Congruence and p-sylow subgroups

[tom.young84]

Homework Statement

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]≡1 mod p.

Homework Equations

Well I believe all we have to do here is conjugate stuff to show membership in various normalizers and then invoke Sylow III.

The Attempt at a Solution

First off we know that for g $$\in$$ N_G(H) such that gPg^-1 and gHg^-1. We also know that there exists a p-sylow subgroup Q=gPg^-1. Since Q is a p-sylow subgroup of H, there exists an h in H such that hPh^-1=Q. Now we see that ghP(gh)^-1=Q. Subbing in something new, j=gh, jPj^-1 is in N_H(P) which is in H. This also shows, by moving stuff around, that N_G(H) is in H. So, now we conclude H is in the Normalizer of H and the Normalizer is in H, so H=N_G(H). The by sylow III, we get that ratio. Is this right, or close?

I was told there is a counting argument that uses group actions and this was more direct but a tad more complicated.

 

[mighty2000]

Your solution is on the right track, but there are a few errors and missing steps. Here is a possible solution:

First, we know that P is a p-sylow subgroup of G, so |P|=p^k for some k. Since P is a subgroup of G, it must also be a subgroup of H. Therefore, P is a p-sylow subgroup of H as well.

Next, we know that N_G(P) is the largest subgroup of G that contains P. Since H contains N_G(P), it follows that N_H(P) is also a subgroup of H. Therefore, N_H(P) is the largest subgroup of H that contains P, which means that N_H(P)=P.

Now, let x be any element of G. Since x is in G, it must also be in H, since H is a subgroup of G. This means that xPx^-1 is also a subgroup of H, and since P is a p-sylow subgroup of H, we must have xPx^-1=P. This shows that x is in N_H(P)=P.

Now, let g be any element of G. We know that gPg^-1 is a p-sylow subgroup of G, so it must also be a subgroup of H. This means that gPg^-1=P. Therefore, g is in N_H(P)=P.

Since g is in P and x is in P, it follows that gx is in P. This means that gx is in N_G(P), and since H contains N_G(P), it follows that gx is in H. Therefore, H contains the subgroup Px, which has order p^k.

Now, we can use Lagrange's theorem to conclude that |H| is divisible by |Px|=p^k, and therefore [G]≡1 mod p, as desired.