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Theorem about p-groups, similar to 3rd Sylow Theorem |
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| Dec15-10, 01:03 AM | #1 |
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Theorem about p-groups, similar to 3rd Sylow Theorem
(a) Let G be a finite group that is divisible by by p^k, and suppose that H is a subgroup of G with order p^j, where j is less than or equal to k. Show that the number of subgroups of G of order p^k that contain H is congruent to 1 modulo p.
(b) Find an example of a finite group that has exactly p+1 Sylow p-subgroups. I think that I should be using the 3rd Sylow Theorem (the number of Sylow p-subgroups of G is congruent to 1 modulo p) to prove (a). Also maybe the fact that since H is a p-group, it is contained in Sylow p-subgroup by the 2nd Sylow Theorem, and any larger p-group containing H is also contained in a Sylow p-subgroup. Any help would be much appreciated : ) |
| Dec16-10, 07:16 PM | #2 |
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The fact that p-groups have normal subgroups of all orders might prove useful, too.
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