒﺄﻥ
( )
G
N
H
ﻫﻲ ﺃﻜﺒﺭ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺘﺤﻭﻱ
H
ﻜﻤﺎ ﺃﻨﻬﺎ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
ﻤﻥ ﺍﻟﻤﺤﺘﻤل ﺃﻥ ﻴﻜﻭﻥ
gS
S
⊂
ﻟﻜﻥ
( )
Stab
g
S
∈
)
ﺍﻨﻅﺭ
1.32
.(
ﺍﻟﺘﺄﺜﻴﺭﺍﺕ ﺍﻟﻤﺘﻌﺩﻴﺔ
(Transitive actions)
ﻗﻀﻴﺔ
7.4
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﺒﺸﻜل ﻤﺘﻌﺩ
ﻷﻱﺫﺌﺩﻨﻋ ،
0
x
X
∈
ﻴﻜﻭﻥ ﺍﻟﺘﻁﺒﻴﻕ،
( )
( )
0
0
0
Stab
:
Stab
g
x
gx
G
x
X
→
a
ﺘﻤﺎﺜل ﻋﻠﻰ
G
-
ﻤﺠﻤﻭﻋﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ،ﻪﻨﻷ ﹰﺍﺩﻴﺠ ﻑﺭﻌﻤ ﻕﻴﺒﻁﺘﻟﺍ ﻥﺇ
( )
0
Stab
h
x
∈
ﺫـﺌﺩﻨﻋ ،
0
0
ghx
gx
=
.
ﻭﻫﻭ ﻤﺘﺒﺎﻴﻥ ﻷﻥ
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١٧
1
0
0
0
0
gx
g x
g g x
x
−
′
′
=
⇒
=
⇒
)
,
′
g g
ﺘﻘﻌ
ﺎﻥ
ـﻓﻲ ﻨﻔﺱ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ ﻟ
( )
0
(Stab x
ﻭﻫﻭ ﻏﺎﻤﺭ ﻷﻥ
G
ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺘﻌﺩ
.
ﻴﻜﻭﻥ ﻭﺍﻀﺤﺎﹰ ﻟﻨﺎ ﺒﺄﻨﻪ،ﹰﺍﺭﻴﺨﺃ
G
-
ﺘﻜﺎﻓﺅ
.
ﻟﻬﺫﺍ ﺇﻥ ﻜل
G
-
ﻤﺠﻤﻭﻋﺔ ﻤﺘﺠﺎﻨﺴﺔ
X
ﺘﻜﻭﻥ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ
G H
ﺭـﻤﺯﻟﺍ ﻥـﻤ ﺽﻌﺒﻟ
ﺍﻟﺠﺯﺌﻴﺔ
H
ﻓﻲ
G
ـ ﻟﻜﻥ ﻫﻜﺫﺍ ﺘﺤﻭﻴل ﻟ،
X
ﻟﻴﺱ ﻗﺎﻨﻭﻨﻴﺎﹰ
:
ﺎﺭـﻴﺘﺨﺍ ﻰﻠﻋ ﺩﻤﺘﻌﻴ
0
x
X
∈
.
ـ ﻴﻜﻭﻥ ﻟ،ﻯﺭﺨﺃ ﺔﻘﻴﺭﻁﺒ ﻙﻟﺫ لﻭﻘﻨﻟ
G
-
ﺍﻟﻤﺠﻤﻭﻋﺔ
G H
،ﺩـﻴﺩﺤﺘﻟﺎﺒﻭ ،ﺔﻠﺼـﻔﻨﻤ ﺔـﻁﻘﻨ
ﺍﻟﻤﺭﺍﻓﻘﺔ
H
ﻹﻋﻁﺎﺀ؛
G
-
ﻤﺠﻤﻭﻋﺔ ﻤﺘﺠﺎﻨﺴﺔ
X
ﻲـﺘﻟﺍ ﹰﺎـﻤﺎﻤﺘ ﻲﻫ ،ﺔﻠﺼﻔﻨﻤﻟﺍ ﺎﻬﻁﺎﻘﻨ ﻊﻤ
ﺘﻌﻁﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻨﺘﻴﺠﺔ
8.4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﻭﻟﻴﻜﻥ،
0
O
Gx
=
ﺍﻟﻤﺩﺍﺭ ﺍﻟﺫﻱ
ﻴﺤﻭ
ﻱ
0
x
.
ﻓﺈﻥ ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲﺫﺌﺩﻨﻋ
O
ﻴﺴﺎﻭﻱ
( )
(
)
0
: Stab
O
G
x
=
ﺇﻥ ﻋﺩﺩ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ،ﹰﻼﺜﻤ
1
gHg
−
ﻟﻠﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﻓﻲ
G
ﻴﺴﺎﻭﻱ
( )
(
)
:
G
G N
H
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
O
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،ﹰﺎﻴﺩﻌﺘﻤ ﻥﻭﻜﻴ
0
g
gx
a
ﺎﹰـﻘﻴﺒﻁﺘ ﻑﺭﻌﻴ
ﺎﺒﻼﹰـﻘﺘ
( )
0
0
Stab
G
x
Gx
→
.
ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻔﻴﺩﺓ ﺩﺍﺌﻤﺎﹰ ﻟﺤﺴﺎﺏ
O
.
ﻗﻀﻴﺔ
9.4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﺒﺸﻜل ﻤﺘﻌﺩ
ﻷﻱ،ﺫﺌﺩﻨﻋ ،
0
x
X
∈
ﺘﻜﻭﻥ،
( )
(
)
Ker
Sym
G
X
→
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻋﻅﻤﻰ ﻤﺤﺘﻭﺍﺓ ﻓﻲ
( )
0
Stab x
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
0
x
X
∈
.
ﻋﻨﺩﺌﺫ
( )
(
)
( )
( )
( )
4.4
1
0
0
Ker
Sym
Stab
Stab
.Stab
x X
g G
G
X
x
gx
g
x g
−
∈
∈
→
=
=
=
I
I
I
ﻓﺈﻥ ﻫﺫﻩ ﺍﻟﻘﻀﻴﺔ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺘﺎﺒﻌﺔ ﻟﻠﺘﻤﻬﻴﺩﻴﺔ،ﻲﻟﺎﺘﻟﺎﺒﻭ
ﺍﻵﺘﻴ
.ﺔ
ﺘﻤﻬﻴﺩﻴﺔ
10.4
ﻷﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
H
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
،
1
g G
gHg
−
∈
I
ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻲﻫ
ﻨﺎﻅﻤﻴﺔ ﻋﻅﻤﻰ ﻤﺤﺘﻭﺍﺓ ﻓﻲ
H
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻼﺤﻅ ﺒﺄﻥ
def
1
0
g G
N
gHg
−
∈
=
I
ﺭﺓـﻤﺯ ﻲـﻬﻓ ،ﺔﻴﺌﺯﺠ ﺭﻤﺯﻟ ﻊﻁﺎﻘﺘ ﺎﻬﻨﺃ ﺎﻤﺒﻭ ،
ﺠﺯﺌﻴﺔ
.
ﻭﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ ﻷﻥ
( ) ( )
1
1
1
1
0
1
0
1
0
−
−
∈
=
=
I
g G
g N g
g g N
g g
N
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٢٧
-
، ﺍﺴﺘﺨﺩﻤﻨﺎ،ﺔﻴﻨﺎﺜﻟﺍ ﺓﺍﻭﺎﺴﻤﻟﺍ لﺠﺃ ﻥﻤ
g
ﺍﻟﺘﻲ ﺘﻤﺴﺢ ﻜل ﻋﻨﺎﺼﺭ
G
ﻭ ﻜﺫﻟﻙ،
1
g g
.
ﻟﻬﺫﺍ
ﻓﺈﻥ
0
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻤﺤﺘﻭﺍﺓ ﻓﻲ
1
eHe
H
−
=
.
ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﺯﻤﺭﺓ ﺃﺨﺭﻯ
ﻋﻨﺩﺌﺫ،ﻩﺫﻬﻜ
1
1
N
gNg
gHg
−
−
=
⊂
ﻟﻜل
g
G
∈
ﻭﻤﻨﻪ،
1
0
g G
N
gHg
N
−
∈
⊂
=
I
ﺼﻔﻭﻑ ﺍﻟﺘﻜﺎﻓﺅ
(The class equation)
ﺇﺫﺍ ﻜﺎﻨﺕ
X
ﻓﺈﻨﻬﺎ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﺠﺘﻤﺎﻉ ﻤﻨﻔﺼل ﻟﻌﺩﺩ ﻤﻨﺘﻪ ﻤﻥ ﺍﻟﻤﺩﺍﺭﺍﺕ،ﺔﻴﻬﺘﻨﻤ
:
1
m
i
i
X
O
=
=
U
ﻭﺒﺎﻟﺘﺎﻟﻲ
:
ﻗﻀﻴﺔ
11.4
ﺇﻥ ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ
X
ﻴﺴﺎﻭﻱ
( )
(
)
1
1
: Stab
,
m
m
i
i
i
i
i
i
X
O
G
x
x
O
=
=
=
=
∈
∑
∑
ﻋﻨﺩﻤﺎ ﺘﺅﺜﺭ ﺍﻟﺯﻤﺭﺓ
G
ﻋﻠﻰ ﻨﻔ
ﺘﺼﺒﺢ ﻫﺫﻩ ﺍﻟﺼﻴﻐﺔ،ﻕﻓﺍﺭﺘﻟﺎﺒ ﺎﻬﺴ
ﻜﻤﺎ ﻴﺄﺘﻲ
:
ﻗﻀﻴﺔ
12.4
)
ﺼﻔﻭﻑ ﺍﻟﺘﻜﺎﻓﺅ
(
( )
(
)
:
G
G
G C
x
=
∑
)
ﺤﻴﺙ
x
ﺘﻤﺴﺢ ﻤﺠﻤﻭﻋﺔ ﻤﻤﺜ
ﻼﺕ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
.(
( )
( )
(
)
:
G
G
Z G
G C
y
=
+
∑
)
ﺇﻥ
y
ﺘﻤﺴﺢ ﻤﺠﻤﻭﻋﺔ ﻤﻤﺜﻼﺕ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﺍﻟ
ﺤﺎﻭﻴﺔ ﻋﻠﻰ ﺃﻜﺜﺭ ﻤﻥ ﻋﻨﺼﺭ ﻭﺍﺤﺩ
.(
ﻤﺒﺭﻫﻨﺔ
13.4
)
ﻜﻭﺸﻲ
(
ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ
p
ﻴﻘﺴﻡ
G
ﺈﻥـﻓ ﺫﺌﺩﻨﻋ ،
G
ﻰـﻠﻋ ﻱﻭـﺤﺘ
ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ
G
.
ﺇﺫﺍ ﻜﺎﻨﺕ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
y
ﻰـﻟﺇ ﻲـﻤﺘﻨﺘ ﻻ
ﻤﺭﻜﺯ
G
،
p
ﺴﻡـﻘﻴ ﻻ
( )
(
)
:
G
G C
y
∑
ﺫـﺌﺩﻨﻋ ،
( )
G
p C
y
ﻕـﺒﻁﻨ ﻥﺃ ﻥـﻜﻤﻴ ﻭ
ﺍﻻﺴﺘﻘﺭﺍﺀ ﻹ
ﻴﺠﺎﺩ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻓﻲ
( )
G
C
y
.
ﻟﺫﻟﻙ ﻴﻤﻜﻥ ﺃﻥ ﻨﻔﺭﺽ ﺃﻥ
p
ﺴﻡـﻘﻴ
ﻜل ﺤﺩﻭﺩ
( )
(
)
:
G
G C
y
ﻓﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟ
ﺼﻔﻭﻑ
)
ﺍﻟﺸﻜل ﺍﻟﺜﺎﻨﻲ
(،
ﻀﺎﹰـﻴﺃ ﻡﺴـﻘﻴ ﻭـﻬﻓ ﻙﻟﺫﻟﻭ
)
ﺍﺠﺘﻤﺎﻉ ﻤﻨﻔﺼل
(
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٣٧
( )
Z G
.
ﻟﻜﻥ
( )
Z G
ﻭ ﻤﻥ ﻤﺒﺭﻫﻨﺔ ﺍﻟﺒﻨﻰ،ﻲﻠﻴﺩﺒﺘ
٣١
ﺄﻥـﺒ ﺩﺠﻨ ﺭﻤﺯﻟﺍ ﻩﺫﻫ لﺜﻤﻟ
( )
Z G
ﻴﺤﻭ
ﻱ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﻨﺘﻴﺠﺔ
14.4
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
p
-
ﺯﻤﺭﺓ ﺇﺫﺍ ﻭ
ﻥـﻤ ﺭﺼﻨﻋ لﻜ ﺔﺒﺘﺭ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓ
ﻋﻨﺎﺼﺭﻫﺎ ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ـ ﻗﻭﻯ ﻟ
p
ﻋﻨﺩﺌﺫ،
(25.1)
لـﻜ ﺔـﺒﺘﺭ ﻥﺄﺒ ﺞﻨﺍﺭﻏﻻ ﺔﻨﻫﺭﺒﻤ ﻥﻴﺒﺘ
ﻋﻨﺼﺭ ﻴﻜﻭﻥ ﻗﻭﻯ ﻟﻠﻌﺩﺩ
p
.
ﻭﺍﻟﻌﻜﺱ ﻴﺒﺭﻫﻥ ﻋﻠﻴﻪ ﻤﻥ ﻤﺒﺭﻫﻨﺔ
ﻜﻭﺸﻲ
.
ﻨﺘﻴﺠﺔ
(15.4)
ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2p
،
p
ﺭﺓـﻤﺯ ﻭﺃ ﺔـﻴﺭﺌﺍﺩ ﻥﻭـﻜﺘ ،ﻱﺩﺭﻓ ﻲﻟﻭﺃ ﺩﺩﻋ
ﺩﻴﻬﻴﺩﺭﺍل
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻌﻠﻡ ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ،ﻲﺸﻭﻜ ﺔﻨﻫﺭﺒﻤ ﻥﻤ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺎﺼﺭ
s
ﻭ
r
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭ
p
ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ
.
ﻟﺘﻜﻥ
H
r
=
.
ﻓﺈﻥ ﺩﻟﻴلﺫﺌﺩﻨﻋ
H
ﻴﺴﺎﻭﻱ
2
ﻭﻟﻬﺫﺍ ﻓﻬﻲ ﻨﺎﻅﻤﻴﺔ،
.
ﻤﻥ
ﺍﻟﻭﺍﻀﺢ ﺃﻥ
s
H
∉
ﻭﻤﻨﻪ،
s
G
H
H
=
U
:
{
}
1
1
1, ,...,
, ,
,...,
p
p
G
r
r
s rs
r
s
−
−
=
ﺎ ﺃﻥــﻤﺒﻭ
H
ﺔــﻴﺌﺯﺠ ﺓﺭــﻤﺯ
ﺔــﻴﻤﻅﺎﻨ
،
1
i
srs
r
−
=
ﺒﻌﺽــﻟ ،
i
.
ﻙــﻟﺫﻭ
ﻷﻥ
( )
2
2
2
2
1
1
1,
i
s
r
s rs
s srs
s
r
−
−
−
=
=
=
=
ﻭﻟﺫﻟﻙ،
2
1mod
i
p
≡
.
ﻷﻥ
p
Z p Z
، ﺤﻘل
ﺘﻜﻭﻥ ﻋﻨﺎﺼﺭﻩ ﻓﻘﻁ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻲ ﺘﺭﺒﻴﻌﻬﺎ ﻴﺴﺎﻭﻱ
1
ﻲـﻫﻭ
1
±
ﺫﻟﻙـﻟﻭ ،
1mod
i
p
≡
ﺃﻭ
1mod
i
p
≡ −
.
ﻓﻲ
ﺍﻟﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ،ﻰﻟﻭﻷﺍ ﺔﻟﺎﺤﻟﺍ
)
ﺃﻱ ﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
ﺍ
ﻟﻤﺘﺒﺎﺩﻟﺔ ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻨﻬﺎ ﺘﺒﺩﻴﻠﻴﺔ
(
ﻭﻓﻲ،
ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ
1
1
srs
r
−
−
=
ﻰـﻠﻋ ﺎﻨﻠﺼﺤ ﺩﻗ ﻥﻭﻜﻨ ﻭ
ﺯﻤﺭﺓ ﺩﻴﻬﻴﺩﺭﺍل
(9.2)
.
p
-
ﺯﻤﺭ
ﻤﺒﺭﻫﻨﺔ
4.16
ﻟﻜل
p
-
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻏﻴﺭ ﺘﺎﻓﻬﺔ ﻤﺭﻜﺯ ﻏﻴﺭ ﺘﺎﻓﻪ
.
13
ﻴﻭﺠﺩ ﻫﻨﺎ ﺒﺭﻫﺎﻥ ﻤﺒﺎﺸﺭ ﻫﻭ ﺃﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻤﺤﻘﻘﺔ ﻤﻥ ﺃﺠل ﺯﻤﺭﺓ ﺁﺒﻠﻴﺔ
Z
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ
Z
.
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ
Z
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻭ ﺍﻟﺘﻲ ﺘﻘﺒل ﺭﺘﺒﺘﻪ ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
ﻭﻯـﻗ ﺽـﻌﺒ ﺫـﺌﺩﻨﻋ ﻪﻨﻷ ،
ﺍﻟﻌﻨﺎﺼﺭ ﺴﻴﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺘﻤﺎﻤﺎﹰ
.
ﻟﻴﻜﻥ
1
g
≠
ﻋﻨﺼﺭ ﻤﻥ
Z
.
ﺇﺫﺍ ﻜﺎﻥ
p
ﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺭﺘﺒﺔ
g
،
ﺴﻴﻘﺴﻡ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓﺫﺌﺩﻨﻋ
Z
g
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻴﻭﺠﺩ ﻓﻴﻬﺎ،
)
ﺒﺎﻻﺴﺘﻘﺭﺍﺀ
(
ﻋﻨﺼﺭ ﻤﻥ
G
ﺒﺤﻴﺙ ﺘﻘﺒل ﺭﺘﺒﺘﻪ ﻓﻲ
Z
g
ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
ﻟﻜﻥ ﺭﺘﺒﺔ ﻤﺜل ﻫﺫﺍ ﺍﻟﻌﻨﺼﺭ ﻴﺠﺏ ﺃﻥ ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
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ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﻤﺎ ﺃﻥ،ﺽﺭﻔﻟﺍ ﻥﻤ
(
)
:1
G
ﻗﻭ
ﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
p
ﺈﻥـﻓ ﻲﻟﺎـﺘﻟﺎﺒﻭ ،
( )
(
)
:
G
G C
y
ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
p
)
0
p
≠
(
ﻟﻜل
y
ﻓﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺼﻔﻭﻑ
)
ﺍﻟﺸﻜل ﺍﻟﺜﺎﻨﻲ
.(
ﺒﻤﺎ ﺃﻥ
p
ﺴﻡـﻘﻴ
ﻜل ﺤﺩ ﻤﻥ
ﻤﻌﺎﺩﻟﺔ
ﺘﺜﻨﺎﺀـﺴﺎﺒ ﻑﻭﻔﺼﻟﺍ
)
ﺎـﻤﺒﺭ
(
( )
(
)
:1
Z G
،
ﺎﻟﻲـﺘﻟﺎﺒﻭ
ﺴﻡـﻘﻴ ﻥﺃ ﺏـﺠﻴ
( )
(
)
:1
Z G
ﺃﻴﻀﺎﹰ
.
ﻨﺘﻴﺠﺔ
17.4
ﺍﻟﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
p
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ
m
p
لـﻜﻟ
m
n
≤
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﺍﻟﺭﻴﺎﻀﻲ ﻋﻠﻰ
n
.
ﺇﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
G
ﻭـﺤﻴ
ﻥـﻤ ﺭﺼـﻨﻋ ﻱ
ﺍﻟﺭﺘﺒﺔ
p
ﻭﻟﺫﻟﻙ،
N
g
=
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﺍﻵﻥ ﻭﻤﻥ ﻓﺭﻀﻴﺔ
ﺍﻻﺴﺘﻘﺭﺍﺀ ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﻔﺭﺽ ﺍﻟﻨﺘﻴﺠﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ
G N
لـﺒﺎﻘﺘﻟﺍ ﺔﻨﻫﺭﺒﻤ ﻥﺈﻓ ﻡﺜ ﻥﻤﻭ ،
(46.1)
ﺘﻌﻁﻴﻨﺎ ﺍﻟﻨﺘﻴﺠﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ
G
.
ﻗﻀﻴﺔ
18.4
ﻜل ﺯﻤﺭﺓ ﺭﺘﺒﺘﻬﺎ
2
p
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ،ﺔﻴﻠﻴﺩﺒﺘ ﺓﺭﻤﺯ ﻲﻫ
p
p
C
C
×
ﺃﻭ
2
p
C
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻌﻠﻡ ﺒﺄﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
Z
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،ﹰﺎﻬﻓﺎﺘ ﺱﻴﻟ
G Z
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
1
ﻥـﻤ ﻭﺃ
ﺍﻟﺭﺘﺒﺔ
p
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،ﺔﻴﺭﺌﺍﺩ ﻥﻭﻜﺘ ﻥﻴﺘﻟﺎﺤﻟﺍ ﻥﻤ لﻜ ﻲﻓ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﺘﻤﻬﻴﺩﻴﺔ
19.4
ﻨﻔﺭﺽ ﺒﺄﻥ
G
ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﻓﻲ ﻤﺭﻜﺯﻫﺎ
)
ﻲـﻬﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ
ﻨﺎﻅﻤﻴﺔ
(
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
G H
ﺩﺍﺌﺭﻴﺔ
.
ﻋﻨﺩﺌﺫ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
a
ﻋﻨﺼﺭﺍﹰ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻭﺍﻟﺫﻱ ﺼﻭﺭﺘﻪ ﻓﻲ
G H
ﻴﻭﻟﺩﻫﺎ
.
لـﻜ ﺫـﺌﺩﻨﻋ
ﻋﻨﺼﺭ ﻤﻥ
G
ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﺘﺏ ﺒﺎﻟﺸﻜل
i
g
a h
=
ﻤﻊ
,
∈
∈
h
H i
.
ﺍﻵﻥ
( )
.
(
.
′
′
′
′
′
′
=
⊂
′
=
′
=
i
i
i
i
i
i
i
i
a h a h
a a hh
H
Z G
a a h h
a h a h
ﻤﻼﺤﻅﺔ
20.4
ﻴﺒﻴﻥ ﺍﻟﺒﺭﻫﺎﻥ ﺍﻟﺴﺎﺒﻕ ﺒﺄﻨﻪ ﺇﺫﺍ ﻜﺎﻥ
( )
H
Z G
⊂
ﻭ
G
ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ
ﻤﻥ ﻤﻤﺜﻼﺕ
G H
ﻓﺈﻥﺫﺌﺩﻨﻋ ،ﺔﻟﺩﺎﺒﺘﻤ ﺎﻫﺭﺼﺎﻨﻋ ﻥﻭﻜﺘ ﺙﻴﺤﺒ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﻤﻥ ﺃﺠل ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ ﺍﻟﻔﺭﺩﻱ
p
ﺭـﻴﻏ ﺓﺭـﻤﺯ ﻱﺃ ﻥﺄـﺒ ﻥﻴﺒﻨ ﻥﺃ ﻥﻵﺍ ﺏﻌﺼﻟﺍ ﻥﻤ ﺱﻴﻟ ،
ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
ﻤﺘﻤﺎﺜﻠﺔ
ﺘﻤﺎﻤﺎﹰ ﻤﻊ ﻭﺍﺤﺩﺓ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﻤﺒﻨﻴﺔ ﻓﻲ
(3.13, 3.14)
)
ﺍﻟﺘﻤﺭﻴﻥ
3-4
.(
ﻴﻭﺠﺩ ﺘﻤﺎﻤﺎﹰ ﺯﻤﺭﺘﻴﻥ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺘﻴﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،لﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ ،ﺍﺫﻬﻟ
3
p
.
)
ﻷن
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ﻤﺜﺎل
21.4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
8
.
ﻴﺠﺏ ﺃﻥ ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﺫﺌﺩﻨﻋ
ﻨﺼﺭ
a
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
1-5
.(
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﺙ ﻻـﻴﺤﺒ
ﻰــﻟﺇ ﻲــﻤﺘﻨﻴ
a
ﺫــﺌﺩﻨﻋ ،
G
a
b
θ
×
ﺙــﻴﺤ
θ
ل ﺍﻟﻭﺤــﺜﺎﻤﺘﻟﺍ ﻭــﻫ
ﺩــﻴ
(
)
2
4
×
→
Z Z
Z 4 Z
ﻭﻟﺫﻟﻙ،
4
G
D
≈
.
ﺃﻱ ﻋﻨﺼﺭ،ﻙﻟﺫﻜ ﻥﻜﻴ ﻡﻟ ﺍﺫﺇ ﺎﻤﺃ
b
ﻥـﻤ
G
ﻻ
ﻴﻘﻊ ﻓﻲ
a
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
ﻭ،
2
2
a
b
=
.
ﺍﻵﻥ
1
bab
−
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺭﺼﻨﻋ ﻭﻫ
4
ﻓﻲ
a
.
ﻭﻻ ﻴﺴﺎﻭﻱ
a
ﻷﻨﻪ ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ،
G
ﺈﻥـﻓ ﺍﺫﻬﻟﻭ ،ﺔﻴﻠﻴﺩﺒﺘ ﻥﻭﻜﺘﺴ
G
ﻲـﻫ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
(17.1, 2.7b)
.
ﺍﻟﺘﺄﺜﻴﺭﺍﺕ
ﻋﻠﻰ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
(Action on the left cosets)
ﺘﺄﺜﻴﺭ ﺍﻟﺯﻤﺭﺓ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
G H
ﻟﻠﺯﻤﺭﺓ
H
ﻓﻲ
G
ﺩﺍﹰـﺠ ﺓﺩﻴﻔﻤ ﺔﻠﻴﺴﻭ
ﻓﻲ ﺩﺭﺍﺴﺔ ﺍﻟﺯﻤﺭ
.
ﺴﻨﻭﻀﺢ ﻫﺫﺍ ﻤﻊ ﺒﻌﺽ ﺍﻷﻤﺜﻠﺔ
.
ﻟﺘﻜﻥ
X
G H
=
.
ﻷﻱ،ﻥﺄﺒ ﺭﻜﺫﺘﻨ
g
G
∈
،
( )
( )
1
1
Stab
Stab
gH
g
H g
gHg
−
−
=
=
ﻭ ﻨﻭﺍﺓ ﺍﻟﺘﻁﺒﻴﻕ
( )
Sym
G
X
→
ﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻋﻅﻤﻰ
1
g G
gHg
−
∈
I
ﻓﻲ
G
ﻤﺤﺘﻭﺍﺓ ﻓﻲ
H
.
ﻤﻼﺤﻅﺔ
22.4
(a)
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ ﻻ
G
ﻏﻴﺭ
1
.
ﻋﻨﺩﺌﺫ
(
)
Sym
G
G H
→
ﻭﻟﻘﺩ ﺘﺤﻘﻘﻨﺎ ﺒﺄﻥ،ﻥﻴﺎﺒﺘﻤ
G
ﻜﺯﻤﺭ
ﻥـﻤ ﺔـﻴﺌﺯﺠ ﺓ
ﺯﻤﺭﺓ ﺘﻨﺎﻅﺭﻴﺔ ﻭﻤﻥ ﺭﺘﺒﺔ ﺃﻗل ﺒﻜﺜﻴﺭ ﻤﻥ
(
)
:1 !
G
.
ﺇﺫﺍ ﻜﺎﻨﺕ،ﹰﻼﺜﻤ
G
ﺒﺴ
ﻴﻥـﺒﺘ ﺫﺌﺩﻨﻋ ،ﺔﻁﻴ
ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
)ﺍ
ﻨﻅﺭ
5
§
(
ﺒﺄﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﻓﻌﻠﻴﺔ
1
H
≠
)
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﺤﺎ
ﻲـﺘﻟﺍ ﺔﻟ
ﺘﻜﻭﻥ ﻓﻴﻬﺎ
G
ﺩﺍﺌﺭﻴﺔ
(
ﻟﻜﻨﻬﺎ،
)
ﻤﻥ ﺍ
ﻟﺘﻌﺭﻴﻑ
(
ﻻ ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻜﻬﺫﻩ
.
(b)
ﺇﺫﺍ ﻜﺎﻨﺕ
(
)
:1
G
ﻻ ﺘﻘﺴﻡ
(
)
:1 !
G
ﻋﻨﺩﺌﺫ،
(
)
Sym
G
G H
→
ﻟﻥ ﻴﻜﻭﻥ ﻤﺘﺒﺎﻴﻨﺎﹰ
)
ﻤﺒﺭﻫﻨﺔ ﻻﻏﺭ
،ﺍﻨﺞ
25.1
(
ﻭﻴﻤﻜﻥ ﺃﻥ ﻨﺴﺘﻨﺞ ﺒﺄﻥ،
H
ﺭﺓـﻤﺯ ﻰـﻠﻋ ﻱﻭـﺤﺘ
ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
1
≠
ﻓﻲ
G
.
ﺇﺫﺍ ﻜﺎﻨﺕ،ﹰﻼﺜﻤ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
99
ﺭﺓـﻤﺯ ﻰـﻠﻋ ﻱﻭﺤﺘ ﺫﺌﺩﻨﻋ ،
ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
N
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
11
)
ﻤﺒﺭﻫﻨﺔ ﻜﻭﺸﻲ
13.4
(
ﻭﻥـﻜﺘ ﻥﺃ ﺏﺠﻴ ﺔﻴﺌﺯﺠﻟﺍ ﺓﺭﻤﺯﻟﺍﻭ ،
ﻨﺎﻅﻤﻴﺔ
.
،ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ
G
N
Q
= ×
.
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٦٧
ﻤﺜﺎل
23.4
ﺇﻥ ﺍﻟﻨﺘﻴﺠﺔ
4.15
ﺘﺒﻴﻥ ﺒﺄﻥ ﻜل ﺯﻤﺭﺓ
G
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
6
ﺭﺓـﻤﺯ ﻭﺃ ﺔـﻴﺭﺌﺍﺩ ﺎـﻤﺇ
ﺩﻴﻬﻴﺩﺭﺍل
.
ﻨﻘﺩﻡ ﻫﻨﺎ ﺘﺭﺘﻴﺏ ﻤﺨﺘﻠﻑ ﻭﺒ
ﺸﻜل ﻤﺒﺴﻁ
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ ﻤﺒﺭﻫﻨﺔ ﻜﻭﺸﻲ
(13.4)
ﺏـﺠﻴ ،
ﺃﻥ ﺘﺤﻭﻱ
G
ﻋﻠﻰ ﻋﻨﺼﺭ
r
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
ﻭﻋﻠﻰ ﻋﻨﺼﺭ
s
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﻋﻼﻭﺓﹰ ﻋﻠﻰ ﺫﻟﻙ
def
N
r
=
ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ ﻻﻥ
6
ﻻ ﺘﻘﺴﻡ
2!
)
ﺴﺎﻭﻱـﻴ ﺎـﻬﻠﻴﻟﺩ ﻥﻷ ﺔﻁﺎﺴﺒﺒ ﻭﺃ
2
.(
ﻟﺘﻜﻥ
H
s
=
.
ﺇﻤﺎ
(a)
ﺘﻜﻭﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺃﻭ،
(b)
H
ﻟﻴﺴﺕ ﺯﻤﺭ
ﺓ
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ
G
.
،ﻰـﻟﻭﻷﺍ ﺔـﻟﺎﺤﻟﺍ ﻲـﻓ
1
rsr
s
−
=
ﺃﻱ ﺃﻥ،
rs
sr
=
ﺫﻟﻙـﻟﻭ ،
2
3
G
r
s
C
C
×
≈
×
.
ﺇﻥ،ﺔﻴﻨﺎﺜﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
(
)
Sym
G
G H
→
ﻭﺒﺎﻟﺘﺎﻟﻲ،ﻥﻴﺎﺒﺘﻤ
ﻭﻤﻨﻪ،ﹰﺍﺭﻤﺎﻏ ﻥﻭﻜﻴ
3
3
G
S
D
≈
≈
.
ﺯﻤﺭ ﺍﻟﺘﺒ
ﺩﻴﻼﺕ
(Permution groups)
ﻨﺄﺨﺫ
( )
Sym X
ﺤﻴﺙ
X
ﺘﺤﻭﻱ ﻋﻠﻰ
n
ﻋﻨﺼﺭ
.
ﺒﻤﺎ ﺃﻥ
)
لـﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ
(
ﺭﺓـﻤﺯﻟﺍ
ﺍﻟﺘﻨﺎﻅﺭﻴﺔ
( )
Sym X
ﺩﺩ ﺍﻟﻌﻨـﻋ ﻰـﻠﻋ ﻁـﻘﻓ ﺩـﻤﺘﻌﺘ
ﻲـﻓ ﺭـﺼﺎ
X
ﺫـﺨﺄﻨ ﻥﺃ ﻥـﻜﻤﻴ ،
{
}
1, 2,...,
X
n
=
ﻭﻫﻜﺫﺍ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭﺓ،
n
S
.
ﺇﻥ ﺍﻟﺭﻤﺯ
(
)
1 2 3 4 5 6 7
2 5 7 4 3 1 6
ﻴﺩل ﻋﻠﻰ ﺍﻟﺘﺒﺩﻴﻠﺔ
ﺍﻟﺘﻲ ﺘﺭﺴل
1
2, 2
5, 3
7
a
a
a
ﺍﻟﺦ،
...
ﻨﻌﺘﺒﺭ ﺍﻟﺘﺒﺩﻴﻠﺔ
.
( ) ( ) ( )
( )
1
2
3
1
2
3
n
n
α
α
α
α
L
L
ﺯﻭﺝـﻟﺍ ﻰﻤﺴﻴ
( )
,
i j
ﺙـﻴﺤ
i
j
<
ﻭ
( )
( )
i
j
α
α
>
ﻭﺱـﻜﻌﻤﺒ
(inversions)
α
،
ﻭﻴﻘﺎل ﺒﺄﻥ
α
ﺕ ﻓﺭـﻨﺎﻜ ﻥﺇ ﺕﺎـﺴﻭﻜﻌﻤﻟﺍ ﺩﺩﻋ ﺏﺴﺤ ﻙﻟﺫﻭ ﻲﺠﻭﺯ ﻭﺃ ﻱﺩﺭﻓ
ﺔـﻴﺩ
(odd)
ﺃﻭ
ﺯﻭﺠﻴﺔ
(even)
..
ﺇﻥ ﺇﺸﺎﺭﺓ
α
،
( )
sign
α
ﺇﻤﺎ ﺘﺴﺎﻭﻱ،
+1
ﺃﻭ
-1
ﺴﺏـﺤ ﻙﻟﺫﻭ
α
ﺇﻥ
ﻜﺎﻨﺕ ﺯﻭﺠﻴﺔ ﺃﻭ ﻓﺭﺩﻴﺔ
.
ﻤﻼﺤﻅﺔ
24.4
ﻟﺤﺴﺎﺏ ﺇﺸﺎﺭﺓ
α
ﻨﺼل،
)
ﺒﺨﻁ
(
ﻜل ﻋﻨﺼﺭ
i
ﻊـﻤ ﻰـﻠﻋﻷﺍ ﺭﻁﺴـﻟﺍ ﻲﻓ
ﻋﻨﺼﺭ
i
ﻭ ﻨﺤﺴﺏ ﻋﺩﺩ ﺍﻟﻤﺭﺍﺕ ﺍﻟﺘﻲ ﺘﻘﺎﻁﻌﺕ ﻓﻴﻬﺎ ﺍﻟﺨﻁﻭﻁ،ﻲﻠﻔﺴﻟﺍ ﺭﻁﺴﻟﺍ ﻲﻓ
:
ﻭﻥـﻜﻴ
α
ﺯﻭﺠﻴﺎﹰ ﺃﻭ ﻓﺭﺩﻴﺎﹰ ﺤﺴﺏ ﻫﺫﺍ ﺍﻟﻌﺩﺩ ﺇﻥ ﻜﺎﻥ ﺯﻭﺠﻴﺎﹰ ﺃﻭ ﻓﺭﺩﻴﺎﹰ
.
،ﻤﺜﻼﹰ
1
2
3
4
5
3 5
1
4
2
ﺇﻨﻪ ﺯﻭﺠﻲ
)
6
ﺘﻘﺎﻁﻌﺎﺕ
.(
ﺒﺴﺒﺏ ﻭﺠﻭﺩ ﺘﻘﺎﻁﻊ ﻭﺍﺤﺩ ﻟﻜل ﻤﻌﻜﻭﺱ،ﻥﻜﻤﻤ ﺍﺫﻫ
.
ﻤﻥ ﺍﺠل ﺍﻟﺘﺒﺩﻴﻠﺔ
α
ﻨﻌﺘﺒﺭ ﺍﻟﺠﺩﺍﺀ،
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٧٧
(
) (
)(
) (
)
(
) (
)
(
)
(
)
1
2 1 3 1 ...
1
3 2 ...
2
...
1
i
j n
V
j
i
n
n
n
n
≤ < ≤
=
− = −
−
−
−
−
−
−
∏
( ) ( )
(
)
( ) ( )
(
)
( ) ( )
(
)
( ) ( )
(
)
( ) ( )
(
)
( ) ( )
(
)
( ) (
)
(
)
1
2
1
3
1 ...
1
3
2 ...
2
...
1 .
i
j n
V
j
i
n
n
n
n
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
≤ < ≤
=
−
=
−
−
−
−
−
−
−
∏
ﺇﻥ ﺍﻟﺤﺩﻭﺩ ﻓﻲ ﻜل ﻤﻥ ﺍﻟﺠﺩﺍﺀﻴﻥ ﻤﺘﺴﺎﻭﻴﺔ ﻤﺎ
ﺎﺭﺓـﺸﺇ ﻲﻁﻌﻴ ﺎﻬﻴﻓ ﺱﻭﻜﻌﻤ لﻜ ﻲﺘﻟﺍ ﺩﻭﺩﺤﻟﺍ ﺍﺩﻋ
ﺴﺎﻟﺒﺔ
14
.
،ﻟﺫﻟﻙ
( )
V
sign
V
α
α
=
ﻤﻥ ﺃﺠل ﺃﻱ ﺘﺒﺩﻴﻠﺔ،ﻪﻨﺄﺒ ﺩﺠﻨ ﺔﻬﺒﺎﺸﻤ ﺔﻨﺭﺎﻘﻤﺒ
β
،
(16)
.
(
)
( )(
)
V
sign
V
α β
α β
=
ﺒﻤﺎ ﺃﻥ
( )
V
sign
V
β
β
=
ﻭ
(
) ( )
( )
V
V
sign
V
α β
αβ
αβ
=
=
ﻫﺫﺍ ﻴﺒﻴﻥ ﻟﻨﺎ،
( )
( )
( )
sign
sign
sign
αβ
α
β
=
،ﻟﺫﻟﻙ
"
"
"
sign
ﻫﻭ ﺍﻟﺘﺸﺎﻜل
{ }
1
n
S
→ ±
.
ﻋﻨﺩﻤﺎ
2
n
≥
ﻭﻥـﻜﺘ ﻲﻟﺎـﺘﻟﺎﺒﻭ ،ﹰﺍﺭﻤﺎﻏ ﻭﻥﻜﻴ ،
ﻨﻭﺍﺘﻪ ﻋﺒﺎﺭﺓ ﻋﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
n
S
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
!
2
n
ﻭ،
ﺔـﺒﻭﺎﻨﺘﻤﻟﺍ ﺓﺭﻤﺯﻟﺎـﺒ ﻰﻤﺴـﺘ
(alternating group)
n
A
.
ﺍﻟﺩﻭﺭﺓ
(cycle)
ﻫﻲ ﺘﺒﺩﻴﻠﺔ ﻤﻥ ﺍﻟﺸﻜل
ﺍﻵﺘﻲ
1
2
3
1
...
r
i
i
i
i
i
a
a
a
a
ﺃﻤﺎ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺘﺒﻘﻴﺔ ﻓﻬﻲ ﺜﺎﺒﺘﺔ
.
14
ﻜل ﻤﻨﻬﺎ ﻫﻭ ﺠﺩﺍﺀ ﻓﻭﻕ ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻤﺅﻟﻔﺔ ﻤﻥ ﻋﻨﺼﺭﻴﻥ ﻤﻥ
{
}
1,..., n
ﻴﻘﺎﺒل ﺍﻟﻌﺎﻤل ﻟﻠﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ،
{ }
,
i
j
ﻭﻫﻭ
(
)
j
i
−
m
.
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٨٧
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ
j
i
ﻤﺨﺘﻠﻔﺎﹰ
.
ﻨﺭﻤﺯ ﻟﻠﺩﻭﺭﺓ ﺒﺎﻟﺭﻤﺯ
(
)
1 2
...
r
i i
i
ﻭ ﻨﺩﻋﻭ،
r
ﺩﻭﺭﺓـﻟﺍ لﻭـﻁﺒ
-
ﻨﻼﺤﻅ ﺃﻥ
r
ﺃﻴﻀﺎﹰ ﻴﺴﺎﻭﻱ ﺭﺘﺒﺔ ﺍﻟﺩﻭﺭﺓ ﻋﻨﻤﺎ ﺘﻌﺘﺒﺭ ﻜﻌﻨﺼﺭ ﻤﻥ
n
S
.
ﻲـﺘﻟﺍ ﺓﺭﻭﺩـﻟﺍ ﻰﻤﺴﺘ
ﻁﻭﻟﻬﺎ ﻴﺴﺎﻭ
ﻱ
2
ﻤﻨﺎﻗﻠﺔ
.
ﺇﻥ ﺍﻟﺩﻭﺭﺓ
( )
i
ﺫﺍﺕ
ﺍﻟﻁﻭل
1
ﻫﻲ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻤﺤﺎﻴﺩ
.
ﺩﻭﺭﺓـﻟﺍ ﺔﻤﺎﻋﺩ
(support of the cycle)
(
)
1
...
r
i
i
ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻌﻨﺎﺼﺭ
{
}
1
,...,
r
i
i
ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﺩﻭﺭﺍﺕ،
ﻤﻨﻔﺼﻠﺔ ﺇﺫﺍ ﻜﺎﻨﺕ ﺩﻋﺎﻤﺎﺘﻬﺎ ﻤﻨﻔﺼﻠﺔ
(disjoint)
.
ﻨﻼ
ﺤﻅ ﺒﺄﻨﻪ ﺘﻜﻭﻥ ﺍﻟﺩﻭﺭﺍﺕ ﺍﻟﻤﻨﻔﺼﻠﺔ ﺘﺒﺩﻴﻠﻴﺔ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
)
ﺩﻭﺭﺍﺕ
ﻤﻨﻔﺼﻠﺔ
(
(
)(
) (
)
1
1
1
...
...
...
...
r
s
u
i
i
j
j
l
l
α
=
ﻋﻨﺩﺌﺫ
)
ﺩﻭﺭﺍﺕ
ﻤﻨﻔﺼﻠﺔ
(
(
) (
) (
)
1
1
1
...
...
...
...
m
m
m
m
r
s
u
i
i
j
j
l
l
α
=
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
α
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
(
)
lcm
, ,...,
r s
u
.
ﻗﻀﻴﺔ
25.4
ﻜل ﺘﺒﺩﻴﻠﺔ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﺘﺏ
)
ﻭﺒ
ﻲـﺴﺎﺴﺃ لﻜﺸـﺒ ﺓﺩـﺤﺍﻭ ﺔﻘﻴﺭﻁ
(
ﺩﺍﺀ ﺩﻭﺭﺍﺕـﺠﻜ
ﻤﻨﻔﺼﻠﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
n
S
α
∈
ﻭﻟﻴﻜﻥ،
{
}
1, 2,...,
O
n
⊂
ـ ﻤﺩﺍﺭ ﻟ
α
.
ﺎﻥـﻜ ﺍﺫﺇ
O
r
=
،
ﻷﻱﺫﺌﺩﻨﻋ
i
O
∈
،
( )
( )
{
}
1
,
,...,
r
O
i
i
i
α
α
−
=
ﻟﺫﻟﻙ ﻓﺈﻥ
α
ﻭ ﺍﻟﺩﻭﺭﺓ
( )
( )
(
)
1
...
r
i
i
i
α
α
−
ﻥـﻤ ﺭﺼﻨﻋ ﻱﺃ ﻰﻠﻋ ﻪﺴﻔﻨ ﺭﻴﺜﺄﺘﻟﺍ ﺎﻤﻬﻟ
O
.
ﻟﻴﻜﻥ
{
}
1
1, 2,...,
m
j
j
n
O
=
=
U
ﺘﺤﻠﻴل
(decomposition)
ﻟﻠﻤﺠﻤﻭﻋﺔ
{
}
1,..., n
ﻻﺠﺘﻤﺎﻉ ﻤﺩﺍﺭﺍﺕ ﻤﻨﻔﺼﻠ
ﺔ
α
ﺘﻜﻥـﻟﻭ ،
j
γ
ﺩﻭﺭﺓ ﻤﺭﺘﺒﻁﺔ
)
ﻜﻤﺎ ﺴﺒﻕ
(
ـﺒ
j
O
.
ﻋﻨﺩﺌﺫ
1
...
m
α γ γ
=
ﻫﻭ ﺘﺤﻠﻴل
α
ﺇﻟﻰ ﺠﺩﺍﺀ ﺩﻭﺭﺍﺕ ﻤﻨﻔﺼﻠﺔ
.
ﻨﻼﺤﻅ ﺃﻥ ﺘﺤﻠﻴل،ﺔﻴﻨﺍﺩﺤﻭﻠﻟ
1
...
m
α γ γ
=
ﺇﻟﻰ ﺠﺩﺍﺀ
ﺩﻭﺭﺍﺕ ﻤﻨﻔﺼﻠﺔ ﻴﺠﺏ ﺃﻥ ﻴﻘﺎﺒل ﺘﺤﻠﻴل
{
}
1,..., n
ﺇﻟﻰ ﻤﺩﺍﺭﺍﺕ
)
ﺩﻭﺭﺍﺕ ﻤﺠﻬﻭﻟﺔ ﻤﻥ ﺍﻟﻁﻭل
1
ﻭ ﻤﺩﺍﺭﺍﺕ ﻤﺅﻟﻔﺔ ﻤﻥ ﻋﻨﺼﺭ ﻭﺍﺤﺩ
.(
ﻴﻤﻜﻥ ﺃﻥ ﻨﺘﺠﺎﻫل ﺍﻟﺩﻭﺭﺍﺕ ﺍﻟﺘﻲ ﻁﻭﻟﻬﺎ ﻴﺴﺎﻭﻱ
1
ﺭـﻴﻐﻨ ،
ﻭ ﻨﻐﻴﺭ ﻜﻴﻔﻴﺔ ﻜﺘﺎﺒﺔ ﻜل ﺩﻭﺭﺓ،ﺕﺍﺭﻭﺩﻟﺍ ﺔﺒﺘﺭ
)
ﺒﺎﺨﺘﻴﺎﺭ ﻋﻨﺎﺼﺭ ﺍﺒﺘﺩﺍ
ﺌﻴﺔ ﻤﺨﺘﻠﻔﺔ
(
لـﻜ ﺍﺫﻫ ﻥﻜﻟ ،
ـﺸﻲﺀ ﻷﻥ ﺍﻟﻤﺩﺍﺭﺍﺕ ﻤﺭﺘﺒﻁﺔ ﺒ
α
.
، ﻤﺜﻼﹰ
PDF created with pdfFactory Pro trial version
٩٧
(
)
( )(
)( )
1
2
3
4
5
6
7 8
15 27634 8
5
7
4
2
1
3
6 8
=
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
( )
lcm 2, 5
10
=
.
ﻨﺘﻴﺠﺔ
26.4
ﺫﺍـﻫ ﻲﻓ ﺕﻼﻗﺎﻨﻤﻟﺍ ﺩﺩﻋ ﻥﺇﻭ ،ﺕﻼﻗﺎﻨﻤ ﺀﺍﺩﺠ لﻜﺸ ﻰﻠﻋ ﺏﺘﻜﺘ ﻥﺃ ﻥﻜﻤﻴ ﺔﻠﻴﺩﺒﺘ لﻜ
ﺍﻟﺠﺩﺍﺀ ﻴﻜﻭﻥ ﺇﻤﺎ ﺯﻭ
ﺠﻴﺎﹰ ﺃﻭ ﻓﺭﺩﻴﺎﹰ ﻭﺫﻟﻙ ﺤﺴﺏ
α
ﺇﻥ ﻜﺎﻨﺕ ﺯﻭﺠﻴﺔ ﺃﻭ ﻓﺭﺩﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺄﺨﺫ ﺍﻟﺩﻭﺭﺓ
(
) ( ) (
)(
)
1
1 2
2
1
1
...
...
,
r
r
r
r
r
i
i
i i
i
i
i
i
−
−
−
=
ﻭﻤﻨﻪ ﻓﺈﻥ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ ﺘﺄﺘﻲ ﻤﻥ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
.
ﻷﻥ
sign
ﻫﻭ ﺘﺸﺎﻜل ﻭ ﺇﺸﺎﺭﺓ ﺍﻟﻤﻨﺎﻗﻠﺔ
-1
،
( ) ( )
sign
1
α
= −
.
ﻨﻼﺤﻅ ﺒﺄﻥ ﺼﻴﻐﺔ ﺍﻟﺒﺭﻫﺎﻥ ﺘﺒﻴﻥ ﺒﺄﻥ ﺇﺸﺎﺭﺓ ﺍﻟﺩﻭﺭﺓ ﺫﻭ ﺍﻟﻁﻭل
r
ﻫﻲ
( )
1
1
r
−
−
، ﺇﻥ،
r
-
ﺩﻭﺭﺓ ﺘﻜﻭﻥ ﺯﻭﺠﻴﺔ ﺃﻭ ﻓﺭﺩﻴﺔﹰ ﺤﺴﺏ
r
ﺇﻥ ﻜﺎﻥ ﻓﺭﺩﻴﺎﹰ ﺃﻭ ﺯﻭﺠﻴﺎﹰ
.
ﻤﻥ ﺍﻟﻤﺤ
ﺩﺩـﻌﻟ ﻥﺎـﻜ ﻥﺇ ﺀﺍﺩـﺠﻟﺍ ﻲﻓ ﺎﻤﻜ ﺔﻴﺩﺭﻓ ﻭﺃ ﺔﻴﺠﻭﺯ ﻥﻭﻜﺘﻟ ﺎﻤ ﺔﻠﻴﺩﺒﺘ ﻑﺭﻌﻨ ﻥﺃ لﻤﺘ
ﺎـﻤﻜ ﺏـﻴﺘﺭﺘﻟﺍ ﺕﻼﻗﺎﻨﻤﻟﺍ ﻯﺩﺤﺇ ﻲﻋﺍﺭﺘ ﻥﺃ ﺏﺠﻴ ﺎﻫﺩﻨﻋ ﻥﻜﻟﻭ ،ﺕﻼﻗﺎﻨﻤﻟﺍ ﻥﻤ ﻱﺩﺭﻓ ﻭﺃ ﻲﺠﻭﺯ
ﺴﺒﻕ ﻟﺘﺒﻴﻥ ﺒﺄﻥ ﻫﺫﺍ ﺍﻟﻤﻔﻬﻭﻡ ﻤﻌﺭﻑ ﺠﻴﺩﺍﹰ
.
ﺘﻨﺹ ﺍﻟﻨﺘﻴﺠﺔ ﻋﻠﻰ ﺃﻥ
n
S
ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﻨﺎﻗﻼﺕ
.
ﺃﻤﺎ
ـ ﺒﺎﻟﻨﺴﺒﺔ ﻟ
n
A
ﻓﺘﻭﺠﺩ ﺍﻟﻨﺘﻴﺠﺔ
ﺍﻵﺘﻴ
.ﺔ
ﻨﺘﻴﺠﺔ
27.4
ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻨﺎﻭﺒﺔ
n
A
ﻤﻭﻟﺩﺓ ﺒﺩﻭﺭﺍﺕ ﺫﺍﺕ ﺍﻟﻁﻭل
3
.
ﺎﻥـﻫﺭﺒﻟﺍ
.
ﺃﻱ
n
A
α
∈
ﻫﻭ ﺠﺩﺍﺀ
)
ﻤﻥ ﺍﻟﻤﺤﺘﻤل ﺃﻥ ﻴﻜﻭﻥ ﺨﺎﻟﻴﺎﹰ
(
،ﺎﻗﻼﺕـﻨﻤﻟﺍ ﻥﻤ ﻲﺠﻭﺯ ﺩﺩﻌﻟ
1 1
...
m m
t t
t t
α
′
′
=
ـ ﻭﻟﻜﻥ ﺠﺩﺍﺀ ﺃﻱ ﻤﻨﺎﻗﻠﺘﻴﻥ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﺘﺏ ﺩﺍﺌﻤﺎﹰ ﻜﺠﺩﺍﺀ ﻟ،
3
-
ﺩﻭﺭﺓ
:
( )( )
( )( )
( )( )( )( ) ( )( )
( ) ( )
,
,
1,
.
ij
jl
j
k
ij
kl
ij
jk
jk
kl
ijk
jkl
ij
kl
=
=
=
=
ﻨﺘﺫﻜﺭ ﺒﺄﻨﻪ ﻴﻘﺎل ﻋﻥ ﺍﻟﻌﻨﺼﺭﻴﻥ
a
ﻭ
b
ﺃﻨﻬﻤﺎ ﻤﺘﺭﺍﻓﻘﺎﻥ
a
b
ﺼﺭـﻨﻋ ﺩـﺠﻭﻴ ﻥﺎﻜ ﺍﺫﺇ
g
G
∈
ﺒ
ﺤﻴﺙ ﺇﻥ
1
b
gag
−
=
ﻭ ﺃﻥ ﺍﻟﺘﺭﺍﻓﻕ ﻫﻭ ﻋﻼﻗﺔ ﺘﻜﺎﻓﺅ،
.
ﺭﺓـﻤﺯﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ
G
ﻥـﻤ ،
ﺍﻟﻤﻔﻴﺩ ﺃﻥ ﻨﺤﺩﺩ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
G
.
ﻤﺜﺎل
28.4
ﻓﻲ
n
S
ﺍﻟﺩﻭﺭﺍﺕ ﺍﻟﻤﺘﺭﺍﻓﻘﺔ ﻤﻊ ﺍﻟﺩﻭﺭﺓ ﺘﻌﻁﻰ ﺒﺎﻟﻌﻼﻗﺔ،
:
.
(
)
( ) ( )
(
)
1
1
1
...
...
k
k
g i
i
g
g i
g i
−
=
ﻣﻨﺎﻗﻼت
#
, , ,
i j k l
ﻤﺨﺘﻠﻔﺔ
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٠٨
ﺒﺎﻟﺘﺎﻟﻲ
(
) (
)
( ) ( )
(
)
( ) ( )
(
)
1
1
1
1
1
...
...
...
...
...
...
r
u
r
u
g i
i
l
l
g
g i
g i
g l
g l
−
=
)
ﻭـﻟ ﻰﺘﺤ
ﻜﺎﻨﺕ ﺍﻟﺩﻭﺭﺍﺕ ﻟﻴﺴﺕ ﻤﻨﻔ
ﻷﻥ ﺍﻟﺘﺭﺍﻓﻕ ﻋﺒﺎﺭﺓ ﻋﻥ ﺘﺸﺎﻜل،ﺔﻠﺼ
.(
ﻰـﻠﻋ لﺼﺤﻨﻟ ،ﻯﺭﺨﺃ ﺕﺎﻤﻠﻜﺒ
1
g g
α
−
ﻨﺴﺘﺒﺩل ﻜل ﻋﻨﺼﺭ ﻓﻲ،
ﻜل ﺩﻭﺭﺓ ﻓﻲ
α
ـ ﺒﺼﻭﺭﺘﻪ ﺒﺎﻟﻨﺴﺒﺔ ﻟ
g
.
ﺴﻭﻑ ﻨﺤﺩﺩ ﺍﻵﻥ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
n
S
.
ﺒﺘﺠﺯﺌﺔ
(partition)
n
ﻥـﻤ ﺔﻴﻟﺎﺘﺘﻤﺒ ﻲﻨﻌﻨ ،
ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ
1
,...,
k
n
n
ﺒ
ﺤﻴﺙ ﺇﻥ
1
2
1
...
k
n
n
n
n
≤ ≤
≤ ≤
≤
ﻭ
1
2
...
k
n
n
n
n
+
+ +
=
ﻴﻭﺠﺩ ﺘﻤﺎﻤﺎﹰ،ﹰﻼﺜﻤ
5
ﺘﺠﺯﺌﺎﺕ ﻟﻠﻌﺩﺩ
4
، ﻭﻫﻲ ﺒﺎﻟﺘﺤﺩﻴﺩ
4 1 1 1 1, 4
1 1 2, 4
1 3, 4
2
2, 4
4
= + + +
= + +
= +
= +
=
ﻭ
1,121,505
ﺘﺠﺯﺌﺔ ﻟﻠﻌﺩﺩ
61
.
ﻨﻼﺤﻅ ﺃﻥ ﺘﺠﺯﺌﺔ
)
ﺍﺠﺘﻤﺎﻉ ﻤﻨﻔﺼل
(
{
}
1
1, 2,...,
...
k
n
O
O
=
U U
ﻟﻠﻤﺠﻤﻭﻋﺔ
{
}
1, 2,..., n
ﺘﺤﺩﺩ ﺘﺠﺯﺌﺔ ﻟﻠﻌﺩﺩ
n
،
1
2
...
,
k
i
i
n
n
n
n
n
O
= + + +
=
ﺒﻤﺎ ﺃﻥ ﻤﺩﺍﺭﺍﺕ ﻋﻨﺼﺭ
α
ﻤﻥ
n
S
ﺘﺸﻜل ﺘﺠﺯﺌﺔ ﻟﻠﻤﺠﻤﻭﻋﺔ
{
}
1, 2,..., n
،
ﺼلـﻨ ﻥﺃ ﻥﻜﻤﻴ
ﺇﻟﻰ ﻜل ﻋﻨﺼﺭ ﻤﺜل
α
ﺒﺘﺠﺯﺌﺔ
n
.
ﺘﺠﺯﺌﺔ ﺍﻟﻌﺩﺩ،ﹰﻼﺜﻤ
8
ــ ﻤﺭﺘﺒﻁﺔ ﺒ
( )(
)( )
15 27634 8
ﻭﻫﻲ
1,2,5
ـ ﻭ ﺍﻟﺘﺠﺯﺌﺔ ﺍﻟﻤﺭﺘﺒﻁﺔ ﺒ
n
ـ ﺘﺭﺘﺒﻁ ﺒ
)
ﺩﻭﺭﺍﺕ
ﻤﻨﻔﺼﻠﺔ
(
(
) (
)
1
1
1
1
...
...
...
, 1
,
k
n
n
i
i
i
i
l
l
n
n
α
+
=
<
≤
ﻭﻫﻲ
1
1,1,...,1,
,...,
k
n
n
(
)
i
n
n
−
∑
ﻗﻀﻴﺔ
29.4
ﻴﻜﻭﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ
α
ﻭ
β
ـ ﻤﺘﺭﺍﻓﻘﺎﻥ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺎ ﻴﻌﺭﻓﺎﻥ ﺍﻟﺘﺠﺯﺌﺔ ﻨﻔﺴﻬﺎ ﻟ
n
.
ﺍﻟﺒﺭﻫﺎﻥ
.
⇒
:
ﺭﺃﻴﻨﺎ ﻓﻲ
(28.4)
ﺃﻥ ﻤﺭﺍﻓﻕ
ﻰ ﺩﻭﺭﺍﺕـﻟﺇ لـﻠﺤﻤﻟﺍ ﺎﻬﻠﻜﺸ ﻰﻠﻋ ﻅﻓﺎﺤﻴ ﺭﺼﻨﻋ
ﻤﻨﻔﺼﻠﺔ
.
⇐
:
ﺒﻤﺎ ﺃﻥ
α
ﻭ
β
ـ ﻴﻌﺭﻓﺎﻥ ﺍﻟﺘﺠﺯﺌﺔ ﻨﻔﺴﻬﺎ ﻟ
n
ﺩﺍﺀ ﺩﻭﺭﺍﺕـﺠ ﻰـﻟﺇ ﺎﻬﻠﻴﻠﺤﺘ ﻥﺈﻓ ﻙﻟﺫﻟ ،
ﻤﻨﻔﺼﻠﺔ ﻟﻪ ﺍﻟﺸﻜل ﻨﻔﺴﻪ
:
(
)(
) (
)
( )(
) ( )
1
1
1
1
1
1
...
...
...
...
,
...
...
...
...
,
r
s
u
r
s
u
i
i
j
j
l
l
i
i
j
j
l
l
α
β
=
′
′
′
′
′
′
=
ﺇﺫﺍ ﻋﺭﻓﻨﺎ
g
ﺒﺎﻟﺸﻜل
ﻣﺮة
PDF created with pdfFactory Pro trial version
١٨
1
1
1
1
1
1
...
...
...
...
,
...
...
...
...
r
s
u
r
s
u
i
i
j
j
l
l
i
i
j
j
l
l
′
′ ′
′
′
′
ﻋﻨﺩﺌﺫ
1
g g
α
β
−
=
ﻤﺜﺎل
30.4
(
)
( )
1
12 3 4...
12 34...
12 3
4...
4...
i j k
i j k
i j k
−
=
.
ﻤﻼﺤﻅﺔ
31.4
ﻟﻜل
1 k
n
< ≤
ﻴﻭﺠﺩ،
(
) (
)
1 ...
1
n n
n
k
k
−
− +
k
-
ﻲـﻓ ﺔـﻔﻠﺘﺨﻤ ﺓﺭﻭﺩ
n
S
.
ﻨﺤﺘﺎﺝ ﺇﻟﻰ
1
k
ﻟﺫﻟﻙ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﻨﺤﺴﺏ
(
) (
)
1 2
1
1
...
...
...
k
k
k
i i
i
i i
i
−
=
=
k
ﻤﺭﺓ
.
ﻲـﻓ ﻕﻓﺍﺭﺘ ﻑﺼ ﻱﺃ ﻲﻓ ﺭﺼﺎﻨﻌﻟﺍ ﺩﺩﻋ ﺏﺴﺤﻨ ﻥﺃ لﻤﺘﺤﻤﻟﺍ ﻥﻤ ،ﻪﺒﺎﺸﻤ لﻜﺸﺒ
n
S
،
ﻟﻜﻥ ﻨﺤﺘﺎﺝ ﺇﻟﻰ ﺍﻟﻘﻠﻴل ﻤﻥ ﺍﻟﺤﺭﺹ ﻋﻨﺩﻤﺎ ﺘﺤﻭﻱ ﺍﻟﺘﺠﺯﺌﺔ ﻋﻠﻰ ﻋﺩﺓ ﺤﺩﻭﺩ ﻤﺘﺴﺎﻭﻴﺔ
.
ﺇﻥ،ﹰﻼﺜـﻤ
ﻋﺩﺩ ﺍﻟﺘﺒﺩﻴﻼﺕ ﻓﻲ
4
S
ﻤﻥ ﺍﻟﺸﻜل
( )( )
ab
cd
ﻴﺴﺎﻭﻱ
1 4 3
2 1
3
2
2
2
×
×
×
=
ﻨﺤﻥ ﺒﺤﺎﺠﺔ ﺇﻟﻰ ﺍﻟﻌﺩﺩ
1
2
ﻟﺫﻟﻙ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﻨﺤﺴﺏ
( )( )
ab
cd
ﻤﺭﺘﻴﻥ
.
ـﺒﺎﻟﻨﺴﺒﺔ ﻟ
4
S
ﻟﺩﻴﻨﺎ
ﺍﻟﺠﺩﻭل
ﺍﻵﺘﻲ
:
ﺍﻟﺘﺠﺯﺌﺔ ﺍﻟﻌﻨﺼﺭ ﻻ ﻴﻘﻊ ﻓﻲ ﺼﻑ ﺍﻟﺘﻜﺎﻓﺅ
ﺍﻟﻨﻭﻉ
1 1 1 1
+ + +
1
1
ﺯﻭﺠﻲ
1 1 2
+ +
( )
ab
6
ﻓﺭﺩﻱ
1 3
+
( )
abc
8
ﺯﻭﺠﻲ
2 2
+
( )( )
ab
cd
3
ﺯﻭﺠﻲ
4
(
)
abcd
6
ﻓﺭﺩﻱ
ﻨﻼﺤﻅ ﺒﺄﻥ
4
A
ﺘﺤﻭﻱ ﻋﻠﻰ
3
ﻋﻨﺎﺼﺭ ﻓﻘﻁ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻥـﻤ ﻲﺘﻟﺍ ﺭﺼﺎﻨﻌﻟﺍ ﺩﻴﺩﺤﺘﻟﺎﺒ ﻲﻫﻭ ،
ﺍﻟﺸﻜل
2+2
ﻭ ﺃﻨﻬﺎ ﺘﺸﻜل ﻤﻊ،
1
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
V
.
ﻫﺫ
ﻩ ﺍﻟﺯﻤﺭﺓ ﻫﻲ ﺍﺠﺘﻤﺎﻉ ﻟﺼﻔ
،ﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
ﻭﻟﺫﻟﻙ ﻓﻬﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴ
ﺔ ﻤﻥ
4
S
.
ﻤﺒﺭﻫﻨﺔ
32.4
(Galois)
ﺇﻥ ﺍﻟﺯﻤﺭﺓ
n
A
ﺒﺴﻴﻁﺔ ﻟﻜل
n
5
≥
.
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٢٨
ﻤﻼﺤﻅﺔ
33.4
ﻤﻥ
ﺃﺠل
2
n
=
ﺘﻜﻭﻥ،
n
A
ﻭ،ﺔﻬﻓﺎﺘ
ﻤﻥ ﺃﺠل
3
n
=
ﺘﻜﻭﻥ،
n
A
ﺯﻤﺭﺓ
ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
ﻤﻥ ﺃﺠل،ﺔﻁﻴﺴﺒ ﻲﻬﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ ،
4
n
=
ﻓﻬﻲ ﻟﻴﺴﺕ ﺘﺒﺩﻴﻠﻴﺔ ﻭﻟﻴﺴﺕ ﺒﺴﻴﻁﺔ
–
ﻭ ﻤﻤﻴﺯﺓ،ﺔﻴﻤﻅﺎﻨ ﺓﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ،
V
)
ﺍﻨﻅﺭ
31.4
.(
ﺘﻤﻬﻴﺩﻴﺔ
34.4
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
n
A
)
5
n
≥
(
ﺇﺫﺍ ﻜﺎﻨﺕ،
N
ﺘﺤﻭﻱ ﻋﻠﻰ
ﺩﻭﺭ ﻁﻭﻟﻪ
3
ﻓﻬﻲ ﺘﺤﻭﻱ ﻋﻠﻰ ﻜل ﺍﻟﺩﻭﺭﺍﺕ ﺍﻟﺘﻲ ﻁﻭﻟﻬﺎﺫﺌﺩﻨﻋ ،
3
ﺴﺎﻭﻱـﺘ ﻲﻬﻓ ﻙﻟﺫﻟﻭ ،
n
A
)
ﻤﻥ
27.4
.(
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
γ
ﺩﻭﺭﺓ ﻁﻭﻟﻬﺎ
3
ﻓﻲ
n
A
ﻭﻟﺘﻜﻥ،
α
ﺩﻭﺭﺓ ﺃﺨﺭﻯ ﻁﻭﻟﻬﺎ
3
ﻓﻲ
n
A
.
ﻨﻌﻠﻡ
ﻤﻥ
(29.4)
ﺒﺄﻥ
1
g g
α
γ
−
=
ﻟﺒﻌﺽ ﺍﻟﻌﻨﺎﺼﺭ
n
g
S
∈
.
