1. ## p-Sylow subgroup congruence

Let $p$ be a prime, $G$ a finite group, and $P$ a $p$-Sylow subgroup of $G$. Let $M$ be any subgroup of $G$ which contains $N_G(P)$. Prove that $[G:M]\equiv 1$ (mod $p$). (Hint: look carefully at Sylow's Theorems.)

2. Originally Posted by Erdos32212
Let $p$ be a prime, $G$ a finite group, and $P$ a $p$-Sylow subgroup of $G$. Let $M$ be any subgroup of $G$ which contains $N_G(P)$. Prove that $[G:M]\equiv 1$ (mod $p$). (Hint: look carefully at Sylow's Theorems.)
The number of conjugates to $P$ is $[G:N(P)]$.
By Sylow's third theorem $[G:N(P)]\equiv 1(\bmod p)$.
But since $[G:M]$ divides $[G:N(P)]$ it means $[G:M]\equiv 1(\bmod p)$ also.

3. Here is essentially a similar proof to that one:

It's important to note though that $N_M(P) = N_G(P)$. By Sylow's Theorem Part 3, we have that $[G: N_G(P)] \equiv 1 \mod{p}$. Then $P$ is a $p$-Sylow group of $M$, and $N_M(P)$ is the normalizer of $P$ in $M$, so $[M: N_M(P)] \equiv 1 \mod{p}$. Thus $[G: M] = \frac {[G: N_G(P)]}{[M: N_M(P)]} \equiv 1 \mod{p}$.