Let p denote a prime number. Let G be a finite group. Let P denote a p-sylow subgroup of G. Finally, let H be any subgroup of G that contains N_G(P). Prove that [G:H]≡1 mod p.
Well here is what I know right away: Nis in H. This means that gPg
. Since it is in H, this also means that N
(P)=hPh^-1 So ghP(gh)^-1. This is about where I'm stuck.
|G:P| is equal to the p-sylow subgroups in G, likewise with |H:
I'm stuck on pulling the G:H ≡1 mod p
This is refining my first post
Lemma 1. Every subgroup H of G which contains the normalizer of a p-Sylow subgroup is its own normalizer.Originally Posted by cmj1988
Let p denote a prime number. Let G be a finite group. Let P denote a p-sylow subgroup of G. Finally, let H be any subgroup of G that contains N_G(P). Prove that [G:H]≡1 mod p.
Well here is what I know right away: N
Lemma 2. If a group H of order(p prime) acts on a finite set S, then
, where
is the number of fixed elements by the given action.
Let S be the set of all left cosets of H in G; let P acts S by conjugation. We see thatfor all
, where K is a left coset of H in G. This condition only holds if and only if
. This further implies that
. We know that
is contained in H and
by lemma 1. Thus
and we conclude that
by lemma 2.
I'm just curious, is this a well known lemma.Lemma 1. Every subgroup H of G which contains the normalizer of a p-Sylow subgroup is its own normalizer.
Lemma 2. If a group H of order
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