ﺇﺫﺍ ﻜﺎﻥ
n
g
A
∈
ﺄﻥـﺒ ﻥﻴﺒﻴ ﺍﺫﻫ ،
α
ﺘﻨﺘﻤﻲ ﺃﻴﻀﺎﹰ ﺇﻟﻰ
N
.
ﻷﻥ،ﻙﻟﺫﻜ ﻥﻜﻴ ﻡﻟ ﺍﺫﺇ
5
n
≥
ﺘﻭﺠﺩ ﻤﻨﺎﻗﻠﺔ،
n
t
S
∈
ﻥـﻋ ﺔﻠﺼﻔﻨﻤ
α
.
ﻋﻨﺩﺌﺫ
n
tg
A
∈
ﻭ
1
1
1
,
t
t
t g g
t
α
α
γ
−
−
−
=
=
ﻭ ﻤﻨﻪ ﻓﺈﻥ
n
A
α
∈
ﺜﺎﻨﻴﺔﹰ
.
ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
ﺍﻵﺘﻴ
ﺔ ﺘﻜﻤل ﺒﺭﻫﺎﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
.
ﺘﻤﻬﻴﺩﻴﺔ
35.4
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
N
ﻓﻲ
n
A
،
,
5,
1
≥
≠
n
N
ﺘﺤﻭﻱ ﺩﻭﺭﺓ ﻁﻭﻟﻬﺎ،
3
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
,
1
N
α
α
∈
≠
.
ﺇﺫﺍ ﻟﻡ ﻴﻜﻥ
α
ﻋﺒﺎﺭﺓ ﻋﻥ
3
-
ﺭـﺨﺁ ﹰﺍﺭﺼﻨﻋ ﻲﻨﺒﻨﺴ ،ﺓﺭﻭﺩ
,
1
N
α
α
′
′
∈
≠
ﺒﺤﻴﺙ ﻨﺜﺒﺕ ﻋﻨ،
ﺎﺼﺭ
{
}
1, 2,..., n
ﺃﻜﺜﺭ ﻤﻤﺎ ﺘﺜﺒﺘﻪ
α
.
ﻥـﻜﻴ ﻡـﻟ ﺍﺫﺇ
α
′
ﻋﺒﺎﺭﺓ ﻋﻥ
3
-
ﻴﻤﻜﻥ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﺒﻨﺎﺀ ﻨﻔﺴﻪﺫﺌﺩﻨﻋ ،ﺓﺭﻭﺩ
.
ﻨﺼل،ﺕﺍﻭﻁﺨﻟﺍ ﻥﻤ ﻲﻬﺘﻨﻤ ﺩﺩﻋ ﺩﻌﺒ
ﺇﻟﻰ
3
-
ﺩﻭﺭﺓ
.
ﺒﻔﺭﺽ ﺃﻥ
α
ﻟﻴﺴﺕ ﻋﺒ
ﺎﺭﺓ ﻋﻥ
3
-
ﺩﻭﺭﺓ
.
ﺇﻤﺎ،ﺔﻠﺼﻔﻨﻤ ﺕﺍﺭﻭﺩﻟ ﺀﺍﺩﺠﻜ ﺎﻬﻨﻋ ﺭﺒﻌﻨ ﺫﺌﺩﻨﻋ
ﺘﺤﻭﻱ ﻋﻠﻰ ﺩﻭﺭﺓ ﻁﻭﻟﻬﺎ
3
≤
ﺃﻭ
ﻨﻘﻭل،ﺕﻼﻗﺎﻨﻤ ﺀﺍﺩﺠ لﻜﺸ ﻰﻠﻋ لﻗﻷﺍ ﻰﻠﻋ ﻥﻭﻜﺘ
(i)
(
)
1 2 3
... ...
i i i
α
=
ﺃﻭ
(ii)
( )( )
1 2
3 4
...
i i
i i
α
=
، ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
α
ﺎـﻤﻬﻨﺄﺒ لﻭﻘﻨ ،ﻥﻴﺩﺩﻋ ﻙﺭﺤﺘ
4
5
,
i i
ﺩﺍـﻋ ﺎـﻤ ،
1
2
3
,
,
i i
i
ﻷﻥ،
(
) (
)
1 2 3
1
4
,
...
i i i
i
i
α
≠
.
ﻟﺘﻜﻥ
(
)
def
1
1
1 2 4
... ...
i i i
N
α
γ α γ
−
=
=
∈
ﻥـﻋ ﺔﻔﻠﺘﺨﻤ ﻲﻫﻭ ،
α
)
ﻷﻥ ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺨﺘﻠﻑ ﻋﻠﻰ
2
i
.(
ﻟﻬﺫﺍ
def
1
1
1
α
α α
−
′ =
≠
ﻥـﻜﻟ
1
1
α γ α γ α
−
−
′ =
ﻴﺜﺒﺕ
2
i
ﻭ ﻜل ﺍﻟﻌﻨﺎﺼﺭ ﻤﺎ ﻋﺩﺍ
1
5
,...,
i
i
ـ ﺘﻜﻭﻥ ﻤﺜﺒﺘﺔ ﺒ
α
-
ﺭـﺼﺎﻨﻋ ﺕـﺒﺜﺘ ﻲﻬﻓ ﻙﻟﺫﻟﻭ
ﺃﻜﺜﺭ ﻤﻥ
α
.
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٣٨
ﻨﺸﻜل،ﺔﻴﻨﺎﺜﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
1
,
,
γ α α
′
ﻜﻤﺎ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ ﻤﻊ
4
i
ﻲـﻓ ﺎﻤﻜ
(ii)
ﻭ
5
i
ﺃﻱ
ﻋﻨﺼﺭ ﻤﺨﺘﻠﻑ ﻋﻥ
1
2
3
4
, , ,
i i i i
.
ﻋﻨﺩﺌﺫ
( )( )
1
1 2
4 5
...
i i
i i
α
=
ﻥـﻋ ﺔـﻔﻠﺘﺨﻤ ﻥﻭﻜﺘ
α
ﻷﻨﻬﺎ ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺨﺘﻠﻑ ﻋﻠﻰ
4
i
.
ﻟﻬﺫﺍ
1
1
1
α α α
−
′ =
≠
ﻟﻜﻥ،
α
′
ﻴﺜﺒﺕ
1
i
ﻭ
2
i
ﻭ ﻜل،
ﺍﻟﻌﻨﺎﺼﺭ
1
5
,...,
i
i
≠
ﻻ ﺘﺜﺒﺕ
ﻤﻥ ﻗﺒل
α
-
ﻭﻟﺫﻟﻙ ﻓﻬﻲ ﺘﺜﺒﺕ ﻋﻨﺎﺼﺭ ﺃﻜﺜﺭ ﻤﻥ
α
ﺩـﺤﺍﻭﺒ
ﻋﻠﻰ ﺍﻷﻗل
.
ﻨﺘﻴﺠﺔ
36.4
ﻤﻥ ﺃﺠل
5
n
≥
ﺇﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﻓﻲ،
n
S
ﻫﻲ
1
،
n
A
ﻭ،
n
S
ﻓﻘﻁ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
n
S
ﻋﻨﺩﺌﺫ،
n
N
A
I
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ
n
A
.
ﺎـﻤﺇ ﻙﻟﺫـﻟ
n
n
N
A
A
=
I
ﺃﻭ
{ }
1
n
N
A
=
I
.
،ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
n
N
A
⊃
ﺴﺎﻭﻱـﻴ ﺎﻬﻠﻴﻟﺩ ﻲﺘﻟﺍ ،
2
ﻲــﻓ
n
S
ﺈﻥــﻓ ﻙﻟﺫــﻟﻭ ،
n
N
A
=
ﺃﻭ
n
S
.
ﻕــﻴﺒﻁﺘﻟﺍ ﻥﻭــﻜﻴ ،ﺔــﻴﻨﺎﺜﻟﺍ ﺔــﻟﺎﺤﻟﺍ ﻲــﻓ
:
n
n
n
x
xA
N
S
A
→
a
ﻭﻤﻨﻪ ﻓﺈﻥ،ﹰﺍﺭﻤﺎﻏ
N
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
ﺃﻭ
2
ﻥ ﺃﻥـﻜﻤﻴ ﻻ ﻥﻜﻟ ،
ﺘﻜﻭﻥ ﺭﺘﺒﺘﻬﺎ
2
ﻷﻨﻪ ﻻ ﻴﻭﺠﺩ ﺼﻑ ﺘﺭﺍﻓﻕ ﻓﻲ
n
S
)
ﻤﺎ ﻋﺩﺍ
{1}
(
ﻴﺘﺄﻟﻑ ﻤﻥ ﻋﻨﺼﺭ ﻭﺍﺤﺩ
.
37.4
ﻴﻭﺠﺩ ﻭﺼﻑ ﻟﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
n
A
لـﺠﺃ ﻥﻤ ﺔﻁﺎﺴﺒﺒ ﺞﺘﻨﺘﺴﺘ ﻥﺃ لﻤﺘﺤﻤﻟﺍ ﻥﻤ ﻲﺘﻟﺍ ،
5
n
≥
.
38.4
ﻴﻘﺎل ﻋﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺃﻨﻬﺎ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﺇﺫﺍ ﻭﺠﺩﺕ ﺯﻤﺭ
ﺍﹰ
ﺠﺯﺌﻴﺔ
{ }
0
1
1
...
...
1
i
i
r
G
G
G
G
G
G
−
=
⊃
⊃ ⊃
⊃
⊃ ⊃
=
ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﻜل ﻤﻥ
i
G
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
1
i
G
−
ﻭ ﻜل ﺯﻤﺭﺓ ﻗﺴﻤﺔ
1
i
i
G
G
−
ﺘﻜﻭﻥ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻟﻬﺫﺍ
ﻓﺈﻥ
n
A
)
ﻭﺃﻴﻀﺎﹰ
n
S
(
ﻟﻴﺴﺕ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﺇﺫﺍ ﻜﺎﻥ
5
n
≥
.
ﻟﻴﻜﻥ
( )
[ ]
f x
x
∈
ﻤﻥ ﺍﻟﺩﺭﺠﺔ
n
.
ـ ﺘﺭﺘﺒﻁ ﺒ،ﺍﻭﻟﺎﺠ ﺔﻴﺭﻅﻨ ﻲﻓ
f
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
f
G
ﺫﻭﺭـﺠ ﺕﻼﻴﺩﺒﺘ ﺭﻤﺯ ﻥﻤ
f
ﻴﻥـﺒﻴﻭ ،
ﺒﺎﻥ ﺠﺫﻭﺭ
f
ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﻤﻥ ﻤﻌﺎﻤﻼﺕ
f
، ﺍﻟﻁﺭﺡ،ﻊﻤﺠﻟﺍ ﻥﻤ ﺔﻴﺭﺒﺠﻟﺍ ﺕﺎﻴﻠﻤﻌﻟﺎﺒ
ﺍﻟﻘﺴ،ﺏﺭﻀﻟﺍ
ﻭ،ﺔﻤ
ﻠـﻟ ﺝﺎﺘﻨﺘﺴﻻﺍ
ﺠﺫﺭ
m
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
f
G
ﻗﺎﺒﻼﹰ ﻟﻠﺤل
)
ﻨﻅﺭﻴﺔ
ﻏ
ﺎﻟﻭﺍ
.(
ﻟﻜل
n
ﻴﻭﺠﺩ ﻋﻠﻰ ﺍﻷﻗل ﻜﺜﻴﺭﺓ ﺤﺩﻭﺩ،
f
ﻤﻥ ﺍﻟﺩﺭﺠﺔ
n
ﺒﺤﻴﺙ
f
n
G
S
≈
ﺎﻟﻲـﺘﻟﺎﺒﻭ
ﺈﻥـﻓ
)
ﻋﻨﺩﻤﺎ
5
n
≥
(
ﺍﻟﻜﺜﻴﺭ ﻤﻥ ﻜﺜﻴﺭﺍﺕ ﺍﻟﺤﺩﻭﺩ ﺘﻜﻭﻥ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﺒﺩﻻﻟﺔ ﺍﻟﺠﺫﻭﺭ
.
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ﺨﻭﺍﺭﺯﻤﻴﺔ
Coxeter
Tood –
ﻟﺘﻜﻥ
G
ﺘﻜﻥـﻟﻭ ،ﻪـﺘﻨﻤ لﻜﺸﺒ لﺜﻤﺘ ﺓﺭﻤﺯ
H
ﺭﺓ ﺠﺯـﻤﺯ
ﺔـﻋﻭﻤﺠﻤﺒ ﺓﺩـﻟﻭﻤ ﺔـﻴﺌ
.
ﺇﻥ
ﺨﻭﺍﺭﺯﻤﻴﺔ
Coxeter
15
-
Tood
ﻫﻲ ﺍﺴﺘﺭﺍﺘﻴﺠﻴﺔ ﻟﻜﺘﺎﺒﺔ ﻓﻴﻤﺎ ﺒﻌﺩ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
ﻟﻠﺯﻤﺭﺓ
H
ﻓﻲ
G
ﻤﻊ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
.
ﻟﻘﺩ ﻭﻀﺤﺘﻬﺎ ﻤﻊ ﻤﺜﺎل
)
ﻤﻥ
Artin 1991, 6.9
ﻪـﻨﺄﺒ ﻅﺤﻼﻨ ﻥﻜﻟ ،ﺭﺜﻜﺃ لﻴﺼﺎﻔﺘﺒ ﺎﻨﺩﻭﺯﻴ ﻱﺫﻟﺍ
ﻴﺭﻜﺏ ﺍﻟﺘﺒﺩﻴﻼﺕ ﻓﻲ ﺍﻻﺘﺠﺎﻩ
ﺍﻟﻤﻌﺎﻜﺱ ﻟﺘﺭﻜﻴﺒﻨﺎ
.(
ﻟﺘﻜﻥ
3
2
2
, ,
,
,
,
G
a b c a b c cba
=
ﻭ ﻟﺘﻜﻥ
H
ـ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻭﻟﺩﺓ ﺒ
c
)
ﻼﻡـﻜﺒ
،ﺃﺩﻕ
H
ﺯﻤﺭﺓ ﺠ
ﺯﺌﻴﺔ ﻤﻭﻟﺩﺓ ﺒﻌﻨﺼﺭ ﻤﻥ
G
ﺍﻟﻤﻤﺜﻠﺔ ﺒﺎﻟﻜﻠﻤﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ
c
.(
ﺔـﻴﻠﻤﻋ ﻑﺼـﺘﺘ
G
ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﺘﺤﻘﻕ ﺍﻟﻘﻭﺍﻨﻴﻥ،ﺕﺍﺩﻟﻭﻤﻟﺍ ﺭﻴﺜﺄﺘﺒ ﺕﺎﻘﻓﺍﺭﻤﻟﺍ ﺔﻋﻭﻤﺠﻤ ﻰﻠﻋ
ﺍﻵﺘﻴ
ﺔ:
(i)
ﻜل ﻤﻭﻟﺩ
)
, ,
a b c
ﻓﻲ ﻤﺜﺎﻟﻨﺎ
(
ﻴﺅﺜﺭ ﻜﺘﺒﺩﻴﻠﺔ
.
(ii)
ﺍﻟﻌﻼﻗﺎﺕ
)
3
2
2
,
,
,
a b c cba
ﻓﻲ ﻤﺜﺎﻟﻨﺎ
(
ﺘﺅﺜﺭ ﺒﺸﻜل ﺘﺎﻓﻪ
.
(iii)
ﻤﻭﻟﺩﺍﺕ
H
)
c
ﻓﻲ ﻤﺜﺎﻟﻨﺎ
(
ﺘﺜﺒﺕ ﺍﻟﻤﺭﺍﻓﻘﺔ
1H
.
(iv)
ﺍﻟﻌﻤﻠﻴﺔ ﻋﻠﻰ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﻤﺘ
ﻌﺩﻴﺔ
.
ﺇﻥ ﺍﻻﺴﺘﺭﺍﺘﻴﺠ
ﻴ،ﺕﺎﻘﻓﺍﺭﻤﻟﺍ ﺝﺎﺘﻨﺘﺴﻻ ﺔﻌﺒﺘﻤﻟﺍ ﺔﻴ
ﺭﻤﺯ ﻟﻬﺎ
1, 2,...
ﺙـﻴﺤ
1 1H
=
ﻭﻥـﻜﻴ ،
ﻀﺭﻭﺭﻴﺎﹰ
.
ﺍﻟﻘﺎﻨﻭﻥ
(iii)
ﻴﺨﺒﺭﻨﺎ ﺒﺒﺴﺎﻁﺔ ﺃﻥ
1
c
c
=
.
ﺴﻨﻁﺒﻕ ﺍﻵﻥ ﺍﻟﻘﺎﻨﻭﻨﻴﻥ ﺍﻷﻭل ﻭ ﺍﻟﺜﺎﻨﻲ
.
ﺒﻤﺎ ﺃﻨﻪ ﻻ
ﻨﻌﻠﻡ ﻤﻥ
ﻫﻭ
1
a
ﻟﻨﺭﻤﺯ ﺒﻤﺎ،
ﻲـﻠﻴ
2 : 1
2
a
=
.
ﻴﻜﻥـﻟ ،ﻪﺒﺎﺸـﻤ لﻜﺸـﺒ ﻭ
2
3
a
=
.
ﺍﻵﻥ
3
3
1
a
a
=
ﻭﻤﻥ،
(ii)
ﻴﺠﺏ ﺃﻥ ﻴﺴﺎﻭﻱ
1
.
ﻨﻜﻭﻥ ﻗﺩ ﺍﺴﺘﻨﺘﺠﻨﺎ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﺜﻼﺜﺔ،ﺍﺫﻬﻟ
1, 2, 3
،
ﺍﻟﺘ
ﺒﺩﻴل ﻭﻓﻕ
a
ﻜﻤﺎ
ﻴﻠﻲ
:
ﻤﺎ ﻫﻲ
1
b
ﺭﻯـﺨﺃ ﺔـﻘﻓﺍﺭﻤ ﺭﺫـﺤﺒ ﺞﻨﺘﺴﻨ ،ﻡﻠﻌﻨ ﻻ ؟
4
1
b
=
.
ﺍﻵﻥ
4 1
b
=
ﻷﻥ
2
1
b
=
،
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ
.
1
4
1
b
b
a a
ﺒﻘﻲ ﻟﺩﻴﻨﺎ ﺍﻟﻌﻼﻗﺔ
cba
.
ﻨﻌﻠﻡ ﺒﺄﻥ
1
2
a
=
ﻟﻜﻥ ﻻ ﻨﻌﻠﻡ ﻤﺎ،
2
b
ﻭﻤﻨﻪ ﻨﻀﻊ،
2
5
b
=
:
15
ﺴﺄﻟﺔـﻤﻟﺍ ﹰﺎﺤﻴﺤـﺼ ﺏﺍﻭـﺠﻟﺍ ﻊـﻤ ﻪـﺘﻨﻤ ﻥـﻤﺯ ﻲـﻓ ﹰﺎﻤﺌﺍﺩ ﺔﻴﻤﺯﺭﺍﻭﺨﻟﺍ ﻲﻬﺘﻨﺘ ﻥﺃ ﺏﺠﻴ ،ﺴﺄﻟﺔﻤﻟﺍ لﺤﻟ
.
ﺇﻥ
ﺨﻭﺍﺭﺯﻤﻴﺔ
Coxeter
Tood –
ﻟﻥ ﺘﺤل ﺍﻟﻤﺴﺄﻟﺔ ﺍﻟﻤﺤﺩﺩﺓ ﻓﻴﻤﺎ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﻤﺜﻴل ﺍﻟﻤﻨﺘﻬﻲ ﻴﻌﺭﻑ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
)
ﻓﻲ
ﻻ ﺘﻭﺠ،ﺔﻘﻴﻘﺤﻟﺍ
ﺩ ﺨﻭﺍﺭﺯﻤﻴﺎﺕ ﻜﻬﺫﻩ
.(
ﻥـﻤ ﺔـﻴﻬﺘﻨﻤﻟﺍ ﺓﺭﻤﺯﻟﺍ ﺔﺒﺘﺭ ﺩﻴﺩﺤﺘ ﺔﻟﺄﺴﻤ لﺤﺒ ،لﺎﺤ ﻱﺃ ﻰﻠﻋ ،ﻡﻭﻘﺘ ﺎﻬﻨﺇ
ﺍﻟﺘﻤﺜﻴل ﺍﻟﻤﻨﺘﻬﻲ ﻟﻠﺯﻤﺭﺓ
)
ﻨﺴﺘﺨﺩﻡ ﺍﻟﺨﻭﺍﺭﺯﻤﻴﺔ ﻤﻊ
H
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺘﺎﻓﻬﺔ
1
.
1
2
3
1.
a
a
a
a a a
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1
2
5
a
b
a a
ﻤﻥ
(iii)
1 1
c
=
ﻭ ﺒﺘﻁﺒﻴﻕ،
(ii)
ﻋﻠﻰ
cba
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ
5 1
c
=
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ،ﻙﻟﺫﻟ
(i)
ﺴﻴﻜﻭﻥ ﻟﻴﻨﺎ
5 1
=
،
ﻨﺴﻘﻁ
5
ﻭﺒﻬﺫﺍ ﻴﻜﻭﻥ ﺍﻵﻥ،
2 1
b
=
.
ﺒﻤﺎ ﺃﻥ
4 1
b
=
ﺴﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ
4
2
=
ﺎ ﺃﻥـﻨﻜﻤﻴ ﻙﻟﺫﻟﻭ ،
ﻨﺴﻘﻁ
4
ﺃﻴﻀﺎﹰ
.
ﻴﻤﻜﻥ ﺃﻥ ﻨﻤﺜل ﻤﺎ ﻋﺭﻓﻨﺎﻩ ﻓﻲ ﺍﻟﻘﺎﺌﻤﺔ
ﺍﻵﺘﻴ
ﺔ:
a a a b b c c a b c
1 2 3 1 2 1 1 1 2 1 1
2 3 1 2 1 2 2 3 2
3 1 2 3 3 3 1 2 3
ﺍﻟﻤﻁﺒﻘﺔ ﻤﻥ،ﻰﻨﻤﻴﻟﺍ ﺔﻴﻠﻔﺴﻟﺍ ﺔﻴﻭﺍﺯﻟﺍ
(ii)
ﺄﻥـﺒ ﺎـﻨﻟ ﻥﻴﺒﺘ ،
2
3
c
=
.
ﻀﺎﹰـﻴﺃ ﻥﻭـﻜﻴ ﻲﻟﺎـﺘﻟﺎﺒﻭ
3
2
c
=
ﻴﺤﺩﺩ ﻫﺫﺍ ﺒﻘﻴﺔ ﺍﻟﻘﺎﺌﻤﺔﺫﺌﺩﻨﻋﻭ ،
:
a a a b b c c a b c
1 2 3 1 2 1 1 1 2 1 1
2 3 1 2 1 2 3 2 3 3 2
3 1 2 3 3 3 2 3 1 2 3
ﻨﻜﻭ
ﻥ
ﺒﺫﻟﻙ
ﻗﺩ ﺤﺼﻠﻨﺎ ﻋﻠﻰ ﺜﻼﺙ ﻤﺭﺍﻓﻘﺎﺕ ﻭﺍﻟﺘﻲ ﻴﺅﺜﺭ ﻋﻠﻴﻬ
ﺎ
, ,
a b c
ﻜﻤﺎ
ﻴﻠﻲ
( )
( )
( )
123
12
23
a
b
c
=
=
=
ﻭ
ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﻜﺘﺏ ﻓﻴﻤﺎ ﺒﻌﺩ ﺍﻟﺘﻁﺒﻴﻕ،ﺭﺜﻜﺃ ﺔﻗﺩﺒ
3
G
S
→
ﺴﺎﺒﻘﺔـﻟﺍ ﻥﻴﻨﺍﻭﻘﻟﺎﺒ ﺔﻔﻟﺅﻤﻟﺍ ﻭ
.
ﻥـﻤ
ﻤﺒﺭﻫﻨﺔ
( Artin 1991, 9.10 )
ﺄﺜﻴﺭـﺘ ﻑﺼﻴ ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ ﺍﺫﻫ ﻥﺃ ﻰﻠﻋ ﺹﻨﺘ ﻲﺘﻟﺍ
G
ﻰـﻠﻋ
G H
.
ﺒﻤﺎ ﺃﻥ ﺍﻟﻌﻨﺎﺼﺭ ﺍﻟﺜﻼﺜﺔ
( )
123
،
( )
12
ﻭ،
( )
23
ﺘﻭﻟﺩ
3
S
ﺄﺜﻴﺭـﺘ ﻥﺄﺒ ﻥﻴﺒﻴ ﺍﺫﻫ ،
G
ﻋﻠﻰ
G H
ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻟﺘﻤﺎﺜل
3
G
S
→
ﻭﺇﻥ،
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﻓﻲ
Artin 1991, 6.9
ﻋﻨﺩﻤﺎ،ﺙﻴﺤﺒ ﺔﻴﻤﺯﺭﺍﻭﺨ ﻰﻟﺇ ﺔﻘﻴﺭﻁﻟﺍ ﻩﺫﻫ لﻴﻭﺤﺘ ﺔﻴﻔﻴﻜ ﻑﺸﺘﻜﺍ ،
ﺴﺘﻜﻭﻥ ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﻗﺩ ﺍﺴﺘﻨﺘﺠﺕ ﺍ،ﺕﺒﺍﻭﺜﻟﺍ ﺔﻤﺌﺎﻗ ﺝﺍﺭﺨﺘﺴﺍ ﻲﻓ ﺢﺠﻨﺘ
،ﻟﻘﺎﺌﻤﺔ ﺍﻟﺼﺤﻴﺤﺔ
ﻫﺫﻩ ﺍﻟﺨﻭﺍﺭﺯﻤﻴﺔ ﻤﻨﺠﺯﺓ ﻓﻲ
Maple
ﺒﺎﺴﺘﺜﻨﺎﺀ ﺘﻠﻙ ﺍﻟﺘﻲ ﺘﺤﺴﺏ ﺍﻟﺘﺄﺜﻴﺭﺍﺕ ﻋﻠﻰ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ،
ﺍﻟﻴﺴﺎﺭﻴﺔ
.
ﻴﻭﺠﺩ ﻫﻨﺎ ﻨﺴﺨﺔ ﻋﻥ ﺫﻟﻙ
:
>with (group) ; [loads the group theory package . ]
>G:=grelgroup ({a,b,c},{[a,a,a],[b,b],c,c],[a,b,c]}) ; [defines G to have
generators a,b,c and relations aaa, bb, cc, abc]
>H:=subgrel ({x=[c]},G) ; [defines H to be the subgroup generated by c]
>permrep (H) ;
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Permgroup (3, a=[[1,2,3],b=[1,2],c=[2,3]])
[computes the acthon of G on the set of right cosets of H in G].
ﺍﻟﺘﺄﺜﻴﺭﺍﺕ ﺍﻻﺒﺘﺩﺍﺌﻴﺔ
(Primitive action)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺍﻟﺘﺄﺜﻴﺭ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻭ ﻟﻴﻜﻥ،
π
ـ ﺘﺠﺯﺌﺔ ﻟ
X
.
ﻨﻘﻭل ﺒﺄﻥ
π
ﺕـﺒﺜﻤ
(stabilized)
G
ﺇﺫﺍ ﻜﺎﻥ
A
g A
π
π
∈ ⇒
∈
ﻤﺜﺎل
39.4
(a)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
(
)
1234
G
=
ﻤﻥ
4
S
ﺘﺜﺒﺕ ﺍﻟﺘﺠﺯﺌﺔ
{ } { }
{
}
1, 3 , 2, 4
ﻤﻥ
{
}
1, 2, 3, 4
.
(b)
ﻨﻔﺭﺽ ﺃﻥ
{
}
1, 2,3, 4
X
=
ﻰـﻠﻋ ﻊـﺒﺭﻤﻟﺍ ﺱﻭﺅﺭ ﺔﻋﻭﻤﺠﻤ ﻊﻤ
4
D
ﺅﺜﺭـﺘ ﻲـﺘﻟﺍ
ﻤﻊ،ﺩﻴﺩﺤﺘﻟﺎﺒ ،ﺔﻴﺩﺎﻴﺘﻋﻻﺍ ﺔﻘﻴﺭﻁﻟﺎﺒ
(
)
( )
1234 ,
24
r
s
=
=
.
ﺫـﺌﺩﻨﻋ
4
D
ﺔـﺌﺯﺠﺘﻟﺍ ﺕـﺒﺜﻴ
{ } { }
{
}
1, 3 , 2, 4
.
(c)
ﻟﺘﻜﻥ
X
ﻤ
ﺠﻤﻭﻋﺔ ﺘﺠﺯﺌﺎﺕ
{
}
1, 2, 3, 4
ﻜل ﻤﻨﻬ،ﻥﻴﺘﻋﻭﻤﺠﻤ ﻰﻟﺇ
ـﻤ
ﻥـﻤ ﻑـﻟﺅﻤ ﺎ
ﻋﻨﺼﺭﻴﻥ
.
ﻓﺈﻥﺫﺌﺩﻨﻋ
4
S
ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﻭ،
( )
(
)
4
Ker
Sym
S
X
→
ﻫﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
V
ﺍﻟﻤﻌﺭﻓﺔ ﻓﻲ
(31.4)
.
ﺇﻥ ﺍﻟﺯﻤﺭﺓ
G
ـ ﺘﺜﺒﺕ ﺩﺍﺌﻤﺎﹰ ﺍﻟﺘﺠﺯﺌﺎﺕ ﺍﻟﺘﺎﻓﻬﺔ ﻟ
X
ﻤﺠﻤﻭﻋﺔ ﻜل ﺍﻟﻤﺠﻤﻭﻋﺎﺕ،ﺩﻴﺩﺤﺘﻟﺎﺒﻭ ،
ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
X
ﻭ
ﻭ،ﺩﺤﺍ ﻭﺭﺼﻨﻋ ﻥﻤ ﺔﻔﻟﺅﻤﻟﺍ
{ }
X
.
ﻨﻘﻭل،ﺕﺎﺌﺯﺠﺘﻟﺍ ﻩﺫﻫ ﻁﻘﻓ ﺕﺒﺜﻨ ﺎﻤﺩﻨﻋ
ﺒﺄﻥ ﺍﻟﺘﺄﺜﻴﺭ ﺍﺒﺘﺩﺍﺌﻲ
(primitive)
ﻭﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﺨﺭﻯ ﻴﻜﻭﻥ ﻏﻴﺭ ﺍﺒﺘﺩﺍﺌﻲ،
(primitive)
ﻴﻘﺎل،
ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
( )
Sym X
)
ﻓﻲ،ﹰﻼﺜﻤ
n
S
(
ﻰـﻠﻋ ﻲﺌﺍﺩﺘﺒﺍ لﻜﺸﺒ ﺕﺭﺜﺃ ﺍﺫﺇ ﺔﻴﺌﺍﺩﺘﺒﺍ
X
.
ﺒﺄﻥ،ﺢﻀﺍﻭﻟﺍ ﻥﻤ
n
S
ﻟﻜﻥ ﺍﻟﻤﺜﺎل،ﺎﻬﺴﻔﻨ ﻰﻠﻋ ﺔﻴﺌﺍﺩﺘﺒﺍ ﺭﻴﺜﺄﺘ ﺓﺭﻤﺯ ﻲﻫ
39.4b
ﻴﺒﻴﻥ ﺒﺄﻥ
4
D
ﺘﻌﺘﺒﺭ ﻜﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ،
4
S
ﻟﻴﺱ ﺍﺒﺘﺩﺍﺌﻴﺎﹰ،ﺔﻨﻴﺒﻤﻟﺍ ﺔﻘﻴﺭﻁﻟﺎﺒ
.
ﻤﺜﺎل
40.4
ﺇﻥ ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻤﺘﻌﺩﻱ ﺍﻟﻤﻀﺎﻋﻑ ﻴﻜﻭﻥ ﺍﺒﺘﺩﺍﺌﻴﺎﹰ
:
ﺇﺫﺍ
ﺜﺒﺕ
{
} { }
{
}
,
,... ,
,... ... ,
x x
y
′
ﻟﻥ ﻴﻜﻭﻥ ﻫﻨﺎﻙ ﻋﻨﺼﺭﺍﹰ ﻴﺭﺴلﺫﺌﺩﻨﻋ
(
)
,
x x
′
ﺇﻟﻰ
(
)
,
x y
.
ﻤﻼﺤﻅﺔ
41.4
ﺇﻥ
G
-
ـ ﻤﺩﺍﺭ ﻴﺸﻜل ﺘﺠﺯﺌﺔ ﻟ
X
ـ ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﺜﺒﺘﺔ ﺒ
G
.
ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﺄﺜﻴﺭ
ﻋﻨﺩﺌﺫ،ﹰﺎﻴﺌﺍﺩﺘﺒﺍ
ﻴﺠﺏ ﺃﻥ ﺘ
ﻜﻭﻥ ﺍﻟﺘﺠﺯﺌﺔ ﺇﻟﻰ ﻤﺩﺍﺭﺍﺕ
ﻫﻲ
ﺇ
ﺤﺩ
ﻯ
ﺍﻟﺘﺠﺯﺌﺎﺕ ﺍﻟﺘﺎﻓﻬﺔ
.
ﺒﺎﻟﺘﺎﻟﻲ
ﺇﺫﺍ ﻜﺎﻥ
ﺍﻟﺘﺄﺜﻴﺭ ﺍﺒﺘﺩﺍﺌﻴﺎﹰ
⇐
ﺍﻟﺘﺄﺜﻴﺭ
ﺃﻭ ﺘﺎﻓﻪﺩﻌﺘﻤ
)
gx
x
=
ﻟﻜل
,
g x
(
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٧٨
ﺒﺎﻟﻨﺴﺒﺔ ﻟﺒ
ﺘﻜﻭﻥ،ﺕﺍﺭﻴﺜﺄﺘﻟﺍ ﻩﺫﻫ ﻲﻗﺎ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﺅﺜﺭ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﺩـﻌﺘﻤ لﻜﺸـﺒ
ﻭﻤﺅﻟﻔﺔ ﻤﻥ ﻋﻨﺼﺭﻴﻥ ﻋﻠﻰ ﺍﻷﻗل
.
ﻗﻀﻴﺔ
42.4
ﺘﺅﺜﺭ ﺍﻟﺯﻤﺭﺓ
G
ﺔـﻴﺌﺯﺠ ﺔﻋﻭﻤﺠﻤ ﺕﺩﺠﻭ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﻲﺌﺍﺩﺘﺒﺍ ﺭﻴﻏ لﻜﺸﺒ
A
ﻤﻥ
X
، ﻭﻤﺅﻟﻔﺔ ﻤﻥ ﻋﻨﺼﺭﻴﻥ ﻋﻠﻰ ﺍﻷﻗل ﺒﺤﻴﺙ ﻴﻜﻭﻥ
(18)
ﻟﻜل
g
G
∈
ﺇﻤﺎ،
gA
A
=
ﺃﻭ
gA
A
φ
=
I
.
ﺍﻟﺒﺭﻫﺎﻥ
.
⇐
:
ﺇﻥ ﺍﻟﺘﺠﺯﺌﺔ
π
ﺍﻟﻤﺜﺒﺘﺔ ﺒﺎﻟﺯﻤﺭﺓ
G
ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﺜل
A
.
⇒
:
ﻤﻥ
A
ﻴﻤﻜﻥ ﺃﻥ ﻨﺸﻜل ﺘﺠﺯﺌﺔ،
{
}
1
2
,
,
,...
A g A g A
ـ ﻟ
X
ـ ﺍﻟﻤﺜﺒﺘﺔ ﺒ،
G
.
ﺘﺴﻤﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺠﺯﺌﻴﺔ
A
ﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﺔـﻗﻼﻌﻟﺍ ﻕـﻘﺤﺘ ﻲﺘﻟﺍ ﻭ
(18)
ﺔـﻴﻠﺨﻟﺎﺒ
(block)
.
ﻗﻀﻴﺔ
4. 43
ﻟﺘﻜﻥ
A
ﺨﻠﻴﺔ ﻓﻲ
X
ﻤﻊ
2
A
≥
ﻭ
A
X
≠
.
ﻷﻱﺫﺌﺩﻨﻋ
x
A
∈
،
( )
( )
Stab
Stab
x
A
G
⊂
⊂
≠
≠
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺩﻴﻨﺎ
( )
( )
Stab
Stab
A
x
⊃
ﻷﻥ
gx
x
gA
A
gA
A
φ
= ⇒
≠ ⇒
=
I
ﻟﻴﻜﻥ
,
y
A y
x
∈
≠
.
ﺒﻤﺎ ﺃﻥ
G
ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺘﻌﺩﻱ ﻋﻠﻰ
X
ﺩـﺠﻭﻴ ،
g
G
∈
ﺙـﻴﺤﺒ
ﻴﻜﻭﻥ
gx
y
=
.
ﻋﻨﺩﺌﺫ
( )
Stab
g
A
∈
ﻟﻜﻥ،
( )
Stab
g
x
∉
.
ﻟﺘﻜﻥ
y
A
∉
.
ﻴﻭﺠﺩ
g
G
∈
ﺒ
ﺤﻴﺙ ﺇﻥ
gx
y
=
ﻭ ﻋﻨﺩﺌﺫ،
( )
Stab
g
A
∉
.
ﻤﺒﺭﻫﻨﺔ
44.4
ﺘﺅﺜﺭ ﺍﻟﺯﻤﺭﺓ
G
ﺒﺸﻜل ﺍﺒﺘﺩﺍﺌﻲ ﻋﻠﻰ
X
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ
ﺼﺭـﻨﻋ لﺠﺃ ﻥﻤ ،
ﻭﺍﺤﺩ
)
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻟﻠﻜل
(
x
ﻤﻥ
X
،
( )
Stab x
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻷﻋﻅﻤﻴﺔ ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻟﻡ ﺘﺅﺜﺭ
G
ﺒﺸﻜل ﺍﺒﺘﺩﺍﺌﻲ ﻋﻠﻰ
X
ﻋﻨﺩﺌﺫ،
)
ﺍﻨﻅﺭ
42.4
(
ﺘ
ﻭﺠﺩ ﺍﻟﺨﻠﻴﺔ
A
X
≠
⊂
ﻭﻤﻥ،لﻗﻷﺍ ﻰﻠﻋ ﻥﻴﺭﺼﻨﻋ ﻥﻤ ﺔﻔﻟﺅﻤﻟﺍﻭ
(43.4)
ﻨﺒﻴﻥ ﺒﺄﻥ
( )
Stab x
ﻟﻥ ﺘﻜﻭﻥ ﺃﻋﻅﻤﻴﺔ ﻷﻱ
x
A
∈
.
ﺒﻔﺭﺽ ﺃﻨﻪ ﻴﻭﺠﺩ،ﺱﻜﻌﻟﺍ
x
ﻤﻥ
X
ﻭﺯﻤﺭﺓ
ﺠﺯﺌﻴﺔ
H
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
Stab x
H
G
≠
≠
⊂ ⊂
ﺃﺩﻋﻲ ﺒﺄﻥﺫﺌﺩﻨﻋ
A
Hx
=
ﺨﻠﻴﺔ
X
≠
ﻤﺅﻟﻔﺔ ﻤﻥ ﻋﻨﺼﺭﻴﻥ ﻋﻠﻰ ﺍﻷﻗل
.
ﻷﻥ
( )
Stab
H
x
≠
،
{ }
Hx
x
≠
ﻭﻤﻨﻪ،
{ }
x
A
X
≠
≠
⊂ ⊂
.
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ﺇﺫﺍ ﻜﺎﻥ
g
H
∈
ﻋﻨﺩﺌﺫ،
gA
A
=
.
ﺃﻤﺎ ﺇﺫﺍ ﻜﺎﻥ
g
H
∉
ﻋﻨﺩﺌﺫ،
gA
ﺼﻠﺔـﻔﻨﻤ ﻥﻭﻜﺘ
ﻥــﻋ
A
:
ﺭﺽــﻔﻨ ﻙﻟﺫــﻟ
ghx
h x
′
=
ﺭــﺼﺎﻨﻋ ﺽﻌﺒــﻟ
h
H
′∈
ﺫــﺌﺩﻨﻋ ،
( )
1
Stab
h
gh
x
H
−
′
∈
⊂
ﻭﻨﻘﻭل ﺒﺄﻥ،
1
h
gh
h
−
′
′′
=
ﻭ،
1
g
h h h
H
−
′ ′′
=
∈
.
ﺘﻤﺎﺭﻴﻥ
1-4
ﻟﺘﻜﻥ
1
H
ﻭ
2
H
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
.
ﺒﻴﻥ ﺒﺄﻥ ﺘﻁﺒﻴﻘﺎﺕ
G
-
ﺔـﻋﻭﻤﺠﻤ
1
2
G H
G H
→
ﺘﻜﻭﻥ ﺘﻁﺒﻴﻘﺎﺕ ﺍﻟﺘﻘﺎﺒل ﺍﻟﻘﺎﻨﻭﻨﻴﺔ ﻤﻊ ﺍﻟﻌﻨﺎﺼﺭ
2
gH
ﻤﻥ
2
G H
ﺒ
ﺤﻴﺙ
ﺇﻥ
1
1
2
H
gH g
−
⊂
.
2-4
(a)
ﺒﺭﻫﻥ ﺒﺄﻥ ﺍﻟ
ﺭـﻤﺯﻟ ﻕـﻓﺍﺭﺘﻟﺍ ﻑﻭﻔﺼـﻟ ﹰﺎﻋﺎﻤﺘﺠﺍ ﻱﻭﺎﺴﺘ ﻥﺃ ﻥﻜﻤﻴ ﻻ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺓﺭﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻌﻠﻴﺔ
.
(b)
ﺔـﻴﻠﻌﻓ ﺔـﻴﺌﺯﺠ ﺔﻋﻭﻤﺠﻤ ﻥﻋ ﹰﻻﺎﺜﻤ ﻁﻋﺃ
S
ﺔـﻴﻬﺘﻨﻤ ﺓﺭـﻤﺯ ﻥـﻤ
G
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ
1
g G
G
gsg
−
∈
=
U
.
3-4
ﺒﺭﻫﻥ ﺃﻥ ﺃﻱ ﺯﻤﺭﺓ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍ
ﻟﺭﺘﺒﺔ
3
p
،
p
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ ﺇﺤﺩﻯ،ﻱﺩﺭﻓ ﻲﻟﻭﺃ ﺩﺩﻋ
ﺍﻟﺯﻤﺭﺘﻴﻥ ﺍﻟﻤﺒﻨﻴﺘﻴﻥ ﻓﻲ
(3.13,3.14)
.
4-4
ﻟﻴﻜﻥ
p
ﺃﺼﻐﺭ ﻋﺩﺩ ﺃﻭﻟﻲ ﻴﻘﺴﻡ
(
)
:1
G
)
ﺒﻔﺭﻀﻪ ﻤﻨﺘﻪ
.(
ﻲـﻓ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻱﺃ ﻥﺄﺒ ﻥﻴﺒ
G
ﺩﻟﻴﻠﻬﺎ
p
ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ
.
5-4
ﻥ ﺒﺄﻥ ﺃﻱ ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔﻴﺒ
2m
،
m
ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ،ﻱﺩﺭﻓ ﺩﺩﻋ
2
.
)
ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﺒﺭﻫﻨﺔ ﻜﺎﻴﻠﻲ
1.21
(
6-4
ﻟﺘﻜﻥ
( )
3
2
GL
G
=
F
.
(a)
ﻥ ﺒﺄﻥﻴﺒ
(
)
:1
168
G
=
.
(b)
ﻟﺘﻜﻥ
X
ـ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﺎﺭﺓ ﻤﻥ ﻤﺒﺩﺃ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﻟ
3
2
F
ﺄﻥـﺒ ﻥﻴﺒ ،
X
ﺘﺤﻭﻱ ﻋﻠﻰ
7
ﻭ ﺃﻨﻪ ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﻤﺘﺒﺎﻴﻥ ﻗ،ﺭﺼﺎﻨﻋ
ﺎﻨﻭﻨﻲ
( )
7
Sym
G
X
S
→
=
.
(c)
ﺍﺴﺘﺨﺩﻡ ﺃﺸﻜﺎل ﺠﻭﺭﺩﺍﻥ ﺍﻟﻘﺎﻨﻭﻨﻴﺔ ﻟﺘﺒﻴﻥ ﺒﺄﻥ
G
ﻊـﻤ ،ﻕـﻓﺍﺭﺘ ﻑﻭﻔﺼ ﺕﺴ ﻰﻠﻋ ﻱﻭﺤﺘ
1,21,42,56,24
ﻭ،
24
ﺏـﻴﺘﺭﺘﻟﺍ ﻰـﻠﻋ ﺭﺼﻨﻋ
] .
ﺕـﻨﺎﻜ ﺍﺫﺇ ﻪـﻨﺄﺒ ﻅـﺤﻼﻨ
M
ﺭﺓـﺤ
[ ]
2
module
≥ −
F
ﻤ
ﻋﻨﺩﺌﺫ،ﺩﺤﺍﻭ ﺔﺒﺘﺭﻤﻟﺍ ﻥ
[ ]
( )
[ ]
2
2
End
M
α
α
=
F
F
[.
(d)
ﺍﺴﺘﻨﺘﺞ ﺒﺄﻥ
G
ﺒﺴﻴﻁﺔ
.
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٩٨
7-4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
.
ﺇﺫﺍ ﻜﺎﻥ
( )
Aut G
ﺒﺭﻫﻥ ﺒﺄﻥ،ﺔﻴﺭﺌﺍﺩ
G
ﻭ،ﺔﻴﻠﻴﺩﺒﺘ
ﺇﺫﺍ ﻜﺎﻨﺕ،ﻪﻨﺃ
G
ﺒﺭﻫﻥ ﺒﺄﻥ،ﺔﻴﻬﺘﻨﻤ
G
ﺩﺍﺌﺭﻴﺔ
.
8-4
ﺄﻥـــﺒ ﻥﻴـــﺒ
n
S
ﺭـــﺼﺎﻨﻌﻟﺎﺒ ﺓﺩـــﻟﻭﻤ
( ) ( ) ( )
12 , 13 ,..., 1n
ﺭـــﺼﺎﻨﻌﻟﺎﺒ ﻭ ،
( ) ( ) (
)
12 , 23 ,...,
1
n
n
−
ﺃﻴﻀﺎﹰ
.
4-9
ﻟﺘﻜﻥ
K
ﺼﻑ ﺘﺭﺍﻓﻕ
ﻤﻥ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
G
ﻤﺤﺘﻭﺍﺓ ﻓﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
H
ﻲـﻓ
G
.
ﺒﺭﻫﻥ ﺃﻥ
K
ــ ﺍﺠﺘﻤﺎﻉ ﻟ
k
ﻲـﻓ ﺔﻴﻭﺎﺴـﺘﻤ ﻡﻭـﺠﺤﺒ ﻕـﻓﺍﺭﺘ ﻑـﺼ
H
ﺙـﻴﺤ ،
( )
(
)
:
.
G
k
G H C
x
=
ﻷﻱ
x
K
∈
.
10-4
(a)
ﻟﻴﻜﻥ
n
A
σ
∈
.
ﻤﻥ ﺍﻟﺘﻤﺭﻴﻥ
4-9
ـ ﻨﻌﻠﻡ ﺒﺄﻥ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻟ
σ
ﻓﻲ
n
S
ﺎـﻤﺇ
ﺘﺒﻘﻰ ﺼﻑ ﺘﺭﺍﻓﻕ ﻭﺍﺤﺩ ﻓﻲ
n
A
ﺃﻭ ﻴﻨﻬﻲ ﻜﺎﺠﺘﻤﺎﻉ ﺼﻔﻲ ﺘﺭﺍﻓﻕ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﺤﺠﻡ
.
ﺄﻥـﺒ ﻥﻴـﺒ
ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻤﺤﻘﻘﺔ
⇔
σ
ﻻ ﺘﺘﺒﺎﺩل ﻤﻊ ﺃﻱ ﺘﺒﺩﻴﻠﺔ ﻓﺭﺩﻴﺔ
⇔
ﺔـﺌﺯﺠﺘ ﻑﺭﻌﺘ
n
ــ ﺒ
σ
ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ ﺃﻋﺩﺍﺩ ﺼﺤﻴﺤﺔ ﻓﺭﺩﻴﺔ ﻤﺨﺘﻠﻔﺔ
.
(b)
ﻟﻜل ﺼﻑ ﺘﺭﺍﻓﻕ
K
ﻓﻲ
7
A
ﺃﻋﻁ ﺸﻜل،
K
ﻭﺤﺩﺩ،
K
.
11-4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻊ ﺍﻟﻤﻭﻟﺩﺍﺕ
,
a b
ﻭ ﺍﻟﻌﻼﻗﺎﺕ
4
2
1
,
a
b aba
bab
= =
=
.
(a)
(4 pts)
ﺍﺴﺘﺨﺩﻡ ﺨﻭﺍﺭﺯﻤﻴﺔ
Coxeter
-
Tood
)
ﻤﻊ
1
H
=
(
ﻭﺭﺓـﺼ ﻑﺭـﻌﺘﻟ
G
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺸﺎﻜل
(
)
,
:1
n
G
S
n
G
→
=
ﺍﻟﻤﻌﻁﻰ ﺒﻤﺒﺭﻫﻨﺔ ﻜﺎﻴﻠﻲ،
1.11
].
ﻻ ﻨﺤﺘ
ﻰ ﺃﻥـﻟﺇ ﺝﺎ
ﺴﻨﻌﻤل ﺒﺸﻜل ﻤﻠﺨﺹ ﻓﻘﻁ،ﺕﺍﻭﻁﺨﻟﺍ لﻜ ﻥﻤﻀﺘﺘ
[.
(b)
(1.pt)
ﺍﺴﺘﺨﺩﻡ
Maple
ﻟﺘﺘﺄﻜﺩ ﻤﻥ ﺇﺠﺎﺒﺘﻙ
.
12-4
ﻥ ﺒﺄﻨﻪ ﺇﺫﺍ ﻜﺎﻥ ﺘﺄﺜﻴﺭﻴﺒ
G
ﻋﻠﻰ
X
ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ﻱﺃ ﺭﻴﺜﺄـﺘ ﺫﺌﺩﻨﻋ ،ﹰﻻﺎﻌﻓ ﻭ ﹰﺎﻴﺌﺍﺩﺘﺒﺍ
ﻨﺎﻅﻤﻴﺔ
1
H
≠
ﻤﻥ
G
ﻴﻜﻭﻥ ﻤﺘﻌﺩﻴﺎﹰ
.
13-4
(a)
ﺘﺄﻜﺩ ﻤﻥ ﺃﻥ
4
A
ﺘﺤﻭﻱ ﻋﻠﻰ
8
ﻋﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
ﻭ،
3
ﻋﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻻ
ﻴﺤﻭ
ﻱ ﺃﻱ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
6
.
(b)
ﺒﺭﻫﻥ ﺒﺄﻥ
4
A
ﻻ ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤ
ﻥ ﺍﻟﺭﺘﺒﺔ
6
(cf. 1.29)
) .
ﺍﺴﺘﺨﺩﻡ
4.23
.(
(c)
ﺒﺭﻫﻥ ﺒﺄﻥ
4
A
ﻫﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ
4
S
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
.
14-4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻊ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ
r
.
ﺒﺭﻫﻥ ﺃﻥ
(a)
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻋﻨﺩﺌﺫ،ﺔﻁﻴﺴﺒ
(
)
:1
G
ﻴﻘﺴﻡ
!
r
.
(b)
ﺇﺫﺍ ﻜﺎﻥ
2, 3
r
=
ﺃﻭ،
4
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥﺫﺌﺩﻨﻋ ،
G
ﺒﺴﻴﻁﺔ
.
(c)
ﻴﻭﺠﺩ ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ
ﻤﻊ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ
5
.
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٠٩
15-4
ﺒﺭﻫﻥ ﺃﻥ
n
S
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
2
n
A
+
.
16-4
ﻟﺘﻜﻥ
H
ﻭ
K
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
.
ﺇﻥ ﺍﻟﻤﺭﺍﻓﻘﺔ ﺍﻟ
ــﺏ ﻟـﻨﺎﺠﻟﺍ ﺔﻴﺌﺎﻨﺜ
H
ﻭ
K
ﻓﻲ
G
ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﺸﻜل
{
}
,
HaK
hak h
H k
K
=
∈
∈
ﻟﺒﻌﺽ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ
a
G
∈
.
)
a
(
ـﻥ ﺒﺄﻥ ﺍﻟﻤﺭﺍﻓﻘﺔ ﺍﻟﺜﻨﺎﺌﻴﺔ ﺍﻟﺠﺎﻨﺏ ﻟﻴﺒ
H
ﻭ
K
ﻓﻲ
G
ﺘﺸﻜل ﺘﺠﺯﺌﺔ ﻟﻠﺯﻤﺭﺓ
G
.
)
b
(
ﻟﺘﻜﻥ
1
H
aKa
−
I
ﺘﺅﺜﺭ ﻋﻠﻰ
H
K
×
ﺒﺎﻟﺸﻜل
(
)
(
)
1
1
,
,
b h k
hb a b ak
−
−
=
.
ــﻲ ﺒﺎﻟــﻫ ﺭﻴﺜﺄــﺘﻟﺍ ﺍﺫــﻫ ﺕﺍﺭﺍﺩــﻤ ﻥﺄــﺒ ﻥﻴــﺒ
ﻕــﻴﺒﻁﺘﻟﺍ ﺕﺎــﺒﻴﻜﺭﺘ ﻁﺒﻀ
(
)
,
:
h k
hak H
K
HaK
× →
a
.
)
c
(
)
ﻴﺤﺴﺏ ﻋﺩﺩ ﺍﻟ
ﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﺜﻨﺎﺌﻴﺔ ﺍﻟﺠﺎﻨﺏ ﺒﺎﻟﺼﻴﻐﺔ
.(
ﺒﺎﺴﺘﺨﺩﺍﻡ
(b)
ﻥ ﺒﺄﻥﻴﺒ
1
H K
HaK
H
aKa
−
=
I
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١٩
ﺍﻟﻔﺼل ﺍﻟﺨﺎﻤﺱ
ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
؛
ﺘﻁﺒﻴﻘﺎﺕ
The Sylow Theorems; Application
ﺴﺘﻜﻭﻥ ﻜل ﺍﻟﺯﻤﺭ،لﺼﻔﻟﺍ ﺍﺫﻫ ﻲﻓ
ﻤﻨﺘﻬﻴﺔ
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻭﻟﻴﻜﻥ
p
ﻋﺩﺩﺍﹰ ﺃﻭﻟﻴﺎﹰ ﻴﻘﺴﻡ
(
)
:1
G
.
ﺘﺩﻋﻰ ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
G
ـ ﺒ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺇﺫﺍ ﻜﺎﻨﺕ ﺭﺘﺒﺘﻬﺎ ﺃﻜﺒﺭ ﻗﻭﺓ ﻟﻠﻌﺩﺩ
p
ﺴﻡـﻘﺘ ﺙﻴﺤﺒ
(
)
:1
G
.
ﺘﻨﺹ ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ ﻋﻠﻰ ﺃﻨﻪ ﻴﻭﺠﺩ
p
-
ﺴﻡـﻘﻴ ﻲﻟﻭﺃ ﺩﺩﻋ لﻜﻟ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺭﻤﺯ
(
)
:1
G
،
ﺒﺤﻴﺙ ﺘﻜﻭﻥ
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﺘﺭﺍﻓﻘﺔ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺜﺒﺕ
p
ﻭﺃﻥ ﻜل،
p
-
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻤﺤﺘﻭﺍﺓ ﻓﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
ﻜﻬﺫﻩ
، ﻋﻠﻰ ﺃﻱ ﺤﺎل،
ﺘﺤﺩﺩ ﺍﻟﻤﺒﺭﻫﻨﺎﺕ ﻋﺩﺩ
p
-
ﺯﻤﺭ
ﺴ
ﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﺤﺘﻤﻠﺔ ﻓﻲ
G
.
ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
ﻨﺴﺘﺨﺩﻡ ﺩﺍﺌﻤﺎﹰ،ﻥﻴﻫﺍﺭﺒﻟﺍ ﻲﻓ
:
ﺇﺫﺍ ﻜﺎﻥ
O
ﻤﺩﺍﺭﺍﹰ ﻟﻠﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﺍﻟﺘﻲ ﺘﺅﺜﺭ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻭ،
0
∈
x
O
ﻤﻥ ﺍﻟﺘﻁﺒﻴﻕﺫﺌﺩﻨﻋ ،
0
,
H
X h
hx
→
a
ﻴﻨﺘﺞ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺘﻘﺎﺒل
( )
0
Stab
;
H
x
O
→
ﺍﻨﻅﺭ
(7.4)
.
ﻟﺫﻟﻙ
( )
(
)
0
: Stab
H
x
O
=
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ،ﺹﺎﺨ لﻜﺸﺒ
H
p
-
، ﺯﻤﺭﺓ
O
ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
p
:
ﻓﺈﻥ
O
ﻥـﻤ ﻑﻟﺄـﺘﻴ
ﺃﻭ،ﺩﺤﺍﻭ ﺭﺼﻨﻋ
O
ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
ﺒﻤﺎ ﺃﻥ
X
،ﺩﺍﺭﺍﺕـﻤﻟﺍ ﻥـﻤ لﺼﻔﻨﻤ ﻉﺎﻤﺘﺠﺍ
ﻨﺴﺘﻨﺘﺞ
:
ﺘﻤﻬﻴﺩﻴﺔ
1.5
ﻟ
ﺘﻜﻥ
H
p
-
ﺯﻤﺭﺓ ﺘﺅﺜﺭ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻨﺘﻬﻴﺔ
X
ﻭ ﻟﺘﻜﻥ،
H
X
ﻤﺠﻤﻭﻋﺔ
ـﺍﻟﻨﻘﺎﻁ ﺍﻟﻤﺜﺒﺘﺔ ﺒ
H
ﻋﻨﺩﺌﺫ،
(
)
mod
H
X
X
p
≡
ﻋﻨﺩﻤﺎ ﺘﻁﺒﻕ
ﺍﻟﺘﻤﻬﻴﺩﻴﺔ ﻋﻠﻰ
p
-
ﺯﻤﺭﺓ
H
ﻨﺠﺩ ﺒﺄﻥ،ﻕﻓﺍﺭﺘﻟﺎﺒ ﺎﻬﺴﻔﻨ ﻰﻠﻋ ﺭﺜﺅﺘ
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٢٩
( )
(
)
(
) (
)
:1
:1
mod
Z H
H
p
≡
ﻭﺒﺎﻟﺘﺎﻟﻲ
( )
(
)
:1
p
Z H
)
ﺍﻨﻅﺭ ﺒﺭﻫﺎﻥ
16.4
.(
ﻤﺒﺭﻫﻨﺔ
2.5
)
ﻴﻠﻭـﺴ
1
(
ﺘﻜﻥـﻟ
G
ﺭﺓ ﻤﻨـﻤﺯ
ﻴﻜﻥـﻟﻭ ،ﺔـﻴﻬﺘ
p
ﺎﹰـﻴﻟﻭﺃ ﹰﺍﺩﺩـﻋ
.
ﺎﻥـﻜ ﺍﺫﺇ
(
)
:1
r
p
G
ﻓﺈﻥﺫﺌﺩﻨﻋ ،
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
r
p
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ
(17.4)
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺃﻥ،
r
p
ﺃﻜﺒﺭ ﻗﻭﻯ
p
ﺘﻘﺴﻡ
(
)
:1
G
ﻥـﻤ ﻭ ،
ﺍﻵﻥ ﻓﺼﺎﻋﺩﺍﹰ ﺴﻨﻔﺭﺽ ﺒﺄﻥ
(
)
:1
r
G
p m
=
ﺤﻴﺙ
m
ﻻ ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
ﻟﺘﻜﻥ
}
ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻤﻊ ﺍﻟﻌﻨﺎﺼﺭ
r
p
{
X
=
ﻤﻊ ﺘﺄﺜﻴﺭ
G
ﺍﻟﻤﻌﺭﻑ ﺒﺎﻟﺸﻜل
(
)
{
}
def
,
,
G
X
X
g A
gA
ga a
A
×
→
=
∈
a
ﻷﻱ
0
0
,
:
a
A h
ha
H
A
∈
→
a
ﺎﻤﺭﺍﹰــﻏ ﻥﻭــﻜﻴ
)
ﺯﺍلــﺘﺨﻻﺍ ﻥﻭﻨﺎــﻗ
(
ﺫﻟﻙــﻟﻭ ،
(
)
:1
r
H
A
p
≤ =
.
ﻓﻲ
ﺍﻟﻤﻌﺎﺩﻟﺔ
(
)
(
)
:1
:1
≤
=
r
r
H
p
، G
p m
ﻭ،
ﺃﻥ
(
)
:
G H
ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ ﻤﺩﺍﺭ
A
.
ﺎـﻨﻨﻜﻤﺃ ﺍﺫﺇ
ﺇﻴﺠﺎﺩ
A
ﺒﺤﻴﺙ ﻻ ﻴﻘﺴﻡ
p
ﻴﻤﻜﻥ ﺃﻥ ﻨﺴﺘﻨﺘﺞ ﺒﺄﻥﺫﺌﺩﻨﻋ ،ﻩﺭﺍﺩﻤ ﻲﻓ ﺭﺼﺎﻨﻌﻟﺍ ﺩﺩﻋ
)
ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ
ﻤﺠﻤﻭﻋﺔ ﻤﺜل
A
(
،
Stab
H
A
=
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
r
p
.
ﺇﻥ ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ
X
ﻫﻲ
( )(
) (
) (
)
(
) (
) (
)
1 ...
...
1
1 ...
...
1
r
r
r
r
r
r
r
r
r
r
r
r
p m
p m
p m
p m
i
p m
p
X
p
p
p
i
p
p
p
−
−
−
+
=
=
−
−
−
+
ﻭﻷﻥ،ﻥﺄﺒ ﻅﺤﻼﻨ
r
i
p
<
ﻗﻭﻯ،
p
ﺍﻟﺘﻲ ﺘﻘﺴﻡ
r
p m
i
−
ﻫﻲ ﻗﻭﻯ
p
ﺍﻟﺘﻲ ﺘﻘﺴﻡ
i
.
ﻨﻔﺱ
ﺍﻟﺤﺎﻟﺔ ﺘﻜﻭﻥ ﺼﺤﻴﺤﺔ ﻤﻥ ﺃﺠل
r
p
i
−
.
ﻟﺫ
ﻰ ﻭـﻠﻋﻷﺍ ﻲﻓ ﺔﻠﺒﺎﻘﺘﻤﻟﺍ ﺩﻭﺩﺤﻟﺍ ﻙﻟ
ﺔـﻠﺒﺎﻗ لﻔـﺴﻷﺍ
ﻟﻠﻘﺴﻤﺔ ﻋﻠﻰ ﺍﻟﻘﻭﻯ ﻨﻔﺴ
ـﻬﺎ ﻟ
p
ﻭﺒﻬﺫﺍ،
p
ﻻ ﻴﻘﺴﻡ
X
.
ـﻷﻥ ﺍﻟﻤﺩﺍﺭ ﻴﺸﻜل ﺘﺠﺯﺌﺔ ﻟ
X
،
ﻭﻤﻨﻪ ﻓﺈﻥ ﺃﺤﺩ
i
O
ﻋﻠﻰ ﺍﻷﻗل ﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
ﻤﺜﺎل
3.5
ﻴﻜﻥـﻟ
p
p
=
Z
Z
F
ﻊـﻤ لـﻘﺤﻟﺍ ،
p
ﻴﻜﻥـﻟﻭ ،ﺭﺼـﻨﻋ
( )
GL
n
p
G
=
F
.
ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﻨﻭﻉ
n n
×
ﻓﻲ
G
،
ﻥـﻤ ﺎﻬﺘﺩﻤﻋﺃ ﺭﺼﺎﻨﻋ ﺫﺨﺄﺘ ﻲﺘﻟﺍ ﻙﻠﺘ ﺭﺜﻜﺃ ﺔﻗﺩﺒ ﻲﻫﻭ
ﻗﺎﻋﺩﺓ
n
p
F
.
ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﻌﻤﻭﺩ ﺍﻷﻭل ﺃﻱ ﻤﺘﺠﻪ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﻓﻲ،ﺍﺫﻬﻟ
n
p
F
ﻴﻭﺠﺩ،
1
n
p
−
i
O
ﻤﺩﺍﺭﺍﺕ ﻤﺨﺘﻠﻔﺔ
,
i
X
O
=
∑
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ﺩـﺠﻭﻴ ،لﻭﻷﺍ ﻪﺠﺘﻤﻟﺍ لﺎﺠﻤ ﻲﻓ ﻊﻘﻴ ﻻ ﻪﺠﺘﻤ ﻱﺃ ﻲﻨﺎﺜﻟﺍ ﺩﻭﻤﻌﻟﺍ ﻥﻭﻜﻴ ﻥﺃ ﻥﻜﻤﻴ ،ﺎﻬﻨﻤ
n
p
p
−
ﻭ ﻫ،ﺎﻬﻨﻤ
ﻜﺫﺍ
.
ﺇﻥ ﺭﺘﺒﺔ،ﻙﻟﺫﻟ
G
ﺘﺴﺎﻭﻱ
(
)(
)(
) (
)
2
1
1
...
n
n
n
n
n
p
p
p
p
p
p
p
−
−
−
−
−
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻗﻭﻯ
p
ﺍﻟﺘﻲ ﺘﻘﺴﻡ
(
)
:1
G
ﻫﻲ
(
)
1 2 ...
1
n
p
+ + + −
.
ﻨﻌﺘﺒﺭ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﺸﻜل
1
0
1
0
0
0
0
1
∗
∗
∗
∗
L
L
L
M
M L M
L
ﺇﻨﻬﺎ ﺘﺸﻜ
ل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
U
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
2
...
n
n
p
p
p
−
−
ﺍﻟﺘﻲ ﺘﻜﻭﻥ،
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
G
.
ﻤﻼﺤﻅﺔ
4.5
ﺘﻌﻁﻲ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺒﺭﻫﺎﻨﺎﹰ ﺁﺨﺭ ﻟﻤﺒﺭﻫﻨﺔ ﻜﻭﺸﻲ
(13.4)
.
ﺇﺫﺍ ﻜﺎﻥ
p
ﺎﹰـﻴﻟﻭﺃ ﹰﺍﺩﺩـﻋ
ﻴﻘﺴﻡ
(
)
:1
G
ﺘﺤﻭﻱـﺴ ﺫﺌﺩﻨﻋ ،
G
ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ﻰـﻠﻋ
H
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
p
ﻭ ﺃﻱ،
,
1
g
H g
∈
≠
ﻫﻭ ﻋﻨﺼﺭ ﻤﻥ،
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﻤﻼﺤﻅﺔ
5.5
ﻴﻤﻜﻥ ﺘﻌﺩﻴل ﺒﺭﻫﺎﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
5.2
ﻭﺓـﻗ لﻜﻟ ﻥﻭﻜﻴ ﻥﺄﺒ ﺭﺸﺎﺒﻤ لﻜﺸﺒ ﺭﻬﻅﻴﻟ
r
p
ﻟﻠﻌﺩﺩ
p
ﺍﻟﺘﻲ ﺘﻘﺴﻡ
(
)
:1
G
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
H
ﻓﻲ
G
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
r
p
.
ﺭﻯـﺨﺃ ﺓﺭـﻤ
ﻨﻜﺘﺏ
(
)
:1
r
G
p m
=
ﻭ ﻨﻌﺘﺒﺭ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻥـﻤ ﺔﻴﺌﺯﺠﻟﺍ ﺕﺎﻋﻭﻤﺠﻤﻟﺍ ﻊﻴﻤﺠ ﻥﻤ ﺔﻔﻟﺅﻤﻟﺍ
ﺍﻟﺭﺘﺒﺔ
r
p
.
،ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
0
r
p
ﺃﻜﺒﺭ ﻗﻭﺓ ﻟﻠﻌﺩﺩ
p
ـ ﻭﺍﻟﻘﺎﺴﻡ ﻟ
X
ﻭﻯـﻗ ﺭـﺒﻜﺃ ﻭﻫ
p
ﻴﻘﺴﻡ
m
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻭﺠﺩ ﻤﺩﺍﺭ ﻓﻲ،
X
ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﺭﺘﺒﺘﻪ ﻻ ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
0 1
r
p
+
.
ﻥـﻤ
ﺃﺠل
A
ﻴﺒﻴﻥ ﺍﻟﺤﺴﺎﺏ ﻨﻔﺴﻪ ﺒﺄﻥ، ﺭﺍﺩﻤﻟﺍ ﻙﻟﺫ ﻲﻓ
( )
Stab A
ﻴﺤﻭﻱ
r
p
ﺼﺭـﻨﻋ
.
ﺼﺢـﻨﻨ
ﺍﻟﻘﺎﺭﺉ ﺒﺄﻥ ﻴﻜﺘﺏ ﺍﻟﺘﻔﺎﺼﻴل
.
ﻤﺒﺭﻫﻨﺔ
6.5
)
ﺴﻴﻠﻭ
II
(
ﻟﺘﻜﻥ
G
، ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
ﺘﻜﻥـﻟﻭ
r
G
p m
=
ﺙـﻴﺤ
m
لـﺒﻘﺘ ﻻ
ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
p
.
(a)
ﺃﻱ
p
-
ﺯﻤﺭﺘﻲ
ﺴﻴﻠﻭ ﺠﺯﺌﻴﺘﻴﻥ ﺘﻜﻭﻨﺎﻥ ﻤﺘﺭﺍﻓﻘﺘﻴﻥ
.
(b)
ﻟﻴﻜﻥ
p
s
ﻋﺩﺩ
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻋﻨﺩﺌﺫ،
1mod
p
s
p
≡
ﻭ
p
s
m
.
(c)
ﻜل
p
-
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺘﻜﻭﻥ ﻤﺤﺘﻭﺍﺓ ﻓﻲ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻨﺘﺫﻜﺭ ﻤﻥ
(4.6, 4.8)
ﺒﺄﻥ ﻤﻨﻅﻡ
H
ﻓﻲ
G
ﻫﻭ
( )
{
}
1
,
G
N
H
g
G gHg
H
−
=
∈
=
ﻭ ﺇﻥ ﻋﺩﺩ ﻤﺭﺍﻓﻘﺎﺕ
H
ﻓﻲ
G
ﻫﻭ
( )
(
)
:
G
G N
H
.
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٤٩
ﺘﻤﻬﻴﺩﻴﺔ
7.5
ﻟﺘﻜﻥ
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﻟﺘﻜﻥ،
H
p
-
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
.
ﺇﺫﺍ
ﻜﺎﻥ
H
ﻤﻨﻅﻡ
P
ﺇﺫﺍ ﻜﺎﻥ،ﻥﺃ ﻱﺃ ،
( )
G
H
N
P
⊂
ﻋﻨﺩﺌﺫ،
H
P
⊂
.
ﻻ،ﺹﺎﺨ لﻜﺸﺒﻭ
ﺘﻭﺠﺩ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺘﻨﻅﻡ
P
ﻏﻴ
ﺭ ﺍﻟﺯﻤﺭﺓ
P
.
ﺎﻥـﻫﺭﺒﻟﺍ
.
ﺒﻤﺎ ﺃﻥ
P
ﻭ
H
ﺯﻤﺭ
ﺘﺎﻥ
ﺠﺯﺌﻴ
ﺘﺎﻥ
ﻤﻥ
( )
G
N
P
ﻭ
P
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
( )
G
N
P
،
ﻭﻥـﻜﻴ
HP
ﻭ،ﺔــﻴﺌﺯﺠ ﺓﺭــﻤﺯ
H H
P
HP P
I
)
ﻕـﻴﺒﻁﺘﺒ
1.45
.(
ﺈﻥــﻓ ﻙﻟﺫــﻟ
(
)
:
HP P
ﻗﻭﺓ ﻟﻠﻌﺩﺩ
p
)
ﺤﻴﺙ ﺇﻨ
ﻨﺎ ﻨﺴﺘﺨﺩﻡ ﻫﻨﺎ
H
p
-
ﺯﻤﺭﺓ
(
ﻟﻜﻥ،
(
) (
)(
)
:1
:
:1 ,
HP
HP P
P
=
ﻭ
(
)
:1
P
ـ ﻫﻲ ﺃﻜﺒﺭ ﻗﻭﺓ ﻟ
p
ﺘﻘﺴﻡ
(
)
:1
G
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺴﺘﻜﻭﻥ ﺃﻜﺒﺭ ﻗﻭﺓ ﺘﻘﺴﻡ،
(
)
:1
HP
،
ﻟﻬﺫﺍ
(
)
0
:
1
HP P
p
=
=
ﻭ،
H
P
⊂
.
ﺒﺭﻫﺎﻥ
)
ﺴﻴﻠﻭ
II
(
(a)
ﻟﺘﻜﻥ
X
ﻤﺠﻤﻭﻋﺔ
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﻟﺘﻜﻥ،
G
ﺘﺅﺜﺭ
ﻋﻠﻰ
X
، ﺒﺎﻟﺘﺭﺍﻓﻕ
(
)
1
,
:
−
×
→
a
g P
gPg
G
X
X
.
ﻟﻴﻜﻥ
O
ﺃﺤﺩ
ﻤﺩﺍﺭﺍﺕ
G
-
ﻤﺩﺍﺭ
:
ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ
O
ﻫﻭ ﻜل
X
.
ﻟﺘﻜﻥ
P
O
∈
ﻭﻟﻴﻜﻥ،
P
ﻴﺅﺜﺭ ﻋﻠﻰ
O
ﻤﻥ ﺨﻼل ﺘﺄﺜﻴﺭ
G
.
ﺇﻥ ﻫﺫﻩ
G
-
ﺩﺓـﺤﺍﻭ ﺭﺍﺩﻤ
ﻤﻥ ﺍﻟﻤﻤﻜﻥ ﺃﻥ ﺘﺘﺠﺯﺃ ﺇﻟﻰ ﺍﻟﻌﺩﻴﺩ ﻤﻥ
P
-
ﺇﺤﺩﺍﻫﺎ ﺴﺘﻜﻭﻥ،ﺭﺍﺩﻤ
{ }
P
.
ﺫﺍـﻫ ﻥﺇ ﺔـﻘﻴﻘﺤﻟﺍ ﻲﻓ
ﺍﻟﻤﺩﺍﺭ ﺍﻟﻤﺅﻟﻑ ﻤﻥ ﻨﻘﻁﺔ ﻭﺍﺤﺩﺓ ﻫﻭ ﺍﻟﻤﺩﺍﺭ ﺍﻟﻭﺤﻴﺩ ﻷﻥ
P
ﻴﻨﻅﻡ
Q
⇔
{ }
Q
ﻫﻭ
P
-
ﻤﺩﺍﺭ
ﺤﻴﺙ ﻨﻌﻠﻡ ﺒﺄﻥ
(7.5)
ﺘﺤﻘﻘﺕ ﻓﻘﻁ ﻋﻨﺩﻤﺎ
Q
P
=
.
لـﻜ ﻲﻓ ﺭﺼﺎﻨﻌﻟﺍ ﺩﺩﻋ ﻥﺈﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ
p
-
ﻤﺩﺍﺭ
ﺒﺎﺴﺘﺜﻨﺎﺀ
{ }
P
ﻗﺎﺒل ﻟﻠﻘﺴﻤﺔ ﻋﻠﻰ
p
ﻭ ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ،
1mod
O
p
≡
.
ﺒﻔﺭﺽ ﺃﻨﻪ ﻴﻭﺠﺩ
P
O
∉
.
ﻭﻟﻴﻜﻥ
P
ﻴﺅﺜﺭ ﻋﻠﻰ
O
ﻟﻜﻥ ﻫﺫﻩ ﺍﻟﻤﺭﺓ ﺍﻟﻤﻨﺎﻗﺸﺔ،ﻯﺭﺨﺃ ﹰﺓﺭﻤ
لـﻜ ﻲـﻓ ﺭﺼﺎﻨﻌﻟﺍ ﺩﺩﻋ ﻥﺈﻓ ﻙﻟﺫﻟﻭ ،ﺓﺩﺤﺍﻭ ﺔﻁﻘﻨ ﻥﻤ ﺔﻔﻟﺅﻤ ﺕﺍﺭﺍﺩﻤ ﺩﺠﻭﻴ ﻻ ﻪﻨﺄﺒ ﻥﻴﺒﺘﺴ
P
-
ﻤﺩﺍﺭ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
P
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻨﺘﺞ ﺒﺄﻨﻪ
#
O
ﻗﺎﺒل ﻟﻠﻘﺴﻤﺔ ﻋﻠﻰ
p
ﺎـﻤ ﺽﻗﺎﻨﻴ ﻱﺫﻟﺍ ،
ﺒﺭﻫﻥ ﻓﻲ ﺍﻟﻔﻘﺭﺓ ﺍﻷﺨﻴﺭﺓ
.
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﻴﻭﺠﺩ ﻜﺎﻟﺯﻤﺭﺓ
P
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،
O
ﻫﻲ ﻜل
X
.
(b)
ﺒﻤﺎ ﺃﻥ
p
s
ﻫﻭ ﺍﻵﻥ ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ
O
ﻴﻤﻜﻥ ﺃﻥ ﻨﺒﻴﻥ ﺃﻴﻀﺎﹰ ﺒﺄﻥ،
1mod
p
s
p
≡
ﻟﻴﻜﻥ
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ
(a)
ﻴﻜﻭﻥ،
p
s
ﺎﺕـﻘﻓﺍﺭﻤ ﺩﺩﻋ
P
ﻭ ﺍﻟﺘﻲ ﺘﺴﺎﻭﻱ،
( )
(
)
(
)
( )
(
)
(
)
( )
(
)
(
)
( )
(
)
:1
:1
:
:1
:
:1
:
G
G
G
G
G
G
m
G N
P
N
P
N
P
P
P
N
P
P
=
=
=
ﻫﺫﺍ ﺘﺤﻠﻴل
m
.
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٥٩
(c)
ﻟﺘﻜﻥ
H
p
-
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭ ﻟﺘﻜﻥ،
H
ﺘﺅﺜﺭ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﺍﻟﻤﺅﻟﻔﺔ
ﻤﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺒﺎﻟﺘﺭﺍﻓﻕ
.
ﻷﻥ
p
X
s
=
ﻰـﻠﻋ ﺔﻤﺴﻘﻟﺍ لﺒﻘﺘ ﻻ
p
ﺏ ﺃﻥـﺠﻴ ،
ﺘﻜﻭﻥ
H
X
ﻏﻴﺭ ﺨﺎﻟﻴﺔ
)
ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
1.5
(
ﺃﺤﺩ،ﻥﺃ ﻱﺃ ،
H
-
ﻤﺩﺍﺭ ﻋﻠﻰ ﺍﻷﻗل ﻴﺘﺄﻟﻑ ﻤﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻭﺍﺤﺩﺓ
.
ﻟﻜﻥ ﻋﻨﺩﺌﺫ
H
ﺘﻨﻅﻡ
P
ﻭﻤﻥ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
7.5
ﻴﻨﺘﺞ ﺒﺄﻥ
H
P
⊂
.
ﻨﺘﻴﺠﺔ
8.5
ﺘﻜﻭﻥ
p
-
ﺕـﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
p
-
ﻴﻠﻭـﺴ ﺓﺭـﻤﺯ
ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ
.
ﺎﻥـﻫﺭﺒﻟﺍ
.
ﻟﺘﻜﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
P
ﻓﻲ
G
.
ﺇﺫﺍ ﻜﺎﻨﺕ
P
ﻋﻨﺩﺌﺫ،ﺔﻴﻤﻅﺎﻨ
(a)
ﻥـﻤ
ﺴﻴﻠﻭ
II
ﻴﻨﺘﺞ ﺒﺄﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ
.
ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﻌﺎﻜﺴﺔ ﺘﻨﺘﺞ ﻤﻥ
(6.3c)
)
ﺤﻴﺙ
ﺒﺄﻥ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ ،ﻥﻴﺒﺘ
P
ﻤﻤﻴﺯﺓ ﺃﻴﻀﺎﹰ
.(
ﻨﺘﻴﺠﺔ
9.5
ﺒﻔﺭﺽ ﺃﻥ
G
ﺯﻤﺭﺓ ﺘﺤﻭﻱ ﻓﻘﻁ ﻋﻠﻰ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻟﻜل ﻋﺩﺩ ﺃﻭﻟﻲ ﻴﻘﺴﻡ
ﺭﺘﺒﺘﻬﺎ
.
ﻓﺈﻥﺫﺌﺩﻨﻋ
G
ـ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟ
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
1
,...,
k
P
P
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﻟﺘﻜﻥ،
i
r
i
i
P
p
=
،
i
p
ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ
ﻤﺨﺘﻠﻔﺔ
.
ﻷﻥ ﻜل ﻤﻥ
i
P
ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻴﻜﻭﻥ ﺍﻟﺠﺩﺍﺀ،
1
...
k
P
P
ﺯﻤ
ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭ
ﻓﻲ
G
.
ﺴﻨﺒﺭﻫﻥ ﺒﺎﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ
k
ﺒﺄﻨﻬﺎ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
1
...
k
r
r
k
p
p
.
ﺇﺫﺍ ﻜﺎﻥ
1
k
=
ﻋﻨﺩﻫﺎ،
ﻭﻟﺫﻟﻙ ﻴﻤﻜﻥ ﺃﻥ ﻨﻔﺭﺽ ﺒﺄﻥ،ﻪﻨﺎﻫﺭﺒﻟ ﺀﻲﺸ ﺩﺠﻭﻴ ﻻ
2
k
≥
ﻭ ﺃﻥ
1
1
...
k
P
P
−
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
1
1
1
1
...
k
r
r
k
p
p
−
−
.
ﻋﻨﺩﺌﺫ
1
1
...
1
−
=
I
k
k
P
P
P
ﻟﺫﻟﻙ ﺘﺒﻴﻥ،
(50.1)
ﺒﺄﻥ
(
)
1
1
...
k
k
P
P
P
−
ﻭـﻫ
ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ
1
1
...
k
P
P
−
ﻭ
k
P
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﻤ،
ﺔـﺒﺘﺭﻟﺍ ﻥ
1
1
...
k
r
r
k
p
p
.
ﻕـﻴﺒﻁﺘﺒ
ﺍﻵﻥ
(51.1)
ﻋﻠﻰ ﻜﺎﻤل ﻤﺠﻤﻭﻋﺔ ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺄﻥـﺒ ﻥﻴﺒﺘﻴ
G
ﺩﺍﺅﻫﺎـﺠ ﻲـﻫ
ﺍﻟﻤﺒﺎﺸﺭ
.
ﻤﺜﺎل
10.5
ﻟﺘﻜﻥ
( )
GL
G
V
=
ﺤﻴﺙ
V
ﻓﻀﺎ
ﺀ ﻤﺘﺠﻬﻲ ﻤﻥ ﺍﻟﺒﻌﺩ
n
ﻰـﻠﻋ
p
F
.
ﺩـﺠﻭﻴ
ﻭﺼﻑ ﻫﻨﺩﺴﻲ ﻟﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺇﻥ ﺍﻟﻤﻌﻠﻡ ﺍﻟﺘﺎﻡ
(full flag)
F
ﻓﻲ
V
ﻫﻭ ﻤﺘﺘﺎﻟﻴﺔ
ﻤﻥ ﺍﻟﻔﻀﺎﺀﺍﺕ ﺍﻟﺠﺯﺌ
ﻴﺔ
{ }
1
1
...
...
0
n
n
i
V
V
V
V
V
−
=
⊃
⊃ ⊃
⊃ ⊃
⊃
ﺤﻴﺙ
( )
dim
i
V
i
=
.
ﺒﺈﻋﻁﺎﺀ ﻤﺜل ﺍﻟﻤﻌﻠﻡ
F
ﻟﺘﻜﻥ،
( )
U F
ﺎﺕـﻘﻴﺒﻁﺘﻟﺍ لـﻜ ﺔـﻋﻭﻤﺠﻤ
ﺍﻟﺨﻁﻴﺔ
:V
V
α
→
ﺒ
ﺤﻴﺙ ﺇﻥ
(a)
( )
i
i
V
V
α
⊂
ﻟﻜل
i
ﻭ،
(b)
ﺍﻟﺘﺸﺎﻜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻋﻠﻰ
1
i
i
V V
−
ـ ﺘﻨﺘﺞ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻤﺤﺎﻴﺩ ﺒﺎﻟﻨﺴﺒﺔ ﻟ
α
.
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٦٩
ﺃﺩﻋﻲ ﺒﺄﻥ
( )
U F
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺩﺓـﻋﺎﻘﻟﺍ ﻲﻨﺒﻨ ﻥﺃ ﻥﻜﻤﻴ
{
}
1
,...,
n
e
e
ـﻟ
V
ﺒﺤﻴﺙ
{ }
1
e
ﻗﺎﻋﺩﺓ
1
V
،
{
}
1
2
,
e e
ﻗﺎﻋﺩﺓ
2
V
ﻭﻫﻜﺫﺍ،
.
ﺇﻥ،ﺩـﻋﺍﻭﻘﻟﺍ ﻩﺫـﻬﻟ ﺔﺒﺴـﻨﻟﺎﺒ
ﻤﺼﻔﻭﻓﺎﺕ ﻋﻨﺎﺼﺭ
( )
U F
ﻫﻲ ﺒﺎﻟﻀﺒﻁ ﻋﻨﺎﺼﺭ ﺍﻟﺯﻤﺭﺓ
U
ﻓﻲ
(5.3)
.
ﻟﺘﻜﻥ
( )
GL
n
g
∈
F
.
{
}
1
,
,...
def
n
n
gF
gV gV
−
=
ﻭ،ﹰﺎﻀـﻴﺃ ﻡﺎـﺘ ﻡـﻠﻌﻤ ﻭﻫ
ﻴﻜﻥـﻟ
( )
( )
1
.
.
U gF
g U F
g
−
=
.
ﻤﻥ
(a)
ﻓﻲ ﺴﻴﻠﻭ
II
ﻨﺭﻯ ﺒﺄﻥ،
p
-
ﻲـﻓ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺭﻤﺯ
G
.
ﻫﻲ ﻭﺒﺩﻗﺔ ﺯﻤﺭ ﻤﻥ ﺍﻟﺸﻜل
( )
U F
ﻟﺒﻌﺽﹴ
ﻤﻥ ﺍﻟﻤﻌﻠﻡ ﺍﻟﺘﺎﻡ
F
.
11.5
ﺘﺴﺘﺨﺩﻡ ﺒﻌﺽ ﺍﻟﻜﺘﺏ ﺘﺭﻗﻴﻤﺎﺕ ﻤﺨﺘﻠﻔﺔ
ﻟﺯﻤﺭ ﺴﻴﻠﻭ
.
ﻟﻘﺩ ﺍﻋﺘﻤﺩﺕ ﻋﻠﻰ ﺍﻟﺘﺭﻗﻴﻤﺎﺕ ﺍﻷﺼﻠﻴﺔ
)
ﺴﻴﻠﻭ
1872
(
ﺒﺸﻜل ﺃﺴﺎﺴﻲ
.
ﺍﻟﺘﻘﺭﻴﺒﺎﺕ ﺍﻟﺒﺩﻴﻠﺔ ﻟﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
ﻨﻨﺴﻰ
ﺒ
ﺄﻨﻨﺎ ﺒﺭﻫﻨﺎ ﻋﻠﻰ ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
.
ﻤﺒﺭﻫﻨﺔ
12.5
ﻟﺘﻜﻥ
G
ﻭﻟﺘﻜﻥ،ﺓﺭﻤﺯ
P
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻟﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴ
ﺔ
H
ﻓﻲ
G
ﻴﻭﺠﺩ،
a
G
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
1
H
a Pa
−
I
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
H
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺘﺫﻜﺭ ﻤﻥ
)
ﻤﻥ ﺍﻟﺘﻤﺭﻴﻥ
4-16
(
ﺒﺄﻥ
G
ﺔـﻴﺌﺎﻨﺜ ﺕﺎﻘﻓﺍﺭﻤﻟ لﺼﻔﻨﻤ ﻉﺎﻤﺘﺠﺍ ﻥﻋ ﺓﺭﺎﺒﻋ
ـﺍﻟﺠﺎﻨﺏ ﻟ
H
ﻭ
P
.
ﻭﻟﺫﻟﻙ
1
a
a
a
H P
G
H P
H
a Pa
−
=
=
∑
∑
I
ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﻤﺠﻤﻭﻉ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﻤﻤﺜﻼﺕ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﺜﻨﺎﺌﻴﺔ ﺍﻟﺠﺎﻨﺏ
.
ﻰـﻠﻋ ﺔﻤﺴﻘﻟﺎﺒ
P
ﻨﺠﺩ ﺒﺄﻥ
1
a
G
H
P
H
a Pa
−
=
∑
I
ﻭ ﻟﺫﻟﻙ ﻴﻭﺠﺩ
a
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
(
)
1
:
H H
a Pa
−
I
ﻰـﻠﻋ ﺔﻤﺴﻘﻟﺍ لﺒﻘﺘ ﻻ
p
.
لـﺠﺃ ﻥـﻤ
ﻋﻨﺼﺭ ﻤﺜل
a
ﺘﻜﻭﻥ،
1
H
a Pa
−
I
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺒﺭﻫﺎﻥ
)
ﺴﻴﻠﻭ
I
(
ﻤﻥ ﻤﺒﺭﻫﻨﺔ ﻜﺎﻴﻠﻲ
(21.1)
ﺘﻜﻭﻥ،
G
ﻤﻐﻤﻭﺭﺓ ﻓﻲ
n
S
ﻭ،
n
S
ﻤﻐﻤﻭﺭﺓ ﻓﻲ
( )
GL
n
p
F
)
ﺍﻨﻅﺭ
1.7
.(
ﻜﻤﺎ ﺃﻥ
( )
GL
n
p
F
ﻴﺤﻭﻱ ﻋﻠﻰ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
)
ﺍﻨﻅﺭ
5.3
(
ﻭ ﻜﺫﻟﻙ ﺃﻴﻀﺎﹰ،
G
.
ﺒﺭﻫﺎﻥ
)
ﺴﻴﻠﻭ
II(a,c)
(
ﻟﺘﻜﻥ
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
ﻓﻲ
G
ﻭﻟﺘﻜﻥ،
P
′
p
-
ﺭﺓـﻤﺯ
ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻓﻲ
G
ﻋﻨﺩﺌﺫ
P
′
ﺘﻜﻭﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ
P
′
ﻭﻟﺫﻟﻙ ﻓﺈﻥ،
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٧٩
ﺍﻟﻤﺒﺭﻫﻨﺔ ﺘﺒﻴﻥ ﺒﺄﻥ
1
aPa
P
−
′
⊃
ﻟﺒﻌﺽ
a
ﺤﻴﺙ
H
P
′
=
.
ﻭﻫﺫﺍ ﻴﻨﺘﺞ
(a)
ﻭ
(c)
ﻤﻥ ﺴﻴﻠﻭ
II
.
ﺃﻤﺜﻠﺔ
ﻨﻁﺒﻕ ﻤﺎ ﺘﻌﻠﻤﻨﺎﻩ ﻟﻨﺤﺼل ﻋﻠﻰ ﻤﻌﻠﻭﻤﺎﺕ ﺠﺩﻴﺩﺓ ﻋﻥ ﺍﻟﺯﻤﺭ ﺒﺭﺘﺏ ﻤﺨﺘﻠﻔﺔ
.
13.5
)
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘ
ﺒﺔ
99
(
ﻟﺘﻜﻥ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
99
.
ﺄﻥـﺒ ﺞﺘﻨﺘﺴـﻨ ﻭﻠﻴﺴ ﺕﺎﻨﻫﺭﺒﻤ ﻥﻤ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺍﺤﺩﺓ
H
ﻋﻠﻰ ﺍﻷﻗل ﻤﻥ ﺍﻟﺭﺘﺒﺔ
11
ﺔـﻘﻴﻘﺤﻟﺍ ﻲـﻓ ﻭ ،
11
99
11
s
ﻭ
11
1mod11
s
≡
.
ﺈﻥـﻓ ﻲﻟﺎﺘﻟﺎﺒ ﻭ
11
1
s
=
ﻭ،
H
ﺔـﻴﻤﻅﺎﻨ
.
،ﺸﺎﺒﻪـﻤ لﻜﺸـﺒ
9
11
s
ﻭ
9
1mod 3
s
≡
ﻭﻤﻨﻪ،
3
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺘﻜﻭﻥ ﺃﻴﻀﺎﹰ ﻨﺎﻅﻤﻴﺔ
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
G
ﺔـﻠﺜﺎﻤﺘﻤ
ﻤﻊ ﺍﻟﺠ
ـﺩﺍﺀ ﺍﻟﻤﺒﺎﺸﺭ ﻟ
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻟﻬﺎ
(9.5)
ﺔـﻴﻠﻴﺩﺒﺘ ﺎﻤﻬﻨﻤ لﻜ ﻲﺘﻟﺍﻭ ،
(18.4)
،
ﻭﻤﻨﻪ ﻓﺈﻥ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﻫﻨﺎ ﺒﺭﻫﺎﻥ ﺒﺩﻴل
.
ﻨﺜﺒﺕ ﻜﻤﺎ ﺴﺒﻕ ﺒﺄﻥ
11
-
ﺔـﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
N
ﻲـﻓ
G
ﻭﻥـﻜﺘ
ﻨﺎﻅﻤﻴﺔ
.
ﺇﻥ
3
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
Q
ﻋﺒﺎﺭﺓ ﻋﻥ ﺘﻁﺒﻴﻘﺎﺕ ﺘﻘﺎﺒل ﻋﻠﻰ
G N
ﺎﻟﻲـﺘﻟﺎﺒ ﻭ ،
G
N
Q
= ×
.
ﺒﻘﻲ ﺃﻥ ﻨﺤﺩﺩ ﺘﺄﺜﻴﺭ
Q
ﻋﻠﻰ
N
ﺒﺎﻟﺘﺭ
ﺍﻓﻕ
.
ﻥـﻜﻟ
( )
Aut N
ﻥـﻤ ﺔـﻴﺭﺌﺍﺩ
ﺍﻟﺭﺘﺒﺔ
10
)
ﺍ
ﻨﻅﺭ
3.4
(
ﻭ ﺒﻬﺫﺍ ﻴﻭﺠﺩ ﺘﺸﺎﻜل ﺘﺎﻓﻪ ﻭﺍﺤﺩ ﻓﻘﻁ،
( )
Aut
Q
N
→
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
G
ـ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟ
N
ﻭ
Q
.
14.5
)
ﺍﻟﺯﻤﺭ
ﻤﻥ ﺍﻟﺭﺘﺏ
pq
،
p
،
q
، ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ
p
q
<
(
ﻟﺘﻜﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺘﻜﻥـﻟ ﻭ ،
P
ﻭ
Q
p
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻭ
q
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
..
ﻋﻨﺩﺌﺫ
(
)
:
G Q
p
=
ﺍﻟﺫﻱ
ﻴﻜﻭﻥ ﺃﺼﻐﺭ ﻋﺩﺩ ﺃﻭﻟﻲ ﻴﻘﺴﻡ
(
)
:1
G
ﻭﻤﻨﻪ،
)
ﺍﻨﻅﺭ ﺍﻟﺘﻤﺭﻴﻥ
4-4
(
ﻓﺈﻥ
Q
ﻨﺎﻅﻤﻴﺔ
.
ﻷﻥ
P
ﺘﻁﺒﻴﻕ ﺘﻘﺎﺒل ﻋﻠﻰ
G Q
ﻭﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ،
,
θ
= ×
G
Q
P
ﻭﺒﻘﻲ ﺃﻥ ﻨﺤﺩﺩ ﺘﺄﺜﻴﺭ
P
ﻋﻠﻰ
Q
ﺒﺎﻟﺘﺭﺍﻓﻕ
.
ﺇﻥ ﺍﻟﺯﻤﺭﺓ
( )
Aut Q
ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
q
−
)
ﺍﻨﻅﺭ
3.4
(
ﺒﺎﺴﺘﺜﻨﺎﺀ،ﻪﻨﻤﻭ ،
1
p q
−
،
G
Q
P
= ×
.
ﺇﺫﺍ ﻜﺎﻥ
1
p q
−
ﻓﺈﻥﺫﺌﺩﻨﻋ ،
( )
Aut Q
)
ﻜﻭﻨﻬﺎ ﺩﺍﺌﺭﻴﺔ
(
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺤﻴﺩﺓ
P
′
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
.
ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﺘﺘﺄﻟﻑ
P
′
ﻤﻥ ﺍﻟﺘﻁﺒﻴﻘﺎﺕ
{
}
,
1
i
p
x
x
i
q
i
∈
=
a
Z
Z
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٨٩
ﻟﺘﻜﻥ
a
ﻭ
b
ﻤﻭﻟﺩﺍﺕ ﻟﻠﺯﻤﺭﺘﻴﻥ
P
ﻭ
Q
ﺄﺜﻴﺭـﺘ ﻥﺃ ﺽﺭﻔﺒﻭ ،ﺏﻴﺘﺭﺘﻟﺍ ﻰﻠﻋ
a
ﻰـﻠﻋ
Q
ﺒﺎﻟﺘﺭﺍﻓﻕ ﻫﻭ
0
0
,
1
i
x
x
i
≠
a
)
ﻓﻲ
q
Z
Z
.(
ﻠﺯﻤﺭﺓـﻟ ﻥﺈﻓ ﺫﺌﺩﻨﻋ
G
ﺍﻟﻤﻭﻟﺩﺍﺕ
a
ﻭ
b
ﻭ
ﺍﻟﻌﻼﻗﺎﺕ
0
1
,
,
i
p
q
a
b
aba
b
−
=
.
ﺒﺎﺨﺘﻴﺎﺭ
0
i
ﻤﺨﺘﻠﻑ ﻴﺘﻭﺍﻓﻕ ﻤﻊ ﺍﺨﺘﻴﺎﺭ ﻤﻭﻟﺩ ﻤﺨﺘﻠﻑ
a
ـ ﻟ
P
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻌﻁﻲ ﺯﻤﺭﺓ ﻤﺘﻤﺎﺜﻠﺔ،
ﻤﻊ
G
.
ﻭﺍﻟﺨﻼﺼﺔ
:
ﺇﺫﺍ ﻜﺎﻥ
p
ﻻ ﺘﻘﺴﻡ
1
q
−
ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔﺫﺌﺩﻨﻋ ،
pq
ﻲـﻫ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺩﺍﺌﺭﻴﺔ
pq
C
ﺇﺫﺍ ﻜﺎﻥ،
1
p q
−
ﺘﻭﺠﺩ ﺃﻴﻀﺎﹰ ﺯﺫﺌﺩﻨﻋ ،
ﻰـﻁﻌﺘ ﺔـﻴﻠﻴﺩﺒﺘ ﺭـﻴﻏ ﺓﺭﻤ
ﺒﺎﻟﻤﻭﻟﺩﺍﺕ ﻭﺍﻟﻌﻼﻗﺎﺕ ﻜﻤﺎ ﺴﺒﻕ
.
15.5
)
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
(
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
.
ﻋﻨﺩﺌﺫ
3
1, 4, 7,10,...
s
=
ﻭﺘﻘﺴﻡ
10
،
5
1, 6,11,...
s
=
ﻭﺘﻘﺴﻡ
6
.
ﻭﺒﺎﻟﺘﺎﻟﻲ
3
1
s
=
ﺃﻭ
10
ﻭ،
5
1
s
=
ﺃﻭ
6
.
ﺃﺤﺩﻫﺎ ﻴﺴﺎﻭﻱ،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
1
ﻭﻤﻥ،لﻗﻷﺍ ﻰﻠﻋ
ﺠﻬﺔ ﺃﺨﺭﻯ ﻨﻜﻭﻥ ﻗﺩ ﺃﻭﺠﺩﻨﺎ
20
ﻤﻥ ﺍﻟﺭﺘﺒﺔًﹰﺍﺭﺼﻨﻋ
3
ﻭ
24
ﻋﻨﺼﺭﺍﹰ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
5
ﻭﺍﻟﺫﻱ،
ﻴﻜﻭﻥ ﻤﺴﺘﺤﻴﻼﹰ
.
،ﻟﺫﻟﻙ
3
–
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
P
ﺃﻭ
5
–
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
Q
ﺘﻜﻭﻥ
ﻭﻤﻨﻪ،ﺔﻴﻤﻅﺎﻨ
H
PQ
=
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﻷﻥ
3
ﻻ ﻴﻘﺴﻡ
5 1
4
− =
ﺘﺒﻴﻥ،
(5.14)
ﺒﺄﻥ
H
، ﺘﺒﺩﻴﻠﻴﺔ
3
5
H
C
C
≈
×
.
ﻭﺒﺎﻟﺘﺎﻟﻲ
(
)
3
5
2
,
G
C
C
C
θ
=
×
×
ﻭﺒﻘﻲ ﺃﻥ ﻨﺤﺩﺩ ﺍﻟﺘﺸﺎﻜﻼﺕ ﺍﻟﻤﺤﺘﻤﻠﺔ
(
)
2
3
5
:
Aut
C
C
C
θ
→
×
.
ﺸﺎﻜلـﺘﻟﺍ ﻥـﻜﻟ
θ
ﻭﻥـﻜﻴ
ﻤﺤﺩﺩﺍﹰ ﺒﺼﻭﺭﺓ ﺍﻟﻌﻨﺼﺭ ﺍﻟﺫﻱ ﻻ ﻴﺴﺎﻭﻱ ﺍﻟﻤﺤﺎﻴﺩ ﻓﻲ
2
C
ﻭﺍﻟﺫﻱ ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻋﻨﺼﺭ،
ﺍﹰ
ﻥـﻤ
ﺍﻟﺭﺘﺒﺔ
2
.
ﻟﺘﻜﻥ
, ,
a b c
ﺘﻭﻟﺩ
3
5
2
,
,
C C C
.
ﻋﻨﺩﺌﺫ
(
)
( )
( )
3
5
3
5
Aut
Aut
Aut
,
C
C
C
C
×
=
×
ﻭ ﻋﻨﺎﺼﺭ
( )
3
Aut C
ﻭ
( )
5
Aut C
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻫﻲ ﻓﻘﻁ
1
a
a
−
a
ﻭ
1
b
b
−
a
.
ﻟﻬﺫﺍ
ﻴﻭﺠﺩ ﺒﺎﻟﻀﺒﻁ
4
ﺘﺸﺎﻜﻼﺕ
θ
ﻭ،
( )
c
θ
ﻫﻭ ﺃﺤﺩ ﺍﻟﻌﻨﺎﺼﺭ
ﺍﻵﺘﻴ
ﺔ:
1
1
1
1
a
a
a
a
a
a
a
a
b
b
b
b
b
b
b
b
−
−
−
−
a
a
a
a
a
a
a
a
.
ﻭ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻘﺎﺒﻠﺔ ﻟﻬﺫﻩ ﺍﻟﺘﺸﺎﻜﻼﺕ ﻟﻬﺎ ﻤﺭﺍﻜﺯ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
،
3
)
ـﻤﻭﻟﺩﺓ ﺒ
a
(
،
5
)
ــﻤﻭﻟﺩﺓ ﺒ
b
(
ﻭ،
1
ﻋ
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ ﻏﻴ،ﺏﻴﺘﺭﺘﻟﺍ ﻰﻠ
ﺭ ﻤﺘﻤﺎﺜﻠﺔ
.
ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻨﻪ
)
ﺘﺤﺕ ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل
(
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٩٩
ﻴﻭﺠﺩ ﺒﺎﻟﻀﺒﻁ
4
ﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
.
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺜﺎﻟﺜﺔ ﻓﻲ ﻗﺎﺌﻤﺘﻨﺎ ﻟﻬﺎ ﺍﻟﻤﻭﻟﺩﺍﺕ،ﹰﻼﺜﻤ
, ,
a b c
ﻭ
ﺍﻟﻌﻼﻗﺎﺕ
3
5
2
1
1
1
,
,
,
,
,
a
b
c
ab
ba cac
a
cbc
b
−
−
−
=
=
=
16.5
)
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
(
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
ﺘﻜﻥـﻟﻭ ،
P
-3
ﻴﻠﻭـﺴ ﺓﺭـﻤﺯ
ﺍﻟﺠﺯﺌﻴﺔ ﻟﻬﺎ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
P
ﻻ ﺘﺤﻭﻱﺫﺌﺩﻨﻋ ،ﺔﻴﻤﻅﺎﻨ
P
ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻏﻴﺭ ﺘﺎﻓﻬﺔ
ﻓﻲ
G
،
ﻭﻤﻨﻪ ﺍﻟﺘﻁﺒﻴﻕ
)
4.2
،
ﺍﻟﺘﺄﺜﻴﺭ ﻋﻠﻰ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﺴﺎﺭﻴﺔ
(
(
)
4
:
Sym
G
G P
S
ϕ
→
≈
ﻭﺼﻭﺭﺘﻪ ﻋﺒﺎﺭﺓ ﻋﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ،ﻥﻴﺎﺒﺘﻤ
4
S
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
.
ﻤﻥ ﺴﻴﻠﻭ
II
ﻨﺭﻯ ﺒﺄﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ
4
ﻤﻥ
-3
ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﺤﻭﻱ،ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
ﺒﺎﻟﻀﺒﻁ ﻋﻠﻰ
8
ﻋﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
.
ﻟﻜﻥ ﺠﻤﻴﻊ ﻋﻨﺎﺼﺭ
4
S
ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
ﻤﺤﺘﻭﺍﺓ ﻓﻲ
4
A
)
ﺍ
ﻲـﻓ لﻭﺩـﺠﻟﺍ ﻰﻟﺇ ﺭﻅﻨ
4.31
(
ﻭﻤﻨﻪ ﻓﺈﻥ،
( )
G
ϕ
ﺘﺘﻘﺎﻁﻊ ﻤﻊ
4
A
ﺒﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺘﺤﻭﻱ ﻋﻠ
ﻰ ﺍﻷﻗل ﻋﻠﻰ
8
ﺭـﺼﺎﻨﻋ
.
ﻤﻥ ﻤﺒﺭﻫﻨﺔ ﻻﻏﺭﺍﻨﺞ ﻨﺠﺩ
( )
4
G
A
ϕ
=
ﻭﻤﻨﻪ،
4
G
A
≈
.
ﻨﻔﺭﺽ ﺍﻵﻥ ﺒﺄﻥ
P
ﻨﺎﻅﻤﻴﺔ
.
ﻋﻨﺩﺌﺫ
G
P Q
= ×
ﺤﻴﺙ
Q
4
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
.
ﺇﺫﺍ
ــﻜﺎﻨ
ﺕ
Q
ﺔــﺒﺘﺭﻟﺍ ﻥــﻤ ﺔــﻴﺭﺌﺍﺩ
4
ﻪــﻓﺎﺘ ﺭــﻴﻏ ﺩــﻴﺤﻭ ﻕــﻴﺒﻁﺘ ﺩــﺠﻭﻴ ﺫــﺌﺩﻨﻋ ،
(
)
( )(
)
4
2
Aut
Q
C
P
C
=
→
=
ﺩﺓـﻴﺤﻭ ﺔـﻴﻠﻴﺩﺒﺘ ﺭـﻴﻏ ﺓﺭـﻤﺯ ﻰـﻠﻋ لﺼﺤﻨ ﻲﻟﺎﺘﻟﺎﺒﻭ ،
3
4
C
C
×
.
ﺎﻥــﻜ ﺍﺫﺇ
2
2
Q
C
C
=
×
ﺎﹰــﻤﺎﻤﺘ ﺩــﺠﻭﻴ ،
3
ﺔــﻬﻓﺎﺘ ﺭــﻴﻏ ﺕﻼﻜﺎﺸــﺘ
( )
:
Aut
Q
P
θ
→
ﻟﻜﻥ ﺜﻼﺜﺎﹰ،
ﻊـﻤ ﺔـﻠﺜﺎﻤﺘﻤ ﻥﻭـﻜﺘ ﺔـﺠﺘﺎﻨﻟﺍ ﺭﻤﺯﻟﺍ ﻥﻤ
3
2
S
C
×
ــ ﺒ
2
Ker
C
θ
=
) .
ـﺍﻟﺘﺸﺎﻜﻼﺕ ﺘﺨﺘﻠﻑ ﺒﺎﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟ
Q
ﻀﺎﹰـﻴﺃ ﻕﺒﻁﻨ ﻥﺃ ﻥﻜﻤﻴ ﻲﻟﺎﺘﻟﺎﺒﻭ ،
ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
17.3
.(
ﺘﻭ،ﻲﻟﺎﺘﻟﺎﺒﻭ
ﺠﺩ
3
ﺯﻤﺭ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
12
ﻭ ﺯﻤﺭﺘﺎﻥ ﺘﺒﺩﻴﻠﻴﺘﺎﻥ
.
17.5
)
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
(
ﻟﺘﻜﻥ
G
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺓﺭﻤﺯ
3
p
،
p
،ﺭﺩﻱـﻓ ﻲـﻟﻭﺃ ﺩﺩـﻋ
ﻭﻟﻨﻔﺭﺽ ﺒﺄﻥ
G
ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻨﻌﻠﻡ ﻤﻥ
(17.4)
ﺒﺄﻥ
G
ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ
N
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
.
ﺇﺫﺍ ﻜﺎﻥ ﻜل ﻋﻨﺼﺭ ﻓﻲ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
)
ﻤﺎ ﻋ
ﺩﺍ
1
(
ﻋﻨﺩﺌﺫ،
p
p
N
C
C
≈
×
ﺩـﺠﻭﻴ ﻭ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
Q
ﻓﻲ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
{ }
1
Q
N
=
I
.
ﺒﺎﻟﺘﺎﻟﻲ
G
N
Q
θ
= ×
ﺸﺎﻜﻼﺕـﺘﻟﺍ ﺽﻌﺒـﻟ ﺔﺒﺴﻨﻟﺎﺒ
:Q
N
θ
→
.
ﺔـﺒﺘﺭ ﻥﺇ
( )
( )
2
Aut
GL
p
N
≈
F
ﺴﺎﻭﻱـﺘ
(
)(
)
2
2
1
p
p
p
−
−
)
ﺍﻨﻅﺭ
3.5
(
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،
p
-
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺎﻬﻴﻓ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺭﻤﺯ
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p
.
ﻭﻤﻥ ﺍﻟﺘﻤﻬﻴﺩﻴ،ﺔﻘﻓﺍﺭﺘﻤ ﻥﻭﻜﺘ ،ﻭﻠﻴﺴ ﺕﺎﻨﻫﺭﺒﻤ ﻥﻤ
ﺔ
18.3
ﺭﺓـﻤﺯ ﻁﺒﻀﻟﺎﺒ ﺩﺠﻭﻴ ﻪﻨﺄﺒ ﻯﺭﻨ
ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﻭﺍﺤﺩﺓ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
.
ﺒﻔﺭﺽ ﺃﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
p
ﻭﻟﺘﻜﻥ،
N
ﺩﺓـﻟﻭﻤ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ
ﺒﻌﻨﺼﺭ ﻤﺜل
a
ﻷﻥ،
(
)
:
G N
p
=
ﺃﺼﻐﺭ
)
ﻭﺍﻟﻭﺤﻴﺩ
(
ﺴﻡـﻘﻴ ﻲـﻟﻭﺃ ﺩﺩـﻋ
(
)
:1
G
،
N
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
)
ﺍﻟﺘﻤﺭﻴﻥ
4-4
.(
ﻨﺒﻴﻥ ﻓﻴﻤﺎ ﺒﻌﺩ ﺒﺄﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻭﻻ
ﻴﻨﺘﻤﻲ ﺇﻟﻰ
N
.
ﻨﻌﻠﻡ ﺒﺄﻥ
( )
1
Z G
≠
ﻭﻷﻥ،
G
، ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ
( )
G Z G
ﻟﻴﺴﺕ ﺩﺍﺌﺭﻴﺔ
(4.19)
.
ﻟﺫﻟﻙ
( )
(
)
:1
Z G
p
=
ﻭ
( )
p
p
G Z G
C
C
≈
×
.
ﺎــﺨ لﻜﺸــﺒ
لــﻜﻟ ﻪــﻨﺄﺒ ﻯﺭــﻨ ،ﺹ
( )
,
p
x
G x
Z G
∈
∈
.
ﻷﻥ
( )
G Z G
ﻴﻜﻭﻥ ﺍﻟﻤﺒﺎﺩل ﻷﻱ ﺯﻭﺝ ﻤﻥ ﻋﻨﺎﺼﺭ،ﺔﻴﻠﻴﺩﺒﺘ
G
ﻴﻨﺘﻤﻲ ﺇﻟﻰ
( )
Z G
ﻭ ﺒﺘﻁﺒﻴﻕ ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻲ ﺍﻟﺒﺴﻴﻁ ﻨﺭﻯ ﺒﺄﻥ،
( )
[
]
(
)
1
2
,
,
1
n n
n
n
n
xy
x y
y x
n
−
=
≥
ﻟﺫﻟﻙ
ﻴﻜﻭﻥ
( )
p
p
p
xy
x y
=
ﻭﻤﻨﻪ،
:
p
x
x
G
G
→
a
ﺘﺸﺎﻜل
.
ﻲـﻓ ﺓﺍﻭﺘﺤﻤ ﻪﺘﺭﻭﺼ
( )
Z G
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻨﻭﺍﺘﻪ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
2
p
ﻋﻠﻰ ﺍﻷﻗل
.
ﺎ ﺃﻥـﻤﺒ
N
ﻰـﻠﻋ ﻁـﻘﻓ ﻱﻭـﺘﺤﺘ
1
p
−
ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻨﺭﻯ ﺒﺄﻨﻪ ﻴﻭﺠﺩ ﻋﻨﺼﺭ،
b
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﺨﺎﺭﺝ
N
.
ﺎﻟﻲـﺘﻟﺎﺒ
2
θ
θ
=
×
≈
×
p
p
G
a
b
C
C
ﻭﺒﻘﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﻤﻼﺤﻅﺔ،
(3.18)
ﺭـﻴﻏ ﺕﻼﻜﺎﺸـﺘﻟﺍ ﻥﺄﺒ
ﺍﻟﺘﺎﻓﻬﺔ
( )
2
1
p
p
p
p
C
Aut C
C
C
−
→
≈
×
ﺘﻌﻁﻲ ﺯﻤﺭﺍﹰ ﻤﺘﻤﺎﺜﻠﺔ
.
ﺇﻥ ﺍﻟﺯﻤﺭ ﻏﻴﺭ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،لﺜﺎﻤﺘﻟﺍ ﻑﻘﺴ ﺕﺤﺘ ،ﺍﺫﻬﻟ
3
p
ﺭـﻤﺯﻟﺍ ﻙﻠﺘ ﻲﻫ
ﻓﻲ
(13.3, 14.3)
.
18.5
)
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺏ
2
n
p
،
4
n
p
ﻭ،
8
n
p
،
p
ﻓ
ﺭﺩﻱ
(
ﻟﺘﻜﻥ
G
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺓﺭﻤﺯ
2
m
n
p
،
1
3
m
≤
≤
،
p
، ﻋﺩﺩ ﺃﻭﻟﻲ ﻓﺭﺩﻱ
1
n
≤
.
ﺴﻨﺭﻯ ﺒﺄ
ﻥ
G
ﺴﻴﻁﺔـﺒ ﺕﺴﻴﻟ
.
ﺘﻜﻥـﻟ
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻭﻟﺘﻜﻥ
( )
G
N
N
P
=
ﻭﻤﻨﻪ،
(
)
:
p
s
G N
=
.
ﻤﻥ ﺴﻴﻠﻭ
II
ﻨﺭﻯ ﺒﺄﻥ،
2 ,
1,
1, 2
1,...
m
p
p
s
s
p
p
=
+
+
.
ﺇﺫﺍ ﻜﺎﻥ
1
p
s
=
ﻭﻥـﻜﺘ ،
P
ﻨﺎﻅﻤﻴﺔ
.
ﺇ
ﻋﻨﺩﻫﺎ ﺘﻭﺠﺩ ﺤﺎﻟﺘﺎ،ﻙﻟﺫﻜ ﻥﻜﻴ ﻡﻟ ﺍﺫ
ﻥ
:
(i)
4
p
s
=
ﻭ
3
p
=
ﺃﻭ
(ii)
8
p
s
=
ﻭ
7
p
=
.
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ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷ
ﺍﻟﺘﺄﺜﻴﺭ ﺒﺎﻟﺘﺭﺍﻓﻕ ﻟﻠﺯﻤﺭﺓ،ﻰﻟﻭ
G
ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ
3
-
ﺔـﻴﺌﺯﺠﻟﺍ ﻭﻠﻴـﺴ ﺭﻤﺯ
٦١
ﺘﻌﺭﻑ ﺍﻟﺘﺸﺎﻜل
4
G
S
→
ﺇﺫﺍ ﻜﺎﻨﺕ، ﻱﺫﻟﺍ ،
G
ـﻭﻥ ﻤﺘﺒﺎﻴﻨـﻜﻴ ﻥﺃ ﺏـﺠﻴ ،ﺔﻁﻴﺴﺒ
ﺎﹰ
.
ﺫﻟﻙـﻟ
(
)
:1 4!
G
ﻭﻤﻨﻪ،
1
n
=
ﻟﺩﻴﻨﺎ،
(
)
:1
2 3
m
G
=
.
ﺍﻵﻥ ﺇﻥ ﺩﻟﻴل
2
-
ﺔـﻴﺌﺯﺠﻟﺍ ﻭﻠﻴـﺴ ﺭﻤﺯ
ﻴﺴﺎﻭﻱ
3
ﻭﻤﻨﻪ ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﺘﺸﺎﻜل،
3
G
S
→
.
ﻨﻭﺍﺘﻪ ﻋﺒﺎﺭﺓ ﻋﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻏﻴﺭ
ﺘﺎﻓﻬﺔ ﻓﻲ
G
.
ﺍﻟﻨﻘﺎﺵ ﻨﻔﺴﻪ ﻴ،ﺔﻴﻨﺎﺜﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓ
ﺄﻥـﺒ ﻥﻴﺒ
(
)
:1 8!
G
ﺎﻟﻲـﺘﻟﺎﺒﻭ ،
1
n
=
ﻀﺎﹰـﻴﺃ
.
ﺫﺍـﻬﻟ
(
)
:1
56
G
=
ﻭ
7
8
s
=
.
ﻟﺫﻟﻙ ﻓﺈﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ
48
ﻋﻨﺼﺭ
ﺍﹰ
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
7
ﺎﻟﻲـﺘﻟﺎﺒﻭ ،
ﻴﻤﻜﻥ ﺃﻥ ﺘﻭﺠﺩ
2
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ،ﻁﻘﻓ ﺓﺩﺤﺍﻭ
.
ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺏ
r
pq
،
p
،
q
ﻋﺩﺩﺍﻥ ﺃﻭﻟﻴﺎ
،ﻥ
p
q
<
ﻷﻥ،ﺔﻁﻴﺴـﺒ ﺕﺴﻴﻟ
ﺍﻟﺘﻤﺭﻴﻥ
4-4
ﻴﺒﻴﻥ ﺒﺄﻥ
p
-
ﺯﻤﺭ
ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
ﺄﻥـﺒ ﻥﻴﺒﻨ ﺕﻻﺎﺤﻟﺍ ﻩﺫﻫ ﺹﺤﻔﺘﻟ
5
A
ﺃﺼﻐﺭ ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻏﻴﺭ ﺩﺍﺌﺭﻴﺔ
.
19.5
)
ﺍ
ﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
60
(
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
60
.
ﺄﻥـﺒ ﻥﻵﺍ ﻥﻴﺒﻨـﺴ
G
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ
5
A
.
ﻭﻷﻥ،ﻥﺄﺒ ﻅﺤﻼﻨ
G
، ﺒﺴﻴﻁﺔ
2
3, 5
s
=
ﺃﻭ
.15
ﺇﺫﺍ ﻜﺎﻨﺕ
P
2
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
ﻭ
( )
G
N
N
P
=
ﻋﻨﺩﺌﺫ،
(
)
2
:
s
G N
=
.
ﺍﻟﺤﺎﻟﺔ
2
3
s
=
ﻷﻥ ﻨﻭﺍﺓ،ﺔﻨﻜﻤﻤ ﺭﻴﻏ ﻥﻭﻜﺘ
(
)
Sym
G
G N
→
ﺴﺘﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
ﻏﻴﺭ ﺘﺎﻓﻬﺔ ﻓﻲ
G
.
ﻓﻲ ﺍﻟﺤﺎﻟﺔ
2
5
s
=
ﻭﺍﺀـﺘﺤﻻﺍ ﻕﻴﺒﻁﺘ ﻰﻠﻋ لﺼﺤﻨ ،
(
)
5
Sym
G
G N
S
→
=
ﺙـﻴﺤ ،
ﻨﻌﺘﺒﺭ
G
ﻜﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ
2
ﻓﻲ
5
S
ﻟﻜﻥ ﻤﻥ،
(36.4)
ﻟﻜل،ﻥﺄﺒ ﻯﺭﻨ
5
n
≥
،
n
A
ﻫﻲ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﺍﻟﺘﻲ ﺩﻟﻴﻠﻬﺎ
2
ﻓﻲ
n
S
.
ﻓﻲ ﺍﻟﺤﺎﻟﺔ
2
15
s
=
ﺍ،
ﻟﻨﻘﺎﺵ ﺫﺍﺘﻪ
)
ﺒﺎﺴﺘﺨﺩﺍﻡ
2
6
s
=
(
ﻥـﻤ ﻥﺎﺘﺭﻤﺯ ﺩﺠﻭﺘ ﻪﻨﺄﺒ ﻯﺭﻨ
2
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
P
ﻭ
Q
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺓﺭﻤﺯﺒ ﻥﺎﻌﻁﺎﻘﺘﺘ
2
.
ﻨﻅﻡـﻤﻟﺍ ﻥﺇ
N
ﺭﺓـﻤﺯﻠﻟ
P
Q
I
ﺘ
ﺤﺘﻭﻱ ﻋﻠﻰ
P
ﻭ
Q
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﻭﻬﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ ،
12
،
20
ﺃﻭ،
60
.
ﺔـﻟﺎﺤﻟﺍ ﻲـﻓ
ﺍﻟﻨﻘﺎﺵ ﺍﻟﺴﺎﺒﻕ ﻴﺒﻴﻥ ﺒﺄﻥ،ﻰﻟﻭﻷﺍ
5
G
A
≈
ـ ﻭ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﺘﺒﻘﻴﺔ ﺘﻨﺎﻗﺽ ﺒﺒﺴﺎﻁﺔ ﻟ،
G
.
ﺘﻤﺎﺭﻴﻥ
16
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻌﺎﺩﻱ،ﺊﻓﺎﻜﻤ لﻜﺸﺒ
(
)
Sym
G
G N
→
.
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٢٠١
1-5
ﺒﻴﻥ
ﺃﻥ ﻜل ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
)
ﻟﻴﺴﺕ ﺒﺎﻟﻀﺭﻭﺭﺓ
ﺃﻥ ﺘﻜﻭﻥ
ﺘﺒﺩﻴﻠﻴ
ﺔ
(
ـﺒ
n
ﻋﻨﺼﺭ ﻋﻠﻰ
ﺍﻷﻜﺜﺭ
ﻭﻤﻥ
ﺭﺘﺒﺔ
ﺘﻘﺴﻡ
n
ﻟﻜل
n
ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ
.
PDF created with pdfFactory Pro trial version
٣٠١
ﺍﻟﻔﺼل ﺍﻟﺴﺎﺩﺱ
ﺍﻟﻤﺘﺴﻠﺴﻼﺕ ﺘ
؛ﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ
ﺍﻟﺯﻤ
ﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل ﻭﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
Subnormal Series; Solvable and Nilpotent Groups
ﺍﻟﻤﺘﺴﻠﺴﻼﺕ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ
(Subnormal Series)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
.
ﺘﺴﻤﻰ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
{ }
0
1
1
...
...
1
i
i
n
G
G
G
G
G
G
+
=
⊃
⊃ ⊃
⊃
⊃ ⊃
=
ﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﻨﺎﻅﻤﻴﺔ
(Subnormal Series)
ﺇﺫﺍ ﻜﺎﻨﺕ
i
G
ﻲـﻓ ﺔﻴﻤﻅﺎﻨ
1
i
G
−
لـﻜﻟ
i
،
ﻭﺘﺴﻤﻰ ﻤﺘﺴﻠﺴﻠﺔ ﻨﺎﻅﻤﻴﺔ
(normal Series)
ﺇﺫﺍ ﻜﺎﻨﺕ
i
G
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻟﻜل
i
٧١
.
ﺎلـﻘﻴ
ﺒﺄﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﻏﻴﺭ ﺘﻜﺭﺍﺭﻴﺔ
(without repetition)
ﻭﺀﺍﺕـﺘﺤﻻﺍ لﻜ ﺕﻨﺎﻜ ﺍﺫﺇ
1
i
i
G
G
−
⊃
ﻓﻌﻠﻴﺔ
) .
،ﺃﻱ ﺃﻥ
1
i
i
G
G
−
≠
.(
ﻨﺩﻋﻭﺫﺌﺩﻨﻋ
n
ﺴﻠﺔـﻠﺴﺘﻤﻟﺍ لﻭﻁ
.
ﺴﻤﺔـﻘﻟﺍ ﺭـﻤﺯ ﻭﻋﺩـﻨ
(quotient groups)
ﺃﻭ
(factor groups)
1
i
i
G
G
−
ﺒﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻟﻠﻤﺘﺴﻠﺴﻠﺔ
.
ﻴﻘﺎل ﻋﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺃﻨﻬﺎ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ
(composition Series)
ﻡـﻟ ﺍﺫﺇ
ﺘﺤﻭ ﻋﻠﻰ ﻤﺘﺴﻠﺴﻠﺔ ﻤﻜﺭﺭﺓ ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﺒﺩﻭﺭﻫﺎ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﻨﺎﻅﻤﻴﺔ ﺃﻴﻀﺎﹰ
.
،ﻭﺒﻜﻠﻤﺎﺕ ﺃﺨﺭﻯ
ﺘﻜﻭﻥ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ ﺇﺫﺍ ﻜﺎﻨﺕ
i
G
ﺃﻋﻅﻤﻴﺔ ﺒﻴﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﻔﻌﻠﻴﺔ
1
i
G
−
لـﻜﻟ
i
.
ﺴﻤﺔـﻗ ﺓﺭﻤﺯ لﻜ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﺔﻴﺒﻴﻜﺭﺘ ﺔﻠﺴﻠﺴﺘﻤ ﺔﻴﻤﻅﺎﻨﻟﺍ ﺕﺤﺘ ﺔﻠﺴﻠﺴﺘﻤﻟﺍ ﻥﻭﻜﺘ ﺍﺫﻬﻟ
ﻟﻬﺎ ﺒﺴﻴﻁﺔ ﻭﻏﻴﺭ ﺘﺎﻓﻬﺔ
.
ﺃﻨﻪ ﻟﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻨﺘﻬ، ﺢﻀﺍﻭﻟﺍ ﻥﻤ
ﻴﺔ ﻓﻲ
G
ﺴﻠﺔـﻠﺴﺘﻤ ﺩـﺠﻭﺘ
ﺘﺭﻜﻴﺒﻴﺔ
)
ﻋﺎﺩﺓ ﻜﺜﻴﺭ
:(
ﻨﺨﺘﺎﺭ
1
G
ﻟﺘﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻌﻠﻴﺔ ﻓﻲ
G
ﺎﺭـﺘﺨﻨ ﻡـﺜ ﻥﻤﻭ ،
2
G
ﻟﺘﻜﻭﻥ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻌﻠﻴﺔ ﻓﻲ
1
G
ﺍﻟﺦ،
..
ﺔ ﺃﻭـﻴﻬﺘﻨﻤ ﺭﻴﻏ ﺓﺭﻤﺯ ﻥﻭﻜﺘ ﻥﺃ ﻥﻜﻤﻴ
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﺤﻭﻱ ﻋﻠﻰ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ ﻤﻨﺘﻬﻴﺔ
.
ﻨﺤﺼل ﻤﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ
{ }
0
1
1
...
...
1
i
i
n
G
G
G
G
G
G
+
=
=
>
>
>
>
>
>
ﻋﻠﻰ ﻤﺘﺘﺎﻟﻴﺔ ﻤﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺘﺎﻤﺔ
17
ﺑﻌﺾ اﻟﻤﺆﻟﻔﯿ
ﻦ ﯾﻜﺘﺒﻮن
"
ﻣﺘﺴﻠﺴﻠﺔ ﻧﺎﻇﻤﯿﺔ
"
ﺣﯿﺚ ﻛﺘﺒﻨ ﺎ
"
ﺳﻠ ﺴﻠﺔ ﺗﺤ ﺖ ﻧﺎﻇﻤﯿ ﺔ
"
و
"
ﺳﻠ ﺴﻠﺔ ﻻ ﻣﺘﻐﯿ ﺮة
"
ﺣﯿ ﺚ ﻛﺘﺒﻨ ﺎ
"
ﺳﻠﺴﻠﺔ ﻧﺎﻇﻤﯿﺔ
."
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٤٠١
1
2
1
2
1
2
2
1
1
0
0
1
1
1
1
1
...
1
1
n
n
n
n
n
n
n
n
G
G
G
G
G
G
G
G
G
G
G G
−
−
−
−
−
−
−
−
→
→
→
→
→
→
→
→
→
→
→
→
ﻟﺫﻟﻙ ﻓﺈﻥ
G
ﺘﺒﻨﻰ ﻭﺤﺘﻰ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
0
1
1
2
1
,
,...,
n
G G G G
G
−
ﺸﻜلـﺒ ﺕﺍﺩـﻴﺩﻤﺘ لﻴﻜﺸﺘﺒ
ﻤﺘﺘﺎﻟﻲ
.
،ﺒﺸﻜل ﺨﺎﺹ
ﺴﻴﻁﺔـﺒ ﺭﻤﺯ ﺝﺭﺎﺨ ﻰﻨﺒﺘ ﺎﻬﻨﺄﻜ ﺎﻫﺭﺒﺘﻌﻨ ﻥﺃ ﻥﻜﻤﻴ ،ﺔﻴﺒﻴﻜﺭﺘ ﺔﻠﺴﻠﺴﺘﻤ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ لﻜﻟ ﻥﺃ ﺎﻤﺒ
.
ﻤﺒﺭﻫﻨﺔ ﺠﻭﺭﺩﺍﻥ
-
، ﻫﻭﻟﺩﺭ
ﻭﺍﻟﺘﻲ ﻫﻲ
ﺭـﻤﺯﻟﺍ ﻩﺫـﻫ ﻥﺄﺒ لﻭﻘﻨ ،ﺓﺭﻘﻔﻟﺍ ﻩﺫﻬﻟ ﻲﺴﻴﺌﺭﻟﺍ ﻉﻭﻀﻭﻤﻟﺍ
ﺍﻟﺒﺴﻴﻁ
ﺔ ﻤﺴﺘﻘﻠﺔ ﻋﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ
ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
)
ﺘﺤﺕ ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل ﻭﺍﻟﺭﺘ
ﺒﺔ
.(
ﺭﺓـــﻤﺯﻠﻟ ﻥﺎـــﻜ ﺍﺫﺇ ﻪـــﻨﺄﺒ ﻅـــﺤﻼﻨ
G
ﺔـــﻴﻤﻅﺎﻨ ﺕـــﺤﺘ ﺔﻠﺴـــﻠﺴﺘﻤ
{ }
0
1
...
1
n
G
G
G
G
=
=
>
> >
ﻋﻨﺩﺌﺫ،
(
)
(
)
(
)
1
1
1
1
:1
:
:1
i n
i
i
i n
i
i
G
G
G
G
G
≤ ≤
−
≤ ≤
−
=
=
∏
∏
ﻤﺜﺎل
1.6
(a)
ﻟﻠﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ
3
S
ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
3
3
1
S
A
>
>
ﻤﻊ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
2
C
ﻭ
3
C
.
(b)
ﻟﻠﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ
4
S
ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
( )( )
4
4
13 24
1,
S
A
V
>
>
>
>
ﺤﻴﺙ
2
2
V
C
C
≈
×
ﻭﺘﺘﺄﻟﻑ ﻤﻥ ﻜل ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻓﻲ
4
A
)
ﺍﻨﻅﺭ
31.4
.(
ﺭـﻤﺯﻭ
ﺍﻟﻘﺴﻤﺔ ﻫﻲ
2
3
2
2
,
,
,
C C C C
.
(c)
ﺃﻱ ﻤﻌﻠﻡ ﺘﺎﻡ ﻓﻲ
p
n
F
،
p
ﻫﻭ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ،ﻲﻟﻭﺃ ﺩﺩﻋ
.
ﻁﻭﻟﻬﺎ
n
ﻭﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ،
ﻟﻬﺎ ﻫﻲ
,
,...,
p
p
p
C C
C
.
(d)
ﻨﺄﺨﺫ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺩﺍﺌﺭﻴﺔ
m
C
a
=
.
ﻤﻥ ﺃﺠل ﺃﻱ ﺘﺤﻠﻴل
1
...
r
m
p
p
=
ـ ﻟ
m
ﻰـﻟﺇ
ﺠﺩﺍﺀ ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ
)
ﻟﻴﺱ ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﺘﻜﻭﻥ ﻤﺨﺘﻠﻔﺔ
(
ﺘﻭﺠﺩ ﺍﻟﻤﺘ،
ﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
1
1 2
...
m
m
m
p
p p
C
C
C
>
>
>
1
1 2
p
p p
a
a
a
ﺍﻟﻁﻭل ﻴﺴﺎﻭﻱ
r
ﻭﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻫﻲ،
1
2
,
,...,
r
p
p
p
C
C
C
.
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٥٠١
(e)
ﺒﻔﺭﺽ ﺃﻥ
G
، ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﺯﻤﺭ ﺒﺴﻴﻁﺔ
1
...
r
G
H
H
=
× ×
.
ﻴﻜﻭﻥ ﻟﻠﺯﻤﺭﺓﺫﺌﺩﻨﻋ
G
ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
2
3
...
...
...
r
r
G
H
H
H
H
× ×
× ×
>
>
>
ﻤﻥ ﺍﻟﻁﻭل
r
ﻤﻊ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
1
2
,
,...,
r
H H
H
.
ﺔـﻠﻴﺩﺒﺘ ﻱﻷ ﻪﻨﺄﺒ ﻅﺤﻼﻨ
π
ﺔـﻋﻭﻤﺠﻤﻠﻟ
{
}
1, 2,..., r
ﺘﻭﺠﺩ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ ﺃﺨﺭﻯ ﻤﻊ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ،
( )
( )
( )
1
2
,
,...,
r
H
H
H
π
π
π
.
(f)
ﻟﻘﺩ ﺭﺃﻴﻨﺎ ﻓﻲ
(36.4)
ﺒﺄﻨﻪ ﻷﺠل
5
n
≥
ﻲـﻓ ﺔﻴﻤﻅﺎﻨﻟﺍ ﺔﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻟﺍ ﻥﺇ ،
n
S
ﻲـﻫ
n
S
ﻭ
n
A
ﻭ
{ }
1
ﻭﻓﻲ،
(32.4)
ﺃﻥ
n
A
ﺒﺴﻴﻁﺔ
.
ﻭﺒﺎﻟﺘﺎﻟﻲ
{ }
1
n
n
S
A
>
>
ﻫﻲ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ
ـﺍﻟﺘﺭﻜﻴﺒﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻟ
n
S
.
ﻤﺒﺭﻫﻨﺔ
(2.6)
)
ﺠﻭﺭﺩﺍﻥ
-
ﻫﻭﻟﺩﺭ
(
(Jordan - Holder)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
{ }
{ }
0
1
0
2
...
1
...
