# Math Help - Self Normalizing groups

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

2. |G: $N_G$P| is equal to the p-sylow subgroups in G, likewise with |H: $N_G$P|. For some p-Sylow subgroup L $\leq$ $N_G$H, there is a g such that L=gPg-1 $\leq$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

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

4. 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 $p^n$ (p prime) acts on a finite set S, then $|S| = |S_0|(\text{mod p})$, where $S_0$ is the number of fixed elements by the given action.
I'm just curious, is this a well known lemma.

5. 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).

6. okay, one more jerk type question

why is S_0=1

7. 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.

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