Thread: Self Normalizing groups

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$\displaystyle _G(P)$ is in H. This means that gPg$\displaystyle ^-1$. Since it is in H, this also means that N$\displaystyle _H$(P)=hPh^-1 So ghP(gh)^-1. This is about where I'm stuck.

|G:$\displaystyle N_G$P| is equal to the p-sylow subgroups in G, likewise with |H:$\displaystyle N_G$P|. For some p-Sylow subgroup L$\displaystyle \leq$$\displaystyle N_G$H, there is a g such that L=gPg-1$\displaystyle \leq$gHg-1=H. We know that's true because there is some theorem that says that if a subgroup contains the normalizer of a p-sylow subgroup it is self normalizing.

I'm stuck on pulling the G:H ≡1 mod p

This is refining my first post

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$\displaystyle _G(P)$ is in H. This means that gPg$\displaystyle ^-1$. Since it is in H, this also means that N$\displaystyle _H$(P)=hPh^-1 So ghP(gh)^-1. This is about where I'm stuck.

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 $\displaystyle p^n$ (p prime) acts on a finite set S, then $\displaystyle |S| = |S_0|(\text{mod p})$, where $\displaystyle S_0$ 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 that $\displaystyle K \in S_0 \leftrightarrow xKx^{-1}=K$ for all $\displaystyle x \in P$, where K is a left coset of H in G. This condition only holds if and only if $\displaystyle P < N_G(K)$. This further implies that $\displaystyle N_G(P) < N_G(N_G(K))$. We know that $\displaystyle N_G(P)$ is contained in H and $\displaystyle N_G(H)=H$ by lemma 1. Thus $\displaystyle S_0=\{H\}$ and we conclude that $\displaystyle [G:H] \equiv 1 (\text{mod p})$ by lemma 2.

Originally Posted by aliceinwonderland

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 $\displaystyle p^n$ (p prime) acts on a finite set S, then $\displaystyle |S| = |S_0|(\text{mod p})$, where $\displaystyle S_0$ is the number of fixed elements by the given action.

I'm just curious, is this a well known lemma.

Originally Posted by cmj1988

I'm just curious, is this a well known lemma.

It seems like you already saw the lemma 1. You can find the lemma 2 in Hungerford's algebra (p93).

okay, one more jerk type question

why is S_0=1

Originally Posted by cmj1988

okay, one more jerk type question

why is S_0=1

It is because P cannot be contained in H and aH (a in G, but not in H) at the same time. So, S_0={H} only and |S_0|=1.

I follow your proof except for K, is K the arbitrary p-sylow subgroup contained in gPg^-1 contained in gHg^-1=H