1
s
t
G
G
G
G
G
H
H
H
=
=
=
=
>
>
>
>
>
>
ﺴﻠﺴﻠﺘﻴﻥ ﺘﺭﻜﻴﺒﻴﺘﻴﻥ ﻟﻠﺯﻤﺭﺓ
G
ﻋﻨﺩﺌﺫ،
s
t
=
ﻭ ﻴﻭﺠﺩ ﺘﺒﺩﻴﻠﺔ
π
ﺔـﻋﻭﻤﺠﻤﻠﻟ
{
}
1, 2,..., r
،
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
( )
1
1
.
i
i
i
i
G G
H
H
π
π
+
+
≈
18
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ ﺭﺘﺒﺔ
G
.
ﺍﻟﺤﺎﻟﺔ
I
:
1
1
H
G
=
.
ﻟﺩﻴﻨﺎ ﺴﻠﺴﻠﺘﺎﻥ ﺘﺭﻜﻴﺒﻴﺘﺎﻥ ﻟﻠﺯﻤﺭﺓ،ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ ﻲﻓ
1
G
ﻥـﻜﻤﻴ ﻲﺘﻟﺍﻭ ،
ﺃﻥ ﻨﻁﺒﻕ ﻋﻠﻴﻬﺎ ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻲ
.
ﺍﻟﺤﺎﻟﺔ
II
:
1
1
H
G
≠
.
ﺒﻤﺎ ﺃﻥ
1
G
ﻭ
1
H
ﻲـﻓ ﻥﺎـﺘﻴﻤﻅﺎﻨ ﻥﺎﺘﻴﺌﺯﺠ ﻥﺎﺘﺭﻤﺯ
G
ﺈﻥـﻓ ،
1
1
G H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﻥـﻤ لـﻜ ﻱﻭﺤﺘ ﺹﺎﺨ لﻜﺸﺒ ﻭ
1
G
ﻭ
1
H
ﻴﻥـﺘﻠﻟﺍ ،
ﺘﻜﻭﻨﺎﻥ ﺃﻋﻅﻤﻴﺘﻴ
ﻥ ﻓﻲ
G
ﻭ ﻤﻨﻪ،
1
1
G H
G
=
.
ﻟﺫﻟﻙ
1
1
1
1
1
1
1
G G
G H G
H G
H
=
I
)
ﺍﻨﻅﺭ
1.45
.(
ﺒﺸﻜل ﻤﺸﺎﺒﻪ
1
1
1
1
G H
G G
H
I
.
ﻟﺘﻜﻥ
2
1
1
K
G
H
=
I
ﻋﻨﺩﺌﺫ،
2
K
ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ
ﻨﺎﻅﻤﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻲ ﻜل ﻤﻥ
1
G
ﻭ
1
H
ﻭ،
(19)
1
1
2
1
1
2
,
G G
H
K
G H
G K
ﻨﺨﺘﺎﺭ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
2
3
...
>
> >
u
K
K
K
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ ﺍﻟﺸﻜل
18
ﻥ ﻫﻭﻟﺩﺭ ﺒﺄﻨﻬﺎ ﻤﺘﻤﺎﺜﻠﺔﻴﺒﻭ ،ﺔﺒﺘﺭﻟﺍ ﺱﻔﻨ ﺎﻬﻟ ﺔﻠﺒﺎﻘﺘﻤﻟﺍ ﺔﻤﺴﻘﻟﺍ ﺭﻤﺯ ﻥﺄﺒ ﻥﺍﺩﺭﻭﺠ ﻥﻴﺒ
.
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٦٠١
1
2
2
1
2
...
...
...
s
u
t
G
G
G
G
K
K
H
H
H
>
>
>
>
>
>
>
>
ﺒﺘﻁﺒﻴﻕ ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻲ ﻋﻠﻰ
1
G
ﻭ
1
H
ﻨﺠﺩ،ﻁﻁﺨﻤﻟﺍ ﻲﻓ ﺔﻴﺒﻴﻜﺭﺘﻟﺍ ﺎﻬﺘﻼﺴﻠﺴﺘﻤ ﻰﻠﻋ ﻭ
(
) {
}
{
}
{
}
{
}
{
}
(
)
1
2
1
1
2
2
3
1
1
2
2
3
1
2
1
2
3
1
1
2
2
3
1
1
2
2
3
1
2
...
,
,
,
...
,
,
, ...
,
,
, ...
,
,
, ...
,
,
,...
...
G
G
G
G G G G G G
G G G K
K
K
H
K G H
K
K
G H
H
K
K
K
G H
H
H
H
H
G
H
H
=
=
>
>
>
>
>
>
ﻨﻼﺤﻅ ﺒﺄﻨﻪ ﺇﺫﺍ ﻁﺒﻘﺕ
ﻫﺫﻩ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻋﻠﻰ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺩﺍﺌﺭﻴﺔ
m
C
ﺼﺤﻴﺢـﻟﺍ ﺩﺩﻌﻟﺍ لﻴﻠﺤﺘ ﻥﺄﺒ ﺞﺘﻨﻴ
ﺇﻟﻰ ﺠﺩﺍﺀ ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ ﻴﻜﻭﻥ ﻭﺤﻴﺩﺍﹰ
.
ﻤﻼﺤﻅﺔ
3.6
ﺘﻭﺠﺩ ﺯﻤﺭ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻤﺘﺴﻠﺴﻼﺘ
ﻬﺎ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ ﻤﻨﺘﻬﻴﺔ
)
ﺭـﻤﺯ ﺩﺠﻭﻴ ﻪﻨﺃ ﻰﺘﺤ
ﺒﺴﻴﻁﺔ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.(
ﻟﺘﻜﻥ،ﺓﺭﻤﺯﻟﺍ ﻩﺫﻫ لﺜﻤﻟ
( )
d G
ﺔـﻴﺒﻴﻜﺭﺘﻟﺍ ﺕﻼﺴـﻠﺴﺘﻤﻠﻟ لﻭﻁ ﺭﻐﺼﺃ
.
ﻭﻷﻥ ﻤﺒﺭﻫﻨﺔ ﺠﻭﺭﺩﺍﻥﺫﺌﺩﻨﻋ
–
ﺎـﻬﻟ ﺔـﻴﺒﻴﻜﺭﺘﻟﺍ ﺕﻼﺴـﻠﺴﺘﻤﻟﺍ لـﻜ ﻥﺄﺒ ﻥﻴﺒﺘﻟ ﻊﺴﻭﺘﺘ ﺭﺩﻟﻭﻫ
ﺍﻟﻁﻭل
( )
d G
ﻭ
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭ ﻗﺴﻤﺔ ﻤﺘﻤﺎﺜﻠﺔ
.
ﻴﺼﻠﺢ ﺍﻟﺒﺭﻫﺎﻥ ﻨﻔﺴﻪ ﻋﺩﺍ ﺃﻨﻪ ﻋﻠﻴﻙ ﺍﺴﺘﺨﺩﺍﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ
( )
d G
ﺒﺩﻻﹰ ﻤﻥ
G
ﻭ
ﺎﻥـﻫﺭﺒ
ﺒﺄﻨﻬﺎ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
ﺯﻤﺭﺓ ﻤﻊ ﻤﺘﺴﻠﺴﻠﺔ
ﺘﺭﻜﻴﺒﻴﺔ ﻤﻨﺘﻬﻴﺔ ﻭﻟﻬﺎ ﺃﻴﻀﺎﹰ
ﻤﺘﺴﻠﺴﻠﺔ
ﺔـﻴﻬﺘﻨﻤ ﺔﻴﺒﻴﻜﺭﺘ
)
ﺍﻟﺘﻤﺭﻴﻥ
1-6
.(
ﺘﺴﻤﻰ ﺃﺤﻴﺎﻨﺎﹰ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻟﻠﻤﺘﺴﻠﺴﻼﺕ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ ﺒﺎﻟﻌﻭﺍﻤل ﺍﻟﺘﺭﻜﻴﺒﻴﺔ
(composition factor)
.
ﺍﻟﺯﻤﺭ ﺍﻟﻘﺎﺒﻠ
ﺔ ﻟﻠﺤل
(Solvable groups)
ﺇﻥ ﺍﻟﻤﺘﺴﻠﺴﻼﺕ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻓﻴﻬﺎ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﺘﺒﺩﻴﻠﻴﺔ ﺘﺴﻤﻰ ﺒﺎﻟﻤﺘﺴﻠﺴﻼﺕ ﺍﻟﻘﺎﺒﻠﺔ
ﻟﻠﺤل
(solvable series)
.
ﻭﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
(solvable)
ﺃﻭ
(soluble)
ﻭﺕـﺤ ﺍﺫﺇ
ﻋﻠﻰ ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ـﺭﺓ ﻗﺎﺒﻠـﻤﺯﻟﺍ ﻥﺄﺒ لﻭﻘﻨ ﻥﺃ ﻥﻜﻤﻴ ،ﺭﺨﺁ لﻜﺸﺒﻭ
ﻥـﻜﻤﺃ ﺍﺫﺇ لـﺤﻠﻟ ﺔ
ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﺒﺘﺸﻜﻴل ﺘﻤﺩﻴﺩﺍﺕ ﻤﺘﺘﺎﻟﻴﺔ ﻟﺯﻤﺭ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻭﺒﻤﺎ ﺃﻥ
:
ﺴﻴﻁﺔـﺒ ﺔﻴﻠﻴﺩﺒﺘﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻥﻭﻜﺘ
ﻨﺭﻯ ﺒﺄﻨﻪ ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ،ﻲﻟﻭﺃ ﺩﺩﻋ ﺎﻬﺘﺒﺘﺭ ﺔﻴﺭﺌﺍﺩ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ
G
ﻗﺎﺒﻠﺔ
ﻁـﻘﻓﻭ ﺍﺫﺇ لﺤﻠﻟ
)
ﺍﻟﺘﻌﺭﻴﻑ
(
)
ﺍﻻﺴﺘﻘﺭﺍﺀ
(
)
ﺒﺘﻁﺒﻴﻕ
(19)
(
)
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
(
)
ﺍﻻﺴﺘﻘﺭﺍﺀ
(
)
ﺍﻟﺘﻌﺭﻴﻑ
(
زﻣﺮ اﻟﻘﺴﻤﺔ
زﻣﺮ اﻟﻘﺴﻤﺔ
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ﺇﺫﺍ ﻜﺎﻥ ﻹﺤﺩﺍﻫﺎ
)
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻟﻜل ﻭﺍﺤﺩﺓ
(
ﻤﺘﺴﻠﺴ
ﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ ﻭﻜل ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻓﻴﻬﺎ ﻫﻲ ﺯﻤﺭ ﺩﺍﺌﺭﻴﺔ
ﺭﺘﺒﺘﻬﺎ ﻋﺩﺩ ﺃﻭﻟﻲ
.
ﻜل ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﺘﻜﻭﻥ ﻗﺎﺒﻠﺔ
ﻜﺫﻟﻙ ﻜل ﺯﻤﺭﺓ ﺩﻴﻬﻴﺩﺭﺍل،لﺤﻠﻟ
.
ﻭ
ﻟﻘﺩ ﺒﻴﻨﺕ ﺍﻟﻨﺘﺎﺌﺞ ﻓﻲ ﺍﻟﻔﺼل
ﺍﻟﺨﺎﻤﺱ ﺃﻥ ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺭﺘﺒﺔ
>
60
ﺘﻜﻭﻥ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
، ﺯﻤﺭﺓ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ ﺒﺴﻴﻁﺔ،ﺱﻜﻌﻟﺎﺒﻭ
ﺇﻥ،ﹰﻼﺜﻤ
n
A
ﻟﻜل
5
n
≥
ﻟﻥ ﺘﻜﻭﻥ ﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل،
.
ﻤﺒﺭﻫﻨﺔ
4.6
)
ﻓﻴﺕ
-
ﻁ
ﻤﺒﺴﻭﻥ
(
(Feit – Thompson)
ﺭﺩﻱـﻓ ﺩﺩﻋ ﺎﻬﺘﺒﺘﺭ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ لﻜ
ﺘﻜﻭﻥ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
19
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻴﺸﻐل ﺍﻟﺒﺭﻫﺎﻥ ﺍﻹﺼﺩﺍﺭﺍﺕ ﺍﻟﻜﺎﻤﻠﺔ ﻟﻤﺠﻠﺔ
Pacific
ﻟﻠﺭﻴﺎﻀﻴﺎﺕ
)
ﻓﻴﺕ
–
ﺴﻭﻥـﺒﻤﻭﻁ
1963
.(
ﺒﻜ
ﻜل ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴ،ﻯﺭﺨﺃ ﺕﺎﻤﻠ
ﻭ،ﺔﻴﺠﻭﺯ ﺔﺒﺘﺭ ﻥﻤ ﻥﻭﻜﺘ ﺔﻴﻬﺘﻨﻤ ﺔ
ﻟﺫﻟﻙ ﻓﻬﻲ
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﺴﻴﻁﺔـﺒﻟﺍ ﺭـﻤﺯﻟﺍ ﻑﻴﻨﺼﺘ ﻲﻓ ﹰﺍﺭﻭﺩ ﺏﻌﻠﺘ ﺔﻨﻫﺭﺒﻤﻟﺍ ﻩﺫﻫ ﻥﺇ
ﺍﻨﻅﺭ،ﺔﻴﻬﺘﻨﻤﻟﺍ
p49
.
ﻤﺜﺎل
5.6
ﻨﺄﺨﺫ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
( )
{ }
0
B
∗ ∗
=
∗
ﻭ
( )
{ }
1
0 1
U
∗
=
ﻤﻥ ﺍﻟﺯﻤﺭﺓ
( )
2
GL F
،
ﺎـــﻤ لـــﻘﺤﻟ
.
ﻭﻥـــﻜﺘ ﺫـــﺌﺩﻨﻋ
U
ﻲـــﻓ ﺔـــﻴﻤﻅﺎﻨ ﺔـــﻴﺌﺯﺠ ﺓﺭـــﻤﺯ
B
ﻭ،
(
)
,
,
B U
F
F
U
F
×
×
×
+
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
B
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ﻗﻀﻴﺔ
6.6
(a)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭ
ﺓ ﻗﺴﻤﺔ ﻟﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻫﻲ ﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
(b)
ﺇﻥ ﺘﻤﺩﻴﺩ ﺍﻟﺯﻤﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل ﻫﻭ ﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
19
ﻜﺘﺏ ﺒﺭﻨﺴﺎﻴﺩ
:(1897, p379)
ﻻ ﺘﻭﺠﺩ ﺯﻤﺭ ﺒﺴﻴﻁﺔ ﺭﺘﺒﺘﻬﺎ ﻋﺩﺩ ﻓﺭ
ﺩﻱ ﺘﻜﻭﻥ ﻓﻲ ﺍﻟﻭﻗﺕ ﺍﻟﺤﺎﻀﺭ ﻤﻌﺭﻭﻓﺔ ﺤﺘﻰ ﺘﺘﻭﺍﺠﺩ
.
ﺭﻱـﺤﺘﻟﺍ ﻭ
ﻟﻭﺠﻭﺩ ﺃﻭ ﻋﺩﻡ ﻭﺠﻭﺩ
ﻴﻤﻜﻥ ﺃﻥ،ﺔﻴﻤﻫﻷﺍ ﻥﻤ ﺔﻴﺎﻏ ﻲﻓ ﺞﺌﺎﺘﻨ ﻰﻟﺇ ،ﺕﺍﺀﺍﻭﺘﺤﻻﺍ ﻥﻜﺘ ﺎﻤﻬﻤ ،ﺩﻴﻜﺃ لﻜﺸﺒ ﺎﻨﺩﻭﻘﻴﺴ ﺭﻤﺯﻟﺍ ﻩﺫﻫ لﺜﻤ
ﻨﻨﺼﺢ ﺍﻟﻘﺎﺭﺉ ﺒﺄﻨﻪ
ﻴﺴﺘﺤﻕ ﺘﺭﻜﻴﺯ ﺍﻨﺘﺒﺎﻫﻪ
.
ﻻ ﺘﻭﺠﺩ ﺯﻤﺭ ﺒﺴﻴﻁﺔ ﻤﻌﻠﻭﻤﺔ ﺒﺤﻴ،ﹰﺎﻀﻴﺃ
ﻼﺙـﺜ ﻥـﻤ لﻗﺃ ﻰﻠﻋ ﺎﻬﺒﺘﺭ ﻱﻭﺤﺘ ﺙ
ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ ﻤﺨﺘﻠﻔﺔ
... .
ﺘﻁﻭﺭ ﻫﺎﻡ ﻓﻲ ﺍﻟﻤﺴﺄﻟﺔ ﺍﻷﻭﻟﻰ ﻟﻡ ﺘﺤل ﺇﻻ ﻓﻲ ﺍﻟﻤﺭﺠﻊ
Suzuki,M.
ﻋﺩﻡ ﻭﺠﻭﺩ ﻨﻤﻁ ﻤﻌﻴﻥ ﻟﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻤﻥ ﺭﺘﺒﺔ
،ﻤﻨﺘﻬﻴﺔ
1957
.
ﺍﻟﺫﻱ ﺒﺭﻫﻥ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻤﻤﻴﺯﺍﺕ ﺃﻥ ﺃﻱ،ﻪﺴﻔﻨ ﺩﻴﺎﺴﻨﺭﺒ لﺒﻗ ﻥﻤ ﺔﺒﻨﺎﺜﻟﺍ ﺔﻟﺄﺴﻤﻟﺍ ﺕﻠﺤ ،لﺎﺤ ﻱﺃ ﻰﻠﻋ
ﺯﻤﺭﺓ
ﺘﺤﻭﻱ ﺭﺘﺒﻬﺎ
ﻋﻠﻰ ﺃﻗل ﻤﻥ ﺜﻼﺙ ﺃﻋﺩﺍﺩ
ﺃ
ﻭﻟﻴﺔ ﻤﺨﺘﻠﻔﺔ ﻫﻲ ﺯﻤﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
)
ﺍﻨﻅﺭ ﺃﻟﺒﻴﺭﻥ ﻭﺒﻴل
1955,p182
.(
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٨٠١
ﺍﻟﺒﺭﻫﺎﻥ
.
(a)
ﻟﺘﻜﻥ
1
...
n
G
G
G
>
>
>
ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻟﻠﺯﻤﺭﺓ
G
ﻭﻟﺘﻜﻥ،
H
ﺯﻤﺭﺓ
ﺠﺯﺌﻴﺔ ﻤﻥ
G
.
ﺇﻥ ﺍﻟﺘﺸﺎﻜل
1
1
:
i
i
i
i
x
xG
H
G
G G
+
+
→
a
I
ﻨﻭﺍﺘﻪ
(
)
1
1
i
i
i
H
G
G
H
G
+
+
=
I
I
I
ﻟﺫﻟﻙ،
1
i
H
G
+
I
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ
i
H
G
I
ﻭ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
1
i
i
H
G
H
G
+
I
I
ﺘﺸﻜل ﺘﻁﺒﻴﻘ
ﺎﹰ
ﻤﺘﺒﺎﻴﻨ
ﺎﹰ
ﺇﻟﻰ
1
i
i
G G
+
ﻲـﺘﻟﺍ ،
ﺘﻜﻭﻥ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻨﻜﻭﻥ ﻗﺩ ﺒﻴﻨﺎ ﺒﺄﻥ
1
...
n
H
H
G
H
G
>
I
> >
I
ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻟﻠﺯﻤﺭﺓ
H
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ ﻟﻠﺯﻤﺭﺓ
G
ﻭﻟﻴﻜﻥ،
i
G
ﺼﻭ
ﺭﺓ
i
G
ﻓﻲ
G
.
ﻋﻨﺩﺌﺫ
{ }
1
...
1
n
G
G
G
=
>
>
>
ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻓﻲ
G
.
(b)
ﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻭﻟﺘﻜﻥ،
G
G N
=
.
ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻨﻪ ﺇﺫﺍ
ﻜﺎﻨﺕ
N
ﻭ
G
ﻗﺎﺒﻠﺘﻴ
ﻓﺈﻥ،لﺤﻠﻟ ﻥ
G
ﺃﻴﻀﺎﹰ ﺯﻤﺭﺓ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ﻟﺘﻜﻥ
{ }
{ }
1
1
...
1
...
1
n
m
G
G
G
N
N
N
=
=
>
>
>
>
>
>
ﻤ
ﺘﺴﻠﺴﻠﺘﻴ
ـﻥ ﻗﺎﺒﻠﺘﻴﻥ ﻟﻠﺤل ﻟ
G
ﻭ
N
ﻭ ﻟﺘﻜﻥ،
i
G
ـ ﺍﻟﺼﻭﺭﺓ ﺍﻟﻌﻜﺴﻴﺔ ﻟ
i
G
ﻓﻲ
G
.
ﻋﻨﺩﺌﺫ
1
1
i
i
i
G G
G G
+
+
)
ﺍﻨﻅﺭ
47.1
(
ﻭ ﻟﺫﻟﻙ،
( )
1
1
...
...
n
m
G
G
G
N
N
N
=
>
>
>
>
>
>
ﻫﻲ ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
ﻓﻲ
G
.
ﻨﺘﻴﺠﺔ
7.6
ﻜل
p
-
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺭﺓـﻤﺯﻟﺍ ﺔـﺒﺘﺭ ﻰـﻠﻋ ﺀﺍﺭﻘﺘـﺴﻻﺍ ﻡﺩﺨﺘﺴﻨ
G
.
ﻰـﻟﺇ ﺓﺩﻭﻌﻟﺎـﺒ
(4.16)
ﻭﻥـﻜﻴ ،
ﺍﻟﻤﺭﻜﺯ
( )
Z G
ﻟﻠﺯﻤﺭﺓ
G
ﻏﻴﺭ ﺘﺎﻓﻪ
ﻭﻤﻥ ﺍﻟﻔﺭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻲ ﻴﻨﺘﺞ ﺒﺄﻥ،
( )
G Z G
ﺔـﻠﺒﺎﻗ
ﻟﻠﺤل
.
ﻷﻥ
( )
Z G
ﻭﻴﺒﻴﻥ،ﺔﻴﻠﻴﺩﺒﺘ
(b)
ﻓﻲ ﺍﻟﻘﻀﻴﺔ ﺒﺄﻥ
G
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
.
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﻤﺒﺎﺩل
ﺍﻟﻌﻨﺼﺭﻴﻥ
,
x y
G
∈
ﻫﻭ
[
]
( )
1
1
1
,
x y
xyx y
xy yx
−
−
−
=
=
ﻟﺫﻟﻙ
[
]
,
1
,
x y
xy
yx
= ⇔
=
ﻭﺘﻜﻭﻥ
G
ﺘﺒﺩﻴﻠﻴﺔ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻜل ﻤﺒﺎﺩل ﻴﺴﺎﻭﻱ
1
.
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٩٠١
ﻤﺜﺎل
8.6
ﻪـﺘﻨﻤ ﺩـﻌﺒ ﻭﺫ ﻲـﻬﺠﺘﻤ ﺀﺎﻀـﻓ ﻱﻷ
V
لـﻘﺤﻟﺍ ﻰـﻠﻋ
K
ﺎﻡـﺘ ﻡـﻠﻌﻤ ﻱﺃ ﻭ
{
}
1
,
,...
n
n
F
V V
−
=
ﻓﻲ
V
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ،
( )
( )
( )
{
}
Aut
,
j
j
B F
V
V
V
j
α
α
=
∈
⊂
∀
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
ﻟﺘﻜﻥ
( )
U F
ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻌﺭﻓﺔ ﻓﻲ ﺍﻟﻤﺜﺎل
5.10
.
ﻋﻨﺩﺌﺫ
( ) ( )
B F
U F
ﺘﺒﺩ
ﻋﻨﺩﻤﺎ،ﻭ ،ﺔﻴﻠﻴ
=
p
K
F
،
( )
U F
ﺘﻜﻭﻥ
p
-
ﺯﻤﺭﺓ
.
ﻫﺫﺍ ﻴﺒﺭﻫﻥ ﺒﺄﻥ
( )
B F
ﺔـﻟﺎﺤ ﻲـﻓ لـﺤﻠﻟ ﺔـﻠﺒﺎﻗ
=
p
K
F
ﺔـﻴﺌﺯﺠﻟﺍ ﺭـﻤﺯﻟﺍ ﺎـﻬﻨﻤ ﺓﺩﺤﺍﻭ ﻑﺭﻌﺘ ﺔﻤﺎﻌﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓﻭ ،
0
1
...
B
B
⊃
⊃
ﻲـﻓ
( )
B F
ﻤﻊ
( )
( )
{
}
,
α
α
−
=
∈
⊂
∀
i
j
j i
B
B F
V
V
j
ﻭ ﻨﻼﺤﻅ ﺒﺄﻥ ﻤﺒﺎﺩل ﻋﻨﺼﺭﻴﻥ ﻓﻲ
i
B
ﻴﻘﻊ ﻓﻲ
1
i
B
+
.
ﻷﻱ ﺘﺸﺎﻜل
:G
H
ϕ
→
[
]
(
)
(
)
( ) ( )
1
1
,
,
,
x y
xyx y
x
y
ϕ
ϕ
ϕ
ϕ
−
−
=
=
،ﺃﻱ ﺃﻥ
ϕ
ﻴﻨﻘل ﻤﺒﺎﺩل ﺍﻟﻌﻨﺼﺭﻴﻥ
,
x y
ﺼﺭﻴﻥـﻨﻌﻟﺍ لﺩﺎـﺒﻤ ﻰﻟﺇ
( ) ( )
,
x
y
ϕ
ϕ
.
ﺸﻜلـﺒ
ﻨﺭﻯ ﺒﺄﻥ،ﺹﺎﺨ
H
ﻓﺈﻥﺫﺌﺩﻨﻋ ،ﺔﻴﻠﻴﺩﺒﺘ
ϕ
ﻴﻨﻘل ﻜل ﺍﻟﻤﺒﺎﺩﻻﺕ ﻓﻲ
G
ﺇﻟﻰ
1
.
ﺭﺓــﻤﺯﻟﺍ ﻥﺇ
( )
1
G
G
′ =
ﻲــﻓ ﺕﻻﺩﺎــﺒﻤﻟﺎﺒ ﺓﺩــﻟﻭﻤﻟﺍ
G
ﺔــﻟﺩﺎﺒﻤﻟﺍ ﺓﺭــﻤﺯﻟﺍ ﻰﻤﺴــﺘ
(commutator)
ﻰـﻟﻭﻷﺍ ﺔـﺠﺭﺩﻟﺍ ﻥـﻤ ﺔﻘﺘﺸـﻤﻟﺍ ﺔـﻴﺌﺯﺠﻟﺍ ﺓﺭـﻤﺯﻟﺍ ﻭﺃ
(first derived
subgroup)
ﻓﻲ
G
.
ﻗﻀﻴﺔ
9.6
ﺍﻟﺯﻤﺭﺓ
ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﺒﺎﺩﻟﺔ
G
′
ﻫﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ ﻓﻲ
G
ﻭﻫﻲ ﺃﺼﻐﺭ ﺯﻤﺭﺓ،
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺒﺤﻴﺙ ﺘﻜﻭﻥ
G G
′
ﺘﺒﺩﻴﻠﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺫﺍﺘﻲ ﻟﻠﺯﻤﺭﺓ
G
ـﻴﻨﻘل ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﻭﻟﺩﺓ ﻟ
G
′
ﺇﻟﻰ
G
′
ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﺘﻨﺘﻘل،
G
′
ﺇﻟﻰ
G
′
.
ـﻭﺒﻤﺎ ﺃﻥ ﻫﺫﺍ ﺼﺤﻴﺢ ﻟﻜل ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﻟ
G
ﻓﺈﻥ،
G
′
ﺯﻤﺭﺓ ﻤﻤﻴﺯﺓ
.
ﻨﻜﺘﺏ
g
g
a
ﻟﻠﺘﺸﺎﻜل
:
g
gG G
G G
′
′
→
a
.
ﻋﻨﺩﺌﺫ
[ ]
,
,
g h
g h
=
ﺍﻟﺫﻱ،
ﻴﺴﺎﻭﻱ
1
ﻷﻥ
[ ]
,
g h
G
′
∈
.
ﻭﺒﺎﻟﺘﺎﻟﻲ
,
1
g h
=
ﻟﻜل
,
g h
G G
′
∈
ﺍﻟﺫﻱ ﻴﺒﻴ،
ﺄﻥـﺒ ﻥ
G G
′
ﺘﺒﺩﻴﻠﻴﺔ
.
ﻟﺘﻜﻥ
N
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻥـﻜﺘ ﺙﻴﺤﺒ
G N
،ﺔـﻴﻠﻴﺩﺒﺘ
ﻭﻤﻨﻪ
[ ]
,
g h
N
∈
.
ﺒﻤﺎ ﺃﻥ ﻫﺫﻩ ﺍﻟﻌﻨﺎﺼﺭ ﺘﻭﻟﺩ
G
′
،
N
G
′
⊃
.
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٠١١
ﻟﻜل
5
n
≥
،
n
A
ﻫﻲ ﺃﺼﻐﺭ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
n
S
ﺔـﻴﻠﻴﺩﺒﺘ ﺔﻤﺴﻗ ﺓﺭﻤﺯ ﻲﻁﻌﺘ
.
ﺒﺎﻟﺘﺎﻟﻲ
( )
n
n
S
A
′ =
.
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌ
ﻴﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ
(second derived subgroup)
ﺭﺓـﻤﺯﻠﻟ
G
ﻫﻲ
( )
G
′
′
ﻭﺍﻟﺜﺎﻟﺜﺔ،
(third)
ﻫﻲ
( )
( )
3
G
G
′
′′
=
ﻭﻫﻜﺫﺍ،
.
ﺯﺓـﻴﻤﻤﻟﺍ ﺔﻴﺌﺯﺠﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻥﺃ ﺎﻤﺒ
ﻟﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ ﺘﻜﻭﻥ ﻤﻤﻴﺯﺓ
(6.3a)
ﺘﻜﻭﻥ ﻜل ﺯﻤﺭ،
ﻲـﻓ ﺓﺯـﻴﻤﻤ ﺓﺭﻤﺯ ﺔﻘﺘﺸﻤ ﺔﻴﺌﺯﺠ ﺓ
G
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ
( )
( )
1
2
...
G
G
G
⊃
⊃
⊃
ﻭ ﺍﻟﺘﻲ ﺘﺩﻋﻰ ﺒﺎﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﻤﺸﺘﻘﺔ
(derived series)
ﻓﻲ
G
.
ﺩﻤﺎـﻨﻋ ،ﹰﻼﺜﻤ
5
n
≥
ﻭﻥـﻜﺘ ،
ﺍﻟﻤﺘﺴﻠﺴ
ﻠﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺯﻤﺭﺓ
n
S
...
n
n
n
n
S
A
A
A
⊃
⊃
⊃
⊃
ﻗﻀﻴﺔ
10.6
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺸﺘﻘﺔـﻤﻟﺍ ﺔﻴﺌﺯﺠﻟﺍ ﺓﺭﻤﺯﻟﺍ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ لﺤﻠﻟ ﺔﻠﺒﺎﻗ
( )
k
G
ﺘﺴﺎﻭﻱ
1
ﻟﺒﻌﺽ ﺍﻷﻋﺩﺍﺩ
k
.
ﺍﻟ
ﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
( )
1
k
G
=
ﺭﺓـﻤﺯﻠﻟ لﺤﻠﻟ ﺔﻠﺒﺎﻗ ﺔﻠﺴﻠﺴﺘﻤ ﻲﻫ ﺔﻘﺘﺸﻤﻟﺍ ﺔﻠﺴﻠﺴﺘﻤﻟﺍ ﻥﺈﻓ ﺫﺌﺩﻨﻋ ،
G
.
ﻟﺘﻜﻥ،ﺱﻜﻌﻟﺎﺒﻭ
...
1
s
G
G
G
G
G
=
=
>
>
> >
ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻟﻠﺯﻤﺭﺓ
G
.
ﺒﻤﺎ ﺃﻥ
1
G G
، ﺘﺒﺩﻴﻠﻴﺔ
1
G
G
′
⊃
.
ﺍﻵﻥ ﺇﻥ
2
G G
′
ﺯﻤﺭﺓ
ﺠﺯﺌﻴﺔ ﻓﻲ
1
G
ﻭﻤﻥ،
2
2
2
1
2
G G
G
G G G
G G
′ ′
′
→
⊂
I
ﻨﺭﻯ ﺒﺄﻥ
1
2
G G
ﺘﺒﺩﻴﻠﻴﺔ
⇐
2
G G
G
′ ′
I
ﺘﺒﺩﻴﻠﻴﺔ
⇐
2
2
G
G
G
G
′′
′
⊂
⊂
I
.
ﻨﺠﺩ ﺒﺄﻥ،ﺔﻘﻴﺭﻁﻟﺍ ﻩﺫﻬﺒ ﺔﻌﺒﺎﺘﻤﻟﺎﺒ
( )
i
i
G
G
⊂
ﻟﻜل
i
ﻭﺒﺎﻟﺘﺎﻟﻲ،
( )
1
s
G
=
.
ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ،ﺍﺫﻬﻟ
G
ﺍﻟﻘﺎﺒﻠﺔ ﻟ
ﺩـﻴﺩﺤﺘﻟﺎﺒ ،ﺔـﻴﻨﻭﻨﺎﻗ لـﺤﻠﻟ ﺔـﻠﺒﺎﻗ ﺔﻠﺴﻠﺴﺘﻤ ﻰﻠﻋ ﻱﻭﺤﺘ لﺤﻠ
ﻭﺍﻟﺘﻲ ﻓﻴﻬﺎ ﻜل ﺍﻟﺯﻤﺭ ﻨﺎﻅﻤﻴﺔ ﻓﻲ،ﺔﻘﺘﺸﻤﻟﺍ ﺔﻠﺴﻠﺴﺘﻤﻟﺍ
G
.
ﺄﻥـﺒ ﻥﻴـﺒﻴ ﺔﻴﻀـﻘﻟﺍ ﻥﺎـﻫﺭﺒ ﻥﺇ
ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻫﻲ ﺃﻗﺼﺭ ﻤﺘﺴﻠﺴﻠﺔ ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻟﻠﺯﻤﺭﺓ
G
.
ﺎﻟﻁﻭل ﺍﻟﻘﺎـﺒ ﺎﻬﻟﻭﻁ ﻰﻤﺴﻴ
لـﺒ
ﻟﻠﺤل ﻟﻠﺯﻤﺭﺓ
G
.
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ﺍﻟﺯﻤﺭ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
(Nilpotent groups)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
.
ﻨﺘﺫﻜﺭ ﺒﺄﻥ
( )
Z G
ﻫﻲ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
G
.
ﺘﻜﻥـﻟ
( )
2
Z
G
G
⊂
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
ـ ﺍﻟﻤﻘﺎﺒل ﻟ
( )
(
)
( )
Z G Z G
G Z G
⊂
.
ﻟﻬﺫﺍ
( )
[
]
( )
2
,
g
Z
G
g x
Z G
∈
⇔
∈
ﻟﻜل
x
G
∈
ﻨﺤﺼل،ﺔﻘﻴﺭﻁﻟﺍ ﻩﺫﻬﺒ ﺔﻌﺒﺎﺘﻤﻟﺎﺒ
ﻋﻠﻰ ﻤﺘﺘﺎﻟﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
)
ﻤﺘﺴﻠﺴﻠﺔ ﻤﺭﻜ
ﺩﺓـﻴﺍﺯﺘﻤ ﺔـﻴﺯ
(
(ascending central series)
{ }
( )
( )
2
1
...
Z G
Z
G
⊂
⊂
⊂
ﺤﻴﺙ
( )
[
]
( )
1
,
i
i
g
Z
G
g x
Z
G
−
∈
⇔
∈
ﻟﻜل
x
G
∈
.
ﺇﺫﺍ ﻜﺎﻥ
( )
m
Z
G
G
=
ﻟﺒﻌﺽ
m
ﺄﻥـﺒ لﺎﻘﻴ ﺫﺌﺩﻨﻋ ،
G
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩﻋ
(nilpotent)
،
ﻭﻴﺴﻤﻰ ﺃﺼﻐﺭ ﻋﺩﺩ ﻤﺜل
m
ﺼﻑ
)
ﺍﻻﻨﻌﺩﺍﻡ
(
(nilpotency)
ﻟﻠﺯﻤﺭﺓ
G
.
لـﻜ ،ﹰﻼﺜـﻤ
p
-
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
)
ﺒﺘﻁﺒﻴﻕ
16.4
.(
{ }
1
ﻫﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻭﺤﻴﺩﺓ ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﺼﻑ
0
ﺼﻑـﻟﺍ ﻥﻤ ﻲﺘﻟﺍ ﺭﻤﺯﻟﺍ ﻭ ،
1
ﺭـﻤﺯﻟﺍ ﻲـﻫ
ﺍﻟ
ﺘﺒﺩﻴﻠﻴﺔ ﺘﻤﺎﻤﺎﹰ
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ
( )
G Z G
ﺘﺒﺩﻴﻠﻴﺔ
-
ﺎلـﻘﻴ ﺎﻤﻜ
ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﻓﻭﻕ ﺁﺒﻠﻴﺔ
(metabelian)
.
ﻤﺜﺎل
11.6
(a)
ﻟﻜﻥ ﺍﻟﻌﻜﺱ ﻟﻴﺱ،لﺤﻠﻟ ﺔﻠﺒﺎﻗ ﺓﺭﻤﺯ ﻲﻫ ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ ﺓﺭﻤﺯﻟﺍ ﻥﺃ ﺢﻀﺍﻭﻟﺍ ﻥﻤ
ﺼﺤﻴﺤﺎﹰ
.
ﻤﻥ ﺃﺠل ﺍﻟﺤﻘل،ﹰﻼﺜﻤ
F
ﻟﻴﻜﻥ،
( )
{
}
, ,
,
0
0
a
b
B
a b c
F ac
c
=
∈
≠
ﻋﻨﺩﺌﺫ
( )
{
}
0
Z B
aI a
=
≠
ﻭ ﻴﻜﻭﻥ،
ﻤﺭﻜﺯ
( )
B Z B
ﺘﺎﻓﻬﺎﹰ
.
ﻟﺫﻟﻙ
( )
B Z B
ﻟﻴﺴﺕ
ﻟﻜﻥ ﻨﻘﻭل ﻓﻲ،ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ
(5.6)
ﺃﻨﻬﺎ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
(b)
ﺍﻟﺯﻤﺭﺓ
1
0
1
0
0 1
G
∗ ∗
=
∗
ﻓﻭﻕ ﺁﺒﻠﻴﺔ
:
ﻤﺭﻜﺯﻫﺎ
1
0
0
1
0
0
0
1
∗
ﻭ،
( )
G Z G
ﺘﺒﺩﻴﻠﻴﺔ
.
(c)
ﺔـﻴﻠﺒﺁ ﺭﻴﻏ ﺓﺭﻤﺯ ﻱﺃ
G
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
3
p
ﺔـﻴﻠﺒﺁ ﻕﻭـﻓ ﻥﻭـﻜﺘ
.
،ﺔـﻘﻴﻘﺤﻟﺍ ﻲـﻓ
( )
G
Z G
′ =
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
)
ﺍﻨﻅﺭ
17.5
(
ﻭ،
G G
′
ﺘﺒﺩﻴﻠﻴﺔ
(18.4)
.
ﺯﻤﺭ،ﺹﺎﺨ لﻜﺸﺒ
ﺩﻴﻬﻴﺩﺭﺍل ﻭﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
8
،
Q
ﻭ
4
D
ﻓﻭﻕ ﺁﺒﻠﻴﺔ،
.
ﺯﻤﺭﺓ ﺩﻴﻬﻴﺩﺭﺍل
2
n
D
ﺔـﻤﻴﺩﻋ
ﺍﻟﻘﻭﻯ
ﻤﻥ ﺍﻟﺼﻑ
n
-
ﺒﺎﺴﺘﺨﺩﺍﻡ ﺃﻥ،ﺀﺍﺭﻘﺘﺴﻻﺎﺒ ﺍﺫﻫ ﻥﻫﺭﺒﻴ ﻥﺃ ﻥﻜﻤﻴ
( )
2
n
Z D
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭ،
( )
1
2
2
2
n
n
n
D
Z D
D
−
≈
.
ﺇﺫﺍ ﻟﻡ ﻴﻜﻥ
n
ﻗﻭﻯ ﻟﻠﻌﺩﺩ
2
ﻋﻨﺩﺌﺫ،
n
D
ﺔـﻤﻴﺩﻋ ﺕﺴﻴﻟ
ﺍﻟﻘﻭﻯ
)
ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻤﺒﺭﻫﻨﺔ
17.6
ﺍﻟﻘﺎﺩﻤﺔ
.(
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ﻗﻀﻴﺔ
12.6
(a)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﻫﻲ ﺯﻤﺭﺓ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
(b)
ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ ﻟﺯﻤﺭﺓ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﻫﻲ ﺯﻤﺭﺓ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
(a)
ﻟﺘﻜﻥ
H
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩﻋ ﺓﺭﻤﺯﻟﺍ ﻥﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
.
،ﺢ ﺃﻥـﻀﺍﻭﻟﺍ ﻥـﻤ
( )
( )
Z H
Z G
H
⊃
I
.
ﺒﻔﺭﺽ
)
ﺒﺎﻻﺴﺘﻘﺭﺍﺀ
(
ﺃﻥ
( )
( )
i
i
Z
H
Z
G
H
⊃
I
ﺫـﺌﺩﻨﻋ ،
( )
( )
1
1
i
i
Z
H
Z
G
H
+
+
⊃
I
ﻷﻥ،
)
ﻟﻜل
h
H
∈
(
( )
[ ]
( )
[ ]
( )
1
,
,
,
,
i
i
i
h
Z
G
h x
Z
G
x
G
h x
Z
H
x
H
+
∈
⇒
∈
∈ ⇒
∈
∈
.
(b)
ﺒﺸﻜل ﻤﺒﺎﺸﺭ
.
ﻤﻼﺤﻅﺔ
13.6
ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﻼﺤ
ﻅ ﺒﺄﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻭﻥـﻜﻴ ﻥﺃ ﻥـﻜﻤﻤﻟﺍ ﻥﻤ ﺫﺌﺩﻨﻋ ،
( )
Z H
ﺃﻜﺒﺭ ﻤﻥ
( )
Z G
.
ﺇﻥ ﻤﺭﻜﺯ،ﹰﻼﺜﻤ
( )
{
}
( )
2
0
0
0
a
H
ab
GL
F
b
=
≠ ⊂
ﻫﻲ
H
ﻟﻜﻥ ﻤﺭﻜ،ﺎﻬﺴﻔﻨ
ﺯ
( )
2
GL F
ﻴﺤﻭﻱ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﻤﺼﻔﻭﻓﺔ ﺍﻟﺴﻠﻤﻴﺔ
.
ﻗﻀﻴﺔ
14.6
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﻤﻥ ﺍﻟﺼﻑ
m
≥
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
[
]
1
2
3
1
...
,
,
,...,
1
m
g g
g
g
+
=
ﻟﻜل
1
2
1
,
,...,
m
g g
g
G
+
∈
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺘﺫﻜﺭ ﺒﺄﻥ
،
( )
[
]
( )
1
,
i
i
g
Z
G
g x
Z
G
−
∈
⇔
∈
ﻟﻜل
x
G
∈
.
ﺒﻔﺭﺽ ﺃﻥ
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﻤﻥ ﺍﻟﺼﻑ
m
≥
ﻋﻨﺩﺌﺫ،
( )
[
]
( )
[
]
( )
[
]
( )
[
]
1
1
2
1
2
2
1
2
3
1
2
3
1
2
3
1
1
2
3
1
1
,
,
,
,
,
,
,
,
......
...
,
,
,...,
,
,...,
...
,
,
,...,
1,
,...,
m
m
m
m
m
m
m
G
Z
G
g g
Z
G
g g
G
g g
g
Z
G
g g
g
G
g g
g
g
Z G
g
g
G
g g
g
g
g
g
G
−
−
+
=
⇒
∈
∈
⇒
∈
∈
⇒
∈
∈
⇒
=
∈
ﻟﻴﻜﻥ،ﺱﻜﻌﻟﺍ لﺠﺃ ﻥﻤ
1
g
G
∈
.
ﻋﻨﺩﺌﺫ
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٣١١
[
]
[
]
( )
[
]
( )
( )
1
2
3
1
1
2
1
1
2
3
1
2
1
2
3
1
1
1
1
1
...
,
,
,...,
,
1,
,
,...,
...
,
,
,...,
,
,...,
...
,
,
,...,
,
,...,
......
,
m
m
m
m
m
m
m
m
g g
g
g
g
g g
g
G
g g
g
g
Z G
g
g
G
g g
g
g
Z
G
g
g
G
g
Z
G
g
G
+
+
−
−
=
∈
⇒
∈
∈
⇒
∈
∈
⇒ ∈
∈
، ﺃﻱ ﺃﻥ،ﻯﻭﻘﻟﺍ ﻡﻴﺩﻋ ﹰﺍﺩﻴﺩﻤﺘ ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ ﺭﻤﺯﻟ ﺩﻴﺩﻤﺘﻟﺍ ﻥﻭﻜﻴ ﻥﺃ ﺓﺭﻭﺭﻀﻟﺎﺒ ﺱﻴﻟ
(20)
N
ﻭ
G N
ﻋﺩﻴﻡ ﺍﻟﻘﻭﻯ
⇐/
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﻤﺜﻼﹰ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ،
U
ﻓﻲ ﺍﻟﺯﻤﺭﺓ
B
ﻓﻲ ﺍﻷﻤﺜﻠﺔ
5.6
ﻭ
11.6
ﺘﺒﺩﻴﻠﻴﺔ ﻭ
B U
، ﺘﺒﺩﻴﻠﻴﺔ
ﻟﻜﻥ
B
ﻟﻴﺴﺕ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﺍﻻﻗﺘﻀﺎﺀ،لﺎﺤ ﻱﺃ ﻰﻠﻋ
(20)
ﻤﺤﻘﻕ ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
N
ﻤﺤﺘﻭﺍﺓ ﻓﻲ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
G
.
ﻓﻲ
ﻟﺩﻴﻨﺎ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻷﻜﺜﺭ ﺘﻔﺼﻴﻼﹰ،ﺔﻘﻴﻘﺤﻟﺍ
ﺍﻵﺘﻴ
.ﺔ
ﻨﺘﻴﺠﺔ
15.6
ﻟﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
N
ﻤﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ
G
،
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘ
ﻭﻯ ﻤﻥ ﺍﻟﺼﻑ
1
m
+ ≥
⇒
G N
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﻤﻥ ﺍﻟﺼﻑ
m
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻜﺘﺏ
π
ﻟﻠﺘﻁﺒﻴﻕ
G
G N
→
.
ﻋﻨﺩﺌﺫ
[
]
]
(
)
[
]
]
1
2
3
1
1
2
3
1
...
,
,
,...,
,
...
,
,
,...,
,
1,
m
m
m
m
g g
g
g
g
g
g
g
g
g
π
π
π
π
π
π
+
+
=
=
1
2
1
,
,...,
m
g g
g
G
+
∈
.
ﺎﻟﻲــﺘﻟﺎﺒ
[
]
]
( )
1
2
3
1
...
,
,
,...,
,
m
m
g g
g
g
g
N
Z G
+
∈
⊂
،
ﻭﻤﻨﻪ
[
]
]
1
2
3
1
2
...
,
,
,...,
,
1
m
m
g g
g
g
g
+
+
=
ﻟﻜل
1
2
,...,
m
g
g
G
+
∈
.
ﻨﺘﻴﺠﺔ
16.6
ﻜل
p
-
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ
G
.
ﻷﻥ
( )
1
Z G
≠
،
( )
G Z G
ﺔـﻤﻴﺩﻋ
ﻭﺍﻟﺘﻲ ﺘﻘﺘﻀﻲ ﺒﺄﻥ،ﻯﻭﻘﻟﺍ
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﺍﻟﺘﻤﺩﻴﺩ
1
1
N
G
Q
ι
π
→
→
→ →
ﻴﻜﻭﻥ ﻤﺭﻜﺯﻴﺎﹰ ﺇﺫﺍ ﻜﺎﻥ
( )
( )
N
Z G
ι
⊂
.
ﻋﻨﺩﺌﺫ
:
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ﺇﻥ ﺍﻟﺯﻤﺭ ﻋﺩﻴﻤﺔ
ﺔـﻴﺯﻜﺭﻤ ﺕﺍﺩـﻴﺩﻤﺘﺒ ﺎﻬﻴﻠﻋ لﻭﺼﺤﻟﺍ ﻥﻜﻤﻴ ﻲﺘﻟﺍ ﺭﻤﺯﻟﺍ ﻙﻠﺘ ﻲﻫ ﻯﻭﻘﻟﺍ
ﻤﺘﺘﺎﻟﻴﺔ
.
ﺒﺎﻟﻤﻘﺎﺒل
:
ﺔـﻴﻠﻴﺩﺒﺘ ﺭـﻤﺯ ﻥـﻤ ﺎـﻬﻴﻠﻋ لﻭﺼﺤﻟﺍ ﻥﻜﻤﻴ ﻲﺘﻟﺍ ﺭﻤﺯﻟﺍ ﻙﻠﺘ ﻲﻫ لﺤﻠﻟ ﺔﻠﺒﺎﻘﻟﺍ ﺭﻤﺯﻟﺍ ﻥﺇ
ﺒﺘﻤﺩﻴﺩﺍﺕ ﻤﺘﺘﺎﻟﻴﺔ
)
ﻟﻴﺱ ﺒﺎﻟﻀﺭﻭﺭﺓ ﺃﻥ ﺘﻜﻭﻥ ﻤﺭﻜﺯﻴﺔ
.(
ﻤﺒﺭﻫﻨﺔ
17.6
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘ
ﺭـﺸﺎﺒﻤﻟﺍ ﺀﺍﺩﺠﻟﺍ ﻱﻭﺎﺴﺘ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ ﺔﻴﻬ
ﻟﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻴﻬﺎ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻥ ﺍﻟﺠﺩﺍﺀ ﺍﻟﻤﺒﺎﺸﺭ ﻟﺯﻤﺭ ﻋﺩﻴ
ﻭ،ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ ﺓﺭﻤﺯ ﻭﻫ ﻯﻭﻘﻟﺍ ﺔﻤ
ﺫﻟﻙـﻟ
ﻓﺈﻥ ﺍﻻﺘﺠﺎﻩ ﺍﻷﻭل ﻴﺄﺘﻲ ﻤﻥ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺴﺎﺒﻘﺔ
.
ﻟﺘﻜﻥ،ﺱﻜﻌﻟﺍ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻋ
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩ
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ
(9.5)
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ ﻜل ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
ﻟﺘﻜﻥ
P
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
ﻤﺎ ﻤﻥ
G
ﻭ ﻟﺘﻜﻥ،
( )
G
N
N
P
=
.
ﺘﺒﻴﻥ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
ﺍﻵﺘﻴﺔ
ﺄﻥـﺒ
( )
G
N
N
N
=
ﻴﻥـﺒﺘﻭ ،
ﺍﻟﺜﺎ
ﻨﻴﺔ ﺒﺄﻥ
N
G
=
، ﺃﻱ ﺃﻥ،
P
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﺘﻤﻬﻴﺩﻴﺔ
18.6
ﻟﺘﻜﻥ
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
.
ﻷﻱ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
H
ﻓﻲ
G
ﺘﺤﺘﻭﻱ ﻋﻠﻰ
( )
G
N
P
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ،
( )
G
N
H
H
=
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
( )
G
g
N
H
∈
ﻭﻤﻨﻪ،
1
gHg
H
−
=
.
ﻋﻨﺩﺌﺫ
1
−
′
⊃
=
H
gPg
P
ﻭﺍﻟﺘﻲ،
ﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ
p
-
ﺭﺓـﻤﺯﻟﺍ ﻲﻓ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺭﺓﻤﺯ
H
.
ﻴﻠﻭـﺴ ﻥـﻤ
II
،
1
hP h
P
−
′
=
ﺒﻌﺽــﻟ
h
H
∈
ﻪــﻨﻤﻭ ،
1
1
hgPg h
P
−
−
⊂
.
ﺎﻟﻲــﺘﻟﺎﺒ
( )
G
hg
N
P
H
∈
⊂
ﻭ،
ﻪــﻨﻤ
g
H
∈
.
ﺘﻤﻬﻴﺩﻴﺔ
19.6
ﻟﺘﻜﻥ
H
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩﻋ ﺔـﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ ﻥﻤ ﺔﻴﻠﻌﻓ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
ﺫـﺌﺩﻨﻋ ،
( )
G
H
N
H
≠
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺫﻟـﻟﻭ ،ﺔـﻴﻠﻴﺩﺒﺘﻟﺍ ﺭﻤﺯﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ ﺔﺤﻴﺤﺼ ﺢﻀﺍﻭ لﻜﺸﺒ ﻥﻭﻜﺘ ﺔﻟﺎﺤﻟﺍ ﻥﺇ
ﻥ ﺃﻥـﻜﻤﻴ ﻙ
ﻨﻔﺭﺽ ﺒﺄﻥ
G
ﻏﻴﺭ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻨﺴﺘﺨﺩﻡ ﺍﻻﺴﺘﻘﺭﺍﺀ ﻋﻠﻰ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺓ
G
.
ﻷﻥ
G
، ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
( )
1
Z G
≠
.
ﺭﺓـﻤﺯﻟﺍ ﺭﺼﺎﻨﻋ ﻥﺇ
( )
Z G
ﻨﻅﻡـﺘ
H
ﺎﻥـﻜ ﺍﺫﺇ ﻪـﻨﻤﻭ ،ﻥﻴـﻌﻤ لﻜﺸـﺒ
( )
Z G
H
⊆
ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ،
( )
( )
.
G
H
Z G H
N
H
≠
⊂
⊂
.
ﺄﻥـﺒ ﺽﺭـﻔﻨ ﻥﺃ ﻥﻜﻤﻴ ﺍﺫﻬﻟ
( )
Z G
H
⊂
.
ﻓﺈﻥ ﻤﻨﻅﻡ ﺍﻟﺯﻤﺭﺓﺫﺌﺩﻨﻋ
H
ﻓﻲ
G
ﺴﺒـﻨﻟﺎﺒ ﻕﺒﺎﻁﺘﻴ
ﺔ
(46.1)
ﻨﻅﻡـﻤ ﻊـﻤ
( )
H Z G
ﻓﻲ
( )
G Z G
ﻭ ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﻔﺭﻭﺽ ﺍﻻﺴﺘﻘﺭﺍﺌﻴﺔ،
.
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٥١١
ﻤﻼﺤﻅﺔ
20.6
ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
ﻨﺴﺘﻨﺘﺞ ﺍﻟﺤﻘﻴﻘﺔ ﺒﺄﻥ
G
ــ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟ
p
-
ﺯﻤﺭ ﺠﺯﺌﻴﺔ ﺃﻭﻟﻴﺔ ﻓﻴﻬﺎ
.
ﻗﻀﻴﺔ
21.6
)
ﻤﻨﺎﻗﺸﺔ ﻓﺭﺍﺘﻴﻨﻲ
(
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
G
ﺘﻜﻥـﻟﻭ ،
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
P
ﻓﻲ ﺍﻟﺯﻤﺭﺓ
H
.
ﻋﻨﺩﺌﺫ
( )
.
G
G
H N
P
=
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
g
G
∈
.
ﻋﻨﺩﺌﺫ
1
1
gPg
gHg
H
−
−
⊂
=
ﻭ ﻜل ﻤﻥ،
1
gPg
−
ﻭ
P
ﻋﺒﺎﺭﺓ
ﻋﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴ
ﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
H
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ ﺴﻴﻠﻭ
II
ﻴﻭﺠﺩ،
h
H
∈
ﺙـﻴﺤﺒ
ﻴﻜﻭﻥ
1
1
gPg
hPh
−
−
=
ﻭﺒﺎﻟﺘﺎﻟﻲ،
( )
1
G
h g
N
P
−
∈
ﻭﻤﻨﻪ
( )
.
G
g
H N
P
∈
.
ﻤﺒﺭﻫﻨﺔ
22.6
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﺇﺫﺍ ﻭ
ﺔـﻴﻠﻌﻓ ﺔﻴﻤﻅﻋﺃ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ لﻜ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓ
ﻨﺎﻅﻤﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻗﻠﻨﺎ ﻓﻲ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
6.19
ﺒﺄﻥ ﻷﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻌﻠﻴﺔ
H
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩﻋ ﺓﺭﻤﺯ ﻥﻤ
G
،
( )
G
H
N
H
≠
⊂
.
،ﺒﺎﻟﺘﺎﻟﻲ
H
ﺃﻋﻅﻤﻴﺔ
( )
G
N
H
G
= ⇐
،ﺃﻱ ﺃﻥ
H
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﺒﻔﺭﺽ ﺃﻥ ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻌﻠﻴﺔ ﻓﻲ،ﺱﻜﻌﻟﺍ
G
ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ
.
ﻥـﻤ ﻕﻘﺤﺘﻨـﺴ
ﺸﺭﻭﻁ ﺍﻟﻤﺒﺭﻫﻨﺔ
17.6
.
ﻟﺘﻜﻥ،ﺍﺫﻬﻟ
P
ﻋﺒﺎﺭﺓ ﻋﻥ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
.
ﺇﺫﺍ ﻟﻡ ﺘﻜﻥ
P
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻌﻠﻴﺔﺫﺌﺩﻨﻋ ،
H
ﻓﻲ
G
ﻭﻱـﺤﺘ
( )
G
N
P
.
ﺈﻥـﻓ ،ﺔـﻴﻤﻅﻋﺃ ﺎﻬﻨﻭﻜﻭ
H
ﺄﻥـﺒ ﻲـﻨﻴﺘﺍﺭﻓ ﺔﺸـﻗﺎﻨﻤ ﻥﻴـﺒﺘ ﻪـﻨﻤﻭ ،ﺔـﻴﻤﻅﺎﻨ
( )
.
G
G
H N
P
H
=
=
-
ﻭﻫﺫﺍ ﺘﻨﺎﻗﺽ
.
ﺍﻟﺯﻤﺭ ﺒﻤﺅﺜﺭﺍﺕ
(Groups with operators)
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ
( )
Aut G
ﻟﻠﺯﻤﺭﺓ
G
ﺭﺓـﻤﺯ ﹰﺎﻀـﻴﺃ ﻲﻫ
.
ﺘﻜﻥـﻟ
A
ﺯﻤﺭﺓ
.
ﻴﺩﻋﻰ ﺍﻟﺯﻭﺝ
(
)
,
G
ϕ
ﺍﻟﻤﺅﻟﻑ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺸﺎﻜلـﺘﻟﺍ ﻊﻤ
( )
:
Aut
A
G
ϕ
→
ـﺒ
A
-
ﺃﻭ ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ،ﺓﺭﻤﺯ
G
ﻤﻊ
A
ﺅﺜﺭﺍﺕـﻤﻟﺍ ﻥـﻤ ﺓﺭﻤﺯﻜ
(Groups with
operators)
ﻟﺘﻜﻥ
G
ﻫﻲ
A
-
ﻭﻟﻨﻜﺘﺏ،ﺓﺭﻤﺯ
x
α
ﻟﻠﺤﺩ
( )
x
ϕ α
ﻋﻨﺩﺌﺫ
( )
( )
( )
( ) ( ) ( ) ( )
( )
1
a
x
x
b
xy
x
y
c
x
x
αβ
α
β
α
α
α
=
=
=
)
ϕ
ﺘﺸﺎﻜل
(
)
( )
ϕ α
ﺘﺸﺎﻜل
(
)
ϕ
ﺘﺸﺎﻜل
(
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٦١١
ﺇﻥ ﺍﻟﺘﻁﺒﻴﻕ،ﺱﻜﻌﻟﺎﺒ
(
)
,
:
x
x A G
G
α
α
× →
a
ﻴﺤﻘﻕ
(a)
،
(b)
،
(c)
ﻥـﻤ ﺞﺘـﻨﻴﻭ
ﺍﻟﺘﺸﺎﻜل
( )
Aut
A
G
→
.
ﻴﺒﻴﻥ ﺍﻟﺸﺭﻁﺎﻥ
(a)
ﻭ
(c)
ﺃﻥ
x
x
α
a
ﺴﻲـﻜﻌﻟﺍ ﻕﻴﺒﻁﺘﻟﺍ ﻭﻫ
ـﻟ
( )
1
x
x
α
−
a
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻜﻭﻥ،
x
x
α
a
ﺘﻘﺎﺒﻼﹰ
G
G
→
.
ﻴﺒﻴﻥ ﺍﻟﺸﺭﻁﺫﺌﺩﻨﻋ
(b)
ﻪـﻨﺄﺒ
ﻰـﻠﻋ ﻲـﺘﺍﺫ لﺜﺎﻤﺘ
G
.
ﻴﻥـﺒﻴ ،ﹰﺍﺭـﻴﺨﺃ
(a)
ﻕـﻴﺒﻁﺘﻟﺍ ﻥﺄـﺒ
( )
α
ϕ α
=
a
x
x
ﺸﺎﻜلـﺘ
( )
Aut
A
G
→
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻊ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﻤﻌﺭﻓﺔ
A
.
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ
H
ﺭﺓـﻤﺯﻟﺍ ﻥـﻤ
G
ﺜﺎﺒﺘﺔ
(admissible)
ﺃﻭ ﻻ ﻤﺘﻐﻴﺭﺓ
(invariant)
ﺇﺫﺍ ﻜﺎﻥ
,
x
H
x
H
A
α
α
∈ ⇒
∈
∈
ﺇﻥ ﺘﻘﺎﻁﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺜﺎﺒﺘﺔ ﻫﻭ ﺯﻤﺭﺓ ﺜﺎﺒﺘﺔ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺎـﻫﺯﻜﺭﻤﻤ ﻭ ﺎﻬﻤﻅﻨﻤ ﻥﻤ لﻜ ﻥﺈﻓ ،ﺔﺘﺒﺎﺜ
ﺯﻤﺭﺓ ﺜﺎﺒﺘﺔ ﺃﻴﻀﺎﹰ
.
A
-
ﺘﺸﺎﻜل
)
ﺃﻭ ﺘﺸﺎﻜل ﺜﺎﺒﺕ
(
(admissible homomorphism)
ـ ﻟ
A
-
ـﺭ ﻫـﻤﺯ
ﻭ
ﺘﺸﺎﻜل
:G
G
γ
′
→
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
( )
g
g
α
α
γ
γ
=
ﻟﻜل
,
A g
G
α
∈
∈
.
ﺎلـﺜﻤ
23.6
(a)
ﻴﻤﻜﻥ ﺃﻥ ﻨﻌﺘﺒﺭ ﺯﻤﺭﺓ
G
ﻜﺯﻤﺭﺓ ﻤﻊ
{ }
1
ﻜﺯﻤﺭﺓ ﻟﻠﻤﺅﺜﺭﺍﺕ
.
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
ﺘﻜﻭﻥ ﻜل ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
ﻭﻱـﺤﺘ ﺕﺍﺭﺜﺅـﻤﻟﺍ ﻊﻤ ﺭﻤﺯﻟﺍ ﺔﻴﺭﻅﻨ ﻥﺈﻓ ﻪﻨﻤﻭ ،ﺔﺘﺒﺎﺜ ﺕﻼﻜﺎﺸﺘﻟﺍ ﻭ
ﻨﻅﺭﻴﺔ ﺍﻟﺯﻤﺭ ﺒﺩﻭﻥ ﻤﺅﺜﺭﺍﺕ
.
(b)
ﻨﻌﺘﺒﺭ
G
ﺒﺎﻋﺘﺒﺎﺭ، ﻱﺃ ،ﻕﻓﺍﺭﺘﻟﺎﺒ ﺎﻬﺴﻔﻨ ﻰﻠﻋ ﺭﻴﺜﺄﺘﻟﺍ ﺓﺭﻤﺯ
G
ﻤﻊ ﺍﻟﺘﺸﺎﻜل
( )
:
Aut
→
a
g
g
i
G
G
ﺍﻟﺯ،ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ ﻲﻓ
ﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ ﻫﻲ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ
.
(c)
ﻨﻌﺘﺒﺭ
G
ﻤﻊ
( )
Aut
A
G
=
ﻜﺯﻤﺭﺓ ﻤﻥ ﻟﻠﻤﺅﺜﺭﺍﺕ
.
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ،ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ ﻲﻓ
ﺍﻟﺜﺎﺒﺘﺔ ﻫﻲ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ
.
ﻟﻘﺩ ﺒﺭﻫﻥ ﻜل ﺸﻲﺀ ﺘﻘﺭﻴﺒﺎﹰ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭ ﻭﺘﺤﻘﻕ ﺃﻴﻀﺎﹰ ﺒ
ﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭ ﻤﻊ ﺍﻟﻤﺅﺜﺭﺍﺕ
.
ﺔـﻟﺎﺤﺒ
ﺇﻥ ﺍﻟﻤﺒﺭﻫﻨﺎﺕ،ﺔﺼﺎﺨ
44.1
،
45.1
ﻭ
46.1
ﺘﺘﺤﻘﻕ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﺭ ﺒﻤﺅﺜﺭﺍﺕ
.
،ﺔـﻟﺎﺤ لﻜ ﻲﻓ
ﻴﻜﻭﻥ ﺍﻟﺒﺭﻫﺎﻥ ﻨﻔﺴﻪ ﻜﻤﺎ ﺴﺒﻕ ﻏﻴﺭ ﺃﻨﻪ ﻴﺠﺏ ﺍﻟﺘﺄﻜﺩ ﻤﻥ ﻗﺎﺒﻠﻴﺔ ﺍﻟﺜﺒﺎﺕ
.
ﻤﺒﺭﻫﻨﺔ
24.6
ﻷﻱ ﺘﺸﺎﻜل ﺜﺎﺒﺕ
:G
G
γ
′
→
ﻥـﻤ
A
-
،ﺭـﻤﺯ
( )
def
Ker
N
γ
=
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﺜﺎﺒﺘﺔ ﻓﻲ
G
،
( )
G
γ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺜﺎﺒﺘﺔ ﻓﻲ
G
′
ﻭ،
ﻴﺘﺤﻠل
γ
ﺔـﻴﻌﻴﺒﻁ ﺔﻘﻴﺭﻁﺒ
ﺘﻤﺎﺜل ﺜ،ﺕﺒﺎﺜ ﺭﻤﺎﻏ ﻕﻴﺒﻁﺘ ﻥﻤ ﺏﻴﻜﺭﺘ ﻰﻟﺇ
ﻭ ﺘﻁﺒﻴﻕ ﻤﺘﺒﺎﻴﻥ ﺜﺎﺒﺕ،ﺕﺒﺎ
:
( )
G
G N
G
G
γ
′
→
→
→
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ﻤﺒﺭﻫﻨﺔ
25.6
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺒﻤﺅﺜﺭﺍﺕ
A
ﻭ ﻟﺘﻜﻥ،
H
ﻭ
N
،ﺎﺒﺘﺘﻴﻥـﺜ ﻥﻴﺘﻴﺌﺯﺠ ﻥﻴﺘﺭﻤﺯ
ﺤﻴﺙ
N
ﻨﺎﻅﻤﻴﺔ
.
ﻋﻨﺩﺌﺫ
H
N
I
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺜﺎﺒﺘﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
H
،
HN
ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ
ﺜﺎﺒﺘﺔ ﻓﻲ
G
ﻭ،
(
)
h H
N
hH
I
a
ﺘﻤﺎﺜل ﺜﺎﺒﺕ
H H
N
HN N
→
I
.
ﻤﺒﺭﻫﻨ
ﺔ
26.6
ﻟﻴﻜﻥ
:G
G
ϕ
′
→
ﺘﺸﺎﻜﻼﹰ ﺜﺎﺒﺘﺎﹰ ﻏﺎﻤﺭﺍﹰ ﻤﻥ
A
-
ﺯﻤﺭ
.
