## A congruence condition on the index of subgroups containing Sylow normalizers

Let $P$ be a Sylow $p$-subgroup of $G$ and let $M$ be any subgroup of $G$ which contains $N_G(P)$. Prove that $[G : M] \equiv 1 \mod p$.

We prove a slightly stronger result by induction. First a definition.

Let $G$ be a finite group and $H \leq G$ a subgroup. The distance from $H$ to $G$ is the length $d$ of the longest chain of proper subgroups $K_i$ such that $H = K_0 < K_1 < \ldots < K_d$.

Lemma: Let $G$ be a finite group and $P \leq G$ a Sylow $p$-subgroup. If $H,K \leq G$ are subgroups such that $N_G(P) \leq H \leq K$, then $[K : H] \equiv 1 \mod p$. Proof: We proceed by induction on the distance $d$ from $N_G(P)$ to $K$. For the base case, if $d = 0$, then $K = N_G(P)$ so that $H = N_G(P)$, and we have $[K : H] = 1$. For the inductive step, suppose that for some $d \geq 0$, if the distance from $N_G(P)$ to $K$ is at most $d$ then for all $H$ such that $N_G(P) \leq H \leq K$ we have $[K : H] \equiv 1 \mod p$. Now let $K \leq G$ be a subgroup containing $N_G(P)$ such that the distance from $N_G(P)$ to $K$ is $d+1$, and suppose further that $N_G(P) \leq H \leq K$. First, if $H = K$, we have $[K : H] = 1$ so that the conclusion holds. Suppose now that the inclusion $H \leq K$ is proper. Note that $[K : N_G(P)] = [K : H] \cdot [H : N_G(P)]$. By the induction hypothesis, we have $[H : N_G(P)] \equiv 1 \mod p$. Recall that $N_K(P) = K \cap N_G(P)$, so that $[K : N_G(P)] = [K : N_K(P)]$ $= n_p(K) \equiv 1 \mod p$ by Sylow's Theorem. Modulo $p$, we thus have $[K:H] \equiv 1$. $\square$

In particular, if $K = G$ and $N_G(P) \leq M \leq G$, we have $[G:M] \equiv 1 \mod p$.

You can do it the other way too, first proving the problem, then extending it to the lemma: Let $N_G(P) \leq M \leq G.$ Then $1 \equiv |G:N_G(P)| = |G:M||M:N_G(P)| = |G:M||M:N_M(P)| \equiv |G:M|.$ For the lemma, since $N_G(P) = N_K(P) \leq H \leq K,$ the same proof applies.