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.

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  • nhanttruong  On July 5, 2011 at 11:34 pm

    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.

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