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﻁﺒﻴﻕ ﺍﻟﺘﻘﺎﺒل
H
H
↔
ﺒﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺍﻟﺤﺎﻭﻴﺔ ﻋﻠﻰ
( )
Ker
ϕ
ﻭ ﻤﺠﻤﻭﻋﺔ
ﺍﻟﺯﻤﺭ
ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
)
ﺍﻨﻅﺭ
64.1
(
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ ﺘﻘﺎﺒل ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ،
.
ﻟﻴﻜﻥ
( )
:
Aut
A
G
ϕ
→
ﺯﻤﺭﺓ ﺒﺎﻟﻤﺅﺜﺭ
A
.
ﺔـﺘﺒﺎﺜﻟﺍ ﺔـﻴﻤﻅﺎﻨﻟﺍ ﺕـﺤﺘ ﺔﻠﺴـﻠﺴﺘﻤﻟﺍ ﻥﺇ
(admissible subnormal series)
ﻫﻲ ﻤﺘ
ﺴﻠﺴﻠﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ ﻓﻲ
G
1
2
...
r
G
G
G
G
⊃
⊃
⊃ ⊃
ﺤﻴﺙ ﻜل
i
G
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
1
i
G
−
.
ﻨﻌﺭﻑ ﺒﺸﻜل ﻤﺸﺎﺒﻪ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ
.
ﺭـﻤﺯ ﻥﻭﻜﺘ
ﺍﻟﻘﺴﻤﺔ ﻓﻲ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﺜﺎﺒﺘ
ﺔ
A
-
ﺔـﻴﺒﻴﻜﺭﺘﻟﺍ ﺔﻠﺴﻠﺴﺘﻤﻟﺍ ﻲﻓ ﺔﻤﺴﻘﻟﺍ ﺭﻤﺯﻭ ،ﹰﺍﺭﻤﺯ
ﺍﻟﺜﺎﺒﺘﺔ
A
-
ﺃﻱ ﺃﻨ،ﺔﻁﻴﺴﺒ ﹰﺍﺭﻤﺯ
ﻪ
ﻻ ﻴﻭﺠﺩ ﺯﻤﺭ ﺠﺯﺌﻴﺔ ﺜﺎﺒﺘﺔ ﻨﺎﻅﻤﻴﺔ ﻏﻴﺭ ﺍﻟﺯﻤﺭﺘﻴﻥ ﺍﻟﺘﺎﻓﻬﺘﻴﻥ،
.
ﺇﻥ ﻤﺒﺭﻫﻨﺔ ﺠﻭﺭﺩﺍﻥ
–
ﻫﻭﻟﺩﺭ ﻤﺴﺘﻤﺭ ﻟﺘﺘﺤﻘﻕ ﻤﻥ ﺃﺠل
A
-
ﺯﻤﺭ
.
ﻭﻥـﻜﺘ ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ ﻲﻓ
ﺍﻟﺘﻤﺎﺜﻼﺕ ﺒﻴﻥ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ ﺍﻟﻤﻘﺎﺒﻠﺔ ﻟﻤﺘﺴﻠﺴﻠﺘﻴﻥ ﺘﺭﻜﻴﺒﻴﺘﻴﻥ ﺜﺎﺒﺘﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ ﻨﻔﺴﻪ ﻜﻤﺎ ﻓﻲ ﺍﻟﻤﺒﺭﻫﻨﺔ
ـ ﺍﻟﺘﻲ ﻗﺩ ﻻﺤﻅﻨﺎ ﺒﺄﻨﻬﺎ ﻤﺤﻘﻘﺔ ﺃﻴﻀﺎﹰ ﺒﺎﻟﻨﺴﺒﺔ ﻟ،ﻁﻘﻓ ﺕﻼﺜﺎﻤﺘﻟﺍ ﻡﺩﺨﺘﺴﺍ ﻪﻨﻷ ،ﺔﻤﺎﻌﻟﺍ
A
-
ﺯﻤﺭ
.
ﻤﺜﺎل
27.6
(a)
ﺒﺎﻋﺘﺒﺎ
ﺭ ﺍﻟﺯﻤﺭﺓ
G
ﺯﻤﺭﺓ ﻜﻤﺅﺜﺭ ﻋﻠﻰ ﻨﻔﺴﻬﺎ ﺒﺎﻟﺘﺭﺍﻓﻕ
.
ﺔ ﺇﻥـﻟﺎﺤﻟﺍ ﻩﺫـﻫ ﻲﻓ
ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
{ }
0
1
2
...
1
s
G
G
G
G
G
⊃
⊃
⊃
⊃ ⊃
=
ﺤﻴﺙ ﺇﻥ
ﻜل
i
G
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺃﻱ ﺃﻨﻬﺎ،
ﻤﺘﺴﻠﺴﻠﺔ ﻨﺎﻅﻤﻴﺔ،
.
ﺇﻥ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
i
G
ﺎﻟﺘﺭﺍﻓﻕـﺒ
ﻟﻴﻌﻁﻲ ﺘﺄﺜﻴﺭ،ﺔﻤﺴﻘﻟﺍ ﺭﻤﺯ ﻰﻟﺇ ﺫﻔﻨﻴ
G
ﻋﻠﻰ
1
i
i
G G
+
.
ﺇﻥ ﺯﻤﺭ
ﻲـﺘ
ﺴﻤﺔـﻘﻟﺍ
ﻟ
ﻤﺘ
ﺴﻠﺘﻴﻥـﻠﺴ
ﺘﺭﻜﻴﺒﻴﺘﻴﻥ ﺜﺎﺒﺘﺘﻴﻥ ﺘﻜﻭﻨﺎﻥ ﻤﺘﻤﺎﺜﻠﺘﻴ
ﻥ ﻜﻤﺎ
ﻓﻲ
G
-
ﺯﻤﺭ
.
(b)
ﺒﺎﻋﺘﺒﺎﺭ
G
ﻤﻊ
( )
Aut
A
G
=
ﻜﺯﻤﺭﺓ ﻤﻥ ﻤﺅﺜﺭﺍﺕ
.
ﺇﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ،ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ ﻲﻓ
ﺘﺤﺕ ﺍﻟﻨﺎﻅﻤﻴﺔ ﺍﻟﺜﺎﺒﺘﺔ ﻫﻲ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ
{ }
0
1
2
...
1
s
G
G
G
G
G
⊃
⊃
⊃
⊃ ⊃
=
ﺤﻴﺙ ﻜل
i
G
ﺯ
ﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻤﻴﺯﺓ ﻓﻲ
G
ﻭ،
ﺎﺒﺘﺘﻴﻥـﺜ ﻥﻴﺘﻴﺒﻴﻜﺭـﺘ ﻥﻴﺘﻠﺴـﻠﺴﺘﻤﻟ ﺔﻤﺴـﻘﻟﺍ ﺭﻤﺯ
ﻤﺘﻤﺎﺜﻠﺘﻴﻥ ﻜﻤﺎ ﻓﻲ
( )
Aut G
-
ﺯﻤﺭ
.
ﻤﺒﺭﻫﻨﺔ ﻜﺭﻭل
ـ
ﺴﺸﻤﻴﺩﺕ
(Krull- Schmidt theorem)
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٨١١
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺘﺤﻠﻴل ﺇﺫﺍ ﻜﺎﻥ
1
G
≠
ﻭ
G
ﺭﺘﻴﻥـﻤﺯﻟ ﺭـﺸﺎﺒﻤ ﺀﺍﺩﺠ لﺜﺎﻤﺘ ﻻ
ﺇﺫﺍ ﻜﺎﻥ،ﻪﻨﺃ ﻱﺃ ،ﻥﻴﺘﻬﻓﺎﺘ ﺭﻴﻏ
1
1
G
H
H
H
H
′
′
≈ ×
⇒
=
=
ﻤﺜﺎل
28.6
(a)
ﺴـﻴﻟ لﻴﻠﺤﺘﻠﻟ ﺔﻠﺒﺎﻘﻟﺍ ﺭﻴﻏ ﺓﺭﻤﺯﻟﺍ ﻥﻜﻟ ،لﻴﻠﺤﺘﻠﻟ ﺔﻠﺒﺎﻗ ﺭﻴﻏ ﺔﻁﻴﺴﺒﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻥﺇ
ﺕ
ﺒﺎﻟﻀﺭﻭﺭﺓ ﺃﻥ ﺘﻜﻭﻥ ﺒﺴﻴﻁﺔ
:
ﻴﻤﻜﻥ
ﺃﻥ ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
،ﻤﺜﻼﹰ
3
S
ﺯﻤﺭﺓ ﻏﻴﺭ
ﻗﺎﺒﻠﺔ ﻟﻠﺘﺤﻠﻴل ﻟﻜﻨﻬﺎ ﺘﺤﻭﻱ ﻋﻠﻰ
3
C
ﻜﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ
.
(b)
ﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺘﺤﻠﻴل ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ
ﻜﺎﻨﺕ
ﺎـﻬﺘﺒﺘﺭ ﺔـﻴﺭﺌﺍﺩ
ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ
.
ﻟﻜﻥ ﻟﻴﺱ ﻤﻥ ﺍﻟﺼﻌﺏ ﺒﺭﻫﺎﻨﻪ ﻤﺒﺎﺸﺭﺓﹰ،ﻑﻴﻨﺼﺘﻟﺍ ﻥﻤ ﹰﺎﺤﻀﺍﻭ ﺍﺫﻫ ﻥﻭﻜﻴ ،ﻊﺒﻁﻟﺎﺒ
.
ﻟﺘﻜﻥ
G
ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
p
ﻭﺒﻔﺭﺽ ﺃﻥ،
G
H
H
′
≈ ×
.
ﻴﺠﺏ ﺃﻥ ﺘﺫﺌﺩﻨﻋ
ﻥـﻤ لﻜ ﻥﻭﻜ
H
ﻭ
H
′
p
-
ﻭﻻ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻨﺎ ﻜﻼﻫﻤﺎ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،ﺓﺭﻤﺯ
m
p
،
m
n
<
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻨﻪ ﻴﺠﺏ
ﺃﻥ ﺘﻜﻭﻥ ﺇﺤﺩﺍﻫﻤﺎ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
p
ﻭﺍﻟﺜﺎﻨﻴﺔ ﺘﺎﻓﻬﺔ،
.
ﺭﺽ ﺃـﻔﺒ ،ﺱﻜﻌﻟﺎـﺒ
ﻥ
G
ﺔـﻴﻠﻴﺩﺒﺘ
ﻭﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺘﺤﻠﻴل
.
ﺒﻤﺎ ﺃﻥ ﻜل ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻨﺘﻬﻴﺔ
)
ﻤﻥ ﺍﻟﻭﺍﻀﺢ
(
ـﻋﺒﺎﺭﺓ ﻋﻥ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟ
p
-
ﺯﻤﺭ ﺤﻴﺙ
p
ﺘﻜﻭﻥ،ﺔﻴﻟﻭﻷﺍ ﺩﺍﺩﻋﻷﺍ لﻜ ﺢﺴﻤﻴ
G
p
-
ﺯﻤﺭﺓ
.
ﺼﺭـﻨﻌﻟﺍ ﻥﺎﻜ ﺍﺫﺇ
g
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺫﻭ ﺍﻟﺭﺘﺒﺔ ﺍﻷﻋﻠﻰ ﻓﻲ
G
ﺇﺤﺩﺍﻫﺎ ﺘﺒﻴﻥ ﺒﺄﻥ،
g
ﻲـﻓ ﺭـﺸﺎﺒﻤ لـﻤﺎﻋ ﻭﻫ
G
،
G
g
H
≈
×
ﻭﻫﺫﺍ ﺘﻨﺎﻗ،
ﺽ
.
(c)
لـﻴﻠﺤﺘﻠﻟ ﺔـﻠﺒﺎﻗ ﺭـﻴﻏ ﺔـﻴﺌﺯﺠ ﺭﻤﺯﻟ ﺭﺸﺎﺒﻤ ﺀﺍﺩﺠﻜ ﺏﺘﻜﺘ ﻥﺃ ﻥﻜﻤﻴ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ لﻜ
)
ﻭﺍﻀﺢ
.(
ﻤﺒﺭﻫﻨﺔ
29.6
(Krull- Schmidt)
ﺒﻔﺭﺽ ﺃﻥ
G
ﺠﺩﺍﺀ
ﺍﹰ
ﻤﺒﺎﺸﺭ
ﺍﹰ
ﺔـﻠﺒﺎﻗ ﺭـﻴﻏ ﺔﻴﺌﺯﺠ ﺭﻤﺯﻟ
ﻟﻠﺘﺤﻠﻴل
1
,...,
s
G
G
ﻭ ﺠﺩﺍﺀ ﻤﺒﺎﺸﺭ ﻟﺯﻤﺭ ﺠﺯﺌﻴﺔ ﻏﻴﺭ ﻗﺎﺒ
ﻠﺔ ﻟﻠﺘﺤﻠﻴل
1
,...,
t
H
H
:
1
1
...
,
...
s
t
G
G
G
G
H
H
× ×
× ×
ﻋﻨﺩﺌﺫ
s
t
=
ﻭ ﻫﻨﺎ ﻨﻌﻴﺩ ﺍﻟﺩﻟﻴل ﺒﺤﻴﺙ ﻴﻜﻭﻥ،
i
i
G
H
≈
.
ﺒﺈﻋﻁﺎﺀ ﻋﺩﺩ ﻤﺎ،ﻙﻟﺫ ﻥﻤ ﺭﺜﻜﺃﻭ
r
ﻴﻤﻜﻥ ﺘﺭﺘﻴﺏ ﺍﻟﺘﺭﻗﻴﻡ ﺒﺤﻴﺙ ﻴﻜﻭﻥ
1
1
...
...
r
r
t
G
G
G
H
H
+
=
× ×
×
× ×
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺍﻨﻅﺭ ﺭﻭﺘﻤﺎﻥ
1995,6.36
.
ﻤﺜﺎل
30.6
ﻟﺘﻜﻥ
p
p
G
=
×
F
F
ﻤﺘﺠﻬﻴﺎﹰ ﻤﻥ ﺍﻟﺒﻌﺩﺀﺎﻀﻓ ﻩﺭﺒﺘﻌﻨ ﻭ ،
2
ﻓﻭﻕ ﺍﻟﺤﻘل
p
F
.
ﻟﺘﻜﻥ
( )
( )
( )
(
)
1
2
1
1
1, 0 ,
0,1 ;
1,1 ,
1, 1 ,
G
G
H
H
=
=
=
=
−
.
ﻋﻨﺩﺌﺫ
1
2
1
2
1
2
,
,
G
G
G
G
H
H
G
G
H
=
×
=
×
=
×
.
أو
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٩١١
ﻤﻼﺤﻅﺔ
(a) 31.6
ﺇ
ﻥ ﻤﺒﺭﻫﻨﺔ
Krull- Schmidt
ﺘﺤﻘﻘﺕ ﺃﻴﻀﺎﹰ
ﺔـﻴﻬﺘﻨﻤ ﺭـﻴﻏ ﺓﺭﻤﺯﻟ ﺔﺒﺴﻨﻟﺎﺒ
ﻭﻗﺩ ﺒﺭﻫﻨﺕ
ﻤﺤﻘﻘﺔﹰ
ﺸﺭﻁﻲ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ
ﺒﺎﻟﻨﺴﺒﺔ
ﺔـﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻟﺍ
ﺩﺓـﻴﺍﺯﺘﻤﻟﺍ ﺕﺎـﻴﻟﺎﺘﺘﻤﻟﺍ ،ﻥﺃ ﻱﺃ ،
ﻭﺍﻟﻤﺘﻨﺎﻗﺼﺔ ﻟﻠﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺘﺼﺒﺢ ﺜﺎﺒﺘﺔ
.
(b)
ﺇﻥ ﻤﺒﺭﻫﻨﺔ
Krull- Schmidt
ﺘﺤﻘﻘﺕ ﺃﻴﻀﺎﹰ
ﺒﺎﻟﻨﺴﺒﺔ ﻟﺯﻤﺭﺓ ﺒﻤﺅﺜﺭﺍﺕ
.
ﻴﻜﻥـﻟ ،ﹰﻼﺜـﻤ
( )
Aut G
ﻤﺅﺜﺭﺍﹰ ﻋﻠﻰ
G
ﻓﺈﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺴﺘﻜﻭﻥ ﺠﻤﻴﻌﻬﺎ ﻤﻤﻴﺯﺓ،
.
(c)
ﺘﺒﻴﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺃﻥ ﺍﻟﺯﻤﺭ،ﺔﻴﻠﺒﻵﺍ ﺔﻴﻬﺘﻨﻤﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻰﻠﻋ ﺎﻨﻘﺒﻁ ﺍﺫﺇ
mi
C
لـﻴﻠﺤﺘﻟﺍ ﻲـﻓ
1
...
=
× ×
r
m
m
G
C
C
ﺤﻴﺙ ﻜل
i
m
ﻋﺒﺎﺭﺓ ﻋﻥ ﻗﻭﻯ ﻟﻌﺩﺩ ﺃﻭﻟﻲ ﻤﺤ
ﺕـﺤﺘ ﺩﻴﺤﻭ لﻜﺸﺒ ﺓﺩﺩ
ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل
)
ﻭﺍﻟﺭﺘﺒﺔ
.(
ﺘﻤﺎﺭﻴﻥ
1-6
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ
)
ﻟﻴﺴﺕ ﺒﺎﻟﻀﺭﻭﺭﺓ ﺃﻥ ﺘﻜﻭﻥ ﻤﻨﺘﻬﻴﺔ
(
ﻤﻊ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﺘﺭﻜﻴﺒﻴﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
0
1
...
1
n
G
G
G
G
=
⊃
⊃ ⊃
=
ﻭﻟﺘﻜﻥ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
.
ﺒﻴﻥ ﺃﻥ
0
1
...
1
n
N
N
G
N
G
N
G
=
⊃
⊃ ⊃
=
I
I
I
ﺘﺘﺤﻭل ﺇﻟﻰ ﻤﺘﺴﻠﺴﻠﺔ ﺘﺭﻜﻴﺒﻴﺔ ﻟﻠﺯﻤﺭﺓ
N
ﺇﺫﺍ ﻟﻡ ﻴﻭﺠﺩ ﺘﻜﺭﺍﺭﺍﺕ
.
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٠٢١
ﺍﻟﻔﺼل ﺍﻟﺴﺎﺒﻊ
ﺘﻤ
ﺜﻴﻼﺕ ﺍﻟﺯﻤﺭ ﺍﻟﻤﻨﺘﻬﻴﺔ
Representations of finite groups
ﻨﻌﺘﺒﺭ،لﺼﻔﻟﺍ ﺍﺫﻫ ﻲﻓ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻭ
F
ﺤﻘﻼﹰ
.
ﻭﻥ ﺫﺍﺕـﻜﺘ ﺔـﻴﻬﺠﺘﻤﻟﺍ ﺕﺍﺀﺎﻀﻔﻟﺍ ﻊﻴﻤﺠ
ﺒﻌﺩ ﻤﻨﺘﻪ
.
F
-
ﺠﺒﺭ ﻫﻭ ﺤﻠﻘﺔ
A
ﺘﺤﻭﻱ
F
ﻓﻲ ﻤﺭﻜﺯﻫﺎ ﻭﺫﺍﺕ ﺒﻌﺩ ﻤﻨﺘﻪ
ﻲـﻓ ﺎـﻤﻜ
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
.
ﻟﻥ ﻨﻌﺘﺒﺭ
A
ﺘﺒﺩﻴﻠﻴﺔ
.
٠٢
ﻜل
A
-
ﺘﻌﺘﺒﺭ ﻭﺫﺌﺩﻨﻋ ﺔﻴﻬﺘﻨﻤ ﺎﻫﺩﺎﻌﺒﺃ ﻥﻭﻜﺘ ﺕﻻﻭﺩﻭﻤ
ﻜﺄﻨﻬﺎ
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
.
ـﺒﺎﻟﻨﺴﺒﺔ ﻟ
A
-
ﻤﻭﺩﻭل
V
ﻴﺭﻤﺯ،
mV
ﻟﻠﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ
m
ﻤﺭﺓ ﻤﻥ
V
.
ﺇﻥ ﺍﻟﻤﻘﺎﺒل
(opposite)
opp
A
ـ ﻟ
F
-
ﺍﻟﺠﺒﺭ
A
ﻫﻭ ﻨﻔﺴﻪ ﻜﻤﺎ ﻓﻲ
A
ﻭﺏـﻠﻘﻤﻟﺍ ﺍﺩﻋ ﺎﻤ
ﻴﻭﺠﺩ ﺘﻁﺒﻴﻕ ﺘﻘﺎﺒل،ﻪﻨﺃ ﻱﺃ ،ﺏﺭﻀﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ
opp
:
a
a A
A
′
↔
a
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻤﺘﻤﺎﺜﻼﹰ
ﻊـﻤ
F
-
ﻓﻀﺎﺀ ﻤ
ﺘﺠﻬﻲ ﻭ ﻴﺤﻘﻕ ﺍﻟﺨﺎﺼﺔ
( )
a b
ba
′
′ ′ =
ﺇﺫﺍ ﻜﺎﻥ
,
a b
A
∈
.
ﺇﻥ
A
-
ﻤﻭﺩﻭل ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ ﻴﻜﻭﻥ ﺒﺴﻴﻁﺎﹰ
(simple)
ﺇﺫﺍ ﻟﻡ ﻴﺤﻭ
ﺔـﻴﺌﺯﺠ ﺕﻻﻭﺩﻭـﻤ ﻰﻠﻋ
ﻭ ﻴﻜﻭﻥ ﻨﺼﻑ ﺒﺴﻴﻁ،ﻱﺭﻔﺼﻟﺍ لﻭﺩﻭﻤﻟﺍ ﺀﺎﻨﺜﺘﺴﺎﺒ ﺔﻴﻠﻌﻓ
(semisimple)
ﺇﺫﺍ ﻜﺎﻥ ﻤﺘ
ﻊـﻤ ﹰﻼﺜﺎﻤ
ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ ﻟﻤﻭﺩﻭﻻﺕ ﺒﺴﻴﻁﺔ
.
ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﻤﺼﻔﻭﻓﻴﺔ
(Matrix representations)
ﺔـﺠﺭﺩﻟﺍ ﻥـﻤ ﺔﻴﻓﻭﻔﺼﻤﻟﺍ ﺕﻼﻴﺜﻤﺘﻟﺍ ﻥﺇ
n
ﺭﺓـﻤﺯﻠﻟ
G
لـﻘﺤﻟﺍ ﻰـﻠﻋ
F
ﺸﺎﻜلـﺘ ﻭـﻫ
( )
GL
n
G
F
→
.
ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﺘﻤﺜﻴل ﺃﻤﻴﻥ
(faithful)
ﺎﹰـﻨﻴﺎﺒﺘﻤ لﻜﺎﺸﺘﻟﺍ ﻥﺎﻜ ﺍﺫﺇ
.
ﺈﻥـﻓ ﻙﻟﺫـﻟ
ﺍﻟﺘﻤﺜﻴل ﺍﻷﻤﻴﻥ ﻴﻌﺭﻑ
G
ﺒﺯﻤﺭﺓ ﻤﻥ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﻨﻭﻉ
n n
×
.
ﻤﺜﺎل
1.7
(a)
ﻴﻭﺠﺩ ﺘﻤﺜﻴل
( )
2
GL
Q
→
ﻟﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ
,
Q
a b
=
ﺍﻟﺘﻲ ﺘﺭﺴل
a
ﺇﻟﻰ
0
1
1
0
−
−
ﻭ
b
ﺇﻟﻰ
0
1
1 0
−
.
،ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ
ﻜﻤﺎ ﻋﺭﻓﻨﺎ
Q
ﺒﺸﻜل ﺃﺴﺎﺴﻲ ﻜﻤﺎ ﻓﻲ
(1.17)
.
20
ﻟﺘﻜﻥ
1
,...,
n
e
e
ـ ﻗﺎﻋﺩﺓ ﻟ
A
ﻤﺜل
F
-
ﻋﻨﺩﺌﺫ،ﻲﻬﺠﺘﻤ ﺀﺎﻀﻓ
ij
k
i j
k
k
e e
a e
=
∑
ﻟﺒﻌﺽ
ij
k
a
F
∈
ﺘﺩﻋﻰ،
ﺜﻭﺍﺒﺕ ﺍﻟﺒﻨﻴﺔ ﻋﻠﻰ
A
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻘﺎﻋﺩﺓ
( )
i i
e
ﺇﻥ ﺍﻟﺠﺒﺭ،ﺕﺭﻴﺘﺨﺍ ﺩﻗ ﺓﺩﻋﺎﻘﻟﺍ ﻥﻭﻜﺘ ﹰﺎﻨﺎﻴﺤﺃ ،
A
ﻴﻜﻭﻥ ﻤﺤﺩﺩﺍﹰ ﺒﺸﻜل
ﻭﺤﻴﺩ ﺒﺜﻭﺍﺒﺕ ﺍﻟﺒﻨﻴﺔ ﻓﻴﻪ
.
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١٢١
(b)
ﻟﺘﻜﻥ
n
G
S
=
.
ﻟﻜل
n
S
σ
∈
ﻟﺘﻜﻥ،
( )
I
σ
ﺼﻔﻭﻓﺔـﻤﻟﺍ ﻥـﻤ ﺔﻠﺼﺎﺤﻟﺍ ﺼﻔﻭﻓﺔﻤﻟﺍ
ﺍﻟﻤﺤﺎﻴﺩﺓ ﺒﺎﺴﺘﺨﺩﺍﻡ
σ
ﻟﺘﺒﺩﻴل ﺍﻷﺴﻁﺭ
.
ﻷﻱ ﻤﺼﻔﻭﻓﺔ،ﺫﺌﺩﻨﻋ
A
ﻤﻥ ﺍﻟﻨﻭﻉ
n n
×
ﺼلـﺤﻨ ،
ﻰــﻠﻋ
( )
I
A
σ
ﻥــﻤ
A
ﺘﺨﺩﺍﻡــﺴﺎﺒ ﻙــﻟﺫﻭ
σ
ﻁﺭــﺴﻷﺍ لﻴﺩــﺒﺘﻟ
.
،ﺎﺹــﺨ لﻜﺸــﺒ
( ) ( ) (
)
I
I
I
σ
σ
σ σ
′
′
=
ﻭﻤﻨﻪ ﻓﺈﻥ،
( )
I
σ
σ
a
ـ ﺘﻤﺜﻴل ﻟ
n
S
.
ﺃﻨﻪ ﺃﻤﻴﻥ،ﺢﻀﺍﻭﻟﺍ ﻥﻤ
.
ﻜﻤﺎ ﺃﻥ
ﻜل ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﺭﺴل ﺇﻟﻰ
n
S
ﻥـﻤ ﺔـﻋﻭﻤﺠﻤ لـﺠﺃ ﻥـﻤ
ﺍﻟﻌﻨﺎﺼﺭ
n
)
ﻤﺒﺭﻫﻨﺔ ﻜﺎﻴﻠﻲ
ﺍﻨﻅﺭ
1
21.
(
لـﻴﺜﻤﺘ ﻰﻠﻋ ﻱﻭﺤﺘ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ لﻜ ﻥﺄﺒ ﻥﻴﺒﻴ ﺍﺫﻫ ،
ﻤﺼﻔﻭﻓﻲ ﺃﻤﻴﻥ
.
(c)
ﻟﺘﻜﻥ
n
G
C
σ
=
=
.
ﺇﺫﺍ ﻜﺎﻥ
F
ﻴﺤﻭﻱ ﻋﻠﻰ
n
ـ ﺠﺫﺭ ﻟ
1
ﻭﻟﻴﻜﻥ،
ξ
ﻋﻨﺩﺌﺫ،
ﻴﻭﺠﺩ ﺍﻟﺘﻤﺜﻴل
( )
1
:
GL
i
i
n
C
F
F
σ
ξ
×
→
=
a
.
ﻴﻜﻭﻥ ﺍﻟﺘﻤﺜﻴل ﺃﻤﻴﻨﺎﹰ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
ξ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﺘﻤﺎﻤﺎﹰ
.
ﺇﺫﺍ ﻜﺎﻥ
n
p
=
ﻋﺩﺩﺍﹰ ﺍﺒﺘﺩﺍﺌﻴﺎﹰ ﻭ ﻜﺎﻥ
F
ﻴﺤﻭﻱ ﻋﻠﻰ ﻤﻤﻴﺯ
p
ﺫـﺌﺩﻨﻋ ،
(
)
1
1
p
p
X
X
− =
−
ﻭﻤﻨﻪ ﻓﺈﻥ،
1
ـ ﻫﻭ ﺍﻟﺠﺫﺭ ﺍﻟﻭﺤﻴﺩ ﻟ
1
ﻤﻥ ﺍﻟﺩﺭﺠﺔ
p
ﻲـﻓ
F
.
ﻲـﻓ
ﻭﻟﻜﻥ ﻴﻭﺠﺩ ﺍﻟﺘﻤﺜﻴل ﺍﻷﻤﻴﻥ،ﹰﺎﻬﻓﺎﺘ ﻥﻭﻜﻴ لﻴﺜﻤﺘﻟﺍ ،ﺔﻟﺎﺤﻟﺍ ﻩﺫﻫ
( )
( )
2
1
:
GL
0
1
i
p
i
C
F
σ
→
a
.
2.7
ﺒﺭﻫﻥ ﺒﺭﻨﺴﺎﻴﺩ ﻋﻠﻰ ﺃﻥ ﻟ
ﻤﺴﺄﻟﺔ ﺒﺭﻨﺴﺎﻴﺩ
)
ﺍﻨﻅﺭ
p33
(
ﻲـﻓ ﺔـﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻠﻟ ﻲﺒﺎﺠﻴﺇ ﺏﺍﻭﺠ
( )
GL
n
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻓﻲ،ﻥﺃ ﻱﺃ ،
( )
GL
n
ﺒﺩﻟﻴل ﻤﻨﺘﻪ ﺘﻜﻭﻥ ﻤﻨﺘﻬﻴﺔ
.
ﻻ ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ ﺒﺩﻟﻴل ﻤﻨﺘﻪ ﻤﻤﺜﻠﺔ ﺒﺸﻜل ﺃﻤﻴﻥ ﻋﻠﻰ،ﻙﻟﺫﻟ
.
ﺠﺫﻭﺭ
1
ﻓﻲ ﺍﻟﺤﻘﻭل
ﺇﻥ ﺘﻤﺜﻴل ﺍﻟﺯﻤﺭﺓ ﻋﻠﻰ ﺍﻟﺤﻘل،ﺭﻴﺨﻷﺍ لﺎﺜﻤﻟﺍ ﻥﻴﺒ ﺎﻤﻜ
F
ﻴﻌﺘﻤﺩ ﻋﻠﻰ
ﺠﺫﻭﺭ
1
لـﻘﺤﻟﺍ ﻲﻓ
.
n
ـﺠﺫﺭ ﻟ
1
ﻴﺸﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
( )
n
F
µ
ﻓﻲ
×
F
ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ،
)
ﺍﻨﻅﺭ
55.1
.(
ﺇﺫﺍ ﻜﺎﻥ ﻤﻤﻴﺯ
F
ﻴﻘﺴﻡ
n
ﻋﻨﺩﺌﺫ،
( )
n
F
n
µ
<
.
،ﺭﻯـﺨﺃ ﺔـﻬﺠ ﻥﻤ
1
n
X
−
ﺎـﻬﻟ
ﺠﺫﻭﺭ ﻤﺨﺘﻠﻔﺔ
)
ﺴﻴﻜﻭﻥ ﺍﻟﺠﺫﺭ ﺍﻟﻤﻀﺎﻋﻑ ﺠﺫﺭﺍﹰ ﻟﻤﺸﺘﻘﻬﺎ
1
n
nX
−
(
ﺏـﺘﺭﻨ ﻥﺃ ﹰﺎـﻤﺌﺍﺩ ﺎﻨﻨﻜﻤﻴﻭ ،
( )
n
F
n
µ
=
ﺒﺘﻤﺩﻴﺩ
F
ﻲـﺌﺯﺠﻟﺍ لﻘﺤﻟﺍ لﻴﺩﺒﺘﺒ ،ﹰﻼﺜﻤ ،
F
ﻲـﻓ
ــ ﺒ
[ ]
F
ξ
ﺙـﻴﺤ
2
/
i n
e
π
ξ
=
ﺃﻭ ﺒﺘﺒﺩﻴل،
F
ـ ﺒ
[ ]
( )
(
)
F X
g X
ﺤﻴﺙ
( )
g X
لـﺒﺎﻗ ﺭـﻴﻏ لـﻤﺎﻋ
ـﻟﻠﺘﺤﻠﻴل ﻟ
1
n
X
−
ﻻ ﻴﻘﺴﻡ
1
m
X
−
ﻷﻱ ﻗﺎﺴﻡ ﻓﻌﻠﻲ
m
ـ ﻟ
n
.
ﻴﺩﻋﻰ ﺍﻟﻌﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻓﻲ
×
F
ﺒﺎﻟﺠﺫﺭ ﺍﻻﺒﺘﺩﺍﺌﻲ
(primitive)
ـ ﻟ
1
.
ـﻟﻜﻲ ﻨﻘ
ﻭل
ﺒﺄﻥ
F
ﺘﺤﻭﻱ ﻋﻠﻰ
n
ـ ﺠﺫﺭ ﺍﺒﺘﺩﺍﺌﻲ ﻟ
1
،
ξ
ﻨﻌﻨﻲ ﺒﺄﻥ ﺍﻟﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
n
ﻭ
ξ
ﻴﻭﻟﺩﻫﺎ
)
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻤﻤﻴﺯ
F
ﻴﺴﺎﻭﻱ
0
ﺃﻭ ﻤﻤﻴﺯ ﺍﺒﺘﺩﺍﺌﻲ ﻻ ﻴﻘﺴﻡ
n
.(
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ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺨﻁﻴﺔ
(Linear representations)
ﻨﺘﺫﻜﺭ
ﻤﻥ
(1.4)
ﺃﻨﻨﺎ
ﻋﺭﻓﻨﺎ ﻤﻔﻬﻭﻡ ﺘﺄﺜﻴﺭ ﺍﻟﺯﻤﺭﺓ
G
ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ
.
ﺔـﻋﻭﻤﺠﻤﻟﺍ ﻥﻭﻜﺘ ﺎﻤﺩﻨﻋ
ﻋﺒﺎﺭﺓ ﻋﻥ
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
V
ﻨﻘﻭل ﺒﺄﻥ ﺍﻟ،
ﺘﺄﺜﻴﺭ ﺨﻁﻲ
(linear)
ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﺘﻁﺒﻴﻕ
ﺨﻁﻴﺎﹰ ﻟﻜل
g
G
∈
.
ﻭﺱـﻜﻌﻤ ﻥﺈـﻓ ﺫـﺌﺩﻨﻋ
gV
ﻲـﻁﺨﻟﺍ ﻕـﻴﺒﻁﺘﻟﺍ ﻭـﻫ
( )
1
−
g
V
ﻭ،
( )
:
GL
g
g v G
V
→
a
ﺘﺸﺎﻜل
.
ﻟ
ﻤﻥ ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﺨﻁﻲ،ﺍﺫﻬ
G
ﻋﻠﻰ
V
ﻰـﻠﻋ لﺼﺤﻨ ،
ﺍﻟﺘﺸﺎﻜل ﺍﻟﺯﻤﺭﻱ
( )
GL
G
V
→
ﺭﺓـﻤﺯﻠﻟ ﹰﺎـﻴﻁﺨ ﹰﺍﺭﻴﺜﺄﺘ ﻑﺭﻌﻴ ﺍﺫﻬﻜ لﻜﺎﺸﺘ لﻜ ،ﺱﻜﻌﻟﺎﺒﻭ ،
G
ﻋﻠﻰ
V
.
ﻨﺩﻋﻭ ﺍﻟﺘﺸﺎ
ﻜل
( )
GL
G
V
→
ﺍﻟﺘﻤﺜﻴل ﺍﻟﺨﻁﻲ
(Linear- representation)
ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ
V
.
ﻨﻼﺤﻅ ﺒﺄﻥ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺨﻁﻲ ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ
n
F
لـﻴﺜﻤﺘ ﹰﺎـﻤﺎﻤﺘ ﻭﻫ
ﻤﺼﻔﻭﻓﻲ ﻤﻥ
ﺍﻟﺩﺭﺠﺔ
n
.
ﻤﺜﺎل
3.7
(a)
ﻟﻴﻜﻥ
n
G
C
σ
=
=
ﻭﻟﻨﻔﺭﺽ ﺒﺄﻥ،
F
ﻴﺤﻭﻱ ﻋﻠﻰ
n
ـ ﺠﺫﺭ ﺍﺒﺘﺩﺍﺌﻲ ﻟ
1
،
ﻭﻟﻴﻜﻥ
ξ
.
ﻟﻴﻜﻥ
( )
GL
G
V
→
ﺘﻤﺜﻴﻼﹰ ﺨ
ﻁﻴﺎﹰ ﻟﻠﺯﻤﺭﺓ
G
.
ﻋﻨﺩﺌﺫ
( )
( )
1
n
n
L
L
σ
σ
=
=
،
ـﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻜﺜﻴﺭﺓ ﺍﻟﺤﺩﻭﺩ ﺍﻷﺼﻐﺭﻴﺔ ﻟ
L
σ
ﺘﻘﺴﻡ
1
n
X
−
.
ﻜﻤﺎ ﺃﻥ
1
n
X
−
ﻟﻬﺎ
n
ﺠﺫﺭ
ﻤﺨﺘﻠﻑ
0
1
,...,
n
ξ
ξ
−
ﻓﻲ
F
ﻴﺘﺤﻠل ﺍﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ ﺇﻟﻰ ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻟﻠﻔﻀﺎﺀﺍﺕ ﺍﻟﺫﺍﺘﻴﺔ،
{
}
def
0
1
,
σ
ξ
≤ ≤ −
= ⊕
=
∈
=
i
i n
i
i
V
V
V
v
V
v
v
ﻜل ﺘﺤﻠﻴل ﺇﻟﻰ ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻜﻬﺫﺍ ﻟﻠﺯﻤﺭﺓ،ﺱﻜﻌﻟﺎﺒ
G
ﻴﻨﺠﻡ ﻋﻥ ﺘﻤﺜﻴل
G
.
(b)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ ﺍﻷﺱ
n
ﻭﻟﻨﻔﺭﺽ ﺒﺄﻥ،
F
ﺘﺤﻭﻱ ﻋﻠﻰ
n
ﺠﺫﺭ ﺍﺒﺘﺩﺍﺌﻲ
ـﻟ
1
.
ﻟﺘﻜﻥ
(
)
( )
(
)
Hom
,
Hom
,
n
G
G F
G
F
µ
∨
×
=
=
ﺇﻥ ﺘﻤﺜﻴل ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ ﺍﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ
V
ﻫﻭ ﻨﻔﺴﻪ ﺍﻟﺘﺤﻠﻴل ﺇﻟﻰ ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ
.
( )
{
}
def
,
G
V
V
V
v
V
v
v
χ
χ
χ
σ
χ σ
∨
∈
= ⊕
=
∈
=
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
G
ﻭ،ﺔﻴﺭﺌﺍﺩ
ﻫﺫﻩ
ﺇﻋﺎﺩﺓ ﻟﻠﺤﺎﻟﺔ
(a)
ﻭﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ ﺘﺄﺘﻲ ﺒﺴﻬﻭﻟﺔ،
)
ﻴﺘﺤﻠل
V
ﻊـﻤ
ﺍﻟﺤﻔﺎ
ﻅ ﻋﻠﻰ ﺘﺄﺜﻴﺭ ﺍﻟﻌﺎﻤل ﺍﻟﺩﺍﺌﺭﻱ ﺍﻷﻭل ﻓﻲ
G
ﻰـﻠﻋ ﻅﺎﻔﺤﻟﺍ ﻊﻤ ﻉﻭﻤﺠﻤ لﻜ لﻠﺤﺘﻴ ﺫﺌﺩﻨﻋ ،
ﺘﺄﺜﻴﺭ ﺍﻟﻌﺎﻤل ﺍﻟﺩﺍﺌﺭﻱ ﺍﻟﺜﺎﻨﻲ ﻓﻲ
G
ﻭﻫﻜﺫﺍ،
.(
,
,
gv
V
V x
gx
= →
a
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٣٢١
ﻤﺒﺭﻫﻨﺔ
Maschke
ﻟﻴﻜﻥ
( )
GL
G
V
→
ﺘﻤﺜﻴﻼﹰ ﺨﻁﻴﺎﹰ ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ
F
-
ﻲـﻬﺠﺘﻤ ﺀﺎﻀﻓ
V
.
ﺄﻥـﺒ لﺎـﻘﻴ
ﺍﻟﻔﻀﺎﺀ ﺍﻟﺠﺯﺌﻲ
W
ﻓﻲ
V
ﻴﻜﻭﻥ
G
-
ﻻ ﻤﺘﻐﻴﺭﺍﹰ
(invariant)
ﺇﺫﺍ ﻜﺎﻥ
gW
W
⊂
لـﻜﻟ
g
G
∈
.
ﻴﻘﺎل ﻋﻥ
F
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
: V
V
α
′
→
ﺭﺓـﻤﺯﻟﺍ ﻰﻠﻋ ﺔﻴﻬﺠﺘﻤﻟﺍ ﺕﺍﺀﺎﻀﻔﻠﻟ
G
ﺍﻟﺘﻲ ﺘﺅﺜﺭ ﻋﻠﻰ ﺒﺨﻁﻴﺔ ﺃﻨﻪ
G
-
ﻻ ﻤﺘﻐﻴﺭ ﺇﺫﺍ ﻜﺎﻥ
( )
( )
g v
g
v
α
α
=
ﻟﻜل
,
g
G v
V
∈
∈
.
ﻴﻘﺎل ﻋﻥ ﺍﻟﺸﻜل ﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ،ﹰﺍﺭﻴﺨﺃ
:V V
F
φ
× →
ﺃﻨﻪ
G
-
ﻻ ﻤﺘﻐﻴﺭ ﺇﺫﺍ ﻜﺎﻥ
(
) (
)
,
,
gv gv
v v
φ
φ
′
′
=
ﻟﻜل
, ,
g
G v v
V
′
∈
∈
.
ﺔـﻨﻫﺭﺒﻤ
4.7
(Maschke)
ﻟﻴﻜﻥ
( )
GL
G
V
→
ﺘﻤﺜﻴﻸ ﺨﻁﻴﺎﹰ ﻟﻠﺯﻤﺭﺓ
G
.
ﺇﺫﺍ ﻜﺎﻥ ﻤﻤﻴﺯ
F
ﻻ ﻴﻘﺴﻡ
G
ﻜل ﻓﻀﺎﺀ ﺠﺯﺌﻲﺫﺌﺩﻨﻋ ،
G
-
ﻻ ﻤﺘﻐﻴﺭ
W
ﻓﻲ
V
ﺔـﻤﻤﺘﻤ ﻰـﻠﻋ ﻱﻭﺤﻴ
G
-
ﻲــﺌﺯﺠ ﺀﺎﻀــﻓ ﺩــﺠﻭﻴ ،ﻪــﻨﺃ ﻱﺃ ،ﺭــﻴﻐﺘﻤ ﻻ
G
-
ﺭــﻴﻐﺘﻤ ﻻ
W
′
ﻭﻥــﻜﻴ ﺙــﻴﺤﺒ
V
W
W
′
=
⊕
.
ﻨﻼﺤﻅ ﺒﺄﻨﻪ ﺘﻁﺒﻕ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺩﺍﺌﻤﺎﹰ ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﻤﻤﻴﺯ
F
ﻫﻭ ﺍﻟﻤﻤﻴﺯ ﺍﻟﺼﻔﺭﻱ
.
ﺇﻥ ﺍﻟﺸﺭﻁ ﻋﻠﻰ ﺍﻟﻤﻤﻴﺯ ﺘﺤﺩﻴﺩﺍﹰ ﻀﺭﻭﺭﻱ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
σ
=
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺔﻴﺭﺌﺍﺩ ﺓﺭﻤﺯ
p
ﺘﺅﺜﺭ ﻋﻠﻰ
2
V
F
=
ﻜﻤﺎ ﺍﻟﻤﺼﻔﻭﻓﺔ
1 1
0 1
)
ﺍﻨﻅﺭ
3.7b
(
ﻋﻨﺩﺌﺫ،
ﻲـﺌﺯﺠﻟﺍ ﺀﺎﻀﻔﻟﺍ ﻥﻭﻜﻴ
0
∗
G
-
ﻟﻜﻥ ﻻ ﻴﻭﺠﺩ ﻓﻀﺎﺀ ﺠﺯﺌﻲ ﻤﺘﻤﻡ،ﺭﻴﻐﺘﻤ ﻻ
F
,
0
a
b
b
≠
ﻴﻜﻭﻥ،
G
-
ﻻ ﻤﺘﻐﻴﺭ
.
ﻨﻘﺩﻡ ﺒﺭﻫﺎﻨﻴﻥ،ﺔﻨﻤﻀﺘﻤﻟﺍ ﺓﺭﻜﻔﻟﺍ ﺔﻴﻤﻫﻷﻭ
ﻟﻤﺒﺭﻫﻨﺔ
Maschke
.
ﺒﺭﻫﺎﻥ ﻤﺒﺭﻫﻨﺔ
Maschke
)
ﻓﻲ ﺍﻟﺤﺎﻟﺔ
=
F
ﺃﻭ
(
ﺘﻤﻬﻴﺩﻴﺔ
5.7
ﻟﺘﻜﻥ
φ
ﺘﻁﺒﻴﻘﺎﹰ ﻤﺘﻨﺎﻅﺭﺍﹰ ﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ ﻋﻠﻰ
V
ﻭ ﻟﻴﻜﻥ،
W
ﻓﻀ
ﺠﺯﺌﻴﺎﹰ ﻋﻠﻰﺀﺎ
V
.
ﻥــﻤ لــﻜ ﻥﺎــﻜ ﺍﺫﺇ
φ
ﻭ
W
G
-
ﻀﺎﹰــﻴﺃ ﻥﻭــﻜﻴ ﺫــﺌﺩﻨﻋ ،ﺭــﻴﻐﺘﻤ ﻻ
(
)
{
}
def
,
0,
W
v
V
w v
w
W
φ
⊥
=
∈
=
∈
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
v
W
⊥
∈
ﻭ ﻟﻴﻜﻥ
g
G
∈
.
(
)
(
)
1
,
,
φ
φ
−
=
w g w
g w v
ﻷﻱ،
∈
w
W
ﻷﻥ
G
φ
-
ﻭ،ﺭﻴﻐﺘﻤ ﻻ
(
)
1
,
0
g w v
φ
−
=
ﻷﻥ
W
ﺃﻴﻀﺎﹰ
G
-
ﻻ ﻤﺘﻐﻴﺭ
.
ﺄﻥـﺒ ﻥﻴﺒﻴ ﺍﺫﻫ
gv
W
⊥
∈
.
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٤٢١
ﻨﺘﺫﻜﺭ ﻤﻥ ﺍﻟﺠﺒﺭ ﺍﻟﺨﻁﻲ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ
φ
ﻟﻴﺱ
ﺸﺎﺫﺍﹰ
ﺫـﺌﺩﻨﻋ ،
⊥
=
⊕
V
W
W
ﺎﻟﻲـﺘﻟﺎﺒﻭ
ﻟﻜﻲ ﻨﺒﺭﻫﻥ ﻋﻠﻰ ﻤﺒﺭﻫﻨﺔ
Maschke
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻨﻪ ﻴﻭﺠﺩ،
G
-
ﺎﺌﻲـﻨﺜ ﻕﻴﺒﻁﺘ ﺭﻴﻐﺘﻤ ﻻ
ﺍﻟﺨﻁﻴﺔ ﻤﺘﻨ
ﺎﻅﺭ
:V V
F
φ
× →
.
ﺘﻤﻬﻴﺩﻴﺔ
6.7
ﺘﻁﺒﻴﻕ ﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ ﻤﺘﻨﺎﻅﺭ
φ
ﻋﻠﻰ
V
،
(
)
(
)
def
,
,
g G
v w
g v g w
φ
φ
∈
=
∑
ﻴﻜﻭﻥ
G
-
ﻻ ﻤﺘﻐﻴﺭ ﺘﻁﺒﻴﻘ
ﺎﹰ
ﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ ﻤﺘﻨﺎﻅﺭ ﻋﻠﻰ
V
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺒﺄﻥ
φ
ﻭﻤﻥ ﺃﺠل،ﺭﻅﺎﻨﺘﻤﻭ ﺔﻴﻁﺨﻟﺍ ﻲﺌﺎﻨﺜ ﻕﻴﺒﻁﺘ
0
g
G
∈
،
(
)
(
)
def
0
0
0
0
,
,
,
g G
g v g w
gg v gg w
φ
φ
∈
=
∑
ﺍﻟﺫﻱ ﻴﺴﺎﻭﻱ
(
)
,
g G
g v g w
φ
∈
∑
، ﻷﻥ
g
ﺘﻤﺴﺢ ﻜل
G
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،
0
g g
ﺘﻤﺴﺢ ﻜل
G
.
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﻨﺘﻭﻗﻊ ﺒﺄﻥ،ﻅﺤﻟﺍ ﺀﻭﺴﻟ
φ
ﻟﻴﺱ
ﺸﺎﺫﺍﹰ
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
φ
ﻜﺫﻟﻙ
)
ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ
ﺍﺴﺘﻁﻌﻨﺎ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ ﺠﻤﻴﻊ
[ ]
F G
-
ﻰـﻠﻋ ﺩﻴﺩﺤﺘﻟﺍ ﻥﻭﺩ ،ﺔﻁﻴﺴﺒ ﻑﺼﻨ ﺕﻻﻭﺩﻭﻤ
F
ﺃﻭ
G
.(
ﺘﻤﻬﻴﺩﻴﺔ
7.7
ﻟﻴﻜﻥ
F
=
.
ﺇﺫﺍ ﻜﺎﻥ
φ
ﺜﻨﺎﺌﻲ ﺍﻟﺨﻁﻴﺔ ﻤﻌﺭﻓﺎﹰ ﻤﻭﺠﺒﺎﹰ ﻋﻠﻰًﺄﻘﻴﺒﻁﺘ
V
ﻋﻨﺩﺌﺫ،
ﻴﻜﻭﻥ
φ
ﻜﺫﻟﻙ ﺃﻴﻀﺎﹰ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
φ
ﻤﻥ ﺃﺠل ﺃﻱﺫﺌﺩﻨﻋ ،ﹰﺎﺒﺠﻭﻤ ﹰﺎﻓﺭﻌﻤ
v
ﻏﻴﺭ ﺼﻔﺭﻱ ﻓﻲ
V
،
( )
(
)
,
,
0
g G
v v
g v g v
φ
φ
∈
=
>
∑
ﻫﺫﺍ ﻴﻜﻤل ﺒﺭﻫﺎﻥ ﻤﺒ
ﺭﻫﻨﺔ
Maschke
ﻋﻨﺩﻤﺎ
F
=
ﻷﻥ ﻴﻭﺠﺩ ﺘﺤﺩﻴﺩﺍﹰ ﺘﻁﺒﻴﻕ،
φ
ﺨﻁﻲ
ﻤﻌﺭﻑ ﻤﻭﺠﺏ ﻋﻠﻰ
V
.
ﺒﺎﻟﻨﻘﺎﺵ ﻨﻔﺴﻪ ﻭ
ﺒﺎﺴﺘﺨﺩﺍﻡ ﺃﺸﻜﺎل ﻫﺭﻤﺘﻴﺔ ﻋﻨﺩﻤﺎ
F
=
)
ﺩﻤﺎـﻨﻋ ﻭﺃ
ﻴﻜﻭﻥ
F
ﺃﻱ ﺤﻘل ﻓﻲ
.(
ﺒﺭﻫﺎﻥ ﻤﺒﺭﻫﻨﺔ
Maschke
)
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ
(
ﻴﺩﻋﻰ ﺍﻟﺘﺸﺎﻜل ﺍﻟﺫﺍﺘﻲ
π
ـ ﻟ
F
-
ﺍﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ
V
ﺇﺴﻘﺎﻁﺎﹰ ﺇﺫﺍ ﻜﺎﻥ
2
π
π
=
.
ﺇﻥ ﻜ
ﺜﻴﺭﺓ
ﺍﻟﺤﺩﻭﺩ ﺍﻷﺼﻐﺭﻴﺔ ﻟﻺﺴﻘﺎﻁ
(projector)
π
ﺘﻘﺴﻡ
(
)
2
1
X
X
X X
−
=
−
ﻭﻤﻨﻪ ﻓﺈﻥ،
V
،ﻴﺘﺤﻠل ﺇﻟﻰ ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻟﻠﺤﻘﻭل ﺍﻟﺫﺍﺘﻴﺔ
( )
( )
0
1
V
V
V
π
π
=
⊕
ﺤﻴﺙ،
( )
{
}
( )
( )
{
}
( )
0
1
0
Im
V
v
V
v
Ker
V
v
V
v
v
π
π
π
π
π
π
= ∈
=
=
= ∈
=
=
.
ﺇﻥ ﺍﻟﺘﺤ،ﺱﻜﻌﻟﺎﺒ
ﻠﻴل
0
1
V
V
V
=
⊕
ﻨﺠﻡ ﻋﻥ ﺍﻟﻤﺴﻘﻁ
(
) (
)
0
1
1
,
0,
v v
v
a
.
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٥٢١
ﻨﻔﺭﺽ ﺍﻵﻥ ﺒﺄﻥ
G
ﺘﺅﺜﺭ ﺒﺨﻁﻴﺔ ﻋﻠﻰ
V
.
ﺇﺫﺍ ﻜﺎﻥ ﺍﻹﺴﻘﺎﻁ
π
ﻴﻜﻭﻥ
G
-
ﺭﺍﹰـﻴﻐﺘﻤ ﻻ
.
ﻓ
ﻟﻜﻲ ﻨﺒﺭﻫﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ،ﻪﻨﺈ
W
ﻫﻲ ﺼﻭﺭﺓ ﺍﻹﺴﻘﺎﻁ
π
G
-
ﻻ ﻤﺘﻐﻴﺭ
.
ﺒﺎﺨﺘﻴﺎﺭ ﻤﺭﺓﹰ ﺃﺨﺭﻯ ﺍﻹﺴﻘﺎﻁ
π
F
-
ﺨﻁﻲ ﻤﻊ ﺍﻟﺼﻭﺭﺓ
W
،ﺩﺩـﺤﻤ لﻜﺸﺒ ﺩﺠﻭﻴ ﻱﺫﻟﺍ ،
ﻭ
ﻋﺩل ﻟﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻤﺴﻘﻁ
π
G
-
ﻻ ﻤﺘﻐﻴﺭ ﻤﻊ ﺍﻟﺼﻭﺭﺓ ﻨﻔﺴﻬﺎ
.
ﻟﻜل
v
V
∈
ﻟﻴﻜﻥ،
( )
( )
(
)
1
1
g G
v
g
g v
G
π
π
−
∈
=
∑
ﺍﻟﺫﻱ ﻴﺤﻭل ﺍﺘﺠﺎﻫﻪ ﻷﻥ
.1
×
∈
G
F
ﻭ ﻴﻌﺭﻑ،
F
-
ﺎﹰـﻴﻁﺨ ﺎﹰـﻘﻴﺒﻁﺘ
:V
V
π
→
.
ﻷﻱ
1
−
∈
g w
W
ﻷﻱ،
∈
w
W
ﻭﻤﻨﻪ
(21)
( )
(
)
1
1
1
g G
g G
w
g g w
w
w
G
G
π
−
∈
∈
=
=
=
∑
∑
ﺇﻥ ﺼﻭﺭﺓ
π
ﺠﺯﺌﻴﺔ ﻓﻲ
W
ﻷﻥ،
( )
Im
W
π
⊂
ﻭ
W
ﻋﺒﺎﺭﺓ ﻋﻥ
G
-
ﻭﻤﻨﻪ،ﺭﻴﻐﺘﻤ ﻻ
( )
( )
(
)
( )
( )
21
def
2
v
v
v
π
π π
π
=
=
ﻷﻱ
v
V
∈
.
ﻴﻜﻭﻥ،ﺍﺫﻬﻟ
π
ﻭﺘﺒﻴﻥ ﺍ،ﹰﺎﻁﺎﻘﺴﺇ
ﻟﻌﻼﻗﺔ
(21)
ﺄﻥـﺒ
( )
Im
W
π
⊃
ﺎﻟﻲـﺘﻟﺎﺒﻭ ،
( )
Im
W
π
=
.
ﺒﻘﻲ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ
π
ﻴﻜﻭﻥ
G
-
ﻻ ﻤﺘﻐﻴﺭ
.
ﻟﻜل
0
g
V
∈
( )
(
)
(
)
(
) (
)
(
)
1
1
1
0
0
0
0
0
1
1
,
π
π
π
−
−
−
∈
∈
=
=
∑
∑
g G
g G
g v
g
g g v
g
g g
g g v
G
G
ﺍﻟﺫﻱ ﻴﺴﺎﻭﻱ
( )
0
g
v
π
، ﻷﻥ
g
ﻴﻤﺴﺢ ﻜل
G
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ،
1
0
g g
−
ﻴﻤﺴﺢ ﻜل
G
.
؛ﺠﺒﺭ ﺍﻟﺯﻤﺭ
ﻨﺼﻑ ﺍﻟﺒﺴﻴﻁﺔ
(The group algebra; semisimplicity)
ﺇﻥ ﺠﺒﺭ ﺍﻟﺯﻤﺭ
(group algebra)
[ ]
F G
ﻤﻌﺭﻑ ﻟﻴﻜﻭﻥ
[ ]
F G
-
ﺩﺓـﻋﺎﻘﺒ ﻲﻬﺠﺘﻤ ﺀﺎﻀﻓ
ﻤﻥ ﻋﻨﺎﺼﺭ
G
ﻤﻊ ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻋﻠﻰ
G
.
،ﻟﻬﺫﺍ
•
ﺃﻱ ﻋﻨﺼﺭ ﻓﻲ
[ ]
F G
ﻫﻭ ﺍﻟﻤﺠﻤﻭﻉ
,
g G
g
g
c g c
F
∈
∈
∑
،
•
ﻴﻜﻭﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ
g G
g
c g
∈
∑
ﻭ
g G
g
c g
∈
′
∑
ﻓﻲ
[ ]
F G
ﻤﺘﺴﺎﻭﻴﻴ
ﻥ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ
ﻜﺎﻥ
g
g
c
c
′
=
ﻟﻜل
g
ﻭ
•
(
)(
)
1 2
1
2
,
g
g G
g
g G
g
g G
g
g
g g
g
g
c g
c g
c g c
c c
=
∈
∈
∈
′
′′
′′
′
=
=
∑
∑
∑
∑
ﻭ
ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﺨﻁﻲ
,
:
g v
gv G V
V
× →
a
ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ
F
-
ﺍﻟﻔﻀﺎﺀ ﺍﻟﻤﺘﺠﻬﻲ ﻴﺘﺤﺩﺩ ﺒﺸﻜل ﻭﺤﻴﺩ ﺇﻟﻰ ﺘﺄﺜﻴﺭ
[ ]
F G
ﻋﻠﻰ
V
،
[ ]
,
:
,
g G
g
g G
g
c g v
c gv F G
V
V
∈
∈
× →
∑
∑
a
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٦٢١
ﺍﻟﺫﻱ ﻴﺤﻭل
V
ﺇﻟﻰ
[ ]
F G
-
ﻤﻭﺩﻭل
.
ﻀﺎﺀﺍﺕـﻔﻟﺍ ﹰﺎﻤﺎﻤﺘ ﻲﻫ ﺭﻴﺜﺄﺘﻟﺍ ﺍﺫﻬﻟ ﺔﻴﺌﺯﺠﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ
ﺍﻟﺠﺯﺌﻴﺔ
G
-
ﻻ ﻤﺘﻐﻴﺭﺓ
.
ﻟﻴﻜﻥ
( )
G
GL V
→
ﺘﻤﺜﻴﻼﹰ ﺨﻁﻴﺎﹰ ﻟﻠﺯﻤﺭﺓ
G
.
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
V
ﺒﺴﻴﻁﺎﹰ
)
ﻨﺼﻑ
ﺴﻴﻁـﺒ
(
ﻭ
ﻜﺫﻟﻙ
[ ]
F G
-
ﻴﻘﺎل ﻋﺎﺩﺓﹰ ﺒﺄﻥ ﺍﻟﺘﻤﺜﻴل ﻏﻴﺭ ﺨﺯﻭل،لﻭﺩﻭﻤ
(irreducible)
)
لـﻴﻠﺤﺘﻠﻟ لـﺒﺎﻗ
ﺒﺸﻜل ﺘﺎﻡ
(completely reducible
.(
ﻋﻠﻰ ﺃﻱ ﺤ
ﺴﻴﻁﺎﹰـﺒ ﹰﻼﻴﺜﻤﺘ ﻡﻬﻴﻤﺴﻨ ﻑﻭﺴ ،لﺎ
)
ﺼﻑـﻨ
ﺒﺴﻴﻁ
.(
ﻗﻀﻴﺔ
8.7
ﺇﺫﺍ ﻜﺎﻥ ﻤﻤﻴﺯ
F
ﻻ
ﻴﻘﺴﻡ
G
ﻜلﺫﺌﺩﻨﻋ ،
[ ]
F G
-
ﻤﻭﺩﻭل ﻫﻭ ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ
ﻟﻤﻭﺩﻭﻻﺕ ﺠﺯﺌﻴﺔ ﺒﺴﻴﻁﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
V
[ ]
F G
-
ﻤﻭﺩﻭل
.
ﺇﺫﺍ ﻜﺎﻥ
V
ﻻ ﻴﻭﺠﺩ ﺃﻱﺫﺌﺩﻨﻋ ،ﹰﺎﻁﻴﺴﺒ
ﺸﻲﺀ ﻟﺒﺭﻫﺎﻨﻪ
.
ﻴﺤﻭﻱ ﻋﻠﻰ ﻤﻭﺩﻭل ﺠﺯﺌﻲ ﻓﻌﻠﻲ ﻏﻴﺭ ﻤﻌﺩﻭﻡ،ﻯﺭﺨﺃ ﺔﻬﺠ ﻥﻤ
W
.
ﻤﻥ ﻤﺒﺭﻫﻨﺔ
Maschke
،
V
W
W
′
=
⊕
ﺤﻴﺙ
W
ﻋﺒﺎﺭﺓ ﻋﻥ
[ ]
F G
-
ﻤﻭﺩﻭل
.
ﻥـﻤ لـﻜ ﻥﺎﻜ ﺍﺫﺇ
W
ﻭ
W
′
ﻴﻤ،ﻯﺭﺨﺃ ﺔﻬﺠ ﻥﻤﻭ ،ﻥﺎﻫﺭﺒﻟﺍ ﻡﺘﻴ ﺫﺌﺩﻨﻋ ،ﹰﺎﻁﻴﺴﺒ
ﺍﻟﺫﻱ ﻴﻨﺘﻬﻲ ﺒ،ﺵﺎﻘﻨﻟﺍ ﻊﺒﺎﺘﻨ ﻥﺃ ﻥﻜ
ﻌﺩﺩ ﻤﻨﺘﻪ ﻤﻥ
ﺍﻟﺨﻁﻭﺍﺕ ﻷﻥ
V
ﺒﺒﻌﺩ ﻤﻨﺘﻪ ﻜﻤﺎ ﺃﻨﻪ
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
.
ﺒﺄﻥ ﺍﻟﺘﻤﺜﻴ،ﺎﻨﻅﺤﻻ ﺩﻗ ﻥﻭﻜﻨ ﻪﻨﺃ ﺎﻤﻜ
ﺭﺓـﻤﺯﻠﻟ ﻲﻁﺨﻟﺍ ل
G
ﺭـﺒﺘﻌﻴ ﻥﺃ ﻥـﻜﻤﻴ
[ ]
F G
-
ﻤﻭﺩﻭل
.
ﺤﺘﻰ ﻨﻔﻬﻡ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺨﻁﻲ ﻟﻠﺯﻤﺭﺓ،ﺍﺫﻬﻟ
G
ﻴﺠﺏ ﺃﻥ ﻨﻔﻬﻡ،
[ ]
F G
-
ﻭ،ﺕﻻﻭﺩﻭـﻤ
ﻟﻬﺫﺍ ﻴﺠﺏ ﺃﻥ ﻨﻔﻬﻡ ﺒﻨﻴﺔ
F
-
ﺍﻟﺠﺒﺭ
[ ]
F G
.
ﻓﻲ ﺍﻟﻤﻘﻁﻊ ﺍﻟﺜﺎﻟﺙ ﺍﻟﺘﺎﻟﻲ ﻨﺩﺭﺱ
F
-
ﺭ ﻭـﺒﺠﻟﺍ
ـ ﻨﺒﺭﻫﻥ ﻤﺒﺭﻫﻨﺎﺕ ﻭﻴﺩﺭﺒﻭﺭﻥ ﺍﻟﻤﺸﻬﻭﺭﺓ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒ،ﺹﺎﺨ لﻜﺸﺒ ،ﺎﻬﺘﻻﻭﺩﻭﻤ
F
-
ﻭﺭـﺒﺠﻟﺍ
ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻭﺩﻭﻻﺘﻬﺎ ﻨﺼﻑ ﺒﺴﻴﻁﺔ
.
ﺍﻟﻤﻭﺩﻭﻻﺕ ﻨﺼﻑ ﺍ
ﻟﺒﺴﻴﻁﺔ
(Semisimple modules)
ﻨﻌﺘﺒﺭ،ﺓﺭﻘﻔﻟﺍ ﺍﺫﻫ ﻲﻓ
A
،
F
-
ﺠﺒﺭﺍﹰ
.
ﻗﻀﻴﺔ
9.7
ﻜل
A
-
ﻤﻭﺩﻭل
V
ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﻜﻌﻼﻗﺔ ﺍﻻﺤﺘﻭﺍﺀ
{ }
0
1
...
0
s
V
V
V
V
=
⊃
⊃ ⊃
=
ﺒﺤﻴﺙ ﺘﻜﻭ
ﻥ ﺯﻤﺭ ﺍﻟﻘﺴﻤﺔ
1
i
i
V V
+
A
-
ﻤﻭﺩﻭﻻﺕ ﺒﺴﻴﻁﺔ
.
ﺇﺫﺍ ﻜﺎﻥ
{ }
0
1
...
0
t
V
W
W
W
=
⊃
⊃ ⊃
=
ﻋﻼﻗﺔ ﺍﻻﺤﺘﻭﺍﺀ ﺍﻟﺜﺎﻨﻴﺔ ﻋﻨﺩﺌﺫ
s
t
=
ﻭ ﺘﻭﺠﺩ ﺘﺒﺩﻴﻠﺔ
σ
ﻟﻠﻤﺠﻤﻭﻋﺔ
{
}
1,..., s
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ
( )
( )
1
1
i
i
i
i
V V
W
W
σ
σ
+
+
≈
ﻟﻜل
i
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ ﻫﺫﺍ ﻤﺨﺘﻠﻑ ﻋﻥ ﻤﺒﺭﻫﻨﺔ ﺠﻭﺭﺩﺍﻥ
-
ﻫﻭﻟﺩﺭ
(2.6)
ﻭﺍﻟﺫﻱ ﻴﺒﺭﻫﻥ ﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ،
.
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٧٢١
ﺘﻤﻬﻴﺩﻴﺔ
10.7
ﺒﻔﺭﺽ
1
1
...
...
s
t
V
V
V
V
W
W
≈ ⊕ ⊕
≈ =
⊕ ⊕
ﻤﻊ ﻜل
A
-
ﻤﻭﺩﻭﻻﺕ
i
V
ﻭ
j
W
ﺍﻟﺒﺴﻴﻁﺔ
.
ﻋﻨﺩﺌﺫ
s
t
=
ﺔـﻠﻴﺩﺒﺘ ﺩﺠﻭﻴ ﻭ
σ
ﺔـﻋﻭﻤﺠﻤﻠﻟ
{
}
1,..., s
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
i
i
V
W
σ
≈
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻭﺍﻟﺘﻲ ﻤﻥ ﺍﻟﻤ،ﺀﺍﻭﺘﺤﺍ ﺔﻗﻼﻋ ﻑﺭﻌﻴ لﻴﻠﺤﺘ لﻜ
ﻤﻜﻥ ﺘﻁﺒﻴﻕ ﺍﻟﻘﻀﻴﺔ ﻋﻠﻴﻬﺎ
.
ﻗﻀﻴﺔ
11.7
ﻟﻴﻜﻥ
V
،
A
-
ﻤﻭﺩﻭل
.
ﺇﺫﺍ ﻜﺎﻥ
V
ﻭلـﻘﻨ ،ﺔﻁﻴﺴﺒ ﺔﻴﺌﺯﺠ ﺕﻻﻭﺩﻭﻤ ﻉﻭﻤﺠﻤ
ﺒﺄﻥ
i I
i
V
S
∈
=
∑
)
ﻟﻴﺱ ﻤﻥ
ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭﺍﹰ
(
ﻷﺫﺌﺩﻨﻋ ،
ﻭﺩﻭلـﻤ ﻱ
ﺠﺯﺌﻲ
W
ﻓﻲ
V
ﺘﻭﺠﺩ ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ،
J
ﻓﻲ
I
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
∈
=
⊕
⊕
i
i J
V
W
S
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺘﻜﻥـﻟ
J
ـﺔ ﺍﻷﻋﻅﻤﻴـﻴﺌﺯﺠﻟﺍ ﺔـﻋﻭﻤﺠﻤﻟﺍ
ﻲـﻓ ﺔ
I
ﻭﻉـﻤﺠﻤﻟﺍ ﻥﻭـﻜﻴ ﺙـﻴﺤﺒ
def
J
j J
j
S
S
∈
=
∑
ﻤﺒﺎﺸﺭﺍﹰ ﻭ
=
I
J
W
S
V
.
ﺃﺩﻋﻲ ﺃﻥ
J
W
S
V
+
=
)
ﺈﻥـﻓ ﻲﻟﺎﺘﻟﺎﺒ
V
ـﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻟ
W
ﻭ
J
S
ﺤﻴﺙ
j
J
∈
.(
ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ ﻜل ﻤﻥ،ﺍﺫﻬﻟ ،ﹰﻻﻭﺃ
i
S
ﻤﺤﺘﻭﺍﺓ ﻓﻲ
J
W
S
+
.
ﻷﻥ
i
S
، ﺒﺴﻴﻁﺔ
(
)
i
J
S
W
S
+
I
ﻴﺴﺎﻭﻱ
i
S
ﺃﻭ
0
.
ﺔـﻟﺎﺤﻟﺍ ﻲـﻓ
،ﺍﻷﻭﻟﻰ
(
)
i
J
S
W
S
⊂
+
ﺔـﻴﻨﺎﺜﻟﺍ ﺔﻟﺎﺤﻟﺍ ﻲﻓﻭ ،
0
J
i
S
S
=
I
ﻭ
(
)
0
J
i
W
S
S
+
=
I
،
ﻭﻫﺫﺍ ﻴﺘﻨﺎﻗﺽ ﻤﻊ ﺘﻌﺭﻴﻑ
I
.
ﻨﺘﻴﺠﺔ
12.7
ﺇﻥ ﺍﻟﺸﺭﻭﻁ
ﺍﻵﺘﻴ
ﺔ ﻋﻠﻰ
A
-
ﻤﻭﺩﻭل
V
ﻤﺘﻜﺎﻓﺌﺔ
:
(a)
V
، ﻨﺼﻑ ﺒﺴﻴﻁ
(b)
V
، ﻤﺠﻤﻭﻉ ﻟﻤﻭﺩﻭﻻﺕ ﺠﺯﺌﻴﺔ ﺒﺴﻴﻁﺔ
(c)
ﻜل ﻤﻭﺩﻭل ﺠﺯﺌﻲ ﻓﻲ
V
ﻟﻪ ﻤﺘﻤﻤﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺘﺒﻴﻥ ﺍﻟﻘﻀﻴﺔ ﺒﺄﻨﻪ ﻤﻥ
(b)
ﻴﻨﺘﺞ
(c)
ﻭﻤﻥ ﺍﻟﻨﻘﺎﺵ ﻓﻲ ﺒﺭ،
ﻫﺎﻥ
(7.8)
ﻥـﻤ
(c)
ﺘﺞـﻨﻴ
(a)
.
ﻭﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻨﻪ ﻤﻥ
(a)
ﻴﻨﺘﺞ
(b)
.
ﻨﺘﻴﺠﺔ
13.7
ﺼﻑـﻨ ﺕﻻﻭﺩﻭـﻤﻠﻟ ﺔﻤﺴﻘﻟﺍ ﺭﻤﺯ ﺕﻻﻭﺩﻭﻤ ﻭ ،ﺔﻴﺌﺯﺠﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ ،ﻉﻭﻤﺠﻤ ﻥﺇ
ﺍﻟﺒﺴﻴﻁﺔ ﻫﻲ ﻨﺼﻑ ﺒﺴﻴﻁﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻜل ﻤﻨﻬﺎ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺠﻤﻭﻉ ﻤﻭﺩﻭﻻﺕ ﺒﺴﻴﻁﺔ
.
F
-
ﺍﻟﺠﺒﻭﺭ ﺍﻟﺒﺴﻴﻁﺔ ﻭ ﻤﻭﺩﻭﻻﺘ
ﻬﺎ
(Simple F-algebra modules)
ﻴﻘﺎل ﻋﻥ
F
-
ﺍﻟﺠﺒﺭ
A
ﺭـﻴﻏ ﺏـﻨﺎﺠﻟﺍ ﺔـﻴﺌﺎﻨﺜ ﺔﻴﻠﻌﻓ ﺕﺎﻴﻟﺎﺜﻤ ﻰﻠﻋ ﻭﺤﻴ ﻡﻟ ﺍﺫﺇ ﻁﻴﺴﺒ ﻪﻨﺃ
0
.
ﺴﻨﺠﻌﻠﻪ ﻴﺘﻜﺭﺭ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺘﻭﻀﻴﺢ
ﺍﻵﺘﻲ
:
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٨٢١
ﺇﻥ ﻨﻭﺍﺓ ﺍﻟﺘﺸﺎﻜل
:
f
A
B
→
ـ ﻟ
F
-
ﺍﻟﺠﺒﻭﺭ ﻫﻭ ﺍﻟﻤﺜﺎﻟﻲ
A
ﻭﻱـﺤﻴ ﻻ ﻱﺫـﻟﺍ
1
،
ﺇﺫﺍ ﻜﺎﻥ،ﻙﻟﺫﻟ
A
، ﺒﺴﻴﻁﺎﹰ
ﻋﻨﺩﺌﺫ
f
ﻤﺘﺒﺎﻴﻥ
.
ﻤﺜﺎل
14.7
ﺔـﻴﺭﺒﺠﻟﺍ ﺔﻓﻭﻔﺼﻤﻟﺍ ﺫﺨﺄﻨ
( )
n
M
F
.
لـﻜﻟ
( )
,
n
A B
M
F
∈
ﻭﺩـﻤﻌﻟﺍ ،
j
،
( )
j
AB
ـ ﻟ
AB
ﻫﻭ
j
AB
ﺤﻴﺙ
j
B
ﺍﻟﻌﻤﻭﺩ
j
ﻓﻲ
B
.
ﻤﻥ،ﻙﻟﺫﻟ
ﺃﺠل ﻤﺼﻔﻭﻓﺔ ﻤﻌﻁﺎﺓ
B
،
( )
( )
0
0
0
j
j
j
j
B
AB
B
AB
= ⇒
=
≠ ⇒
ـﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻟﻤﺜﺎﻟﻲ ﺍﻟﻴﺴﺎﺭﻱ ﻟ
( )
n
M
F
ﺸﻜلـﻟﺍ ﻥـﻤ ﺔﻋﻭﻤﺠﻤ ﻭﻫ
( )
L I
ﺙـﻴﺤ
I
ﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻤﻥ
{
}
1, 2,..., n
ﻭ
( )
L I
ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ ﺍﻷﻋﻤﺩﺓ
j
ﺘﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ ﻟﻜل
j
I
∉
.
ﺎﺕـﻋﻭﻤﺠﻤﻟﺍ ﻲـﻫ ﺔﻴﺭﺎﺴﻴﻟﺍ ﺔﻴﺭﻐﺼﻷﺍ ﺎﺕﻴﻟﺎﺜﻤﻟﺍ ،ﺔﺼﺎﺨ ﺔﻟﺎﺤﺒ
{ }
( )
L
j
.
،ﻤﺜﻼﹰ
{ }
( )
0
0
0
0
0
0
0
0
0
0
0
0
3
L
∗
∗
∗
∗
=
ﻭ
{ }
(
)
0
0
0
0
0
0
0
0
2, 4
L
∗
∗
∗
∗
∗
∗
∗
∗
=
ﺍﻟﻤﺜﺎﻟﻲ ﺍﻟﻴﺴﺎﺭﻱ ﻭ ﺍﻟﻤﺜﺎﻟﻲ ﺍﻟﻴﺴﺎﺭﻱ ﺍﻷﺼﻐﺭﻱ،ﺏﻴﺘﺭﺘﻟﺍ ﻰﻠﻋ ،ﻲﻫ
.
ﺸﻜلـﻟﺍ ﺔﻴﻨﻴﻤﻴﻟﺍ ﺕﺎﻴﻟﺎﺜﻤﻠﻟ
ﻲـﻓ ﻡﻭﺩـﻌﻤ ﺭـﻴﻏ ﺏـﻨﺎﺠﻟﺍ ﻲﺌﺎـﻨﺜ ﻲﻟﺎﺜﻤ ﻱﺃ ﻥﺈﻓ ﻪﻨﻤﻭ ،ﺭﻁﺴﻷﺍ ﻲﻓ ﺩﻭﺩﺤﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ ﻪﺴﻔﻨ
( )
n
M
F
ﻴﺸﻜل ﺤﻠﻘﺔ ﻜﺎﻤﻠﺔ
.
( )
n
M
F
ﻋﺒﺎﺭﺓ ﻋﻥ
F
-
ﺠﺒﺭ ﺒﺴﻴﻁ
.
ﻤﺜﺎل
15.7
ﻴﻘﺎل ﻋﻥ
F
-
ﺠﺒﺭ ﺃﻨﻪ ﺠﺒﺭ ﺘﻘﺴﻴﻡ ﺇﺫﺍ ﻜﺎﻥ ﻟﻜل ﻋﻨﺼﺭ ﻏﻴﺭ ﻤﻌﺩﻭﻡ
a
،ﺎﹰـﺴﻭﻜﻌﻤ
ﻴﻭﺠﺩ،ﻪﻨﺃ ﻱﺃ
b
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
1
ab
ba
= =
.
ﺴﻠﻤﺎﺕـﻤﻟﺍ لـﻜ ﻕـﻘﺤﻴ ﻡﻴﺴﻘﺘﻟﺍ ﺭﺒﺠ ﻥﺈﻓ ﺍﺫﻬﻟ
ﻟﻴﻜﻭﻥ
ﺤﻘﻼﹰ ﻤﺎ ﻋﺩﺍ ﺍﻟﻘﺎﻨﻭﻥ ﺍﻟﺘﺒﺩﻴﻠﻲ
)
ﻼﹰـﻘﺤ ﹰﺎﻨﺎﻴﺤﺃ ﻰﻋﺩﻴ ﺏﺒﺴﻟﺍ ﺍﺫﻬﻟﻭ
ﺩﻴﻠﻲـﺒﺘ ﺭـﻴﻏ
.(
ﻥـﻤ
،ﺏـﻨﺎﺠﻟﺍ ﺔـﻴﺌﺎﻨﺜ ﻭﺃ ،ﺔﻴﻨﻴﻤﻴ ،ﺔﻴﺭﺎﺴﻴ ،ﺔﻴﻠﻌﻓ ﺕﺎﻴﻟﺎﺜﻤ ﻰﻠﻋ ﻱﻭﺤﻴ ﻻ ﻡﻴﺴﻘﺘﻟﺍ ﺭﺒﺠ ﻥﺃ ،ﺢﻀﺍﻭﻟﺍ
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺒﺴﻴﻁﺔ
.
ﺇﺫﺍ ﻜﺎﻥ
D
ﺠﺒﺭ ﺘﻘﺴﻴﻡ
(division algebra)
ﻜﻤﺎ ﻓﻲ،ﺫﺌﺩﻨﻋ ،
(7.14)
ﺍﻟﻤﺜﺎﻟﻲ ﺍﻟﻴﺴﺎﺭﻱ،
ﻓﻲ
( )
n
M
D
ﻫﻭ ﻤﺠﻤﻭﻋﺎﺕ ﻤﻥ ﺍﻟﺸﻜل
( )
L I
ﻭ
( )
n
M
D
ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﺒﺴﻴﻁﺔ
.
ﻤﺜﺎل
16.7
ﻟﻜل
,
×
∈
a b
F
ﻭﻟﺘﻜﻥ،
( )
,
H a b
ﻥـﻋ ﺓﺭﺎـﺒﻋ
F
-
ﺩﺓـﻋﺎﻘﻟﺍ ﻊـﻤ ﺭـﺒﺠ
1, , ,
i j k
)
F
-
ﻓﻀﺎﺀ ﻤﺘﺠﻬﻲ
(
ـﻭﺍﻟﻀﺭﺏ ﺍﻟﻤﺤﺩﺩ ﺒ
2
2
,
,
i
a
j
b ij
k
ji
=
=
= = −
ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﺍﺨﺘﻴﺎﺭﻴﺎﹰ
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٩٢١
)
ﻭﻤﻨﻪ
ik
iij
aj
=
=
ﺍﻟﺦ
.
.(
ﻋ
ﻨﺩﺌﺫ
( )
,
H a b
ﻋﺒﺎﺭﺓ ﻋﻥ
F
-
ﻴﺩﻋﻰ ﺍﻟﺠﺒﺭ ﺍﻟﺭﺒﺎﻋﻲ،ﺭﺒﺠ
(quaternion algebra)
ﻋﻠﻰ
F
.
ﺇﺫﺍ ﻜﺎﻥ،ﹰﻼﺜﻤ
F
=
ﺫـﺌﺩﻨﻋ ،
(
)
1, 1
H
− −
ﺭـﺒﺠﻟﺍ
ﺍﻟﺭﺒﺎﻋﻲ ﺍﻟﻌﺎﻡ
.
ﻴﻤﻜﻨﻨﺎ
ﺃﻭﻻﹰ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻥ
( )
,
H a b
ﻊـﻤ لﺜﺎﻤﺘﻴ ﻭﺃ ﻡﻴﺴﻘﺘ ﺭﺒﺠ ﺎﻤﺇ
( )
2
M
F
.
ﻴﻜﻭﻥ ﺒﺴﻴﻁﺎﹰ،ﺹﺎﺨ لﻜﺸﺒ
.
17.7
ﺍﻟﻜﺜﻴﺭ ﻤﻥ ﺍﻟﺠﺒﺭ ﺍﻟﺨﻁﻲ ﻻ ﻴﺘﻁﻠﺏ ﺒﺄﻥ ﻴﻜﻭﻥ ﺍﻟﺤﻘل ﺘﺒﺩﻴﻠﻴﺎﹰ
.
ﻴﻥـﺒﺘ ﺔﻤﺎﻌﻟﺍ ﺕﺎﺸﺎﻘﻨﻟﺍ ،ﹰﻼﺜﻤ
ﺒﺄﻥ ﺍﻟﻤﻭﺩﻭل ﺍﻟﻤﻨﺘﻬﻲ ﺍﻟﺘﻭﻟﻴﺩ
V
ﻓﻭﻕ ﺠﺒﺭ ﺍﻟﺘﻘﺴﻴﻡ
D
ﻭﺃﻥ ﻜل ﻗﻭﺍﻋﺩﻩ ﺘﺤﻭﻱ ﻋﻠﻰ، ﺩﻋﺍﻭﻗ ﻪﻟ
ﺍﻟﻌﺩﺩ ﻨﻔﺴﻪ
n
-
ﻨﺴﻤﻲ
n
ﺒﻌﺩ
(dimension)
V
.
ﻊـﻴﻤﺠ ،ﺔﺼﺎﺨ ﺔﻟﺎﺤﺒ
D
-
ﻭﺩﻭﻻﺕـﻤ
ﺍﻟﻤﻨﺘﻬﻴﺔ ﺍﻟﺘﻭﻟﻴﺩ ﺘﻜﻭﻥ ﺤﺭﺓ
.
18.7
ﻟﻴﻜﻥ
A
F
-
ﻭ ﻟﻨﻌﺘﺒﺭ،ﺭﺒﺠ
A
A
ـ ﺭﻤﺯﺍﹰ ﻟ
A
-
ﻤﻭﺩﻭل ﻴﺴﺎﺭﻱ
.
ﻥـﻤ ﺀﺍﺩـﺠﻟﺍ
ﺍﻟﻴﻤﻴﻥ
x
xa
a
ﻋﻠﻰ
A
A
ﺒﺎﻟﻌﻨﺼﺭ
a
ﻓﻲ
A
ﻴﻜﻭﻥ
A
-
ـ ﺘﺸﺎﻜﻼﹰ ﺫﺍﺘﻴﺎﹰ ﺨﻁﻴﺎﹰ ﻟ
A
A
.
ﻜل،ﻙﻟﺫ ﻰﻠﻋ ﹰﺓﻭﻼﻋ
A
-
ﺘﻁﺒﻴﻕ ﺨﻁﻲ
:
A
A
A
A
ϕ
→
ﺸﻜلـﻟﺍ ﺍﺫﻫ ﻥﻤ ﻭﻫ
( )
1
a
ϕ
=
.
،ﻟﻬﺫﺍ
)
ﻨﻌﺘﺒﺭﻩ
F
-
ﻓﻀﺎﺀ
ﻤﺘﺠﻬﻴ
(ﺎﹰ
( )
End
A
A
A
A
ﻟﻴﻜﻥ
a
ϕ
ﺍﻟﺘﻁﺒﻴﻕ
x
xa
a
.
ﻋﻨﺩﺌﺫ
(
)( )
( )
(
)
( )
( )
def
1
1
1
a
a
a
a
a
a a
a
a a
ϕ ϕ
ϕ ϕ
ϕ
ϕ
′
′
′
′
′
=
=
=
=
o
ﻭﻤﻨﻪ
)
ﻨﻌﺘﺒ
ﺭﻩ
F
-
ﺠﺒﺭ
(
( )
opp
End
A
A
A
A
،ﻭﺒﺎﻟﺘﻌﻤﻴﻡ
( )
opp
End
A
V
A
ﻷﻱ
A
-
ﻤﻭﺩﻭل
V
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﺤﺭﺍﹰ ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ
(rank)
1
ﻭ،
( )
( )
opp
End
A
n
V
M
A
ﻷﻱ
A
-
ﻤﻭﺩﻭل ﺤﺭ
V
ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ
n
(cf. 7.30)
.
ﺍﻟﻤﻤﺭﻜﺯﺍﺕ
(Centralizers)
ﻟﻴﻜﻥ
A
،
F
-
ﺠﺒﺭﺍﹰ ﺠﺯﺌﻴﺎﹰ ﻓﻲ
F
-
ﺍﻟﺠﺒﺭ
B
.
ﺇﻥ ﻤﻤﺭﻜﺯ
(centralizer)
A
ﻓﻲ
B
ﻫﻭ
( )
{
}
,
B
C
A
b
B ba
ab
a
A
= ∈
=
∀ ∈
ﻭﻫﻭ ﺃﻴﻀﺎﹰ
F
-
ﺠﺒﺭ ﺠﺯﺌﻲ ﻓﻲ
B
.
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ﻤﺜﺎل
19.7
ﻓﻲ ﺍﻷﻤﺜﻠﺔ
ﺍﻵﺘﻴ
ﺘﺅﺨ،ﺔ
ﺫ ﺍﻟﻤﻤﺭﻜﺯﺍﺕ ﻤﻥ
( )
n
M
F
.
)
a
(
ﻟﺘﻜﻥ
A
ﻲـﻓ ﺔﻴﻤﻠﺴﻟﺍ ﺕﺎﻓﻭﻔﺼﻤﻟﺍ ﻥﻤ ﺔﻋﻭﻤﺠﻤ
( )
n
M
F
، ﺃﻱ ﺃﻥ،
n
A
FI
=
.
،ﻤﻥ ﺍﻟﻭﺍﻀﺢ
( )
( )
n
C A
M
F
=
.
)
b
(
ﻟﺘﻜﻥ
( )
n
A
M
F
=
.
ﻓﺈﺫﺌﺩﻨﻋ
ﻥ
( )
C A
ﻫﻭ ﻤﺭﻜﺯ
( )
n
M
F
ﺴﺒﻪـﺤﻨ ﻱﺫﻟﺍ ،
ﺍﻵﻥ
.
ﻟﺘﻜﻥ
ij
e
ﺍﻟﻤﺼﻔﻭﻓﺔ ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻓﻴﻬﺎ
1
ﻓﻲ ﺍﻟﻤﻭﻀﻊ
( )
,
i j
ﻭﺍﻟﺼﻔﺭ ﻓﻲ ﺒﻘﻴﺔ
ﻟﺫﻟﻙ،ﻊﻀﺍﻭﻤﻟﺍ
,
0,
=
=
≠
i m
i j
l m
e
j
l
e e
j
l
ﺘﻜﻥــــﻟ
( )
( )
i j
n
a
M
F
α
=
∈
.
ﺫــــﺌﺩﻨﻋ
,
i j
i j
i j
a e
α
=
∑
ﻪــــﻨﻤﻭ
lm
i
il
im
e
a e
α
=
∑
ﻭ
l m
j
m j
l j
e
a
e
α
=
∑
.
ﺇﺫﺍ
ﺎﻥــﻜ
α
ﺯــﻜﺭﻤ ﻲــﻓ
( )
n
M
F
ﺫــﺌﺩﻨﻋ ،
l m
l m
e
e
α
α
=
،
ﻪــﻨﻤﻭ
0
il
a
=
ﻟﻜل
i
l
≠
،
0
m j
a
=
ﻟﻜل
j
m
≠
ﻭ،
l l
m m
a
a
=
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻤﺭﻜﺯ
( )
n
M
F
ﻫﻭ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﺼﻔ
ﻭﻓﺎﺕ ﺍﻟﺴﻠﻤﻴﺔ
.
n
F I
.
ﻟﻬﺫﺍ
( )
.
n
C A
F I
=
.
)
c
(
ﻟﺘﻜﻥ
A
ﻲـﻓ ﺔﻴﺭﻁﻘﻟﺍ ﺕﺎﻓﻭﻔﺼﻤﻟﺍ لﻜ ﺔﻋﻭﻤﺠﻤ
( )
n
M
F
.
،ﺔـﻟﺎﺤﻟﺍ ﻩﺫـﻫ ﻲـﻓ
( )
C A
A
=
.
،ﻨﻼﺤﻅ ﺃﻨﻪ ﻓﻲ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺜﻼﺙ
( )
(
)
C C A
A
=
.
ﻤﺒﺭﻫﻨﺔ
20.7
)
ﻤﺒﺭﻫﻨﺔ ﺍﻟﻤﻤﺭﻜﺯﺍﺕ ﺍﻟﻤﺯﺩﻭﺠﺔ
(
ﻴﻜﻥــﻟ
A
،
F
-
ﻴﻜﻥــﻟﻭ ،ﹰﺍﺭــﺒﺠ
V
،
A
-
ﻭﺩﻭلــﻤ
ــﺴﻴﻁ ﺃﻤﻴﻨــﺒ ﻑﺼــﻨ
ﺎﹰ
.
ﺫــﺌﺩﻨﻋ
( )
(
)
C C A
A
=
)
ﺍﻟﻤﺭﺍ
ﻜﺯ ﻤﺄﺨﻭﺫﺓ ﻤﻥ
( )
End
F
V
.(
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
( )
D
C A
=
ﻭ ﻟﻴﻜﻥ
( )
B
C D
=
.
ﺢ ﺃﻥـﻀﺍﻭﻟﺍ ﻥﻤ
A
B
⊂
ﺄﺘﻲـﻴﻭ ،
ﺍﻻﺤﺘﻭﺍﺀ ﺍﻟﻤﻌﺎﻜﺱ ﻤﻥ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
ﺍﻵﺘﻴ
ﺔ ﻋﻨﺩﻤﺎ ﻨﺄﺨﺫ
1
,...,
n
v
v
ﻟﺘﻭﻟﺩ
V
ﺒﺎﻋﺘﺒﺎﺭﻩ
F
-
ﻀﺎﺀـﻓ
ﻤﺘﺠﻬﻲ
.
ﺘﻤﻬﻴﺩﻴﺔ
21.7
ﻷﻱ
1
,...,
n
v
v
V
∈
ﻭ
b
B
∈
ﻴﻭﺠﺩ،
a
A
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
1
1
2
2
,
,...,
n
n
av
bv
av
bv
av
bv
=
=
=
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺒﺭﻫﻥ ﺫﻟﻙ ﺃﻭﻻﹰ ﻤﻥ ﺃﺠل
1
n
=
.
ﻨﻼﺤﻅ ﺒﺄﻥ
1
Av
ﻋﺒﺎﺭﺓ ﻋﻥ
A
-
ﻤﻭﺩﻭل ﺠﺯﺌﻲ
ﻲــﻓ
V
ﻪــﻨﻤﻭ ،
)
ﺭــﻅﻨﺍ
12.7
(
ﺩــﺠﻭﻴ
A
-
ﻲــﺌﺯﺠ لﻭﺩﻭــﻤ
W
ﻲــﻓ
V
ﺙــﻴﺤﺒ
1
V
Av
W
=
⊕
.
ﻟﻴﻜﻥ
:V
V
π
→
ﺍﻟﺘﻁﺒﻴﻕ
(
) (
)
1
1
,
, 0
a v w
av
a
)
ﺍﻹﺴﻘﺎﻁ ﺍﻟﻐﺎﻤﺭ
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١٣١
ﻋﻠﻰ
1
Av
.(
ﺇﻨﻪ
A
-
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﻘﻊ ﻓﻲ،ﻲﻁﺨ
D
ﻭ ﻴﺤﻘﻕ ﺍ،
ﻟﺨﺎﺼﺔ
( )
v
v
π
=
ﺇﺫﺍ ﻭﻓﻘﻁ
ﺇﺫﺍ ﻜﺎﻥ
1
v
Av
∈
.
ﺍﻵﻥ
( ) ( )
1
1
1
,
bv
b
v
bv
π
π
=
=
ﻭﻤﻨﻪ
1
1
bv
Av
∈
.
ﻜﻤﺎ ﻫﻭ ﻤﻁﻠﻭﺏ
.
ﻨﺒﺭﻫﻥ ﺍﻵﻥ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ
.
ﻟﻴﻜﻥ
W
ـ ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ ﻟ
V
،
n
ﻤﺭﺓ ﺒﺎﻟﺘﺄﺜﻴﺭ ﺍﻟﻘﻁﺭﻱ
A
، ﺃﻱ ﺃﻥ،
(
) (
)
1
1
,...,
,...,
,
,
n
n
i
a v
v
a v
a v
a
A v
V
=
∈
∈
ﻓﺈﻥﺫﺌﺩﻨﻋ
W
ﻋﺒﺎﺭﺓ ﻋﻥ
A
-
ﻤﻭﺩﻭل
ﻀﺎﹰـﻴﺃ ﻁﻴﺴـﺒ ﻑﺼﻨ
(13.7)
.
ﺯـﻜﺭﻤﻤ ﻥﺇ
A
ﻲـﻓ
( )
End
F
W
ﺼﻔﻭﻓﺎﺕـﻤﻟﺍ ﻥﻤ ﻑﻟﺄﺘﺘ
( )
( )
1
,
,
End
i j
i j
F
i j n
V
γ
γ
≤
≤
∈
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ ،
( ) ( )
i j
i j
a
a
γ
γ
=
لـﻜﻟ
a
A
∈
، ﺃﻱ ﺃﻥ،
i j
D
γ
∈
(cf. 30.7)
.
ﺇﻥ،ﻯﺭـﺨﺃ ﺕﺎـﻤﻠﻜﺒ
ﻤﻤﺭﻜﺯ
A
ﻓﻲ
( )
End
F
A
ﻫﻲ
( )
n
M
D
.
ﺎلـﺜﻤﻟﺍ ﻲﻓ ﺎﻤﻜ ﺵﻗﺎﻨﻨ
7.19(b)
ﺘﺨﺩﺍﻡـﺴﺎﺒ ،
ﺍﻟﻤﺼﻔﻭﻓﺎﺕ
( )
i j
e
δ
ﺤﻴﺙ ﻴﺘﻭﻀﻊ
δ
ﻓﻲ ﺍﻟﻤﻭﻀﻊ
ij
ﺄﻥـﺒ ﻥﻴﺒﻴ ،ﻊﻀﺍﻭﻤﻟﺍ ﺔﻴﻘﺒﺒ ﺭﻔﺼﻟﺍ ﻭ
ﻤﻤﺭﻜ
ﺯ
( )
n
M
D
ﻓﻲ
( )
End
F
W
ﻴﺘﺄﻟﻑ ﻤﻥ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻘﻁﺭﻴﺔ
0
0
0
0
0
0
β
β
β
L
K
M
M
O
M
L
ﺤﻴﺙ
B
β
∈
.
ﻨﻁﺒﻕ ﺍﻵﻥ ﺍﻟﺤﺎﻟﺔ
1
n
=
ـ ﻤﻥ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﻟ
A
،
W
،
b
ﻭ ﺍﻟﺸﻌﺎﻉ،
(
)
1
,...,
n
v
v
ﻟﻴﺘﻡ ﺍﻟﺒﺭﻫﺎﻥ
.
ﻤﺒﺭﻫﻨﺔ
22.7
ﻜل
F
-
ﺠﺒﺭ ﺒﺴﻴﻁ ﻤﺘﻤﺎﺜل ﻤﻊ
( )
n
M
D
ﻟﺒﻌﺽ ﻤﻥ
n
ﺽـﻌﺒ ﻭ
F
-
ﺠﺒﺭ ﺘﻘﺴﻴﻡ
D
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﺨﺘﺎﺭ
A
-
ﻤﻭﺩﻭل ﺒﺴﻴﻁ
S
ﻲـﻓ ﻱﺭﻐﺼﺃ ﻱﺭﺎﺴﻴ ﻲﻟﺎﺜﻤ ﻱﺃ ،ﹰﻼﺜﻤ ،
A
.
ﺫـﺌﺩﻨﻋ
ﻴﺅﺜﺭ
A
ﻋﻠﻰ
S
ـ ﻷﻥ ﺍﻟﻨﻭﺍﺓ ﻟ،ﻥﻴﻤﺃ لﻜﺸﺒ
( )
End
F
A
S
→
ﺴﺘ
ﻜﻭﻥ ﻤﺜﺎﻟﻲ ﺜﻨﺎﺌﻲ ﺍﻟﺠﺎﻨﺏ
ﻓﻲ
A
ﻻ ﻴﺤﻭﻱ ﻋﻠﻰ
1
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﺴﺎﻭﻱ،
0
.
ﻟﻴﻜﻥ
D
ﻤﻤﺭﻜﺯ
A
ﻓﻲ
F
-
ﺠﺒﺭ
( )
End
F
S
ـ ﻟ
F
-
ﻲـﻁﺨﻟﺍ ﻕﻴﺒﻁﺘﻟﺍ
S
S
→
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ ﻤﺒﺭﻫﻨﺔ ﺍﻟﻤﻤﺭﻜﺯﺍﺕ ﺍﻟﻤﺯﺩﻭﺠﺔ
(20.7)
ﻓﺈﻥ ﻤﻤﺭﻜﺯ،
D
ﻓﻲ
( )
End
F
S
ﻲـﻫ
A
ﺃﻱ ﺃﻥ،
،
( )
End
D
A
S
=
.
ﺇﻥ ﺘﻤﻬﻴﺩﻴﺔ ﺸﻭﺭ
(23.7)
ﺘﺒﻴﻥ ﺒﺄﻥ
D
ﺠﺒﺭ ﺘﻘﺴﻴﻡ
.
ﻟﺫﻟﻙ ﻓﺈﻥ
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٢٣١
S
D
-
ﻤﻭﺩﻭل ﺤﺭ
(17.7)
، ﻨﻘﻭل،
n
S
D
≈
ﻭﻤﻨﻪ،
( )
( )
opp
End
D
n
S
M
D
≈
)
ﺍﻨﻅﺭ
7.18
.(
ﺘﻤﻬﻴﺩﻴﺔ
23.7
)
ﺘﻤﻬﻴﺩﻴﺔ
ﻭﺭـﺸ
(
لـﻜﻟ
F
-
ﺭـﺒﺠ
A
ﻭ
A
-
ﺴﻴﻁـﺒ لﻭﺩﻭـﻤ
S
ﺇﻥ،
( )
End
A
S
ﺠﺒﺭ ﺘﻘﺴﻴﻡ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
γ
،
F
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
S
S
→
.
ﻋﻨﺩﺌﺫ
( )
Ker
γ
A
-
ﻲـﺌﺯﺠ لﻭﺩﻭﻤ
ﻓﻲ
S
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻭ ﻴﺴﺎﻭﻱ،
S
ﺃﻭ ﻴﺴﺎﻭﻱ
0
.
،ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
γ
ﻲـﻓﻭ ،ﺭﻔﺼﻟﺍ ﻱﻭﺎﺴﻴ
ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻴﺸﻜل ﺘﻤﺎﺜﻼﹰ
ﻟﻪ ﻤﻌﻜﻭﺱ ﻭﻫﻭ ﺃﻴﻀﺎﹰ،ﻥﺃ ﻱﺃ ،
A
-
ﺨﻁﻲ
.
ﺍﻟﻤﻭﺩﻭﻻﺕ ﻋﻠﻰ
F
-
ﺍﻟﺠﺒﻭﺭ ﺍﻟﺒﺴﻴﻁﺔ
ﻷﻱ
F
-
ﺍﻟﺠﺒﺭ
A
ﻲـﻓ ﺔﻴﺌﺯﺠﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ ،
A
A
ﻲـﻓ ﺔﻴﺭﺎﺴـﻴ ﺕﺎـﻴﻟﺎﺜﻤ ﻲـﻫ
A
ﻭ،
ﺍﻟﻤﻭﺩﻭﻻﺕ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺒﺴﻴﻁﺔ ﻓﻲ
A
A
ﻫﻲ ﻤﺜﺎﻟﻴﺎﺕ ﺃﺼﻐﺭﻴﺔ ﺒﺴﻴﻁﺔ
.
ﻗﻀﻴﺔ
24.7
ﺇﻥ ﺃﻱ ﻤﺜﺎﻟﻴﻴﻥ ﻴﺴﺎﺭﻴﻴﻥ ﺃﺼﻐﺭﻴﻴﻥ ﻤﻥ
F
-
ــﺎﺜﻼﻥ ﻜـﻤﺘﻤ ﻁﻴﺴﺒﻟﺍ ﺭﺒﺠﻟﺍ
A
-
ﻭ،ﺔﻴﺭﺎﺴﻴﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ
A
A
ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻟﻤﺜﺎﻟﻴﺎﺘﻪ ﺍﻟﻴﺴﺎﺭﻴﺔ ﺍﻷﺼﻐﺭﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﻌﺩ ﺍﻟﻤﺒﺭﻫﻨﺔ
22.7
ﻴﻤﻜﻥ ﺃﻥ ﻨﻔﺭﺽ ﺃﻥ،
( )
n
A
M
D
=
ﺴﻴﻁﺔـﺒﻟﺍ ﺭﻭﺒﺠﻟﺍ ﺽﻌﺒﻟ
D
.
ﺫﻜﺭﻨﺎ ﻓﻲ
(15.7)
ﺃﻥ ﺍﻟﻤﺜﺎﻟﻴﺎﺕ ﺍﻷﺼﻐﺭﻴﺔ ﻓﻲ
( )
n
M
D
ﻤﻥ ﺍﻟﺸﻜل
{ }
( )
L
j
.
ﻥـﻤ
ﺍﻟﻭﺍﻀﺢ
{ }
( )
1 j n
A
L
j
≤ ≤
= ⊕
ﻭ ﻜل
{ }
( )
L
j
ﻊـﻤ لﺜﺎﻤﺘﻤ
n
D
ﻲـﻓ ﻲـﻌﻴﺒﻁ ﺭﻴﺜﺄـﺘﺒ
( )
n
M
D
.
ﻤﺒﺭﻫﻨﺔ
25.7
ﻟﻴﻜﻥ
F
-
ﺍﻟﺠﺒﺭ ﺍﻟﺒﺴﻴﻁ
A
ﻭ ﻟﻴﻜﻥ،
A
-
ﺍﻟﻤﻭﺩﻭل ﺍﻟﺒﺴﻴﻁ
S
.
لـﻜ ﺫﺌﺩﻨﻋ
A
-
ﻤﻭﺩﻭل ﻴﻜﻭﻥ ﻤﺘﻤﺎﺜﻼﹰ ﻤﻊ ﺍﻟﻤﺠﻤﻭﻉ
ﺍﻟﻤﺒﺎﺸﺭ ﻟﻌﺩﺩ ﻤﻥ
S
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
0
S
ﺍﻟﻤﺜﺎﻟﻲ ﺍﻷﺼﻐﺭﻱ ﺍﻟﻴﺴﺎﺭﻱ ﻓﻲ
A
.
ﺘﺒﻴﻥ ﺍﻟﻘﻀﻴﺔ ﺒﺄﻥ
0
n
A
A
S
≈
ﻟﺒﻌﺽ
n
.
ﻟﺘﻜﻥ
1
,...,
r
e
e
ﻤ
ﺠﻤﻭﻋﺔ ﻤﻥ ﻤﻭﻟﺩﺍﺕ
V
ﻜﺫﻟﻙ
A
-
ﻤﻭﺩﻭل
.
ﺍﻟﺘﻁﺒﻴﻕ
(
)
1
,...,
r
i
i
a
a
a e
∑
a
ﻨﻌﺘﺒﺭ
V
ﻜﺯﻤﺭﺓ ﻗﺴﻤﺔ ﻤﻥ ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ
r
ﻤﺭﺓ ﻤﻥ
A
A
ﻭﺒﺎﻟﺘﺎﻟ،
ﺴﻤﺔـﻗ ﺓﺭﻤﺯ ﺎﻬﻨﺈﻓ ﻲ
ﻤﻥ
0
n r S
.
ﺇﻥ،ﺍﺫﻬﻟ
V
لـﺜﺎﻤﻴ ﺎﻬﻨﻤ لﻜ ﺔﻁﻴﺴﺒ ﺔﻴﺌﺯﺠ ﺕﻻﻭﺩﻭﻤﻟ ﺭﺸﺎﺒﻤ ﻉﻭﻤﺠﻤ ﻥﻋ ﺓﺭﺎﺒﻋ
0
S
ﻥـﻤﻭ ،
ﺍﻟﻘﻀﻴﺔ
11.7
ﻨﺠﺩ ﺒﺄﻥ
0
V
mS
≈
ﻟﺒﻌﺽ
m
.
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٣٣١
ﻨﺘﻴﺠﺔ
26.7
ﻟﻴﻜﻥ
F
-
ﺍﻟﺠﺒﺭ ﺍﻟﺒﺴﻴﻁ
A
.
ﺃﻱﺫﺌﺩﻨﻋ
A
-
ﻤﻭﺩﻭﻟﻴﻥ ﺠﺯﺌﻴﻴﻥ ﺒﺴﻴﻁﻴﻥ ﻴﻜﻭﻨﺎﻥ
ﻭﺃﻱ،ﻥﻴﻠﺜﺎﻤﺘﻤ
A
-
ﻤﻭﺩﻭﻟﻴﻥ ﻟﻬﻤﺎ ﺍﻟﺒﻌﺩ ﻨﻔﺴﻪ ﻋﻠﻰ
F
ﻤﺘﻤﺎﺜﻠ
ﻴ
ﻥ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻭﺍﻀﺢ ﻤﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
.
ﻨﺘﻴﺠﺔ
27.7
ﺇﻥ ﺍﻟﻌﺩﺩ
n
ﻓﻲ ﺍﻟﻤﺒﺭﻫﻨﺔ
7.22
ـ ﻤﺤﺩﺩ ﺒ
A
ﻭ،ﺩﻴﺤ ﻭلﻜﺸﺒ
D
ﻤﺤﺩﺩ ﺒﺸﻜل
ﻭﺤﻴﺩ ﺘﺤﺕ ﺴﻘﻑ ﺍﻟﺘﻤﺎﺜل
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﺫﺍ ﻜﺎﻥ
( )
n
A
M
D
≈
ﻋﻨﺩﺌﺫ،
( )
End
A
D
S
≈
ﻷﻱ
A
-
ﺴﻴﻁـﺒﻟﺍ لﻭﺩﻭـﻤﻟﺍ
S
.
ﻭﻤﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ،ﺍﺫﻬﻟ
25.7
ﻴﺘﻡ ﺍﻟﻤﻁﻠﻭﺏ
.
ﺘﺼﻨﻴﻑ ﺠﺒﻭﺭ ﺍﻟﺘﻘﺴﻴﻡ ﻋﻠﻰ
F
ﺒﻌﺩ ﺍﻟﻤﺒﺭﻫﻨﺔ
22.7
ﻟﻜﻲ ﻨﺼﻨﻑ ﺍﻟﺠﺒﻭﺭ ﺍﻟﺒﺴﻴﻁﺔ ﻋﻠﻰ،
F
ﻰـﻠﻋ ﻡﻴﺴﻘﺘﻟﺍ ﺭﻭﺒﺠ ﻑﻴﻨﺼﺘ ﻲﻘﺒ ،
F
.
ﻗﻀﻴﺔ
28.7
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
F
ﻤﻐﻠﻕ
ﺇﻥ ﺠﺒﺭ ﺍﻟﺘﻘﺴﻴﻡ ﺍﻟﻭﺤﻴﺩ ﻋﻠﻰ،ﹰﺎﻴﺭﺒﺠ
F
ﻫﻭ
F
ﻨﻔﺴﻪ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
D
ﺠﺒﺭ ﺘﻘﺴﻴﻡ ﻋﻠﻰ
F
.
ﻷﻱ
D
α
∈
ﺇﻥ،
F
-
ﺍﻟﺠﺒﺭ ﺍﻟﺠﺯﺌﻲ
[ ]
F
α
ﻓﻲ
D
ـ ﺍﻟﻤﻭﻟﺩ ﺒ
α
ﻫﻭ ﺤﻘل
)
ﻷﻨﻪ ﻤﻨﻁﻘﺔ ﺘﻜ
ﺎﻤﻠﻴﺔ ﻤﻥ ﺩﺭﺠﺔ ﻤﻨﺘﻬﻴﺔ ﻋﻠﻰ
F
.(
ﻟﺫﻟﻙ
F
α
∈
.
29.7
ﺇﻥ ﺘﺼﻨﻴﻑ ﺼﻔﻭﻑ ﺍﻟﺘﻤﺎﺜل ﻟﺠﺒﻭﺭ ﺍﻟﺘﻘﺴﻴﻡ ﻋﻠﻰ ﺍﻟﺤﻘل ﻫﻭ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻤﺴﺎﺌل ﺍﻟﻬﺎﻤﺔ ﻭ
ﺍﻟﺼﻌﺒﺔ ﻓﻲ ﺍﻟﺠﺒﺭ ﻭﻨﻅﺭﻴﺔ ﺍﻷﻋﺩﺍﺩ
.
ﻤﻥ ﺃﺠل
F
=
ﻴﻜﻭﻥ ﺠﺒﺭ ﺍﻟﺘﻘﺴﻴﻡ ﺍ،
ﺭـﺒﺠﻟﺍ ﻭﻫ ﺩﻴﺤﻭﻟ
ﺍﻟﺭﺒﺎﻋﻲ ﺍﻟﻌﺎﺩﻱ
.
ﺇﺫﺍ ﻜﺎﻥ
F
ﺯـﻜﺭﻤﺒ ﺩـﻴﺤﻭﻟﺍ ﻡﻴﺴﻘﺘﻟﺍ ﺭﺒﺠ ﻥﺈﻓ ،ﹰﺎﻴﻬﺘﻨﻤ
F
ﻭـﻫ
F
ﺴﻪـﻔﻨ
)
ﻤﺒﺭﻫﻨﺔ ﻭﻴﺩﺭﺒﻭ
ﺭﻥ
.(
ﻴﺴﻤﻰ ﺠﺒﺭ ﺍﻟﺘﻘﺴﻴﻡ ﻋﻠﻰ
F
ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ
F
ﻤﺭﻜﺯﻴﺎﹰ
(central)
)
ﻨﺎﻅﻤﻲ ﺴﺎﺒﻘﺎﹰ
(normal)
.(
ﺒﻴﻥ ﺒﺭﻭﻓﻴﺭ ﺒﺄﻥ ﻤﺠﻤﻭﻋﺔ ﺼﻔﻭﻑ ﺍﻟﺘﻤﺎﺜﻼﺕ ﻟﺠﺒﻭﺭ ﺍﻟﺘﻘﺴﻴﻡ ﺍﻟﻤﺭﻜﺯﻴﺔ ﻋﻠﻰ ﺤﻘل ﻴﺸﻜل
،ﺭﺓـﻤﺯ
ﺴﻤﻴﺕ
)
ـﺒ
ﻫﺎﺱ ﻭ ﻨﻭﺜﻴﺭ
(
ﺯﻤﺭﺓ ﺒﺭﻭﺍﻴﺭ
21
(Brauer group)
ﻟﻠﺤﻘل
.
ﺇﻥ ﻤﻌﻨﻰ ﻤﺎ ﺠﺎﺀ ﻓﻲ
21
ﺇﻥ ﺍﻟﺠﺩﺍﺀ ﺍﻟﺘﻨﺴﻭﺭﻱ
(tensor product)
F
D
D
′
⊗
ﻟﺠﺒﺭﻴﻥ ﺒﺴﻴﻁﻴﻥ ﻤﺭﻜﺯﻴﻴﻥ ﻋﻠﻰ
F
ﺴﻴﻁـﺒ ﺭـﺒﺠ ﻭﻫ
ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻴﻜﻭﻥ ﻤﺘﻤﺎﺜﻼﹰ ﻤﻊ،ﹰﺎﻀﻴﺃ ﻱﺯﻜﺭﻤ
( )
r
M
D
′′
ﻟﺒﻌﺽ ﻤﻥ ﺍﻟﺠﺒﻭﺭ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﺭﻜﺯﻴﺔ
D
′′
.
ﻨﻌﺭﻑ
D
D
D
′
′′
=
ﺼﻑ ﺘﻤﺎﺜل،ﺔﻴﻌﻴﻤﺠﺘ ﺔﻴﺭﻭﺴﻨﺘﻟﺍ ﺕﺍﺀﺍﺩﺠﻟﺍ ﻥﻷ ﻲﻌﻴﻤﺠﺘ ﺀﺍﺩﺠﻟﺍ ﺍﺫﻫ ﻥﺇ
F
ﻭ،ﻱﺩﺎﻴﺤﻟﺍ ﺭﺼﻨﻌﻟﺍ ﻭﻫ
opp
D
ﻫﻭ
ﻤﻌﻜﻭﺱ
D
.
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٤٣١
ﺍﻟﻔﻘﺭﺓ ﺍﻷﺨﻴﺭﺓ ﻫﻭ ﺃﻥ ﺯﻤﺭ ﺒﺭﻭﺍﻴﺭ ﻟﻠﺤﻘ
ﺇﻥ،ﺭﻔﺼـﻟﺍ ﻲـﻫ ﺔﻴﻬﺘﻨﻤﻟﺍ لﻭﻘﺤﻟﺍﻭ ﹰﺎﻴﺭﺒﺠ ﺔﻘﻠﻐﻤﻟﺍ لﻭ
ﺯﻤﺭﺓ ﺒﺭﻭﺍﻴﺭ ﻓﻲ
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﺃﺤﺼﻴﺕ ﺯﻤﺭ ﺒﺭﻭﺍﻴﺭ ﻓﻲ
ﻥـﻤ ﺔـﻴﻬﺘﻨﻤﻟﺍ ﺎﻬﺘﺍﺩﻴﺩﻤﺘ ﻭ
ﻭ ﻨﻭﺜﻴﺭ ﻓﻲ ﻋﺎﻡ،ﺱﺎﻫ ،ﺭﻴﺍﻭﺭﺒ ،ﺕﺭﻴﺒﻟﺃ لﺒﻗ
1930
ﻜﻨﺘﻴﺠﺔ ﻟﺼﻑ ﻨﻅﺭﻴﺔ ﺍﻟﺤﻘﻭل
.
F
-
ﺎـﻬﺘﻻﻭﺩﻭﻤﻭ ﺔﻁﻴﺴـﺒﻟﺍ ﻑﺼﻨ ﺭﻭﺒﺠﻟﺍ
(Semisimple F - algebras
and their modules)
ﻴﻘﺎل ﺇﻥ
F
-
ﺍﻟﺠﺒﺭ
A
لـﻜ ﻥﺎـﻜ ﺍﺫﺇ ﻁﻴﺴـﺒ ﻑﺼـﻨ ﻪﻨﺃ
A
-
ﺴﻴﻁـﺒ ﻑﺼـﻨ لﻭﺩﻭـﻤ
(semisimple)
.
ﺘﺒﻴﻥ ﻟﻨﺎ
ﺍﻟﻤﺒﺭﻫﻨﺔ
7.25
ﺃﻥ
F
-
ﻭ،ﺔﻁﻴﺴـﺒ ﻑﺼﻨ ﻥﻭﻜﺘ ﺔﻁﻴﺴﺒﻟﺍ ﺭﻭﺒﺠﻟﺍ
ﺘﺒﻴﻥ ﻤﺒﺭﻫﻨﺔ ﻤﺎﺴﺸﻙ ﺒﺄﻥ ﺠﺒﺭ ﺍﻟﺯﻤﺭﺓ
[ ]
F G
ﺔـﺒﺘﺭ ﻥﻭﻜﺘ ﺎﻤﺩﻨﻋ ﻁﻴﺴﺒ ﻑﺼﻨ
G
لـﺒﻘﺘ ﻻ
ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﻤﻤﻴﺯ
F
)
ﺍﻨﻅﺭ
8.7
.(
ﻨﺘﺫﻜﺭ ﺒﻌﻀﺎﹰ ﻤﻥ ﻤﺒﺭﻫﻨﺎﺕ ﺍﻟﻤﻭﺩﻭﻻﺕ ﺍﻷﺴﺎﺴﻴﺔ،ﺓﺭﻘﻔﻟﺍ ﻩﺫﻬﻟ ﺔﻴﺴﻴﺌﺭﻟﺍ ﺔﺠﻴﺘﻨﻟﺍ ﺽﺭﻋ لﺒﻗ
.
30.7
ﻟﻴﻜﻥ
F
-
ﺍﻟﺠﺒﺭ
A
ﻭﻟﻨﺄﺨﺫ ﺍﻟﻤﻭﺩﻭﻻﺕ،
1
1
n
n
M
M
M
N
N
N
=
⊕
⊕
=
⊕
⊕
L
L
ﻭﻟﻴﻜﻥ
A
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
: M
N
α
→
.
ﻟﻜل
j
j
x
M
∈
ﻭﻟﻴﻜﻥ،
(
)
(
)
1
0,..., 0,
, 0,..., 0
,...,
j
m
x
y
y
α
=
ﻓﺈﻥﺫﺌﺩﻨﻋ
j
i
x
y
→
ﻋﺒﺎﺭﺓ ﻋﻥ
A
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
j
i
M
N
→
ﻪـﻟ ﺯـﻤﺭﻨ ﻱﺫـﻟﺍﻭ ،
ﺒﺎﻟﺭﻤﺯ
i j
α
.
ﻓﺈﻥ،ﺍﺫﻬﻟ
α
ﻴﻌﺭﻑ ﺍﻟﻤﺼﻔﻭﻓﺔ
m n
×
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻓﻴﻬﺎ ﺍﻟﻤﻌﺎﻤل
ij
ﻋﺒﺎﺭﺓ ﻋﻥ
A
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
j
i
M
N
→
.
ﺫﻩـﻬﻜ ﺔﻓﻭﻔﺼﻤ لﻜ ،ﺱﻜﻌﻟﺎﺒﻭ
( )
i j
α
ﺭﻑـﻌﺘ
A
-
ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﺨﻁﻲ
M
N
→
، ﺘﺤﺩﻴﺩﺍﹰ،
( )
( )
( )
( )
( )
( )
11
1
1
11
1
1
1
1
1
1
1
1
1
1
j
n
n
n
def
j
i
ij
jn
j
i
in
n
n
m
mj
mn
n
m
mn
n
x
x
x
x
x
x
x
x
x
x
x
x
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
+ +
=
+ +
+ +
L
L
L
M
M
M
M
M
M
L
L
a
L
M
M
M
M
M
M
L
L
L
ﻨﺭﻯ،ﺍﺫﻬﺒﻭ
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٥٣١
(22)
(
)
(
)
(
)
1
,1
Hom
,
Hom
,
A
A
j
i
j n
i m
M N
M
N
≤ ≤
≤ ≤
)
ـﺘﻤﺎﺜل ﻟ
F
-
ﺍﻟﻔﻀﺎﺀﺍﺕ ﺍﻟﻤﺘ
ﺠﻬﻴﺔ
.(
ﺇﺫﺍ ﻜﺎﻥ
M
N
=
ـ ﻴﺼﺒﺢ ﻫﺫﺍ ﺍﻟﺘﻤﺎﺜل ﺘﻤﺎﺜﻼﹰ ﻟ،
F
-
ﺠﺒﻭﺭ
.
ﺇﺫﺍ ﻜﺎﻨﺕ،ﹰﻼﺜﻤ
M
ـ ﻤﺠﻤﻭﻋﺎﹰ ﻤﺒﺎﺸﺭﺍﹰ ﻟ
m
ﻤﺭﺓ ﻤﻥ
0
M
ﻋﻨﺩﺌﺫ،
(23)
( )
( )
(
)
0
End
End
A
m
A
M
M
M
)
ﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻟﻨﻭﻉ
m m
×
ﺤﻴﺙ ﺇﻥ
ﻤﻌﺎﻤﻼﺘﻬﺎ ﻤﺄﺨﻭﺫﺓ ﻤﻥ ﺍﻟﺤﻠﻘﺔ
( )
0
End
A
M
.(
ﻤﺒﺭﻫﻨﺔ
31.7
ﻜل
F
-
ﺠﺒﺭ ﻨﺼﻑ ﺒﺴﻴﻁ
A
ﻴﻜﻭﻥ ﻤﺘﻤﺎﺜﻼﹰ ﻤﻊ ﺠﺩﺍﺀ
F
-
ﺠﺒﺭ ﺒﺴﻴﻁ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﺎﺨﺘﻴﺎﺭ
A
-
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺨﻁﻲ
A
i
i
i
A
r S
→ ⊕
ﺤﻴﺙ
i
S
A
-
، ﻤﻭﺩﻭﻻﺕ ﺒﺴﻴﻁﺔ
ﺎـﻤﻬﻨﻤ ﻥﻴـﺘﻨﺜﺍ ﻱﺃ ﻥﻴﺒ لﺜﺎﻤﺘ ﺩﺠﻭﻴ ﻻﻭ
.
ﺫـﺌﺩﻨﻋ
( )
(
)
End
End
A
A
A
i
i
i
A
r S
≈
⊕
.
ﻷﻥ
(
)
Hom
,
0
A
j
i
S S
=
ﻟﻜل
i
j
≠
،
(
)
(
)
( )
End
End
i
A
i
i
i
A
i
i
i
r
i
i
r S
r S
M
D
⊕
∏
∏
ﺤﻴﺙ
( )
End
i
A
i
D
S
=
.
،ﻭﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ
( )
opp
End
A
A
A
ﻤﻥ
(7.18)
ﻭ ﻟﺫﻟﻙ،
( )
opp
i
r
i
i
A
M
D
≈
∏
ﻭﺒﺘﻁﺒﻴﻕ ﺍﻟﺠﺒﺭ ﺍﻟﻤﻌﺎﻜﺱ
ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﺘﻤﺎﺜل،
( )
opp
i
r
i
i
A
M
D
≈
∏
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ ﺘﻤﻬﻴﺩﻴﺔ ﺸﻭﺭ
(23.7)
ﻴﻜﻭﻥ،
i
D
ﻥـﻤﻭ ،ﻡﻴﺴـﻘﺘ ﺭـﺒﺠ
(7.15)
ﺄﻥـﺒ ﺩـﺠﻨ
( )
opp
i
r
i
M
D
ﻋﺒﺎﺭﺓ ﻋﻥ
F
-
ﺠﺒﺭ ﺒﺴﻴﻁ
.
ﻤﻭﺩﻭﻻﺕ ﻋﻠﻰ
F
-
ﺠﺒﻭﺭ ﻨﺼﻑ ﺒﺴﻴﻁﺔ
(Modules over semisimple
F- Algebras)
ﻟﻴﻜﻥ
A
B C
= ×
ـ ﺠﺩﺍﺀ ﻟ
F
-
ﺠﺒﻭﺭ
.
ﺇﻥ
B
-
ﻤﻭﺩﻭل
M
ﻰـﻟﺇ لﻭﺤﺘﻴ
A
-
ﻭﺩﻭلـﻤ
ﺒﺎﻟﺘﺄﺜﻴﺭ
ﻣﻦ
)
(22
ﻣﻦ
(23)
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٦٣١
( )
,
b c m
bm
=
ﻤﺒﺭﻫﻨﺔ
32.7
ﻟﻴﻜﻥ
F
-
ﻭ ﻟﻴﻜﻥ،ﻁﻴﺴﺒﻟﺍ ﻑﺼﻨ ﺭﺒﺠ
1
...
t
A
A
A
= × ×
ﺤﻴﺙ ﺇﻥ
ﻥـﻤ لﻜ
i
A
ﻋﺒﺎﺭﺓ ﻋﻥ
F
-
ﺠﺒﺭ ﺒﺴﻴﻁ
.
ﻟﻜل
i
A
ﻟﻴﻜﻥ،
i
A
-
ﻤﻭﺩﻭل ﺒﺴﻴﻁ
i
S
(cf. 26.7)
.
(a)
ﻜل
i
S
ﻋﺒﺎﺭﺓ ﻋﻥ
A
-
ﻭ ﻜل،ﻁﻴﺴﺒ لﻭﺩﻭﻤ
A
-
ﻊـﻤ لﺜﺎﻤﺘﻤ ﻁﻴﺴﺒ لﻭﺩﻭﻤ
i
S
ﺘﻤﺎﻤﺎﹰ
.
(b)
ﻜل
A
-
ﻤﻭﺩﻭل ﻴﺘﻤﺎﺜل ﻤﻊ
i
i
i
r S
⊕
ﻟﺒﻌﺽ
i
r
∈
ﻭ ﻴﻜﻭﻥ ﺍﻟﻤﻭﺩﻭﻻﻥ،
i
i
i
r S
⊕
ﻭ
i
i
i
r S
′
⊕
ﻤﺘﻤﺎﺜﻠﻴ
ﻥ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ
i
i
r
r
′
=
ﻟﻜل
i
.
ﺍﻟﺒﺭﻫﺎﻥ
.
(a)
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺒﺄﻥ ﻜل ﻤﻥ
i
S
ﺒﺴﻴﻁ
ﻋﻨﺩﻤﺎ ﻨﻌﺘﺒﺭﻩ
A
-
ﺩـﺠﻭﻴ ﻻ ﻭ ،لﻭﺩﻭـﻤ
ﺘﻤﺎﺜل ﺒﻴﻥ ﺃﻱ ﺍﺜﻨﻴﻥ ﻤﻨﻬﻤﺎ
.
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻤﻥ
(7.24)
ﻨﺠﺩ ﺒﺄﻥ
A
i
i
A
r S
≈ ⊕
ﺒﻌﺽـﻟ
i
r
∈
.
ﻟﻴﻜﻥ
A
-
ﻤﻭﺩﻭل ﺒﺴﻴﻁ
S
ﻭ ﻟﻴﻜﻥ،
x
ﻋﻨﺼﺭ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﻓﻲ
S
.
ﻕـﻴﺒﻁﺘﻟﺍ ﻥﺈـﻓ ﺫﺌﺩﻨﻋ
:
A
a x
A
S
→
ﺔـﻋﻭﻤﺠﻤ ﻰﻠﻋ ﻩﺭﻭﺼﻘﻤ ﻥﺈﻓ ﻲﻟﺎﺘﻟﺎﺒﻭ ،ﺭﻤﺎﻏ
i
S
ﻲـﻓ
A
A
ﺴﺎﻭﻱـﻴ ﻻ
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻭ ﺘﻤﺎﺜل،ﺭﻔﺼﻟﺍ
.
(b)
ﻴﺒﺭﻫﻥ ﻋﻠﻰ ﺍﻟﺠﺯﺀ ﺍﻷﻭل ﻤﻥ
(a)
ﻭ
ﺯﺀـﺠﻟﺍ ﻰﻠﻋﻭ ،ﺔﻁﻴﺴﺒﻟﺍ ﻑﺼﻨ ﺔﻘﻠﺤﻟﺍ ﻑﻴﺭﻌﺘ ﻥﻤ
ﺍﻟﺜﺎﻨﻲ ﻤﻥ
(10.7)
.
ﺘﻤﺜﻴﻼﺕ
G
(The representation of G)
ﻗﻀﻴﺔ
33.7
ﺇﻥ ﺒﻌﺩ ﻤﺭﻜﺯ
[ ]
F G
ﺒﺎﻋﺘﺒﺎﺭﻩ
F
-
ﻕـﻓﺍﺭﺘﻟﺍ ﻑﻭﻔﺼ ﺩﺩﻋ ﻭﻫ ﻲﻬﺠﺘﻤ ﺀﺎﻀﻓ
ﻓﻲ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
1
,...,
t
C
C
ﻲـﻓ ﻕـﻓﺍﺭﺘﻟﺍ ﻑﻭﻔﺼ
G
لـﻜﻟ ،ﻭ ،
i
ﻴﻜﻥـﻟ ،
i
c
ﺼﺭـﻨﻌﻟﺍ
i
a C
a
∈
∑
ﻓﻲ
[ ]
F G
.
ﺴﻨﺒﺭﻫﻥ ﻋﻠﻰ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻗﻭﻯ
:
(24)
ﺇﻥ ﻤﺭﻜﺯ
[ ]
F G
ﻫﻭ
1
...
t
c
c
F
F
⊕ ⊕
ﺒﻤﺎ ﺃﻥ
1
,...,
t
c
c
ﻟﺫﻟﻙ ﻴﻜﻔﻲ ﺃﻥ ﻨﺒﻴﻥ ﺒﺄﻨﻬﺎ ﺘﻭﻟﺩ ﺍﻟﻤﺭﻜﺯ،ﹰﺎﻴﻁﺨ ﺔﻠﻘﺘﺴﻤ
.
ﻷﻱ
g
G
∈
ﻭ
[ ]
a
a G
m a
F G
∈
∈
∑
،
(
)
1
1
a
a
a G
a G
g
m a g
m gag
−
−
∈
∈
=
∑
∑
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٧٣١
ﺇﻥ ﻤﻌﺎﻤل
a
ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻓﻲ ﺍﻟﻤﺠﻤﻭﻉ ﻫﻭ
1
g
ag
m
−
ﻭﻤﻨﻪ،
(
)
1
1
a
a G
a G
g
ag
g
m a g
m
a
−
−
∈
∈
=
∑
∑
ﻴﺒﻴﻥ ﻫﺫﺍ ﺒﺄﻥ
a
a G
m a
∈
∑
ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﻤﺭﻜﺯ
[ ]
F G
ﺔـﻟﺍﺩﻟﺍ ﺕﻨﺎﻜ ﺍﺫﺇ ﻁﻘﻓﻭ ﺍﺫﺇ
a
ma
a
ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ،ﻥﺃ ﻱﺃ ،ﻕﻓﺍﺭﺘﻟﺍ ﻑﻭﻔﺼ ﻲﻓ ﺔﺘﺒﺎﺜ
a
i
a G
i
m a
Fc
∈
∈
∑
∑
.
ﺔــﻅﺤﻼﻤ
34.7
ﻥ ﺃــﻜﻤﻴ
ﺼﺭــﻨﻌﻟﺍ ﻥــﻋ ﺭــﺒﺘﻌﻨ ﻥ
a
a G
m a
∈
∑
ﻲــﻓ
[ ]
F G
ﺎﻟﺘﻁﺒﻴﻕــﺒ
:
a
a
m G
F
→
a
.
،ﺒﻬﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ
[ ]
(
)
Map
,
F G
G F
≈
.
ﺇﻥ ﺘﺄﺜﻴﺭ
G
ﻋﻠﻰ
[ ]
F G
ﻴﺘﻤﺜل ﺒﺎﻟﺘﺄﺜﻴﺭ
( )( )
( )
1
g f
a
f
g a
−
=
ﻜلـﻟ
g
G
∈
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺩﺍﻟﺔ
:
f G
F
→
.
ﻲـﻓ
ﺒﻴﻨﺎ ﺒﺄﻥ ﻋﻨﺎﺼﺭ ﻤﺭﻜﺯ،ﻕﺒﺎﺴﻟﺍ ﻥﺎﻫﺭﺒﻟﺍ
[ ]
F G
ﺔـﻟﺍﺩﻟﺎﺒ ﹰﺎﻤﺎﻤﺘ لﺜﻤﺘﺘ
:
f G
F
→
ﻭ
ﻲـﺘﻟﺍ
ﺘﻜﻭﻥ ﺜﺎﺒﺘﺔ ﻋﻠﻰ ﻜل ﺼﻑ ﺘﺭﺍﻓﻕ
.
ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﺩﻭﺍل
ﺒﺩﻭﺍل ﺍﻟﺼﻑ
(class function)
.
ﻨﻔﺭﺽ ﺃﻥ،لﺼﻔﻟﺍ ﺍﺫﻫ ﻲﻓ ﻰﻘﺒﺘ ﺎﻤ ﻲﻓ
F
ﻭﻥـﻜﻴ ﺙـﻴﺤﺒ ﺭﻔﺼـﻟﺍ ﻱﻭﺎﺴﻴ ﻩﺯﻴﻤﻤ لﻘﺤ
[ ]
F G
ﻴﺘﻤﺎﺜل ﻤﻊ ﺠﺩﺍﺀ ﺠﺒﻭﺭ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻋﻠﻰ
F
.
ﻴﻤﻜﻥ ﺃﻥ ﻨﺄﺨﺫ،ﹰﻼﺜﻤ
F
ﻼﹰـﻘﺤ ﻥﻭﻜﻴﻟ
، ﻤﺜﻼﹰ، ﺭﻔﺼﻟﺍ ﻱﻭﺎﺴﻴ ﻩﺯﻴﻤﻤ ﹰﺎﻴﺭﺒﺠ ﻕﻠﻐﻤ
22
.
ﻴﺩﻋﻰ ﺍﻟﺘﻤﺜﻴل
[ ]
( )
(
)
GL
F G
G
F G
→
ﺒﺎﻟﺘﻤﺜﻴل ﺍﻟﻤﻨﻅﻡ
(regular representation)
.
ﻤﺒﺭﻫﻨﺔ
35.7
(a)
ﺇﻥ ﻋﺩﺩ ﺼﻔﻭﻑ ﺍﻟﺘﻤﺎﺜل ﻓﻲ
[ ]
F G
-
ﻤﻭﺩﻭﻻﺕ ﺒ
ﺴﻴﻁﺔ ﻤﺘﺴﺎﻭﻴﺔ ﻤﻊ ﻋﺩﺩ
ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
G
.
(b)
ﺇﻥ ﺠﺩﺍﺀ ﺃﻱ ﺘﻤﺜﻴل ﺒﺴﻴﻁ
S
ﻓﻲ ﺘﻤﺜﻴل ﻤﻨﺘﻅﻡ ﻴﺴﺎﻭﻱ ﺩﺭﺠﺘﻪ
dim
F
i
S
.
(c)
ﻟﺘﻜﻥ
1
,...,
t
s
s
ﻤﺠﻤﻭﻋﺔ ﺘﻤﺜﻴﻼﺕ ﻟﺼﻔﻭﻑ ﻤﺘﻤﺎﺜﻠﺔ ﻤﻥ
G
−
F
،ﺴﻴﻁﺔـﺒ ﺕﻻﻭﺩﻭﻤ
ﻭﻟﻴﻜﻥ
dim
=
i
F
i
f
S
ﻋﻨﺩﺌﺫ
2
1
i
i t
f
G
≤ ≤
=
∑
ﺍﻟﺒﺭﻫﺎﻥ
.
(a)
، ﻤﻥ ﺍﻟﻔﺭﺽ
[ ]
( )
( )
1
...
t
f
f
F G
M
F
M
F
≈
× ×
ﺩﺍﺩـﻋﻷﺍ ﻥـﻤ ﺔﻋﻭﻤﺠﻤﻟ
ﺍﻟﺼﺤﻴﺤﺔ
1
,...,
t
f
f
.
22
ﻤﻥ ﻤﺒﺭﻫﻨﺔ ﻤﺎﺸﻙ
(7.8)
ﻴﻜﻭﻥ ﻋﻨﺩﺌﺫ،
F G
ﻨﺼﻑ ﺒ
ﻤﺒﺎﺸﺭﺍﹰ ﻟﺠﺒﻭﺭ ﺒﺴﻴﻁﺔﺀﺍﺩﺠ ﻥﻭﻜﻴ ﻲﻟﺎﺘﻟﺎﺒﻭ ،ﻁﻴﺴ
(7.32)
.
ﻜل ﻤﻨﻬﺎ ﻋﺒﺎﺭﺓ ﻋﻥ ﺠﺒﺭ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﻋﻠﻰ ﺠﺒﺭ ﺘﻘﺴﻴﻡ
(7.22)
ﻟﻜﻥ ﺠﺒﺭ ﺍﻟﺘﻘﺴﻴﻡ ﺍﻟﻭﺤﻴﺩ ﺍﻟﺫﻱ ﻴﻜﻭﻥ،
ﺤﻘﻼﹰ ﻤﻐﻠﻕ ﺠﺒﺭﻴﺎﹰ ﻫﻭ ﺍﻟﺤﻘل ﻨﻔﺴﻪ
(28-7)
.
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٨٣١
ﻤﻥ ﺍﻟﻤﺒﺭﻫﻨﺔ
32.7
ﻴﻜﻭﻥ ﻋﺩﺩ ﺼﻔﻭﻑ ﺍﻟﺘﻤﺎﺜل ﻤﻥ،
[ ]
F G
-
ﺩﺩـﻋ ﻭﻫ ﺔﻁﻴﺴﺒﻟﺍ ﺕﻻﻭﺩﻭﻤﻟﺍ
ﺍﻟﻌﻭﺍﻤل
t
.
ﺇﻥ ﻤﺭﻜﺯ ﺍﻟﺠﺩﺍﺀ
F
-
ﺯـﻜﺭﻤ ﻥﺇ ﻲﻟﺎـﺘﻟﺎﺒ ﻭ ،ﺎـﻫﺯﻜﺍﺭﻤ ﺀﺍﺩـﺠ ﻱﻭﺎﺴﻴ ﺭﻭﺒﺠﻟﺍ
[ ]
F G
ﻤﺘﻤﺎﺜل ﻤﻊ
t F
.
ﻟﺫﻟﻙ ﻓﺈﻥ
t
ﻴﺴﺎﻭﻱ ﺒﻌﺩ ﻤﺭﻜﺯ
F
ﺩﺩـﻋ ﻱﻭﺎﺴﻴ ﻪﻨﺄﺒ ﻡﻠﻌﻨ ﻱﺫﻟﺍ ،
ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ
G
.
(b)
ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺭﻤﻭﺯ ﻓﻲ
(14.7)
،
( )
{ }
( )
{ }
( )
1
...
f
M
F
L
L
r
⊕ ⊕
.
(c)
ﺒﺒﺴﺎﻁﺔ ﻨﺤﺼل
ﻋﻠﻰ ﺍﻟﻤﺴﺎﻭﺍﺓ
[ ]
( )
1
dim
dim
i
F
F
f
i t
F G
M
F
≤ ≤
=
∑
ﻤﻴﺯﺍﺕ
G
(The characters of G)
ﻨﺘﺫﻜﺭ ﺒﺄﻥ ﺍﻷﺜﺭ
( )
Tr
V
α
ﻟﻠﺘﺸﺎﻜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ
: V
V
α
→
ﻭـﻫ ﻲـﻬﺠﺘﻤﻟﺍ ﺀﺎﻀـﻔﻟﺍ ﻲـﻓ
i j
a
∑
ﺤﻴﺙ
( )
i j
a
ﻤﺼﻔﻭﻓﺔ
α
ـ ﺒﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﺽ ﺍﻟﻘﻭﺍﻋﺩ ﻟ
V
.
ﺎﺭـﻴﺘﺨﺍ ﻥﻋ ﺔﻠﻘﺘﺴﻤ ﺎﻬﻨﺇ
ﺍﻟﻘﻭﺍﻋﺩ
)
ﺇﻥ ﺃﺜﺎ
ﺭ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻤﺘﺭﺍﻓﻘﺔ ﻤﺘﺴﺎﻭﻴﺔ
.(
ﻨﺤﺼل ﻤﻥ ﻜل ﺘﻤﺜﻴل
( )
:
GL
g
gv G
V
→
a
ﻋﻠﻰ ﺍﻟﺩﺍﻟﺔ،
χV
ﻋﻠﻰ
G
،
( )
( )
χ
=
V
V g
Tr
gv
ﻴﺩﻋﻰ ﺒﺎﻟﻤﻴﺯﺓ
(character)
ρ
.
ﻨﻼﺤﻅ ﺒﺄﻥ
χ V
ــ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺼﻑ ﺍﻟﺘﻤﺎﺜل ﻟ
[ ]
F G
-
ﻤﻭﺩﻭل
V
ﻭﺒﺄﻥ،
χV
ﻋﺒﺎﺭﺓ ﻋﻥ ﺩﺍﻟﺔ ﺼﻔﻴﺔ
.
ﺯﺓـﻴﻤﻟﺍ ﻥﺄﺒ لﺎﻘﻴ
χ
ﺴﻴﻁﺔـﺒ
(simple)
)
ﺃﻭ
ﻏﻴﺭ ﺨﺯﻭﻟﺔ
(
(irreducible)
ﺇﺫﺍ ﻜﺎﻨﺕ ﻤﻌﺭﻓﺔ ﻋﻠﻰ
FG
-
ﻤﻭﺩﻭل ﺒﺴﻴﻁ
.
ﺇﻥ ﺍﻟﻤﻴﺯﺓ ﺍﻷﺴﺎﺴﻴﺔ
(principal character)
1
χ
ﺭﺓـﻤﺯﻠﻟ ﻪـﻓﺎﺘﻟﺍ لـﻴﺜﻤﺘﻟﺎﺒ ﺔﻓﺭﻌﻤﻟﺍ ﺓﺯﻴﻤﻟﺍ ﻙﻠﺘ ﻲﻫ
G
)
ﻪـﻨﻤﻭ
( )
1
1
g
χ
=
ﻟﻜل
g
G
∈
(
ﻭ ﺍﻟﻤﻴﺯﺓ ﺍﻟﻤﻨﺘﻅﻤﺔ،
(regular character)
reg
χ
ﻙـﻠﺘ ﻲـﻫ
ﺍﻟﻤﻴﺯﺓ ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺘﻤﺜﻴل ﺍﻟﻤﻨﺘﻅﻡ
.
ﺒﺤﺴﺎﺏ
( )
reg
g
χ
ﺭـﺼﺎﻨﻋ ﻡﺍﺩﺨﺘـﺴﺎﺒ
G
ــﺩ ﻟـﻋﺍﻭﻘﻜ
[ ]
F G
ﻓﻲ ﺍﻟﺒﺩﺍﻴﺔ ﻨﺭﻯ ﺒﺄﻥ،
( )
reg
g
χ
ﻫﻭ ﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ
a
ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻥـﻜﻴ ﺙﻴﺤﺒ
ga
a
=
ﻭﻤﻨﻪ،
( )
,
0
,
reg
G
g
e
g
g
e
χ
=
=
≠
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٩٣١
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ
V
ﻤﻥ ﺍﻟﺒﻌﺩ
1
ﻴﻘﺎل ﺒﺄﻥ ﺍﻟﻤﻴﺯﺓﺫﺌﺩﻨﻋ ،
χ V
ﺨﻁﻴﺔ
(linear)
.
،ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
( )
GL V
F
×
ﻭ ﻤﻨﻪ،
( )
( )
χ
ρ
=
V g
g
.
،ﻟﺫﻟﻙ
χ V
ﻫﻭ ﺘﺸﺎﻜل
×
→
G
F
ﻭﻤﻨﻪ،
ﻓﺈﻥ ﻫﺫﺍ ﺍﻟﺘﻌﺭﻴﻑ
"
ﺍﻟﻤﻴﺯﺓ ﺍﻟﺨﻁﻴﺔ
"
ﻴﺘﻭﺍﻓﻕ ﺒﺸﻜل ﺃﺴﺎﺴﻲ ﻤﻊ ﺍﻟﺘﻌﺭﻴﻑ ﺍﻷﻭل
.
ﺘﻤﻬﻴﺩﻴﺔ
36.7
ﻷﻱ
G
-
ﻤﻭﺩﻭل
V
ﻭ
V
′
،
χ
χ
χ
′
′
⊕
=
⊕
V
V
V
V
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺒﺤﺴﺎﺏ ﻤﺼﻔﻭﻓﺔ
L
g
ﻤﻊ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻘﺎﻋﺩﺓ
V
V
′
⊕
ﺍﻟﺘﻲ ﺘﺸﻜﻠﺕ ﺒﺘﺭﻜﻴﺏ ﻗﺎﻋﺩﺓ
V
ﻤﻊ ﻗﺎﻋﺩﺓ
V
′
.
ﻟﺘﻜﻥ
1
,...,
t
S
S
ﻤﺠﻤﻭﻋﺔ ﻜل ﺘﻤﺜﻴﻼ
ﺕ ﺼﻔﻭﻑ ﺍﻟﺘﻤﺎﺜل ﻤﻥ
FG
-
ﺴﻴﻁﺔـﺒ ﺕﻻﻭﺩﻭـﻤ
ﺒﺤﻴﺙ ﻨﺨﺘﺎﺭ
1
S
ﻭﻟﺘ،ﻪﻓﺎﺘﻟﺍ لﻴﺜﻤﺘﻟﺍ ﻥﻭﻜﻴﻟ
ﻜﻥ
1
,...,
t
χ
χ
ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﻤﻘﺎﺒﻠﺔ
.
ﻗﻀﻴﺔ
37.7
ﺇﻥ ﺍﻟﺩﻭﺍل
1
,...,
t
χ
χ
ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎﹰ ﻋﻠﻰ
F
ﺎﻥـﻜ ﺍﺫﺇ ،ﻪﻨﺃ ﻱﺃ ،
1
,...,
t
c
c
F
∈
ﺒﺤﻴﺙ ﻴﻜﻭﻥ
( )
0
i
i
i
c
g
χ
=
∑
ﻟﻜل
g
G
∈
ﻜلﺫﺌﺩﻨﻋ ،
i
c
ﺘﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻨﻜﺘﺏ
[ ]
( )
( )
1
...
t
f
f
F G
M
F
M
F
≈
× ×
ﻭﻟﻴﻜﻥ،
(
)
0,...,0,1, 0,..., 0
i
e
=
.
ﻋﻨﺩﺌﺫ
i
e
ﺘﺅﺜﺭ ﻜﻤﺎ ﻴﺅﺜﺭ
1
ﻋﻠﻰ
i
S
ﻭ ﻜﺫﻟﻙ
0
ﻋﻠﻰ
j
S
ﻟﻜل
i
j
≠
ﻭﻤﻨﻪ،
(25)
( )
dim
,
0
,
i
F
i
j
i
f
S
j
i
e
j
i
χ
=
=
=
≠
ﻟﺫﻟﻙ
( )
,
i
i
i
i
i
i
c
e
c f
χ
=
∑
ﻭ ﺍﻟ
ﺘﻲ ﻤﻨﻬﺎ ﻴﻨﺘﺞ ﻤﺎﻴﻠﻲ
.
ﻗﻀﻴﺔ
38.7
ﻴﻜﻭﻥ
F G
-
ﺍﻟﻤﻭﺩﻭ
ﻻﻥ ﻤﺘﻤﺎﺜﻠﻴ
ﻥ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ﻤﻴﺯﺍﺘﻬﻤﺎ ﻤﺘﺴﺎﻭﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻘﺩ ﺒﺭﻫﻨﺎ
ﺘﻤﺎﻤﺎﹰ ﺒﺄﻥ ﻤﻴﺯﺓ ﺍﻟﺘﻤﺜﻴل ﺘﻌﺘﻤﺩ ﻓﻘﻁ ﻋﻠﻰ ﺼﻑ ﺍﻟﺘﻤﺎﺜل ﻟﻬﺎ
.
ﺇﺫﺍ ﻜﺎﻥ،ﺱﻜﻌﻟﺎﺒ
1
≤ ≤
=
⊕
i
i
i t
V
c S
،
i
c
∈
ﻓﺈﻥ ﻤﻴﺯﺘﻬﺎ ﻫﻲﺫﺌﺩﻨﻋ ،
1
χ
χ
≤ ≤
=
∑
i
i
i t
V
c
ﻭ ﻤﻥ،
(25)
ﻴﺘﺒﻴﻥ ﺒﺄﻥ
( )
χ
=
i
i
i
c
V e
f
.
ﻟﺫﻟﻙ ﻓﺈﻥ
χ V
لـﻜﺒ ﺀﺍﺩﺠﻟﺍ ﺩﺩﺤﻴ
i
S
ﻲـﻓ ﺭـﻬﻅﺘ
V
،
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻨﻪ ﻴﺤﺩﺩ ﺼﻑ ﺍﻟﺘﻤﺎﺜل
ـ ﻟ
V
.
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٠٤١
39.7
ﺘﻜﻭﻥ ﺍﻟﻘﻀﻴﺔ ﺨﺎﻁﺌﺔ ﺇﺫﺍ ﻜﺎﻥ
F
ﺯـﻴﻤﻤﻟﺍ ﻰﻠﻋ ﻱﻭﺤﻴ
0
p
≠
.
لـﻴﺜﻤﺘﻟﺍ ﻥﺇ ،ﹰﻼﺜـﻤ
( )
( )
2
1
:
GL
0
1
i
p
i
C
F
σ
→
a
ﻓﻲ
(c1.7)
ﻟﻜﻨﻪ ﻴﺤﻭﻱ ﻋﻠﻰ ﺍﻟﻤﻴﺯﺓ ﻨﻔﺴﻬﺎ،ﹰﺎﻬﻓﺎﺘ ﺱﻴﻟ
ﻜ
ﺎﻟﺘﻤﺜﻴل ﺍﻟﺘﺎﻓﻪ
.
ﺤﺘﻰ ﺃﻥ
ﺍﻟﻘﻀﻴﺔ ﺨﺎﻁﺌﺔ ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﻤﻤﻴﺯ
F
ﻷﻱ،ﻪﻨﻷ ،ﺓﺭﻤﺯﻟﺍ ﺔﺒﺘﺭ ﻡﺴﻘﻴ ﻻ
ﺘﻤﺜﻴل
( )
GL
G
V
→
ﻴﻜﻭﻥ ﻤﻤﻴﺯ ﺘﻤﺜﻴل،
G
ﻋﻠﻰ
pV
ﻴﺘﻁﺎﺒﻕ ﻤﻌﻪ ﺍﻟﺼﻔﺭ
.
ﺃﻱ ﺩﺍﻟﺔ
G
F
→
ـ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﻴﻌﺒﺭ ﻋﻨﻬﺎ ﻜ
Z
-
ﺴﻤﻰـﺘ ﺕﺍﺯﻴﻤﻟﺍ ﻥﻤ ﻲﻁﺨ ﺏﻴﻜﺭﺘ
ﺒﺎﻟﻤﻴﺯﺓ ﺍﻟﺘﻘﺩﻴﺭﻴﺔ
23
(virtual character)
.
ﻗﻀﻴﺔ
40.7
ﺇﻥ ﺍﻟﻤﻴﺯﺓ ﺍﻟﺒﺴﻴﻁﺔ ﻟﻠﺯﻤﺭﺓ
G
ﺘﺸﻜل
Z
-
ﻗﺎﻋﺩﺓ ﻟﻤﻴﺯﺍﺕ
G
ﺍﻟﺘﻘﺩﻴﺭﻴﺔ
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﺘﻜﻥ
1
,...,
t
χ
χ
ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺒﺴﻴﻁﺔ ﻓﻲ
G
.
ﻓﺈﻥ ﻤﻴﺯﺍﺕﺫﺌﺩﻨﻋ
G
ﻫﻲ ﺘﻤﺎﻤﺎﹰ ﺍﻟﺩﻭﺍل
ﺍﻟﺼﻔﻴﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﻴﻌﺒﺭ ﻋﻨﻬﺎ ﺒﺎﻟﺸﻜل
,
i
i
i
m
m
χ
∈
∑
ﻭﻤﻨﻪ ﻓﺈﻥ ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺘﻘ،
ﺔـﻴﺭﻴﺩ
ﻫﻲ ﺘﻤﺎﻤﺎﹰ ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﻴﻌﺒﺭ ﻋﻨﻬﺎ ﺒﺎﻟﺸﻜل
,
i
i
i
m
m
χ
∈
∑
Z
.
ﺈﻥـﻓ ﻙﻟﺫـﻟ
ـﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺒﺴﻴﻁﺔ ﺘﻭﻟﺩ ﺒﺎﻟﺘﺤﺩﻴﺩ ﺍﻟ
Z
-
ﻀﻴﺔـﻘﻟﺍ ﻥﻴـﺒﺘ ﻭ ، ﺔﻴﺭﻴﺩﻘﺘﻟﺍ ﺕﺍﺯﻴﻤﻠﻟ لﻭﺩﻭﻤ
7.37
ﺒﺄﻨﻬﺎ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎﹰ ﻋﻠﻰ
Z
)
ﻭ ﺤﺘﻰ ﻋﻠﻰ
F
.(
ﻗﻀﻴﺔ
41.7
ﺇﻥ ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺒﺴﻴﻁﺔ ﻓﻲ
G
ﺘﺸﻜل
F
-
ﻗﺎﻋﺩﺓ ﻟﻠﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﻋﻠﻰ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺇﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺼﻔﻭﻑ ﺍﻟﻤﺘﺭ
ﺍﻓﻘﺔ ﻓﻲ
G
ﻰـﻟﺇ
F
.
ﻜﻤﺎ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺘﺤﻭﻱ ﻋﻠﻰ
t
ﺘﺸﻜل،ﺭﺼﻨﻋ
F
-
ﻓﻀﺎﺀ
ﻤﺘﺠﻬﻴ
ﺎﹰ
ﺒﻌﺩﻩ
t
.
ﻜﻤﺎ ﺃﻥ
ﺍﻟﻤﻴﺯﺍﺕ ﺍﻟﺒﺴﻴﻁﺔ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻌﻨﺎﺼﺭ ﺍﻟﻤ
ﻴﺠﺏ ﺃﻥ ﺘﺸﻜل،ﻙﻟﺫﻟ ﻲﻬﺠﺘﻤﻟﺍ ﺀﺎﻀﻔﻠﻟ ﹰﺎﻴﻁﺨ ﺔﻠﻘﺘﺴ
ﻗﺎﻋﺩﺓ
.
ﻨﻔﺭﺽ ﺍﻵﻥ ﺒﺄﻥ
F
ﺤﻘل ﺠﺯﺌﻲ ﻓﻲ
ﺜﺎﺒﺕ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﺭﺍﻓﻕ ﺍﻟﻤﺭﻜﺏ
c
c
a
.
ﻟﻠﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ
1
f
ﻭ
2
f
ﻋﻠﻰ
G
ﺘﻌﺭﻑ،
(
)
( ) ( )
1
2
1
2
1
∈
=
∑
a G
f
f
f
a f
a
G
ﺘﻤﻬﻴﺩﻴﺔ
42.7
ﺇﻥ ﺍﻟﺯﻭﺝ
)
|
(
ﻫﻭ ﺠﺩﺍﺀ ﺩﺍﺨﻠﻲ ﻋﻠﻰ
F
-
ﻓﻀﺎﺀ ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﻋﻠﻰ
G
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺒﺭﻫﻥ ﺒﺄﻥ
23
ﻟﻜﻥ ﺤﺘﻰ ﻴﻜﻭﻥ ﻫﺫﺍ ﻤﺘﺠﻨﺒﺎﹰ،ﺔﻤﺎﻌﻟﺍ ﺕﺍﺯﻴﻤﻟﺍ ﺎﻬﻨﻭﻋﺩﻴ ﻥﻴﻔﻟﺅﻤﻟﺍ ﺽﻌﺒ
:
ﻴﻭﺠﺩ ﺃﻜﺜﺭ ﻤﻥ
ﻴﻡـﻤﻌﺘﻟ ﺓﺩـﺤﺍﻭ ﺔﻘﻴﺭﻁ
ﻤﻔﻬﻭﻡ ﺍﻟﻤﻴﺯﺓ
.
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١٤١
•
(
) (
) (
)
1
2
1
2
+
=
+
f
f
f
f
f
f
f
ﻟﻜل ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ
1
f
،
2
f
،
f
،
•
(
)
(
)
1
2
1
2
,
=
cf
f
c f f
ﻟﻜل
c
f
∈
ﻭ ﺍﻟﺩﺍﻟﺘﻴﻥ ﺍﻟﺼﻔﻴﺘﻴﻥ
1
f
،
2
f
،
•
(
) (
)
2
1
1
2
=
f
f
f
f
ﻟﻜل ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ
1
f
،
2
f
،
•
(
)
0
>
f f
ﻟﻜل ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻤﺔ
f
.
ﺇﻥ ﻜل ﺫﻟﻙ ﻭﺍﻀﺢ ﻤﻥ ﺍﻟﺘﻌﺭﻴﻑ
.
ﻟﻜل
[ ]
F G
-
ﺍﻟﻤﻭ
ﺩﻭل
V
ﻴﺭﻤﺯ،
[ ]
F G
ــﺔ ﺒـﺘﺒﺜﻤﻟﺍ ﺭﺼﺎﻨﻌﻟﺍ ﻥﻤ ﻲﺌﺯﺠﻟﺍ لﻭﺩﻭﻤﻠﻟ
G
:
{
}
,
;
G
V
v
V gv
v g
G
= ∈
=
∈
ﺘﻤﻬﻴﺩﻴﺔ
7.43
ﻟﻴﻜﻥ
π
ﺍﻟﻌﻨﺼﺭ
1
a G
a
G
∈
∑
ﻓﻲ
[ ]
F G
.
ﻷﻱ
[ ]
F G
-
ﺍﻟﻤﻭﺩﻭل
V
،
v
π
ﻫﻭ ﺇﺴﻘﺎﻁ ﺒﺼﻭﺭﺓ
G
V
.
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻷﻱ
g
G
∈
،
(26)
1
1
a G
a G
g
ga
a
G
G
π
π
∈
∈
=
=
=
∑
∑
ﻭﺍﻟﺘﻲ ﻤﻨﻬﺎ ﻴﻜﻭﻥ
π π π
=
)
ﻓﻲ
F
-
ﺍﻟﺠﺒﺭ
[ ]
F G
.(
ﻷﻱ،ﻙﻟﺫﻟ
[ ]
F G
-
ﺍﻟﻤﻭﺩﻭل
V
،
2
v
v
π
π
=
ﻭﻤﻨﻪ
v
π
ﻴﻜﻭﻥ
ﺇﺴﻘﺎﻁﺎﹰ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
v
ﻨﻘﻭل ﺒﺄﻥ،ﺎﻬﺘﺭﻭﺼ
0
v
v
π
=
،
ﻋﻨﺩﺌﺫ
def
0
0
g v
g
v
v
v
π
π
=
=
=
ﺈﻥــﻓ ﻪــﻨﻤﻭ
v
ﻰــﻟﺇ ﻲــﻤﺘﻨﻴ
G
V
.
ﺎﻥــﻜ ﺍﺫﺇ ،ﺱﻜﻌﻟﺎــﺒﻭ
G
v
V
∈
ﻥ ﺍﻟﻭﺍــﻤ ،
ﺢ ﺃﻥــﻀ
1
a G
v
a v
v
G
π
∈
=
=
∑
ﻭﻤﻨﻪ ﻓﺈﻥ،
v
ـ ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺼﻭﺭﺓ ﻟ
π
.
ﻗﻀﻴﺔ
44.7
ﻷﻱ
[ ]
F G
-
ﻤﻭﺩﻭل
V
،
( )
1
dim
χ
∈
=
∑
G
F
a G
V
V a
G
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٢٤١
ﺍﻟﺒﺭﻫﺎﻥ
.
ﻟﻴﻜﻥ
π
ﻜﻤﺎ ﻓﻲ ﺍﻟﺘﻤﻬﻴﺩﻴﺔ
7.43
.
ﻷﻥ
v
π
ﻴﻜﻭﻥ،ﻁﺎﻘﺴﺇ
V
ﺭـﺸﺎﺒﻤﻟﺍ ﻉﻭﻤﺠﻤﻟﺍ
ـﻟ
0
-
ﻓﻀﺎﺀﺍﺘﻬﺎ ﺍﻟﺫﺍﺘﻴﺔ ﻭ
1
-
ﻭـﻫ ﻑﺭـﺤﻟﺍ ﻥﺄـﺒ ﻥﻴﺒﻨ ﻭ ،ﺔﻴﺘﺍﺫﻟﺍ ﺎﻬﺘﺍﺀﺎﻀﻓ
G
V
.
،ﺫﻟﻙـﻟ
(
)
Tr
dim
G
F
V
V
V
π
=
.
ﻤﻥ
، ﻷﻥ ﺍﻷﺜﺭ ﻫﻭ ﺩﺍﻟﺔ ﺨﻁﻴﺔ،ﻯﺭﺨﺃ ﺔﻬﺠ
(
)
( )
( )
1
1
Tr
Tr
a G
a G
v
V
V a v
V a
G
G
π
χ
∈
∈
=
=
∑
∑
ﻤﺒﺭﻫﻨﺔ
45.7
ﻷﻱ
[ ]
F G
-
ﺍﻟﻤﻭﺩﻭﻟﻴﻥ
V
،
W
،
[ ]
(
)
(
)
dim Hom
,
F
F G
V W
V
W
χ
χ
=
ﺍﻟﺒﺭﻫﺎﻥ
.
ﺘﺅﺜﺭ ﺍﻟﺯﻤﺭﺓ
G
ﻋﻠ
ﻰ ﺍﻟﻔﻀﺎﺀ
(
)
Hom
,
F
V W
ــ ﻟ
F
-
ﺔـﻴﻁﺨﻟﺍ ﺕﺎـﻘﻴﺒﻁﺘﻟﺍ
V
W
→
، ﻟﻘﺎﻋﺩﺓ
( )( )
( )
(
)
(
)
,
,
Hom
,
,
F
g
v
g
v
g
G
V W
v
V
ϕ
ϕ
ϕ
=
∈
∈
∈
ﻭ
(
)
(
)
Hom
,
Hom
,
G
F
F G
V W
V W
=
.
ﻨﺘﻴﺠﺔ
46.7
ﺇﺫﺍ ﻜﺎﻥ
χ
ﻭ
χ
′
ﻋﻨﺩﺌﺫ،ﻥﻴﺘﻁﻴﺴﺒ ﻥﻴﺘﺯﻴﻤ
(
)
1,
0,
χ χ
χ χ
χ χ
′
=
′ =
′
≠
ﻟﺫﻟﻙ ﻓﺈﻥ ﺍﻟﻤﻤﻴﺯﺍﺕ ﺍﻟﺒﺴﻴﻁﺔ ﺘﺸﻜل ﻗﺎﻋﺩﺓ ﻤﺘﻌﺎﻤﺩﺓ ﻟﻔﻀﺎﺀ ﺍﻟﺩﻭﺍل ﺍﻟﺼﻔﻴﺔ ﻋﻠﻰ
G
.
ﻗﺎﺌﻤﺔ ﺨﻭﺍﺹ ﺍﻟﺯﻤﺭﺓ
ﻟﻠﻜﺘﺎﺒﺔ
ﺍﻷﻤﺜﻠﺔ
ﻟﻠﻜﺘﺎﺒﺔ
.
ﻴﺅﻜﺩ،ﺔﻴﻬﺘﻨﻤﻟﺍ ﺭﻤﺯﻟﺍ ﺔﻴﺭﻅﻨ لﻴﺜﻤﺘﻟ ﻲﺨﻴﺭﺎﺘﻟﺍ ﺩﻬﻌﻟﺍ ﻥﺄﺒ ﻅﺤﻼﻨ
ﺍﻟﻌﻤل
"
ﻴﻴﻥـﺴﺎﺴﻷﺍ ﻥﻴﻤﻫﺎﺴﻤﻟﺍ
ﺍﻷﺭﺒﻌﺔ ﻟﻬﺫﻩ ﺍﻟﻤﺒﺭﻫﻨﺔ ﻓﻲ ﻤﺭﺍﺤل ﺘﺸﻜﻴﻠﻬﺎ
:
،ﺴﺎﻴﺩـﻨﻴﺭﺒ ﻡﺎﻴﻠﻴﻭ ،ﺱﻭﻴﻨﻴﺒﻭﺭﻓ ﺝﺭﻭﺠ ﺩﻨﺎﻨﻴﺩﺭﻴﻓ
ﺍﻨﻅﺭ،ﺭﻴﺍﻭﺭﺒ ﺩﺭﺎﺸﺘﻴﺭ ،ﺭﻭﺸﻴﺴ ﻱﺎﺴﻏ
Curtis 1999
"
ﻟﻠﻜﺘﺎﺒﺔ
ﺍﻟﺯﻤﺭ ﺍﻟﺠﺒﺭﻴﺔ ﺍﻟﺨﻁﻴﺔ
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ﺘﻁﻭﺭ
ﺍﻟﻤﺒﺭﻫﻨﺔ ﺒﺸﻜل ﻜﺎﻑ
ﻟﺘﻜﻭﻥ ﻗﺎﺩﺭﺓ
ﻋﻠﻰ
ﺘﺼﻨﺒﻑ
ﺴﻴﻁﺔ ﺍـﺒﻟﺍ ﺔﻴﺭﺒﺠﻟﺍ ﺭﻤﺯﻟﺍ
ﺼﻠﺔ ﻭـﻔﻨﻤﻟ
ﺘﻤﺜﻴﻼﺘﻬﺎ ﻓﻲ ﺤﺩﻭﺩ ﻤﺨﻁﻁﺎﺕ ﺩﻨﻜﻴﻥ
.
ﻨﺫﻜﺭ ﻜﻴﻑ ﻨﺸﺄﺕ ﺯﻤﺭ ﻜﻭﺴﺘﻴﺭ
.
ﺴﺠﻴلـﺘ ﺔـﻴﻔﻴﻜ ﺵﻗﺎﻨﻨ
ﺯﻤﺭ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﺒﺴﻴﻁﺔ ﺍﻟﻤﻨﺘﻬﻴﺔ
.
ﻁﺒﻭﻟﻭﺠﻴﺔ ﺍﻟﺯﻤﺭ
ﺘﻁﻭﺭ ﺍﻟﻤﺒﺭﻫﻨﺔ ﺍﻷﺴﺎﺴﻴﺔ ﻭ ﺘﻨﺎﻗﺵ ﺍﻟﻤﻭﺍﻀﻴﻊ
ﺍﻵﺘﻴ
ﺔ
:
ﺔـﺼﺍﺭﺘﻤﻟﺍ ﺭﻤﺯﻟﺍ ﻑﻴﻨﺼﺘ ،ﺭﺎﻫ ﺕﺎﺴﺎﻴﻗ
ﺘﻤﺜﻴل ﺍ،ﻥﻴﻐﻴﺭﺘﻨﻭﺒ ﺔﻴﺌﺎﻨﺜ ،ﺔﻴﻠﺒﻷﺍ ﹰﺎﻴﻌﻀﻭﻤ
ﻟﺯﻤﺭ ﺍﻟﻤﺘﺭﺍﺼﺔ
.
ﺯﻤﺭ ﻟﻲ
ﺘﺸﺭﺡ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﺯﻤﺭ،ﻲﻟ ﺭﻤﺯ ﻲﻫ ﺔﻴﺠﻭﻟﻭﺒﻁﻟﺍ ﺭﻤﺯﻟﺍ ﻡﻅﻌﻤ ﻥﺇ ،ﺔﻴﺴﺎﺴﻷﺍ ﺔﻨﻫﺭﺒﻤﻟﺍ ﺭﻭﻁﺘ
ﺘﺫﻜﺭﺓ ﺒﺎﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺤﺴﺎﺒﻴﺔ،ﺔﻁﻴﺴﺒﻟﺍ ﺭﻤﺯﻟﺍ ﻑﻴﻨﺼﺘ ﺞﺘﻨﺘﺴﺘ ﻭ ،ﺔﻴﺭﺒﺠﻟﺍ
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٤٤١
ﺍﻟﻤﻠﺤﻕ
A
ﺘﻤﺎﺭﻴﻥ ﻤﺤﻠﻭﻟﺔ
ﺘﺘﺭﺍﻭﺡ ﻫﺫﻩ ﺍﻟﺤﻠﻭل ﻤﺎ ﺒﻴﻥ ﺍﻟﺘﻠﻤﻴﺢ ﺃﺤﻴﺎﻨﺎﹰ ﻭ
ﺒﻴﻥ ﺍﻟﺤﻠﻭل ﺍﻟﻜﺎﻤﻠﺔ ﺃﺤﻴﺎﻨﺎﹰ ﺃﺨﺭﻯ
.
ﺃﺘﻭﻗﻊ ﻤﻥ ﺍﻟﻁﻼﺏ
ﺃﻥ ﻴﻜﺘﺒﻭﺍ ﺍﻟﺤل ﺒﺸﻜل ﻜﺎﻤل
.
1-1
.
ﺇﻥ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻭﺤﻴﺩ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،ﺭﺎﺒﺘﺨﻻﺎﺒ
2
ﻫﻭ
2
2
c
a
b
=
=
.
ﺎ ﺃﻥـﻤﺒ
1
gcg
−
ﻭـﻫ
ﺃﻴﻀﺎﹰ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻴﺠﺏ ﺃﻥ ﻴﺴﺎﻭﻱ،
c
، ﺃﻱ ﺃﻥ،
1
gcg
c
−
=
ﻟﻜل
g
Q
∈
.
ﻟﻬﺫﺍ
ﻓﺈﻥ
c
ﺘﺘﺒﺎﺩل ﻤﻊ ﺠﻤﻴﻊ ﻋﻨﺎﺼﺭ
Q
ﻭ،
{ }
1,c
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
Q
.
ﺭـﻤﺯﻟﺍ ﻥﺇ
ﺍﻟﺠﺯﺌﻴﺔ
ﺍﻟﻤﺘﺒﻘﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺏ
1
،
4
ﺃﻭ،
8
،
ﻭﻫﻲ ﻨﺎﻅﻤﻴﺔ ﺒﺸﻜل ﺘﻠﻘﺎﺌﻲ
)
ﺍﻨﻅﺭ
35.1a
.(
1
-
2
.
ﺇﻥ ﺍﻟﻌﻨﺼﺭ
( )
1 1
0 1
ab
=
ﻭ،
( ) ( )
1 1
1
0 1
0
1
n
n
=
.
1
-
3
.
ﻨﺄﺨﺫ ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺍﻟﺠﺯﺌﻴﺔ
{
}
1
,
g g
−
ﻤﻥ
G
.
ﺼﺭـﻨﻋ ﻰﻠﻋ ﻱﻭﺤﺘ ﺔﻋﻭﻤﺠﻤ لﻜ
ﻴﻥ
ﻓﻘﻁ ﻤﺎ ﻋﺩﺍ
g
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
1
ﺃﻭ
2
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﺘﺤﻭﻱ ﻋﻠﻰ،
1
ﻋﻨﺼﺭ
.
ﺒﻤﺎ ﺃﻥ
G
ﻴﺠﺏ ﺃﻥ ﻴﻭﺠﺩ،ﺕﺎﻋﻭﻤﺠﻤﻟﺍ ﻩﺫﻬﻟ لﺼﻔﻨﻤﻟﺍ ﻉﺎﻤﺘﺠﻻﺍ ﻱﻭﺎﺴﺘ
)
ﺼﻔﺭـﻟﺍ ﻱﻭﺎﺴﻴ ﻻ
(
ﺩﺩـﻋ
ﺯﻭﺠﻲ
ﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ
1
ﻭﺒﺎﻟﺘﺎﻟﻲ، ﺭﺼﻨﻋ
ﻥـﻤ لـﻗﻷﺍ ﻰﻠﻋ ﺩﺤﺍﻭ ﺭﺼﻨﻋ ﺩﺠﻭﻴ
ﺍﻟﺭﺘﺒﺔ
2
.
1
-
4
.
ﺒﻤﺎ ﺃﻥ ﺍﻟﺯﻤﺭﺓ
G N
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
n
،
( )
1
n
gN
=
لـﻜﻟ
g
G
∈
)
ﺔـﻨﻫﺭﺒﻤ
ﻻﻏﺭﺍ
ﻨﺞ
.(
ﻟﻜﻥ
( )
n
n
gN
g N
=
ﻪـﻨﻤﻭ ،
n
g
N
∈
.
ﺭـﺒﺘﻌﻨ ،ﺔـﻴﻨﺎﺜﻟﺍ ﺔـﻟﺎﺤﻠﻟ ﺔﺒﺴـﻨﻟﺎﺒ
{ }
3
1,
N
D
τ
=
⊂
.
ﺩﻟﻴلـﻟﺍ ﻥﻤ ﻲﻫﻭ
3
ﺼﺭـﻨﻌﻟﺍ ﻥـﻜﻟ ،
τ σ
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
2
ﻪـﻨﻤﻭ ،
( )
3
N
τ σ
τ σ
=
∉
.
1
-
5
.
ﻟﻴﻜﻥ
,
a b
G
∈
.
ﻭ ﻟﺩﻴﻨﺎ
( )
2
2
2
a
b
ab
e
=
=
=
.
،ﺔـﺼﺎﺨ ﺔـﻟﺎﺤﺒ
abab
e
=
.
ﺒﺎﻟﻀﺭﺏ ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﺒﺎﻟﻌﻨﺼﺭ
ba
ﻨﺠﺩ ﺒﺄﻥ،
ab
ba
=
.
1
-
6
.
ﺎـﻬﻨﺄﺒ ﻥﻫﺭـﺒﻨ ﻥﺃ ﻲﻔﻜﻴ ،ﺔﻴﺭﻅﺎﻨﺘﻭ ﺔﻴﺴﺎﻜﻌﻨﺍ ﺔﻴﻠﻤﻋ ﻲﻫ ﺱﺎﻴﻘﻟﺍ ﺔﻴﻠﺒﺎﻗ ﻥﺃ ﺢﻀﺍﻭﻟﺍ ﻥﻤ
ﻤﺘﻌﺩﻴﺔ
.
ﺴﻭﻑ ﻨﺴﺘﺨﺩﻡ ﻋﻠﻰ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻲـﻓ ﻪـﺘﻨﻤ لﻴﻟﺩـﺒ
G
،
ﻋﻨﺩﺌﺫ
H
G
′
I
ﻤﻥ ﺩﻟﻴل ﻤﻨﺘﻪ ﻓﻲ
G
′
ﻷﻱ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
G
′
ﻓﻲ
G
)
ﻷﻥ ﺍﻟﺘﻁﺒﻴﻕ ﺍﻟﻁﺒﻴﻌﻲ
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٥٤١
G H
G
G H
′
′ →
I
ﻏﺎﻤﺭ
.(
ﺕـﻨﺎﻜ ﺍﺫﺇ ﻪـﻨﺄﺒ ﺞﺘﻨﻴ ،ﻕﻴﺒﻁﺘﻟﺍ ﺍﺫﻫ ﻡﺍﺩﺨﺘﺴﺎﺒ
1
H
ﻭ
3
H
ﻗﺎﺒﻠﺔ ﻟﻠﻘﻴﺎﺱ ﻤﻊ
2
H
ﻋﻨﺩﺌﺫ،
1
2
3
H
H
H
I
I
ﻲـﻓ ﻪـﺘﻨﻤ لـﻴﻟﺩ ﻥﻤ
1
2
H
H
I
ﻲـﻓ ﻭ
2
3
H
H
I
)
ﻭ
ﺒﺎﻟﺘﺎﻟﻲ ﻓﻲ
1
H
ﻭ
3
H
ﺃﻴﻀﺎﹰ
.(
ﺎ ﺃﻥـﻤﻜ
1
3
1
2
3
H
H
H
H
H
⊃
I
I
I
،
ﻭﻫﻲ ﺃﻴﻀﺎﹰ ﻤﻥ ﺩﻟﻴل ﻤﻨﺘﻪ ﻓﻲ ﻜل ﻤﻥ
1
H
ﻭ
3
H
.
2
-
1
.
،ﺔـﻴﻠﻴﺩﺒﺘ ﻥﻭﻜﺘ ﻥﺃ ﺏﺠﻴ ﺭﺼﺎﻨﻌﻟﺍ ﻥﻤ ﺔﻴﻠﻴﺩﺒﺘ ﺔﻋﻭﻤﺠﻤﺒ ﺓﺩﻟﻭﻤ ﺓﺭﻤﺯ ﻱﺃ ﻥﺃ ﹰﻻﻭﺃ ﻅﺤﻼﻨ
ﻭﻤﻨﻪ ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺘﺒﺩﻴﻠﻴﺔ ﻓﻲ ﺍﻟﻤﺴﺄﻟﺔ
.
ﻤﻥ
(2.8)
ﻨﺠﺩ ﺒﺄﻥ ﺃﻱ ﺘﻁﺒﻴﻕ،
{
}
1
,...,
n
a
a
A
→
ﺤﻴﺙ
A
ﺘﺒﺩﻴﻠﻴﺔ ﻴﺘ
ﻤﺩﺩ ﺒﺸﻜل ﻭﺤﻴﺩ ﺇﻟﻰ ﺍﻟﺘﺸﺎﻜل
G
A
→
ﺭﺓـﻤﺯﻠﻟ ﻥﺈﻓ ﻪﻨﻤﻭ ،
G
ﺔـﺼﺎﺨﻟﺍ
ﺍﻟﻌﺎﻤﺔ ﺍﻟﺘﻲ ﺘﻤﻴﺯ ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ ﺍﻵﺒﻠﻴﺔ ﻋﻠﻰ ﺍﻟﻤﻭﻟﺩﺍﺕ
i
a
.
2
-
2
.
(a)
ﺇﺫﺍ ﻜﺎﻥ
a
b
≠
ﻓﺈﻥ ﺍﻟﻜﻠﻤﺔﺫﺌﺩﻨﻋ ،
1
1
...
...
a
a b
b
−
−
ﻤﺨﺘﺯﻟﺔ ﻭ
1
≠
.
ﺇﺫﺍ،ﻙﻟﺫـﻟ
ﻜﺎﻥ
1
n
n
a b
−
=
ﻋﻨﺩﺌﺫ،
a
b
=
.
( )
b
ﺸﺎﺒﻪـﻤ لﻜﺸـﺒ
.
(c)
ــﺯل ﻟـﺘﺨﻤﻟﺍ لﻜﺸـﻟﺍ ﻥﺇ
,
1
n
x
x
≠
ﻤﻥ ﺍﻟﻁﻭل،
n
ﻋﻠﻰ ﺍﻷﻗل
.
2
-
3
.
(a)
ﺒﺸﻜل ﻋﺎﻡ
.
(b)
C
C
∞
∞
×
ﻭ ﺍﻟﺯﻤﺭ ﺍﻟﺤﺭﺓ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﻫﻲ،ﺔﻴﻠﻴﺩﺒﺘ
1
ﻭ
C
∞
ﻁـﻘﻓ
.
(c)
ﺒﻔﺭﺽ ﺃﻥ
a
ﻲـﻓ ﺔﻴﻟﺎﺨ ﺭﻴﻏ ﺔﻟﺯﺘﺨﻤ ﺔﻤﻠﻜ
1
,...,
n
x
x
ﺄﻥـﺒ لﻭـﻘﻨ ،
...
i
a
x
=
)
ﺃﻭ
1
...
i
x
−
.(
ﻟﻜل
j
i
≠
ــ ﺇﻥ ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺯل ﻟ،
def
1
1
,
−
−
=
j
j
j
x
a
x a x a
ﻥ ﺃﻥـﻜﻤﻴ ﻻ
ﻭﻤﻨﻪ ﻓﺈﻥ،ﹰﺎﻴﻟﺎﺨ ﻥﻭﻜﻴ
a
ﻭ
j
x
ﻻ ﻴﺘﺒﺎﺩﻻﻥ
.
2
-
4
.
ﺇﻥ ﺍﻟﻌﻨﺼﺭ
ﺍﻟﻭﺤﻴﺩ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻫﻭ
2
b
.
ﺭﺓـﻤﺯﻠﻟ
2
n
Q
b
ﺩﺍﻥـﻟﻭﻤﻟﺍ
a
ﻭ
b
،
ﻭﺍﻟﻌﻼﻗﺎﺕ
2
2
2
1
1
1,
1,
n
a
b
b a b
a
−
−
−
=
=
=
ﺭﺓـﻤﺯﻠﻟ لﻴﺜﻤﺘ ﻭﻫ ﻱﺫﻟﺍﻭ ،
2
2
n
D
−
)
ﺭـﻅﻨﺍ
2.9
.(
2
-
5
.
(a)
ﺒﻤﻘﺎﺭﻨﺔ ﺘﻤﺜﻴل
4
2
4
,
,
1
D
r
s
s r s r
=
=
لـﻴﺜﻤﺘ ﻊﻤ
G
ﻊـﻀﻭ ﺡﺭـﺘﻘﻨ
r
a b
=
ﻭ
s
a
=
.
ﺭﺍﺠﻊ
)
ﺒﺎﺴﺘﺨﺩﺍﻡ
2.8
(
ﺒﺄﻨﻪ ﺘﻭﺠﺩ ﺍﻟﺘﺸﺎﻜﻼﺕ
:
1
4
4
,
,
,
,
,
D
G
r
a b
s
a
G
D
a
s
b
s r
−
→
→
a
a
a
a
ﺇﻥ ﻜل ﻤﻥ ﺍﻟﻤﺭﻜﺒﺎﺕ
4
4
D
G
D
→ →
ﻭ
4
G
D
G
→
→
ﻫﻲ ﺍﻟﺘﻁﺒﻴﻘﺎﺕ ﺍﻟﻤﺤﺎﻴﺩﺓ ﻋﻠﻰ
ﻭﻟﺫﻟﻙ،ﺓﺩﻟﻭﻤﻟﺍ ﺭﺼﺎﻨﻌﻟﺍ
)
2.8
ﺃﻴﻀﺎﹰ
(
ﻓﻬﻲ ﺍﻟﺘﻁﺒﻴﻘﺎﺕ ﺍﻟﻤﺤﺎﻴﺩﺓ
.
(b)
ﺘﻬﻤل
.
2
-
6
.
ﺒﻤﻼﺤﻅﺔ ﺃﻥ
3
1
3
1
a b a
b c b
−
−
=
.
ﻟﻜﻥ
3
1
b
=
ﻭ
3
1
c
=
.
ﻟﺫﻟ
ﻙ
1
a
=
.
ﻤﻥﺫﺌﺩﻨﻋ
ﺍﻟﻌﻼﻗﺔ ﺍﻷﺨﻴﺭﺓ ﻨﺠﺩ ﺒﺄﻥ
1
b
=
.
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٦٤١
2
-
8
.
ﺭــﺼﺎﻨﻌﻟﺍ ﻥﺇ
2
x
،
x y
،
2
y
ﺄﻥــﺒ ﻯﺭــﻨ ﺔﻟﻭﻬﺴــﺒﻭ ،ﺓﺍﻭــﻨﻟﺍ ﻰــﻟﺇ ﻲــﻤﺘﻨﺘ
2
2
,
,
,
x y x
x y y
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
)
ﻋﻠﻰ ﺍﻷﻜﺜﺭ
(
2
ﻭﺒﺎﻟﺘﺎﻟﻲ،
ﻴﺠﺏ ﺃﻥ ﺘﻭﻟﺩ ﺍﻟﻨﻭﺍﺓ
)
ﺒﺎﻋﺘﺒﺎﺭﻫﺎ
ﺯﻤﺭﺓ ﻨﺎﻅﻤﻴﺔ ﻋﻠﻰ ﺍﻷﻗل
–
ﺍﻟﻤﺴﺄﻟﺔ ﻏﻴﺭ ﻭﺍﻀﺤﺔ
.(
ﺄﻥـﺒ ﻥﻫﺭـﺒﻨ ﻥﺃ ﹰﺓﺭﺸﺎﺒﻤ ﺎﻨﻨﻜﻤﻴ ﺔﻴﺍﺩﺒﻟﺍ ﻲﻓ
ﺃﻭ ﻨﻁﺒﻕ ﻤﺒﺭﻫﻨﺔ ﻨﻠﺴﻭﻥ،ﺓﺭﺤ ﺭﺼﺎﻨﻌﻟﺍ ﻩﺫﻫ
-
ﺴﺸﺭﻴﺭ
.(2.6)
ﻨﻼﺤﻅ ﺒﺄﻥ ﺍﻟﺼﻴﻐﺔ ﻓﻲ
p. 30
)
ﺒﺸﻜل ﺩﻗﻴﻕ
(
ﺘﺘﻨﺒﺄ ﺒﺄﻥ ﺍﻟﻨﻭﺍﺓ
ﺤﺭﺓ ﻭﻤﻥ ﺍﻟﺭﺘﺒﺔ
2.2 2 1 3
− + =
.
2
-
8
.
ﺼﺭﺍﻥـﻨﻌﻟﺍ ﻥﺎـﻜ ﺍﺫﺇ ﻪـﻨﺄﺒ ﻥﻴﺒﻨ ﻥﺃ ﺎﻨﻴﻠﻋ
s
ﻭ
t
ﺎﻥـﻘﻘﺤﻴ ﻭ ﺔـﻴﻬﺘﻨﻤ ﺓﺭـﻤﺯ ﻥـﻤ
1
3
5
t s t
s
−
=
ﻓﺈﻥ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻤﻌﻁﻰﺫﺌﺩﻨﻋ ،
g
ﻴﺴﺎﻭﻱ
١
.
،ﻟﺫﻟﻙ
1
n
s
=
ﻟﺒﻌﺽ
n
.
ﻭﻥـﻜﺘ
ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺸﻭﻗﺔ ﻋﻨﺩﻤﺎ
( )
3,
1
n
=
.
،ﻟﻜﻥ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
3r
s
s
=
ﻟﺒﻌﺽ ﻤﻥ
r
.
ﺎﻟﻲـﺘﻟﺎﺒ ﻭ
(
)
1
3
1
3
5
r
r
r
t s t
t s t
s
−
−
=
=
.
،ﺍﻵﻥ
(
) (
)
1
1
1
1
1
5
5
1;
r
r
g
s
t s t s t s t
s s
s s
−
−
−
−
−
−
=
=
=
ﺍﻨﻅﺭ
]
ﺒﺤﺜﺕ ﻋﻥ ﻨﻤﻭﺫﺝ ﻟﻠﺘﻔﺼﻴل،ﻩﺫﻬﻜ ﺔﻟﺄﺴﻤ ﻲﻓ
.
ﺎـﻤ ﻥـﻜﻟ ،ﹰﺎﻀﻴﺃ ﻙﻟﺫ ﻯﺭﻷ ﹰﺎﺘﻗﻭ ﺕﺫﺨﺃ
ﺤﺩﺙ ﺃﺨﻴﺭﺍﹰ ﺃﻨﻪ ﻜﺎﻥ ﻟﻠﻌﻨﺼﺭ
g
ﻤﺭﺍﻓﻘﺎ
ــﺴﺒﺔ ﻟـﻨﻟﺎﺒ ﺕﺎﻗﻼﻌﻟﺍ ﻪﺘﻠﻌﻓ ﺎﻤﻜ ،ﺎﻬﻴﻓ ﻥ
G
ﺫﻟﻙـﻟ
ﺤﺎﻭﻟﺕ ﺃﻥ ﺃﺠﺩ ﺍﻟﻌﻼ
ﻗﺔ ﺒﻴﻨﻬﻡ
.[
3
-
1
.
ﺇﻥ ﺍﻟﻨﻘﺎﻁ ﺍﻟﻤﻭﺼﻠﺔ ﻟﻠﺤل ﻫﻲ ﺃﻥ
2
n
a
a
a
=
×
.
ﻨﻁﺒﻕ
(1.49)
ﺄﻥـﺒ ﻯﺭـﻨﻟ
2n
D
ﺘﺘﺤﻠل ﻜﻤﺎ ﻓﻲ ﺍﻟﺠﺩﺍﺀ
.
3
-
2
.
ﻟﺘﻜﻥ
N
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻓﻲ
G
.
ﻋﻨﺩﺌﺫ
G N
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
4
،
ﻟﻜﻥ ﻻ ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
Q
G
⊂
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
ﺤﻴﺙ
1
Q
G
=
I
)
ﻥـﻤ ﺓﺭـﻤﺯ لﻜ ﻥﻷ
ﺍﻟﺭﺘﺒﺔ
4
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
(
ﻭﻤﻨﻪ،
G
N
Q
ϕ
≠ ×
ﻷﻱ
Q
.
ﺎﺵـﻘﻨﻟﺍ ﺱﻔﻨـﺒ
ﻨﻁﺒﻘﻪ ﻋﻠﻰ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
.
3
-
3
.
ﻷﻱ
g
G
∈
،
1
gMg
−
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
m
ﻭ ﻟﺫﻟﻙ ﻓﻬﻲ ﺘﺴﺎﻭﻱ،
M
.
ﻟﻬﺫﺍ
ﻓﺈﻥ
M
)
ﻭﺒﺸﻜل ﻤ
ﺸﺎﺒﻪ
N
(
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻭ،
M N
ﻲـﻓ ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
.
ﺇﻥ ﺭﺘﺒﺔ ﺃﻱ ﻋﻨﺼﺭ ﻓﻲ
M
N
I
ﻴﻘﺴﻡ
(
)
gcd
,
1
m n
=
ﻭﻤﻨﻪ ﻓﻬﻲ ﺘﺴﺎﻭﻱ،
1
.
ﺍﻵﻥ
ﺘﺒﻴﻥ
(1.50)
ﺃﻥ
M
N
M N
×
≈
ﻭﺍﻟﺘﻲ ﺒﻌﺩ ﺫﻟﻙ ﺴﺘﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
m n
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ،
ﺘﺴﺎﻭﻱ
G
.
3
-
4
.
ﺒﻴﻥ ﺃﻥ
( )
2
2
GL
F
ﻲـﻓ ﺔـﻤﻭﺩﻌﻤﻟﺍ ﺭﻴﻏ ﺔﺜﻼﺜﻟﺍ ﺕﺎﻬﺠﺘﻤﻟﺍ لﺩﺎﺒﺘ
2
2
×
F
F
)
ﻀﺎﺀـﻓ
ﻤﺘﺠﻬﻲ ﻤﻥ ﺍﻟﺒﻌﺩ
2
ﻋﻠﻰ
2
F
.(
3
-
5
.
ﺇﻥ ﺍﻟﺤﻠﻭل
ﺍﻵﺘﻴ
ﺔ ﻤﻘﺘﺭﺤﺔ ﻤﻥ ﻗﺒل ﺍﻟﻘﺎﺭﺌﻴﻥ
.
ﻨﻜﺘﺏ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ ﺒﺎﻟﺸﻜل
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٧٤١
{
}
1,
,
,
Q
i
j
k
= ± ± ± ±
(A)
ﻨﺄﺨﺫ ﻤﻜﻌﺒﺎﹰ
.
ﻨﻜﺘﺏ ﺴﺘﺔ ﻋﻨﺎﺼﺭ ﻤﻥ
Q
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
4
ﻋﻠﻰ ﺍﻷﻭﺠﻪ
ﺍﻟﺴﺘﺔ ﺤﻴﺙ
i
ﻫﻭ
ﻤﻌﺎﻜﺱ
i
−
ﺍﻟﺦ،
..
ﻜل ﺩﻭﺭﺍﻥ ﻟﻠﻤﻜﻌﺏ ﻴﻨﺘﺞ ﺘﻤﺎﺜﻼﹰ
ـ ﺫﺍﺘﻴ
ﺎﹰ
ﺭﺓـﻤﺯﻠﻟ
Q
ﻭ،
( )
Aut Q
ﻲـﻫ
،ﺍﻟﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﻟﻠﻤﻜﻌﺏ
4
S
.
(B)
ﺘﺤﻭﻱ ﺍﻟﺯ
ﻤﺭﺓ
Q
ﻋﻠﻰ ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻭﻫﻭ ﺘﺤﺩﻴﺩﺍﹰ،
-1
ﻥـﻤ ﺭﺼﺎﻨﻋ ﺔﺘﺴ ﻭ ،
ﺍﻟﺭﺘﺒﺔ
4
ﻭﻫﻲ ﺘﺤﺩﻴﺩﺍﹰ
,
,
i
j
k
± ± ±
.
ﺃﻱ ﺘﻤﺎﺜل
α
ﻟﻠﺯﻤﺭﺓ
Q
لـﺴﺭﻴ ﻥﺃ ﺏﺠﻴ
-1
ﻰـﻟﺇ
ﻨﻔﺴﻪ ﻭ ﻭﻴﺒﺎﺩل ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒ
ﺔ
4
.
ﻨﻼﺤﻅ ﺒﺄﻥ
,
,
i j
k j k
i k i
j
=
=
=
ﻟﺫﻟﻙ ﻓﺈﻥ،
α
ﺩـﺤﺃ ﻰﻟﺇ ، ﻱﺃ ،ﺎﻬﺴﻔﻨ ﺕﺎﻋﻭﻤﺠﻤﻟﺍ ﻰﻟﺇ ﻱﺭﺌﺍﺩ لﻜﺸﺒ ﺔﺒﺘﺭﻤﻟﺍ ﺕﺎﻋﻭﺠﻤﻟﺍ لﺴﺭﺘ ﻥﺃ ﺏﺠﻴ
ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺍﻟﺜﻤﺎﻨﻴﺔ ﺍﻟﺘﻲ ﻓﻲ ﺍﻟﻘﺎﺌﻤﺔ
:
i
j
k
i
j
k
i
j
k
i
j
k
i
k
j
i
k
j
i
k
j
i
k
j
−
−
−
−
−
−
−
−
−
−
−
−
ﻭﻷﻥ
( )
1
1
α
− =
ﻴﻤ،
ﻜﻥ ﺃﻥ ﺘﺒﺎﺩل
α
لـﻜ ﻥﺄﺒ ﻯﺭﻨ ﻥﺃ ﺏﻌﺼﻟﺍ ﻥﻤ ﺱﻴﻟﻭ ،ﺔﻤﺌﺎﻘﻟﺍ ﺭﻁﺴﺃ
ﺍﻟﺘﺒﺎﺩﻴل ﻤﻤﻜﻨﺔ
.
3
-
6
.
ﺇﻥ ﺍﻟﺯﻭﺝ
1
0
0
1
0
0
1
b
N
c
=
ﻭ
0
0
0
0
0
0
a
Q
a
d
=
ﻴﺤﻘﻕ ﺍﻟﺸﺭﻭﻁ
(i)
ﻭ
(ii)
ﻭ
(iii)
ﻓﻲ
(3.7)
.
ﻤﻥ ﺃﺠل،ﹰﻼﺜﻤ
(i)
)
ﻴﻘﻭل
Maple
ﺒﺄﻥ
(
(
)
(
)
1
1
1
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
0
1
0
0
0
0
1
b
b
ab
d
d
a
b
b
a
b
c
a
c
c
a
c
c
ac
d
d
d
d
−
− +
+
=
− +
+
ﻤﺒﺎﺸﺭﺍﹰ ﻟﻠﺯﻤﺭﺘﻴﻥ ﻷﻨﻬﺎ ﻟﻴﺴﺕ ﺘﺒﺩﻴﻠﻴﺔﺀﺍﺩﺠ لﻜﺸﺘ ﻻ
.
4
-
1
.
ﻟﻴﻜﻥ
g
ﻤﻭﻟﺩ
ﺍﹰ
ﻟﻠﺯﻤﺭﺓ
C
∞
.
ﻓﺈﻥ ﺍﻟﻤﻭﻟﺩ ﺍﻵﺨﺭ ﺍﻟﻭﺤﻴﺩ ﻫﻭﺫﺌﺩﻨﻋ
1
g
−
ﻭ ﺍﻟﺘﻤﺎﺜل ﻏﻴﺭ،
ﺍﻟﺘﺎﻓﻪ ﺍﻟﻭﺤﻴﺩ ﻫﻭ
1
g
g
−
a
.
ﺒﺎﻟﺘﺎﻟﻲ
( ) { }
Aut
1
∞
= ±
C
.
ﺇﻥ ﺍﻟﺘﺸﺎﻜل
( )
3
3
Aut
→
S
S
ﻤﺘﺒﺎﻴﻥ ﻷﻥ
( )
3
1
Z S
=
ﻟﻜﻥ،
3
S
ﻴﺤﻭﻱ ﻋﻠﻰ ﺜﻼﺜﺔ ﻋﻨﺎﺼﺭ ﻫﻲ
1
a
،
2
a
ﻭ
3
a
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
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٨٤١
2
ﻭ ﻋﻠﻰ ﻋﻨﺼﺭﻴ
ﻥ ﻭﻫﻤﺎ
b
ﻭ
2
b
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
.
ﺇﻥ ﺍﻟﻌﻨﺼﺭﻴ
ﻥ
1
a
ﻭ
b
ﺩﺍﻥـﻟﻭﻴ
3
S
ﻭ،
ﻴﻭﺠﺩ ﻓﻘﻁ ﺴﺘﺔ ﺍﺤﺘﻤﺎﻻﺕ ﻤﻥ ﺃ
ﺠل
( )
1
a
α
،
( )
b
α
ﻭﻤﻨﻪ،
( )
3
3
Aut
→
S
S
ﻏﺎﻤﺭ ﺃﻴﻀﺎﹰ
.
4
-
2
.
ﻟﺘﻜﻥ
H
ﺭﺓـﻤﺯﻟﺍ ﻲﻓ ﺔﻴﻠﻌﻓ ﺔﻴﺌﺯﺠ ﺭﺓﻤﺯ
G
ﺘﻜﻥـﻟﻭ ،
( )
G
N
N
H
=
.
ﺩﺩـﻋ ﻥﺇ
ﻤﺭﺍﻓﻘﺎﺕ
H
ﻴﺤﻘﻕ
(
) (
)
:
:
G N
G H
≤
)
ﺍ
ﻨﻅﺭ
4.8
.(
ـﺒﻤﺎ ﺃﻥ ﻜل ﻤﺭﺍﻓﻕ ﻟ
H
ﻴﺤﻭﻱ
ﻋﻠﻰ
(
)
:1
H
ﻋﻨﺼﺭ ﻭ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺘﺘﺩﺍﺨل
)
ﻋﻠﻰ ﺍﻷﻗل
(
ﻓﻲ
{ }
1
ﻨﺭﻯ ﺒﺄﻥ،
.
(
)(
) (
)
1
:
:1
:1
gHg
G H
H
G
−
<
=
U
ﻨﺨﺘﺎﺭ،ﻲﻨﺎﺜﻟﺍ ﺀﺯﺠﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ
S
ﻟﺘﻜﻭﻥ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺘﻤﺜﻴﻼﺕ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
.
4
-
3
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ
4.17
ﻭ
4.18
ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ،
N
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ
2
p
ﻲـﺘﻟﺍﻭ ،
ﺘﻜﻭﻥ ﺘﺒﺩﻴﻠﻴﺔ
.
ﻨﺒﻴﻥ ﺍﻵﻥ ﺃﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ
c
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻰـﻟﺇ ﻲﻤﺘﻨﻴ ﻻ ﻭ
N
،
ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
G
N
c
ϕ
= ×
ﺍﻟﺦ،
..
.
4
-
4
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﻟﻴﻠﻬﺎ
p
ﻭﻟﺘﻜﻥ،
N
ﻨﻭﺍﺓ ﺍﻟﺘﻁﺒﻴﻕ
(
)
Sym
→
G
G H
-
ﻭﻫﻲ ﺃﻜﺒﺭ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﻭﻤﺤﺘﻭﺍﺓ ﻓﻲ
H
)
ﺍﻨﻅﺭ
4.22
.(
ﺇﺫﺍ ﻜﺎﻥ
N
H
≠
،
ﻋﻨﺩﺌﺫ
(
)
:
H N
ﺘﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ
q
p
≥
ﻭ،
(
)
:
G N
ﻰـﻠﻋ ﺔﻤﺴﻘﻟﺍ لﺒﻘﺘ
pq
.
ﻟﻜﻥ
pq
ﻻ ﺘﻘﺴﻡ
!
p
-
ﻭﻫﺫﺍ ﺘﻨﺎﻗﺽ
.
4
-
5
.
ﻨﺜﺒﺕ
G
ﻓﻲ
2m
S
ﻭﻟﻴﻜﻥ،
2m
N
S
G
=
I
.
ﻋﻨﺩﺌﺫ
2
2
2
m
m
G N
S
A
C
→
=
،
ﻭﻤﻨﻪ
(
)
:
2
G N
≤
.
ﻟﻴﻜﻥ
a
ﻋﻨﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
ﻓﻲ
G
ﻭﻟﻴﻜﻥ،
1
,...,
m
b
b
ﺔـﻋﻭﻤﺠﻤ
ﻤﻥ ﺍﻟﻤﺭﺍﻓﻘﺎﺕ ﺍﻟﻴﻤﻴﻨﻴﺔ ﻟﻠﺯﻤﺭﺓ
a
ﻓﻲ
G
ﺈﻥـﻓ ﻙﻟﺫﻟ ،
{
}
1
1
,
,...,
,
m
m
G
b a b
b
a b
=
.
ﺇﻥ
ﺼﻭﺭﺓ
a
ﻓﻲ
2m
S
ﻫﻭ ﺠﺩﺍﺀ
m
ﻤﻨﺎﻗ
ﻠﺔ
(
) (
)
1
1
,
,...,
,
m
m
b a b
b
a b
ﻭﺒﻤﺎ ﺃﻥ،
m
، ﻓﺭﺩﻱ
ﻴﻨﺘﺞ ﺒﺄﻥ
a
N
∉
.
4
-
6
.
(a)
ﺇﻥ ﻋﺩﺩ ﺍﻷﺴﻁﺭ ﺍﻷﻭﻟﻰ ﺍﻟﻤﻤﻜﻨﺔ ﻴﺴﺎﻭﻱ
3
2
1
−
ﺔـﻴﻨﺎﺜﻟﺍ ﺭﻁﺴﻷﺍ ﻭ ،
3
2
2
−
ﻭ،
ﺍﻷﺴﻁﺭ ﺍﻟﺜﺎﻟﺜﺔ
3
2
2
2
−
ﻭﺒﺎﻟﺘﺎﻟﻲ،
(
)
:1
7 6 4
168
G
= × × =
.
(b)
ﻟﻴﻜﻥ
3
2
V
=
F
.
ﻋﻨﺩﺌﺫ
3
2
8
V
=
=
.
ﻰـﻠﻋ ﻱﻭﺤﻴ ﺕﺎﻴﺜﺍﺩﺤﻹﺍ ﺃﺩﺒﻤ ﻥﻤ ﺭﺎﻤ ﻁﺨ لﻜ
ﻭﻤﻨﻪ،ﺕﺎﻴﺜﺍﺩﺤﻹﺍ ﺃﺩﺒﻤ ﻱﻭﺎﺴﺘ ﻻ ﹰﺎﻤﺎﻤﺘ ﺓﺩﺤﺍﻭ ﺔﻁﻘﻨ
7
X
=
.
(c)
ﻨﺸﻜ
ل ﻗﺎﺌﻤﺔ ﺒﺎﻟﻤﻤﻴﺯﺍﺕ ﺍﻟﻤﻤﻜﻨﺔ ﻭ ﻜﺜﻴﺭﺍﺕ ﺍﻟﺤﺩﻭﺩ ﺍﻷﺼﻐﺭﻴﺔ
:
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٩٤١
ﺭﺘﺒﺔ ﺍﻟﻌﻨﺼﺭ ﻓﻲ ﺍﻟﺼﻑ ﺍﻟﺤﺠﻡ
ﺍﻟﺤﺩﻭﺩﻴﺔ ﺍﻷﺼﻐﺭﻴﺔ ﺍﻟﻤﻤﻴﺯﺍﺕ
1
1
1
X
+
3
2
1
X
X
X
+
+
+
2
21
(
)
2
1
X
+
3
2
1
X
X
X
+
+
+
4
42
(
)
3
1
X
+
3
2
1
X
X
X
+
+
+
3
56
ﻨﻔﺴﻪ
(
)
(
)
3
2
1
1
1
X
X
X
X
+ =
+
+
+
7
24
ﻨﻔﺴﻪ
)
ﻏﻴﺭ ﻗﺎﺒل ﻟﻠﺘﺤﻠﻴل
(
2
1
X
X
+
+
7
24
ﻨﻔﺴﻪ
)
ﻏﻴﺭ ﻗﺎﺒل ﻟﻠﺘﺤﻠﻴل
(
3
2
1
X
X
+
+
ﺇﻥ ﺍﻟﺤﺠﻡ ﻫﻨﺎ ﻴﺭﻤﺯ ﺇﻟﻰ ﻋﺩﺩ ﻋﻨﺎﺼﺭ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
.
ﺍﻟﺤﺎﻟﺔ
5
:
ﻟﻴﻜﻥ
α
ﺯﺓـﻴﻤﻤﻟﺍ ﺩﻭﺩﺤﻟﺍ ﺓﺭﻴﺜﻜ ﻊﻤ ﻲﺘﺍﺫﻟﺍ لﻜﺎﺸﺘﻟﺍ
3
2
1
X
X
+
+
.
ﻥـﻤ ﺩـﺠﻨ
ﻜﻭﻨﻬﺎ ﻜﺜﻴﺭﺓ ﺤﺩﻭﺩ ﺃﺼﻐﺭﻴﺔ ﺃﻥ
7
1
α
=
ﻭﻤﻨﻪ ﻓﺈﻥ،
α
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
7
.
ﻨﻼﺤﻅ ﺃﻥ
V
ﺎﺭﺓـﺒﻋ
ﻋﻥ
[ ]
2
α
F
-
ﺯـﻜﺭﻤﻤ ﻥﺈـﻓ ﻪـﻨﻤﻭ ،ﺓﺩـﺤﺍﻭ ﺔـﺒﺘﺭ ﻥﻤﻭ ﺭﺤ لﻭﺩﻭﻤ
α
ﻲـﻓ
G
ﻭـﻫ
[ ]
2
G
α
α
=
I
F
.
ﻟﻬﺫﺍ
( )
7
G
C
α
=
ﻭﻋﺩﺩ ﺍﻟﻌﻨﺎﺼﺭ ﻓﻲ ﺼﻑ ﺘﺭﺍﻓﻕ،
α
ﺴﺎﻭﻱـﻴ
168 7
24
=
.
ﺍﻟﺤﺎﻟﺔ
6
:
ﺘﻤﺎﻤﺎﹰ ﻜﻤﺎ ﻓﻲ ﺍﻟﺤﺎﻟﺔ
5
.
ﺍﻟﺤﺎﻟﺔ
4
:
ﻫﻨﺎ
1
2
V
V
V
= ⊕
ﻜﻤﺎ ﺃﻨﻪ
[ ]
2
α
F
-
ﻭ،لﻭﺩﻭﻤ
[ ]
( )
[ ]
( )
[ ]
( )
2
2
2
1
2
End
End
End
α
α
α
=
⊕
V
V
V
F
F
F
ﻴﻨﺘﺞ ﺃﻥ
( )
3
G
C
α
=
ﻭﻤﻨﻪ ﻋﺩﺩ ﻤﺭﺍﻓﻘﺎﺕ،
α
ﻴﺴﺎﻭﻱ
168
56
3
=
.
ﺍﻟﺤﺎﻟﺔ
3
:
ﻴﻜﻭﻥ ﻫﻨﺎ
( )
[ ]
2
G
C
G
α
α
=
= ≥
I
F
ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،
4
.
ﺍﻟﺤﺎﻟﺔ
1
:
α
ﻫﻭ ﺍﻟﻌﻨﺼﺭ ﺍﻟﺤﻴﺎﺩﻱ ﻫﻨﺎ
.
ﺍﻟﺤﺎﻟﺔ
2
:
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻜﻭﻥ
1
2
V
V
V
= ⊕
ﻋﺒﺎﺭﺓ ﻋﻥ
[ ]
2
α
F
-
ﺅﺜﺭـﻴ ﺙﻴﺤ ،لﻭﺩﻭﻤ
α
ﻜﺘﺄﺜﻴﺭ
1
ﻋﻠﻰ
1
V
ﻭﻴﻤﻠﻙ ﻜﺜﻴﺭﺓ ﺍﻟﺤﺩﻭﺩ
2
1
X
+
ﻋﻠﻰ
2
V
.
ﺴﻴﻁﺔـﺒ ﺎﻬﻨﺇ ﻭﺃ ،لﻠﺤﺘﺘ ﺎﻤﺇ
ﻨﻼﺤﻅ ﺒﺄﻥ ﺼﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻫﺫﺍ ﻴﺤﻭﻱ ﻋﻠﻰ ﺍﻟﻌﻨﺎﺼﺭ ﺍﻟﺒﺎﻗﻴﺔ
.
(d)
ﺒﻤﺎ ﺃﻥ
3
168
2
3 7
= × ×
ﺴﺘﻜﻭﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺨﺎﺼﺔ ﻏﻴﺭ ﺍﻟﺘﺎﻓﻬﺔ ﻤﻥ،
ﺍﻟﺭﺘﺒﺔ
2, 4, 8, 3, 6, 12, 24, 7, 14, 28, 56, 21, 24
ﺃﻭ،
84
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ـ ﺴﺘﻜﻭﻥ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﺠﺘﻤﺎﻉ ﻤﻨﻔﺼل ﻟ،ﺔﻴﻤﻅﺎﻨ
{ }
1
ﻭ ﺒﻌﺽ ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
ﻭﻤﻨﻪ،ﻯﺭﺨﻷﺍ
(
)
:1
1
i
N
c
= +
∑
ﺤﻴﺙ
i
c
ﺴﺎﻭﻱـﺘ
21
،
24
،
42
ﺃﻭ،
56
ﻥـﻟ ﻥـﻜﻟ ،
ﻴﺤﺩﺙ ﻫﺫﺍ
.
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٠٥١
4
-
7
.
ﺒﻤﺎ ﺃﻥ
( )
( )
Aut
→
G Z G
G
ﻨﺭﻯ ﺃﻥ،
( )
G Z G
ﻭ ﻤﻥ،ﺔﻴﺭﺌﺍﺩ
(4.19)
ﻨﺠﺩ
ﺒﺄﻥ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻓﻬﻲ ﺘﺤﻭﻱ ﺍﻟﻌﺎﻤل،ﺔﻴﺭﺌﺍﺩ ﺭﻴﻏ ﻭ ﺔﻴﻬﺘﻨﻤ
r
s
p
p
C
C
×
ﺍﻟﺦ
..
4
-
8
.
ﻤﻥ ﺍﻟﻭﺍﻀﺢ ﺃﻥ
( ) ( )( )( )
1
1
1
i j
j
i
j
=
.
ﻰـﻠﻋ ﻱﻭـﺤﺘ ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻱﺃ ﻲﻟﺎﺘﻟﺎﺒ
( ) ( )
12 , 13 ,...
ﻭ ﻨﻌﻠﻡ ﺒﺄﻥ،ﺕﻼﻗﺎﻨﻤﻟﺍ لﻜ ﻰﻠﻋ ﻱﻭﺤﺘ
n
S
ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﻨﺎﻗﻼﺕ
.
4
-
9
.
ﺄﻥــــﺒ ﻅــــﺤﻼﻨ
( )
( )
G
H
C
x
H
C
x
=
I
ﻭ،
ﻴﻜﻭﻥــــﺴ ﻲﻟﺎــــﺘﻟﺎﺒ
( )
( )
( )
.
H
G
G
H C
x
H C
x
C
x
≈
.
ﺒﺭﻫﻥ ﺃﻥ ﻜل ﺼﻑ ﻴﺤﻭﻱ ﻋﻠﻰ ﻨﻔﺱ ﺍﻟﻌﺩﺩ
c
ﻤﻥ
ﺍﻟﻌﻨﺎﺼﺭ
.
ﻋﻨﺩﺌﺫ
( )
(
)
( )
(
)
( )
( )
(
)
:
:
.
.
:
G
G
G
G
K
G C
x
G H C
x
H C
x
C
x
kc
=
=
=
4
-
10
.
(a)
ﺍﻟﻤﺴﺎﻭﺍﺓ ﺍﻷﻭﻟﻰ ﺘﺄﺘﻲ ﻤﻥ ﺍﻟﻤﺴﺄﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ
.
ﻨﻼﺤﻅ ﺃﻥ،ﺔﻴﻨﺎﺜﻠﻟ ﺔﺒﺴﻨﻟﺎﺒ
σ
ﺎﺩلـﺒﺘﺘ
ﻭﻤﻨﻪ ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﺯﻭﺠﻴ،ﺎﻬﻠﻴﻠﺤﺘ ﻲﻓ ﺭﺍﻭﺩﻷﺍ لﻜ ﻊﻤ
ﺔ
)
،ﺃﻱ ﺃﻥ
ﺭﺩﻱـﻓ ﺎﻬﻟﻭﻁ
(
ﺎﻥـﻜ ﺍﺫﺇ ،
ﻟﺩﻭﺭﻴﻥ ﺍﻟﻁﻭل ﺍﻟﻔﺭﺩﻱ ﻨﻔﺴﻪ
k
ﻴﻤﻜﻥ ﺃﻥ ﻨﺠﺩ ﺃﻭﻻﹰ ﺠﺩﺍﺀ،
k
ﻤﻨﺎﻗﻠﺔ ﺘﺘﺒﺎﺩل ﻤﻌﻬﻡ
.
ﻭ ﺘﺘﺒﺎﺩل ﻤﻊ
σ
ﻨﺒﻴﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ ﺘﺠﺯﺌﺔ،ﺱﻜﻌﻟﺎﺒﻭ ،
n
ـ ﺍﻟﻤﻌﺭﻓﺔ ﺒ
σ
ﺤﻴﺤﺔـﺼ ﺩﺍﺩـﻋﺃ ﻥـﻤ ﻑﻟﺄﺘﺘ
ﺘﺘﺒﺎﺩلﺫﺌﺩﻨﻋ ،ﺔﻔﻠﺘﺨﻤ
σ
ﻓﻘﻁ ﻤﻊ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻭﻟﺩﺓ ﺒﺎﻷﺩﻭﺍﺭ ﻓﻲ ﺘﺤﻠﻴﻠﻬﺎ ﺍﻟﺩﻭﺭﻱ
.
(b)
ﻨﺴﺠل ﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ
ﻓﻲ
7
S
ﺃﺤﺠﺎﻤﻬﺎ،
ﻭ،لﺜﺎﻤﺘﻟﺍ ،
)
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﺍ
ﺎﹰـﻴﺠﻭﺯ ﺅﻓﺎﻜﺘﻟ
(
ﻓﻴﻤﺎ ﺇﺫﺍ ﻜﺎﻥ ﻤﻨﻔﺼﻼﹰ ﻓﻲ
7
A
.
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١٥١
ﺘﺤﻭﻱ
( )
7
C
σ
ﻤﻨﻔﺼﻠﺔ ﻓﻲ
7
A
ﺘﻤﺎﺜل
ﺍﻟﺤﺠﻡ
ﺍﻟﺩﻭﺭ
( )
( )
( )
( )
(
)
(
)
( )
(
)
(
)
( )( )
( )
( )( )
( )(
)
( )
( )(
)
( )( )
( )( )( )
( )(
)
( )( )( )
( )( )( )
( )
1
1
1
2
12
21
3
123
70
67
4
1234
210
5
12345
504
67
6
123456
840
7
1234567
720
2520
720
8
12 34
105
67
9
12 345
420
10
12 3456
630
12
11
12 3456
504
12
123 456
280
14
25 36
13
123 4567
420
14
12 34 56
105
15
12 34 567
210
12
E
N
O
E
N
O
E
N
O
E
Y
E
N
O
E
N
O
E
N
O
O
E
N
4
-
11
.
ﺒﺎﻟﻌﻭﺩﺓ ﺇﻟﻰ
Maple
،
( )( )( )
( )( )( )
6,
13 26 45 ,
12 34 56
n
a
b
=
a
a
.
4
-
12
.
ﺒﻤﺎ ﺃﻥ
( )
( )
1
0
0
Stab
Stab
−
=
gx
g
x
g
ﺎﻥـﻜ ﺍﺫﺇ ،
( )
0
Stab
⊂
H
x
ﺫـﺌﺩﻨﻋ
( )
0
Stab
⊂
H
x
ﻟﻜل
x
ﻭﻤﻨﻪ،
1
H
=
ﺭﺽـﻔﻟﺍ ﺽﻘﻨـﺒ ،
.
ﻭﻥ ﺍﻵﻥـﻜﺘ
( )
0
Stab x
ﻭﻤﻨﻪ،ﺔﻴﻤﻅﻋﺃ
( )
0
.Stab
=
H
x
G
ﺍﻟﺘﻲ ﺘﺒﻴﻥ ﺒﺄﻥ،
H
ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺘﻌﺩ
.
5
-
1
.
ﻟﻴﻜﻥ
p
ﻋﺩﺩﺍﹰ ﺃﻭﻟﻴﺎﹰ ﻴﻘﺴﻡ
G
ﻭ
ﻟﺘﻜﻥ
p
-
ﺔـﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
S
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ،
m
p
.
ﺘﺤﻭﻱﺫﺌﺩﻨﻋ
S
ﻋﻠﻰ ﺍﻷﻜﺜﺭ
1
1
1
...
1
m
m
m
p
p
p
p
p
−
−
+ + +
=
<
−
ﻋﻨﺼﺭ ﺭﺘﺒﺔ ﻜل ﻤﻨﻬﺎ ﺃﻗل ﻤﻥ
m
p
ﻭﺒ،
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺭﺼﻨﻋ ﻰﻠﻋ ﻱﻭﺤﺘ ﻥﺃ ﺏﺠﻴ ﻲﻟﺎﺘﻟﺎ
m
p
.
ﻭﻤﻨﻪ ﻓﻬﻲ ﺘﺤﻭﻱ ﻋﻠﻰ،ﺔﻴﺭﺌﺍﺩ ﻥﻭﻜﺘ ،ﻙﻟﺫﻟ
1
m
m
p
p
−
−
ﺔـﺒﺘﺭﻟﺍ ﻥﻤ ﺭﺼﻨﻋ
m
p
.
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺜﺎﻨﻴﺔ ﺴﺘﺤﻭﻱ ﺃﻴﻀﺎﹰ ﻋﻠﻰ
1
m
m
p
p
−
−
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﹰﺍﺭﺼﻨﻋ
m
p
ﻭ،
ﺴﺘﻜﻭﻥ ﻤﺨﺘﻠﻔﺔ ﻋﻥ ﻋﻨﺎﺼﺭ
S
ﻭﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﻤﺜﻠﻬﺎ،
.
ﻟﺫﻟﻙ ﻓﺈﻥ
G
ﻰـﻠﻋ ﻱﻭﺤﺘ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻭﺍﺤﺩﺓ ﻓﻘﻁ ﻟﻜل
p
ﻭ ﻜل ﻤﻨﻬﺎ ﺘﻜﻭﻥ ﺩﺍﺌﺭﻴﺔ،ﺎﻬﺘﺒﺘﺭ ﻡﺴﻘﻴ
.
ﺍﻵﻥ ﺘﺒﻴﻥ
ﻻ ﺗﻘﺴﻢ
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(5.9)
ﺃﻥ
G
ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ،ﹰﺎﻴﺒﺴﻨ ﺔﻴﻟﻭﺃ ﺩﺍﺩﻋﺃ ﺎﻬﻨﻤ لﻜ ﺔﺒﺘﺭ ﺔﻴﺭﺌﺍﺩ ﺭﻤﺯﻟ ﺀﺍﺩﺠ ﻥﻋ ﺓﺭﺎﺒﻋ
ﻫﻲ ﺃﻴﻀﺎﹰ ﺩﺍﺌﺭﻴﺔ
.
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٣٥١
ﺍﻟﻤﻠﺤﻕ
B
ﻤﺴﺎﺌل
ﻏﻴﺭ
ﻤﺤﻠﻭﻟﺔ
34
.
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
ﺏ ﺃﻥـﺠﻴﻭ ﻁﻘﻓ ﺓﺩﺤﺍﻭ ﺔﻴﻤﻅﻋﺃ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﻰﻠﻋ ﻱﻭﺤﺘ
ﺘﻜﻭﻥ
p
-
، ﺯﻤﺭﺓ ﺩﺍﺌﺭﻴﺔ
p
ﻋﺩﺩ ﺃﻭﻟﻲ
.
35
.
ﻟﻴﻜﻥ
a
ﻭ
b
ﻋﻨﺼﺭﻴﻥ ﻤﻥ
67
S
.
ﻥـﻤ لـﻜ ﻥﺎﻜ ﺍﺫﺇ
a
ﻭ
b
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ
146
ﻭ
a b
b a
=
ﻤﺎ ﻫﻲ ﺍﻟﺭﺘﺏ ﺍﻟﻤﻤﻜﻨﺔ ﻟﻠﺠﺩﺍﺀ،
a b
؟
36
.
ﺒﻔﺭﺽ ﺃﻥ
G
ﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔ
X
.
(a)
ﺒﻴﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ
1
gxg
X
−
∈
ﻟﻜل
,
x
X
g
G
∈
∈
ﺔـﻴﺌﺯﺠﻟﺍ ﺓﺭـﻤﺯﻟﺍ ﺫـﺌﺩﻨﻋ ،
ﺍﻟﻤﺒﺎﺩﻟﺔ ﻓﻲ
G
ﻤﻭﻟﺩﺓ ﺒﻤﺠﻤﻭﻋﺔ ﻜل ﺍﻟﻌﻨﺎﺼﺭ
1
1
xyx y
−
−
ﻟﻜل
,
x y
X
∈
.
(b)
ﺒﻴﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ
2
1
x
=
ﻟﻜل
x
X
∈
ﺔـﻴﺌﺯﺠﻟﺍ ﺓﺭﻤﺯﻟﺍ ﻥﺈﻓ ﺫﺌﺩﻨﻋ ،
H
ﻲـﻓ
G
ﻤﻭﻟﺩﺓ ﺒﻤﺠﻤﻭﻋﺔ ﻜل ﺍﻟﻌﻨﺎﺼﺭ
x y
ﻟﻜل
,
x y
X
∈
ﻤﻥ ﺍﻟﺩﻟﻴل
١
ﺃﻭ
2
.
37
.
ﺒﻔﺭﺽ
3
p
≥
ﻭ
2
1
p
−
ﻋﺩﺩﻴﻥ ﺃﻭﻟﻴﻴﻥ
)
،ﻤﺜﻼﹰ
3, 7,19, 31,...
p
=
.(
ﺃﻭ ﻻ، ﻥﻫﺭﺒ
ﺃﻥ ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ،لﺎﺜﻤﺒ ﻥﻫﺭﺒﺘ
(
)
2
1
p
p
−
ﺘﺒﺩﻴﻠﻴﺔ
.
38
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
.
ﺒﺭﻫﻥ ﺃﻭ ﻻ ﺘﺒﺭﻫﻥ ﻤﺎﻴﻠﻲ
:
(a)
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻤﻨﺘﻬﻴﺔ ﻭ
P
ﻋﺒﺎﺭﺓ ﻋﻥ
p
-
ﻋﻨﺩﺌﺫ،ﺔﻴﺌﺯﺠ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
H
P
I
ﻫﻲ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻓﻲ
H
.
(b)
ﺇﺫﺍ ﻜﺎﻨﺕ
G
، ﻤﻨﺘﻬﻴﺔ
P
ﻋﺒﺎﺭﺓ ﻋﻥ
p
-
ﻭ،ﺔﻴﺌﺯﺠ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
( )
G
H
N
P
⊃
،
ﻋﻨﺩﺌﺫ
( )
G
N
H
H
=
.
(c)
ﺇﺫﺍ ﻜﺎﻥ
g
ﻋﻨﺼﺭ
ﺍﹰ
ﻤﻥ
G
ﺤﻴﺙ ﺇﻥ
1
gHg
H
−
⊂
ﻋﻨﺩﺌﺫ،
( )
G
g
N
H
∈
.
39
.
ﺒﺭﻫﻥ ﺃﻨﻪ ﻻ ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
616
.
40
.
ﻟﻴﻜﻥ
n
ﻭ
k
ﻋﺩﺩﻴﻥ ﺼﺤﻴﺤﻴﻥ
1
k
n
≤ ≤
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
n
S
ﺩﺓـﻟﻭﻤ
ﺒﺎﻟﺩﻭﺭ
(
)
1
,...,
k
a
a
.
ﺃﻭﺠﺩ ﺭﺘﺒﺔ ﻤﻤﺭﻜﺯ
H
ﻓﻲ
n
S
.
ﻨﻅﻡـﻤ ﺔﺒﺘﺭ ﺩﺠﻭﺃ ﺫﺌﺩﻌﺒ
H
ﻲـﻓ
n
S
] .
ﻤﻨﻅﻡ
H
ﻫﻭ ﺍﻟﻤﺠﻤﻭﻋﺔ
g
G
∈
ﺤﻴﺙ
1
ghg
h
−
=
ﻟﻜل
h
H
∈
.
ﻀﺎﹰـﻴﺃ ﻲﻫ ﻭ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
[.
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٤٥١
41
.
ﺒﺭﻫﻥ ﺃﻭ ﻻ ﺘﺒﺭﻫﻥ ﺍﻟﺤﺎﻟﺔ
ﺍﻵﺘﻴ
ﺔ
:
ﺇﺫﺍ ﻜﺎﻨﺕ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
G
،
ﻟﻜلﺫﺌﺩﻨﻋ
x
G
∈
ﻴﺘﺤﻘﻕ،
1
1
xHx
H
x Hx
H
−
−
⊂
⇒
⊂
.
42
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻨﺘﻬﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻴﻜﻥـﻟﻭ ،
g
ﻥـﻤ ﺭﺼـﻨﻋ
G
.
ﺒﻔﺭﺽ ﺃﻥ
g
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﻭ
ﺃﻥ ﺍﻟﻌﻨﺼﺭ ﺍﻟﻭﺤﻴﺩ ﻤﻥ
H
ﺍﻟﺫﻱ ﻴﺘﺒﺎﺩل ﻤﻊ
g
ﻫﻭ
١
.
ﺒﻴﻥ
ﺃﻥ
:
(a)
ﺍﻟﺘﻁﺒﻴﻕ
1
1
h
g h g h
−
−
a
ﺘﻘﺎﺒل ﻤﻥ
H
ﺇﻟﻰ
H
.
(b)
ﺍﻟﻤﺭﺍﻓﻕ
gH
ﻴﺘﺄﻟﻑ ﻤﻥ ﻋﻨﺎﺼﺭ
G
ﺫﻭﺍﺕ ﺍﻟﺭﺘﺏ
n
.
43
.
ﺒﻴﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ ﺘﺒﺩﻴﻠﺔ ﻤﺎ ﻓﻲ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
G
ﻤﻥ
n
S
ﺘﺭﺴل
x
ﻰـﻟﺇ
y
ﺫـﺌﺩﻨﻋ ،
ﻟﻤﻨﻅﻤﺎﺕ ﻜل ﻤﻥ ﺍﻟﻤﺜﺒﺘﺎﺕ
( )
Stab x
ﻭ
( )
Stab y
ـ ﻟ
x
ﻭ
y
ﺍﻟﺭﺘﺒﺔ ﻨﻔﺴﻬﺎ
.
44
.
ﺒﺭﻫﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ ﻜل ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
G
ﻨ
ﺫـﺌﺩﻨﻋ ،ﺔﻴﻠﺒﺁ ﻭ ﺔﻴﻤﻅﺎ
ﻓﺈﻥ ﺍﻟﺯﻤﺭﺓ ﺁﺒﻠﻴﺔ
.
45
.
ﺼﺭﻴﻥــﻨﻌﻟﺎﺒ ﺓﺩــﻟﻭﻤ ﺎــﻤ ﺓﺭــﻤﺯ ﻥﺃ ﺽﺭــﻔﺒ
a
ﻭ
b
ﺎﺕــﻗﻼﻌﻟﺍ ﻕــﻘﺤﺘ ﻭ
:
3
2
,
1,
1
m
n
a
b
a
b
=
=
=
ﺤﻴﺙ
m
ﻭ
n
ﺃﻋﺩﺍﺩ
ﺍﹰ
ﺼﺤﻴﺤﺔ ﻤ
ﻭﺠﺒﺔ
.
ــﻤﻥ ﺃﺠل ﺃﻱ ﻗﻴﻤﺔ ﻟ
m
ﻭ
n
ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ
G
ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.
46
.
ﺕـﻨﺎﻜ ﺍﺫﺇ ﻪﻨﺃ ﻥﻴﺒ
G
ﺼﺭﻴﻥـﻨﻌﻟﺎﺒ ﺓﺩـﻟﻭﻤ ﺓﺭـﻤﺯ
x
ﻭ
y
ﺔـﻓﺭﻌﻤﻟﺍ ﺕﺎـﻗﻼﻌﻟﺍ ﻭ
( )
4
2
3
1
x
y
xy
=
=
=
ﻭ ﺃﻭﺠﺩ ﺭﺘﺒﺔ،لﺤﻠﻟ ﺔﻠﺒﺎﻗ ﺓﺭﻤﺯ ﻲﻫ
G
ﻭ ﺯﻤﺭﻫﺎ ﺍﻟﻤﺸﺘﻘﺔ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ
,
,
G G
G
′ ′′ ′′′
.
47
.
G
ﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﺎﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﺎﻅﻤﻴﺔ
X
ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ ﺍﻟﻌﻨﺎﺼﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
.
ﻴﻥ ﺃﻥـﺒ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﺒﺩﻟﺔ
G
′
ﻓﻲ
G
ﻤﻭﻟﺩﺓ ﺒﻜل ﻤﺭﺒﻌﺎﺕ ﺍﻟﺠﺩﺍﺀ
x y
ﺭـﺼﺎﻨﻋ ﻥﻤ ﺝﺍﻭﺯﻷ
X
.
48
.
ﺤﺩﺩ ﺍﻟﻤﻨﻅﻡ
N
ﻓﻲ
( )
GL
n
F
ﺍﻟﻤﺅﻟﻑ ﻤﻥ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ
H
، ﻟﻠﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻘﻁﺭﻴﺔ
ﻭ ﺒﺭﻫﻥ ﺃﻥ
N H
ﻤﺘﻨﺎﻅﺭﺓ ﻤﻊ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﺘﻨﺎﻅﺭﺓ
n
S
.
49
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻭﻟﺩﺓ ﺒﺎ
ﻟﻌﻨﺼﺭﻴﻥ
x
ﻭ
y
ﺔـﻓﺭﻌﻤﻟﺍ ﺕﺎﻗﻼﻌﻟﺍ ﻭ
( )
4
2
5
,
,
x
y
xy
.
ﻤﺎ ﻫﻭ ﺍﻟﺩﻟﻴل ﻓﻲ
G
ﻟﻠﺯﻤﺭﺓ ﺍﻟﻤﺒﺩﻟﺔ
G
′
ﻓﻲ
G
.
50
.
ﻟﺘﻜﻥ
G
ﻭ،ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻭﻟﺩﺓ ﺒﺎﻟﻌﻨﺎﺼﺭ ﺫﺍﺕ ﺍﻟﺭﺘﺏ ﺍﻟﻔﺭﺩﻴﺔ
.
ﻴﻥـﺒ
ﺃﻥ
H
ﻭ ﺃﻥ ﺭﺘﺒﺔ،ﺔﻴﻤﻅﺎﻨ
G H
ﻗﻭﻯ ﻟﻠﻌﺩﺩ
2
.
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51
.
ﻟﺘﻜﻥ
G
ﻭ ﻟﺘﻜﻥ،ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ
P
.
ﺕـﻨﺎﻜ ﺍﺫﺇ ﻪﻨﺃ ﻥﻴﺒ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺤﻴﺙ ﺇﻥ
( )
G
N
P
H
G
⊂
⊂
ﻋﻨﺩﺌﺫ
(a)
ﺇﻥ ﻤﻨﻅﻡ
H
ﻓﻲ
G
ﻴﺴﺎﻭﻱ
H
،
(b)
(
) (
)
:
1 mod
G H
p
≡
.
52
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
33 . 25
.
ﺒﻴﻥ ﺃﻥ
G
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
) .
ﻫﻨﺕ
:
ﺍﻟﺨﻁﻭﺓ ﺍﻷﻭﻟﻰ ﻫﻲ
ﺇﻴﺠﺎﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻥ ﺍﻟﺭﺘ
ﺒﺔ
11
ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ
(.
53
.
ﺒﻔﺭﺽ ﺃﻥ
α
ﺘﺸﺎﻜل ﺫﺍﺘﻲ ﻟﻠﺯﻤﺭﺓ
G
ﻥـﻤ ﻕﻴﺒﻁﺘ ﻭﻫ ﻱﺫﻟﺍ
G
ﻰـﻟﺇ
G
ﺎ ﻭـﻬﻠﻤﺎﻜﺒ
ﻴﺘﺒﺎﺩل ﻤﻊ ﻜل ﺍﻟﺘﻤﺎﺜﻼﺕ ﺍﻟﺫﺍﺘﻴﺔ ﺍﻟﺩﺍﺨﻠﻴﺔ ﻟﻠﺯﻤﺭﺓ
G
.
ﺒﻴﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ ﺘﺴﺎﻭﻱ ﺯﻤﺭﺘﻬﺎ ﺍﻟﺠﺯﺌﻴﺔ
ﻋﻨﺩﺌﺫ،ﺔﻟﺩﺎﺒﻤﻟﺍ
x
x
α
=
ﻟﻜل
x
ﻓﻲ
G
.
54
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻤﻊ ﺍﻟﻤﻭﻟﺩﺍﺕ
s
ﻭ
t
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﺎﻤﻬﻨﻤ لﻜ ﻲﺘﻟﺍ ﻭ
2
.
ﺘﻜﻥـﻟ
(
)
:1 2
n
G
=
.
(a)
ﺒﻴﻥ ﺃﻥ
G
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
.
ﺒﻔﺭﺽ ﺃﻥ
n
ﻋﺩﺩ ﻓﺭﺩﻱ
.
(b)
ﺼﻑ ﺠ
ﻤﻴﻊ ﺼﻔﻭﻑ ﺘﺭﺍﻓﻕ
G
.
(c)
ﻲــﻓ ﺔــﻴﺌﺯﺠﻟﺍ ﺭــﻤﺯﻟﺍ ﻊــﻴﻤﺠ ﻑــﺼ
G
ﺸﻜلــﻟﺍ ﻥــﻤ ﻥﻭــﻜﺘ ﻲــﺘﻟﺍ
( )
{
}
,
C x
y
G x y
y x
x
G
=
∈
=
∈
.
(d)
ﺼﻑ ﺠﻤﻴﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺩﺍﺌﺭﻴﺔ ﻓﻲ
G
.
(e)
ﺼﻑ ﺠﻤﻴﻊ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
ﻓﻲ ﺤﺎﻟﺘﻲ
(b)
ﻭ
(d)
.
(f)
ﺒﺭﻫﻥ ﺃﻥ ﺃﻱ
p
-
ﺯﻤﺭﺘﻴﻥ ﺠﺯﺌﻴﺘﻴﻥ ﻓﻲ
G
ﻤﺘﺭﺍﻓﻘﺘﺎ
ﻥ
)
p
ﺃﻭﻟﻲ
.(
55
.
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺘﻌﺩ
ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋ
ﺔ
X
.
ﻭ ﻟﺘﻜﻥ
N
ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
ﻓﻲ
G
ﻭ ﻟﺘﻜﻥ،
Y
ﻤﺠﻤﻭﻋﺔ ﻤﺩﺍﺭﺍﺕ
N
ﻓﻲ
X
.
ﺒﺭﻫﻥ ﺃﻥ
:
(a)
ﻴﻭﺠﺩ ﻤﺅﺜﺭ ﻁﺒﻴﻌ
ﻲ ﻟﻠﺯﻤﺭﺓ
G
ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
Y
ﺍﻟﺫﻱ ﻴﻜﻭﻥ ﻤﺘﻌﺩﻴ
ﺎﹰ ﻭ
لـﻜ ﻥﺃ ﻥﻴﺒ
ﻤﺩﺍﺭ ﻤﻥ ﻤﺩﺍﺭﺍﺕ
N
ﻋﻠﻰ
X
ﻟﻬﺎ ﺍﻟﻘﺩﺭﺓ ﻨﻔﺴﻬﺎ
.
(b)
ﻭﻀﺢ ﺒﻤﺜﺎل ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﻓﺈﻨﻪ ﻟﻴﺱ ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺃﻥ ﺘﻜﻭﻥﺫﺌﺩﻨﻋ ﺔﻴﻤﻅﺎﻨ ﺭﻴﻏ
ﻟﻤﺩﺍﺭﺍﺘﻬﺎ ﺍﻟﻘﺩﺭﺓ ﻨﻔﺴﻬﺎ
.
56
.
ﺒﺭﻫﻥ ﺃﻥ ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻤﻥ
p
-
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ ﺒﺩﻟﻴل ﺃﻭﻟﻲ
)
p
ﺃﻭﻟﻲ
.(
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57
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻓﻭﻕ ﺩ
ﺍﺌﺭﻴﺔ
(metacyclic)
ﺇﺫﺍ ﺤﻭﺕ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﺩﺍﺌﺭﻴﺔ
N
ﻤﻊ ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ ﺍﻟﺩﺍﺌﺭﻴﺔ
G N
.
ﺭـﻤﺯﻟﺍ ﻥﻤ ﺔﻤﺴﻘﻟﺍ ﺭﻤﺯ ﻭ ﺔﻴﺌﺯﺠﻟﺍ ﺭﻤﺯﻟﺍ ﻥﺃ ﻥﻫﺭﺒ
ﺘﺤﺕ ﺍﻟﺩﺍﺌﺭﻴﺔ ﻫﻲ ﺯﻤﺭ ﻓﻭﻕ ﺩﺍﺌﺭﻴﺔ
.
ﺭـﻤﺯﻟ ﺓﺭﺸﺎﺒﻤﻟﺍ ﺕﺍﺀﺍﺩﺠﻟﺍ ﻥﺃ ﻥﻫﺭﺒﺘ ﻻ ﻭﺃ ﻥﻫﺭﺒ
ﻭﻕـﻓ
ﺩﺍﺌﺭﻴﺔ ﻫﻲ ﻓﻭﻕ ﺩﺍﺌﺭﻴﺔ
.
58
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﺅﺜﺭﺓ
ﺒﺸﻜل ﻤﺘﻌﺩ
ﻭ
ﻤﻀﺎﻋﻑ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ
X
ﻴﻜﻥـﻟﻭ ،
x
X
∈
.
ﺒﺭﻫﻥ ﺃﻥ
:
(a)
ﺍﻟﻤﺜﺒﺕ
x
G
ـ ﻟ
x
ﻫﻭ ﺯ
ﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻲ
G
.
(b)
ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺇﻤﺎﺫﺌﺩﻨﻋ ،
N
ﺠﺯﺌﻴﺔ ﻓﻲ
x
G
ﺃﻭ ﺘﺅﺜﺭ
ﺒﺸﻜل ﻤﺘﻌﺩ
ﻋﻠﻰ
X
.
59
.
ﻟﻴﻜﻥ
x
ﻭ
y
ﻋﻨﺼﺭﻴ
ﻥ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺤﻴﺙ ﺇﻥ
1
5
x y x
y
−
=
،
x
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
،
ﻭ
1
y
≠
ﻤﻥ ﺭﺘﺒﺔ ﻓﺭﺩﻴﺔ
.
ﺍﻭﺠﺩ
)ﺒ
ﺩﻭﻥ ﺒﺭﻫﺎﻥ
(
ﺭﺘﺒﺔ
y
.
60
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻟﺘﻜﻥ،
A
ﻲـﻓ ﺔـﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
ﺍﻟﺯﻤﺭﺓ
H
ﺤﻴﺙ ﺇﻥ
ﻤﺭﺍﻓﻘﺎﺕ
A
ﻓﻲ
G
ﺘﻭﻟﺩﻫﺎ
.
(a)
ﺒﺭﻫﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﻲـﻓ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
ﺎـﻤﺇ ﺫـﺌﺩﻨﻋ ،
N
H
⊂
ﺃﻭ
G
NA
=
.
(b)
ﻟﺘﻜﻥ
M
ﺘﻘﺎﻁﻊ ﻤﺭﺍﻓﻘﺎﺕ
H
ﻓﻲ
G
.
ﺒﺭﻫﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﺎـﻬﺘﺭﻤﺯ ﻱﻭﺎﺴـﺘ
ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻤﺒﺎﺩﻟﺔ ﻭ
A
ﻋﻨﺩﺌﺫ،ﺔﻴﻠﺒﺁ
G M
ﺯﻤﺭﺓ ﺒﺴﻴﻁﺔ
.
61
.
(a)
ﺒﺭﻫﻥ ﺃﻥ ﻤﺭﻜﺯ ﺍﻟﺯﻤﺭﺓ ﻏﻴﺭ ﺍﻵﺒﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
p
،
p
ﺔـﺒﺘﺭﻟﺍ ﻥـﻤ ﻥﻭﻜﻴ ،ﻲﻟﻭﺃ
p
.
(b)
ﻤﺎ ﻫﻲ ﺍﻟﺯﻤﺭﺓ ﻏﻴﺭ ﺍﻵﺒﻠﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
١
6
ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﺭﻜﺯﻫﺎ ﻏﻴﺭ ﺩﺍﺌﺭﻱ
.
62
.
ﺒﻴﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﻤﻊ ﺍﻟﻤﻭﻟﺩﺍﺕ
α
ﻭ
β
ﻭ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﻤﻌﺭﻓﺔ
( )
3
2
2
1
α
β
α β
=
=
=
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ ﺯﻤﺭﺓ ﺍﻟﺘﻨﺎﻅﺭﺍﺕ
3
S
ﻤﻥ ﺍﻟﺩﺭﺠﺔ
3
ﺍﻟﺘﻤﺎﺜل ﺍﻟﺼﺭﻴﺢ،ﻥﺎﻫﺭﺒﻟﺍ ﻊﻤ ،ﺀﺎﻁﻋﺈﺒ
.
63
.
ﺒﺭﻫﻥ ﺃﻭ ﺃﻋﻁ ﻤﺜﺎل ﻤﻌﺎﻜﺱ
:
(a)
ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
١
5
.
(b)
ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
30
ﺘﻜﻭﻥ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
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64
.
ﻴﻜﻥــﻟ
t
∈
Z
ﺘﻜﻥــﻟﻭ ،
G
ﺩﺍﺕــﻟﻭﻤﻟﺍ ﻊــﻤ ﺓﺭــﻤﺯ
x
ﻭ
y
ﺎﺕــﻗﻼﻌﻟﺍ ﻭ
1
3
,
1
t
x y x
y
x
−
=
=
.
(a)
ﻤﺎ ﻫﻲ ﺍﻟﺸﺭﻭﻁ ﺍﻟﻼﺯﻤﺔ ﻭ ﺍﻟﻜﺎﻓﻴﺔ ﻋﻠﻰ
t
ﻟﺘﻜﻭﻥ
G
ﻤﻨﺘﻬﻴﺔ
.
(b)
ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻓﻴﻬﺎ
G
ﺤﺩﺩ ﺭﺘﺒﺘﻬﺎ،ﺔﻴﻬﺘﻨﻤ
.
65
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭ
ﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p q
،
p
q
≠
ﻋﺩﺩﺍﻥ ﺃﻭﻟﻴﺎ
ﻥ
.
(a)
ﺒﺭﻫﻥ ﺃﻥ
G
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
(b)
ﺒﺭﻫﻥ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
⇔
G
ﺁﺒﻠﻴﺔ
⇔
G
ﺩﺍﺌﺭﻴﺔ
.
(c)
ﻫل
G
؟ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
)
ﺒﺭﻫﻥ ﺃﻭ ﺃﻭﺠﺩ ﻤﺜﺎل ﻤﻌﺎﻜﺱ
.(
66
.
ﻟﺘﻜﻥ
X
ﻤﺠﻤﻭﻋﺔ ﻤﺅﻟﻔﺔ ﻤﻥ
n
p
، ﻋﻨﺼﺭ
p
ﻭ ﻟﺘﻜﻥ،ﻲﻟﻭﺃ
G
ﺅﺜﺭـﺘ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ
ﻋﻠﻰ
X
ﺒﺸﻜل ﻤﺘﻌﺩ
.
ﺒﺭﻫﻥ ﺃﻥ ﻜل
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻓﻲ
G
ﺘﺅﺜﺭ ﻋﻠﻰ
X
ﺸﻜلـﺒ
ﻤﺘﻌﺩ
.
67
.
ﻟ
ﺘﻜﻥ
4
2
2
1
1
, ,
,
1,
,
G
a b c b c
c b a
b
c
a c a
c a b a
b c
−
−
=
=
=
=
=
=
=
.
ﺤﺩﺩ ﺭﺘﺒﺔ
G
ﻭ ﺃﻭﺠﺩ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺯﻤﺭﺓ
G
.
68
.
ﻟﺘﻜﻥ
N
ﻭﻯـﻘﻟﺍ ﺔـﻤﻴﺩﻋ ﺓﺭﻤﺯﻟﺍ ﻥﻤ ﺔﻬﻓﺎﺘ ﺭﻴﻏ ﺔﻴﻤﻅﺎﻨ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ
G
.
ﺭﻫﻥ ﺃﻥـﺒ
( )
1
N
Z G
≠
I
.
69
.
ﻻ ﺘﻔﺭﺽ ﻤﺒﺭﻫﻨﺎﺕ ﺴﻴﻠﻭ ﻓﻲ ﺍﻟﻤﺴﺄﻟﺔ
.
(a)
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
G
ﻭ ﻟﺘﻜﻥ،
P
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺒﺭﻫﻥ ﺃﻨﻪ ﻴﻭﺠﺩ
x
G
∈
ﺤﻴﺙ ﺇﻥ
1
x P x
H
−
I
ﻫﻲ
p
-
ﻲـﻓ ﺔﻴﺌﺯﺠﻟﺍ ﻭﻠﻴﺴ ﺓﺭﻤﺯ
H
.
(b)
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﺅﻟﻔﺔ ﻤﻥ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ
1
0
1
0
1
∗
L
L
L
ﻤﻥ ﺍﻟﻨﻭ
ﻉ
n n
×
ﻲـﻫ
p
-
ﺯﻤﺭﺓ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻓﻲ
( )
n
p
GL
F
.
70
.
ﺒﻔﺭﺽ ﺃﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻓﻲ ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ
G
ﺙ ﺇﻥـﻴﺤ
G H
ﺩﺍﺌﺭ
ﺔـﻴ
ﻭﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
ﺤﻴﺙ،
n
ﻭ
(
)
:1
G
ﺃﻭﻟﻴﺎﻥ ﻨﺴﺒﻴﺎﹰ
.
ﺒﺭﻫﻥ ﺃﻥ
G
ﻴﺴﺎﻭﻱ ﺍﻟﺠﺩﺍﺀ ﺸﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ
H
S
θ
×
ﺤﻴﺙ
S
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺩﺍﺌﺭ
ﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
.
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71
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﺃﺼﻐﺭﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﺍﻟﻤﻨﺘﻬﻴﺔ ﻭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل
.
ﺭﻫﻥـﺒ
ﺃﻥ
H
ﻤﺘﻤﺎﺜﻠﺔ ﻤﻊ ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭ ﺍﻟﺩﺍﺌﺭﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
p
ﻟﺒﻌﺽ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ
p
.
72
.
(a)
ﺒﺭﻫﻥ ﺃﻨﻪ ﺘﻜﻭﻥ ﺭﺘﺒﺔ ﺍﻟﺯﻤﺭﺘﻴﻥ ﺍﻟﺠﺯﺌﻴﺘﻴﻥ
A
ﻭ
B
ﺭﺓـﻤﺯﻟﺍ ﻥﻤ
G
ﺔ ﺇﺫﺍـﻴﻬﺘﻨﻤ
ﻜﺎﻨﺕ
A
B
I
ﻤﻥ ﺭﺘﺒﺔ ﻤﻨﺘﻬﻴﺔ
.
(b)
ﻴﻘﺎل ﻋﻥ ﺍﻟﻌﻨﺼﺭ
x
ﻓﻲ ﺍﻟﺯﻤﺭﺓ
G
ﺃﻨﻪ
FC
-
ﻲـﻓ ﻪﺘﻨﻤ لﻴﻟﺩ ﻥﻤ ﻪﺘﺒﺜﻤ ﻥﺎﻜ ﺍﺫﺇ ﺭﺼﻨﻋ
G
.
ﺒﺭﻫﻥ ﺃﻥ ﻤﺠﻤﻭﻋﺔ ﺠﻤﻴﻊ
FC
ﻋﻨﺎﺼﺭ ﻓﻲ
G
ﺘﻜﻭﻥ ﻨﺎﻅﻤﻴﺔ
.
73
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
2
2
p q
ﻟﻠﻌﺩﺩﻴﻥ ﺍﻷﻭﻟﻴﻴﻥ
p
q
>
.
ﺒﺭﻫﻥ ﺃﻥ
G
ﺘﺤﻭﻱ ﻋﻠﻰ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
n
p
ﻟﺒﻌﺽ
1
n
≥
.
74
.
(a)
ﻟﺘﻜﻥ
K
ﻭﻟﺘﻜﻥ،ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ ﺔﻴﻬﺘﻨﻤ ﺓﺭﻤﺯ
L
ﻲـﻓ ﺔـﻴﺌﺯﺠ ﺓﺭﻤﺯ
K
ﺙـﻴﺤﺒ
.
L K
K
δ
=
ﺤﻴﺙ،
K
δ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﺸﺘﻘﺔ
.
ﺒﺭﻫﻥ ﺃﻥ
L
K
=
] .
ﻴﺠﺏ ﺃﻥ ﺘﻔﺭﺽ ﺒﺄﻥ
ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻨﺘﻬﻴﺔ ﺘﻜﻭﻥ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ﻜ
ﻭﻥـﻜﺘ ﺔـﻴﻤﻅﻋﺃ ﺔـﻴﺌﺯﺠ ﺓﺭـﻤﺯ ل
ﻨﺎﻅﻤﻴﺔ
.[
(b)
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.
ﺇﺫﺍ ﺤﻭﺕ
G
ﻋﻠﻰ
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ
H
ﻥـﻤ لـﻜ ﻥﻭـﻜﻴ ﺙﻴﺤﺒ
G
H
δ
ﻭ
H
ﺒﺭﻫﻥ ﺃﻥ،ﻯﻭﻘﻟﺍ ﺔﻤﻴﺩﻋ
G
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
75
.
ﻟﺘﻜﻥ
G
p
-
ﺯﻤﺭﺓ ﻏﻴﺭ ﺩﺍﺌﺭﻴﺔ
ﻤﻨﺘﻬﻴﺔ
.
ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺸﺭﻭﻁ
ﺍﻵﺘﻴ
ﺔ ﻤﺘﻜﺎﻓﺌﺔ
:
(a)
( )
(
)
2
:
G Z G
p
≤
.
(b)
ﻜل ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﺃﻋﻅﻤﻴﺔ ﻓﻲ
G
ﺘﻜﻭﻥ ﺁﺒﻠﻴﺔ
.
(c)
ﺘﻭﺠﺩ ﻋﻠﻰ ﺍﻷﻗل ﺯﻤﺭﺘﺎﻥ ﺠﺯﺌﻴﺘﺎﻥ ﺃﻋﻅﻤﻴﺘﺎ
ﻥ ﺒﺤﻴﺙ ﺘﻜﻭﻨﺎﻥ ﺁﺒﻠﻴﺘﻴﻥ
.
76
.
ﺒﺭﻫﻥ ﺃﻥ ﻜل ﺯﻤﺭﺓ
G
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
56
ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﺘﺏ
)
ﺒﺸﻜل ﻏﻴﺭ ﺘﺎﻓﻪ
(
ﻋﻠﻰ ﺸﻜل ﺠﺩﺍﺀ
ﺸﺒﻪ ﻤﺒﺎﺸﺭ
.
ﺃﻭﺠﺩ
)
ﺒﺩﻭﻥ ﺒﺭﻫﺎﻥ
(
ﺯﻤﺭﺘﻴﻥ ﻏﻴﺭ ﻤﺘﻤﺎﺜﻠﺘﻴﻥ ﻭ ﻏﻴﺭ ﺁﺒﻠﻴﺘﻴﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
56
.
77
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻭ
:
ϕ
→
G
G
ﺘﺸﺎﻜل
(a)
ﺒﺭﻫﻥ ﺃﻨﻪ ﻴﻭﺠﺩ ﻋﺩﺩ ﺼ
ﺤﻴﺢ
0
n
≥
ﺤﻴﺙ ﺇﻥ
( )
( )
n
m
G
G
ϕ
ϕ
=
ﺩﺍﺩـﻋﻷﺍ لﻜﻟ
ﺍﻟﺼﺤﻴﺤﺔ
m
n
≥
.
ﻟﺘﻜﻥ
n
α ϕ
=
.
(b)
ﺒﺭﻫﻥ ﺃﻥ
G
ﺠﺩﺍﺀ ﺸﺒﻪ ﻤﺒﺎﺸﺭ ﻟﻠﺯﻤﺭﺘﻴﻥ ﺍﻟﺠﺯﺌﻴﺘﻴﻥ
( )
Ker
α
ﻭ
( )
Im
α
.
(c)
ﺒﺭﻫﻥ ﺃﻥ
( )
Im
α
ﻨﺎﻅﻤﻴﺔ ﻓﻲ
G
ﺃﻭ ﺃﻋﻁ ﻤﺜﺎل ﻤﻌﺎﻜﺱ
.
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78
.
ﻟﺘﻜﻥ
S
ﻤﺠﻤﻭﻋﺔ ﺘﻤﺜﻴﻼﺕ ﻟﺼﻔﻭﻑ ﺍﻟﺘﺭﺍﻓﻕ ﻓﻲ ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
G
ﻭ ﻟﺘﻜﻥ
H
ﺭﺓـﻤﺯ
ﺠﺯﺌﻴﺔ ﻓﻲ
G
.
ﺒﺭﻫﻥ ﺃﻥ
S
H
H
G
⊂
⇒
=
.
79
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.
(a)
ﺒﺭﻫﻥ ﺃﻨﻪ ﺘﻭﺠﺩ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻭﺤﻴﺩﺓ
K
ﻓﻲ
G
ﺒﺤﻴﺙ
(i)
G K
ﻗﺎﺒﻠﺔ ﻟﻠﺤل ﻭ
(ii)
ﺇﺫﺍ ﻜﺎﻨﺕ
N
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅﻤﻴﺔ ﻭ
G N
ﻋﻨﺩﺌﺫ،لﺤﻠﻟ ﺔﻠﺒﺎﻗ
N
K
⊃
.
(b)
ﺒﺭﻫﻥ ﺃﻥ
K
ﻤﺘﻤﻴﺯﺓ
.
(c)
ﺒﺭﻫﻥ ﺃﻥ
[
]
,
K
K K
=
ﻭ ﺃﻥ
1
K
=
ﺃﻭ
K
ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
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٠٦١
ﺍﻟﻤﻠﺤﻕ
C
ﺍﻤﺘﺤﺎﻥ ﻟﺴﺎﻋﺘﻴﻥ
1
.
ﺃﻱ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻵﺘﻴﺔ ﺼﺤﻴﺤﺔ
)
ﻥـﻤ لـﻜﻟ ﹰﺍﺯﺠﻭﻤ ﹰﻼﻴﻠﻌﺘ ﻁﻋﺃ
(a)
ﻭ
(b)
ﻭ
(c)
ﻭ
(d)
ﺃﻋﻁ ﻤﺠﻤﻭﻋﺔ ﺼﺤﻴﺤﺔ ﻤﻥ ﻤﺤﺘﻭﻴﺎﺕ،
(e)
.(
)
a
(
ﻥــﻤ لــﻜ ﻥﺎــﻜ ﺍﺫﺇ
a
ﻭ
b
ﺫــﺌﺩﻨﻋ ،ﺓﺭــﻤﺯ ﻥــﻤ ﻥﻴﺭﺼــﻨﻋ
( )
6
2
3
1,
1
1
a
b
a b
=
= ⇒
=
.
)
b
(
ﺍﻟﻌﻨﺼﺭﺍﻥ
ﺍﻵﺘﻴ
ﺎﻥ ﻤﺘﺭﺍﻓﻘﺎﻥ ﻓﻲ
7
S
:
(
) (
)
1
2
3
4
5
6
7
1
2
3
4
5
6
7
,
3
4
5
6
7
2
1
2
3 1
5
6
7
4
)
c
(
ﺇﺫﺍ ﻜﺎﻨﺕ
G
ﻭ
H
ﻴﻥ ﻭـﺘﻴﻬﺘﻨﻤ ﻥﻴﺘﺭـﻤﺯ
59 4
59 4
G
A
H
A
×
≈
×
،
ﻋﻨﺩﺌﺫ
G
H
≈
.
)
d
(
ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻓﻲ
5
A
ﺘﺤﻭﻱ
( )
123
ﻫﻲ
5
A
ﻨﻔﺴﻬﺎ
.
)
e
(
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
⇐
ﺩﺍﺌﺭﻴﺔ
⇐
ﺘﺒﺩﻴﻠﻴﺔ
⇐
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
)
ﻤﻥ ﺃﺠل
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ
.(
2
.
ﻤﺎ ﻋﺩﺩ
11
-
ﺯﻤﺭ ﺴﻴﻠﻭ ﺍﻟﺠﺯﺌﻴﺔ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
110
2.5.11
=
ﺼﻨﻑ؟
ﺍﻟﺯﻤﺭ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
110
ﺍﻟﺘﻲ ﺘﺤﻭﻱ
ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
.10
ﻫل ﻜل ﺯﻤﺭﺓ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
110
ﺘﺤﻭﻱ ﻋﻠﻰ ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
10
.
3
.
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﻤﻨﺘﻬﻴﺔ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
ﻲـﻓ ﺔـﻴﻠﻴﺩﺒﺘ ﺔﻤﺴﻗ ﺓﺭﻤﺯ لﻜ ﺕﻨﺎﻜ ﺍﺫﺇ ﻪﻨﺃ ﻥﻴﺒ
G
ﻓﺈﻥﺫﺌﺩﻨﻋ ،ﺔﻴﺭﺌﺍﺩ
G
ﺘﺒﺩﻴﻠﻴﺔ
.
ﻫل ﺍﻟﺤﺎﻟﺔ ﺼﺤﻴﺤﺔ ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭ ﻏﻴﺭ ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
.
(a) .4
ﻟﺘﻜﻥ
G
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻤﻥ
( )
Sym X
ﺤﻴﺙ،
X
ﻤﺠﻤﻭﻋﺔ ﻤﻥ
n
ﺼﺭـﻨﻋ
.
ﺇﺫﺍ
ﻜﺎﻨﺕ
G
ﺘﺒﺩﻴﻠﻴﺔ ﻭ ﺘﺅﺜﺭ ﺒﺸﻜل ﻤﺘﻌﺩ
ﻋﻠﻰ
X
ﺒﻴﻥ ﺃﻥ ﻜل ﻋﻨﺼﺭ،
1
g
≠
ﻤﻥ
G
ﺭﻙـﺤﻴ
ﻜل ﻋﻨﺼﺭ ﻤﻥ
X
.
ﻭﺍﺴﺘﻨﺘﺞ ﺒﺄﻥ
(
)
:1
G
n
≤
.
)
b
(
ﻟﻜل
1
m
≥
،
ﺃﻭﺠﺩ ﺍﻟﺯﻤﺭ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ
3m
S
ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3m
.
)
c
(
ﺒﻴﻥ ﺃﻥ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺠﺯﺌﻴﺔ ﺍﻟﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ
n
S
ﺘﻜﻭﻥ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
3
n
≥
.
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١٦١
5
.
ﻟﺘﻜﻥ
H
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻨﺎﻅ
ﻤﻴﺔ ﻤﻥ ﺍﻟﺯﻤﺭﺓ
G
ﻭﻟﺘﻜﻥ،
P
ﺯﻤﺭﺓ ﺠﺯﺌﻴﺔ ﻓﻲ
H
.
ﺍﻓﺭﺽ
ﺃﻥ ﻜل ﺘﻤﺎﺜل ﺫﺍﺘﻲ ﻓﻲ
H
ﻴﻜﻭﻥ
ﺩﺍﺨﻠﻴﺎﹰ
.
ﺒﺭﻫﻥ ﺃﻥ
( )
.
G
G
H N
P
=
.
(a) .6
ﺃﻋﻁ ﻭﺼﻔﺎﹰ ﻟﺯﻤﺭﺓ ﻤﻊ ﺍﻟﻤﻭﻟﺩ
ﺍﺕ
x
ﻭ
y
ﻭ ﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﻼﻗﺎﺕ
1
1
y x y
x
−
−
=
.
)
b
(
ﺩﺍﺕـــﻟﻭﻤﻟﺍ ﻊـــﻤ ﺓﺭـــﻤﺯﻟ ﹰﺎﻔـــﺼﻭ ﻁـــﻋﺃ
x
ﻭ
y
ﺔــــﻓﺭﻌﻤ ﻭ
ﺒﺎﻟﻌﻼﻗﺎﺕ
1
1
1
1
,
y x y
x
x y x
y
−
−
−
−
=
=
.
ﻴﻤﻜﻨﻙ ﺃﻥ ﺘﺴﺘﺨﺩﻡ
ﻟﻜﻥ ﻴﻨﺒﻐﻲ ﺃﻥ ﺘﺸﻴﺭ ﺒﻭﻀﻭﺡ،ﺕﺎﻅﺤﻼﻤﻟﺍ ﻭﺃ ﻑﺼﻟﺍ ﻲﻓ ﺕﻨﻫﺭﺒ ﻲﺘﻟﺍ ﺞﺌﺎﺘﻨﻟﺍ
ﺇﻟﻰ ﻤﺎ ﺘﺴﺘﺨﺩﻤﻪ
.
ﺍﻟﺤل
1
.
(a)
ﺨﺎﻁﺌﺔ
:
ﻓﻲ
2
3
,
,
a b a b
،
a b
ﻤﻥ ﺭﺘﺒﺔ ﻏﻴﺭ ﻤﻨﺘﻬﻴﺔ
.
(b)
ﺍﻟﺘﺤﻠﻴل ﺍﻟﺩﻭﺭﻱ ﻫﻭ،ﺔﺤﻴﺤﺼ
(
)( ) ( )(
)
1357
246 , 123 4567
.
(c)
ﻨﺴﺘﺨﺩﻡ ﻤﺒﺭﻫﻨﺔ ﻜﺭﻭل،ﺔﺤﻴﺤﺼ
-
ﺴﻴﺸﻤﻴﺩﺕ
.
(d)
ﺍﻟﺯﻤﺭﺓ ﺍﻟﻤﻭﻟﺩﺓ ﺘﻜﻭﻥ ﻓﻌﻠﻴﺔ،ﺔﺌﻁﺎﺨ
.
(e)
ﺩﺍﺌﺭﻴﺔ
⇐
ﺘﺒﺩﻴﻠﻴﺔ
⇐
ﻋﺩﻴﻤﺔ ﺍﻟﻘﻭﻯ
⇐
ﻗﺎﺒﻠﺔ ﻟﻠﺤل
.
2
.
ﺇﻥ ﻋﺩﺩ
11
-
ﺯﻤﺭ ﺴﻴﻠ
ﻭ ﺍﻟﺠﺯﺌﻴﺔ
11
1,12,...
s
=
ﻭ ﺘﻘﺴﻡ
10
.
ﺒﺎﻟﺘﺎﻟﻲ ﺘﻭﺠﺩ
11
-
ﺭـﻤﺯ
ﺴﻴﻠﻭ ﺠﺯﺌﻴﺔ ﻭﺍﺤﺩﺓ ﻓﻘﻁ
P
.
ﻭﻟﻬﺎ
11
10
5
,
,
,
G
P
H
P
C
H
C
H
D
θ
= ×
=
=
=
ﻋﻠﻴﻨﺎ ﺍﻵﻥ ﺃﻥ ﻨﻨﻅﺭ ﺇﻟﻰ ﺍﻟﺘﻁﺒﻴﻕ
( )
11
10
:
Aut
θ
→
=
H
C
C
.
ﺸﻭﺭـﻴﺴ ﺔﺌﻁﻭﺘ ﻥﻤ ،ﻡﻌﻨ
-
ﺯﺍﺴﻴﻨﻬﺎﻭﺱ
.
3
.
ﺒﻔﺭﺽ ﺃﻥ
G
ﻴﻤﻠﻙ ﺼﻔﺎﹰ
1
<
.
ﺘﺤﻭﻱﺫﺌﺩﻨﻋ
G
ﻋﻠﻰ ﺯﻤﺭﺓ ﺍﻟﻘﺴﻤﺔ
H
ﻤﻥ ﺍﻟﺼﻑ
2
.
ﻨﻌﺘﺒﺭ
( )
( )
1
1
Z H
H
H Z H
→
→
→
→
ﻋﻨﺩﺌﺫ
H
ﺘﺒﺩﻴﻠﻴﺔ ﻤﻥ
(4.17)
ﻭ ﻫﺫﺍ،
ﺘﻨﺎﻗﺽ
.
ﻟﺫﻟﻙ ﻓﺈﻥ
G
ﺘﺒﺩﻴﻠﻴﺔ ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺩﺍﺌﺭﻴﺔ
.
ﺍﻟﺫﻱ ﻴﺒﻴﻥ ﺒﺄﻥ،ﺀﺍﺭﻘﺘﺴﻻﺎﺒ ،لﺩﺎﺒﺘﻟﺎﺒ
( )
G Z G
ﺩﺍﺌﺭﻴﺔ
.
ﻻ
!
ﻟﻴﺱ ﺼﺤﻴﺤﺎﹰ ﻤﻥ ﺃﺠل ﺍﻟﺯﻤﺭ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺤل،ﺔﻘﻴﻘﺤﻟﺍ ﻲﻓ
)
،ﻤﺜﻼﹰ
3
S
.(
4
.
(a)
ﺇﺫﺍ ﻜﺎﻥ
g x
x
=
ﻋﻨﺩﺌﺫ،
g h x
h g x
h x
=
=
.
ﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ
g
لـﻜ ﺕـﺒﺜﻴ
ﻋﻨﺼﺭ ﻤﻥ
X
ﻭﻤﻨﻪ،
1
g
=
.
ﻨﺜﺒﺕ
x
X
∈
ﺫـﺌﺩﻨﻋ ،
:
g
g x G
X
→
a
ﺎﻤﺭـﻏ
.
]
ﻨﻼﺤﻅ ﺃﻥ ﻤﺒﺭﻫﻨﺔ ﻜﺎﻴﻠﻲ ﺘﻌﻁﻲ ﺍﻟﺘﺸﺎﻜل ﺍﻟﺫﺍﺘﻲ
(
)
,
:1
n
G
S
n
G
→
=
.[
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٢٦١
(b)
ﺒﺘﺠﺯﺌﺔ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺇﻟﻰ ﻤﺠﻤﻭﻋﺎﺕ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﺭﺘﺒﺔ
3
ﻭ ﻟﺘﻜﻥ،
1
...
m
G
G
G
=
× ×
.
(c)
ﻟﺘﻜﻥ
1
,...,
r
O
O
ﻤﺩﺍﺭﺍﺕ
G
ﻭ ﻟ،
ﺘﻜﻥ
i
G
ﺼﻭﺭﺓ
G
ﻲـﻓ
( )
Sym
i
O
.
ﺫـﺌﺩﻨﻋ
1
...
r
G
G
G
→ × ×
ﻭﻤﻨﻪ،
)
ﺒﺎﻻﺴﺘﻘﺭﺍﺀ
(،
(
)
(
) (
)
1
3
3
3
1
:1
:1 ...
:1
3
... 3
3
r
n
n
n
r
G
G
G
≤
≤
=
5
.
ﻟﺘﻜﻥ
g
G
∈
ﻭ ﻟﺘﻜﻥ،
h
H
∈
ﺤﻴﺙ
ﺇﻨ
ﻬﺎ ﻤﺭﺍﻓﻘﺔ
h
ﻋﻠﻰ
H
ﺘﺘﻭﺍﻓﻕ ﻤﻊ ﻤﺭﺍﻓﻘﺔ
g
.
ﻋﻨﺩﺌﺫ
1
1
gPg
hPh
−
−
=
ﻭﻤﻨﻪ،
( )
1
G
h g
N
P
−
∈
.
6
.
(a)
ﺇﻨﻬﺎ ﺍﻟﺯﻤﺭﺓ
G
x
y
C
C
θ
θ
∞
∞
=
×
=
×
ﻤﻊ
( )
:
Aut
1
θ
∞
∞
→
= ±
C
C
.
ﺸﻜلـﻟﺎﺒ ﺩﻴﺤﻭ ﺸﻜلﺒ ﺭﺼﺎﻨﻌﻟﺍ ﺏﺘﻜﺘ ﻥﺃ ﻥﻜﻤﻴ ،لﺩﺎﺒﺘﻟﺎﺒ
, ,
i
j
x y
i j
∈
Z
ﻭ،
1
y x
x y
−
=
.
(b)
ﺇﻨﻬﺎ ﺍﻟﺯﻤﺭﺓ ﺍﻟﺭﺒﺎﻋﻴﺔ
.
ﻤﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ ﻨﺤﺼل ﻋﻠﻰ
1
1
,
y x
x y
y x
x y
−
−
=
=
ﻭﻤﻨﻪ
2
2
x
y
=
.
ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺜﺎﻨﻴﺔ
2
1
2
2
,
,
x y x
y
y
−
−
=
=
ﻭﻤﻨﻪ
4
1
y
=
.
ﺘﺒﻴﻥ ﺨﻭﺍﺭﺯﻤﻴﺔ ﺘﻭﺩ،لﺩﺎﺒﺘﻟﺎﺒ
-
ﻥـﻤ ﺔﻴﺌﺯﺠ ﺓﺭﻤﺯ ﺎﻬﻨﺃ ﺭﻴﺘﻴﺴﻜﻭﻜ
8
S
ﺭـﺼﺎﻨﻌﻟﺎﺒ ﺓﺩـﻟﻭﻤ
(
)(
)
1287 3465
ﻭ
(
)(
)
1584
2673
.
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٣٦١
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SERRE, J.-P. 1980. Trees. Springer-Verlag, Berlin. Translated from the French
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