Let P be Sylow p-subgroup of G and let M be any subgroup of G which contains N_G(P). Prove that |G:M|=1 mod p.

Question

Could anyone help me out please? This is from Dummit and Foote Exercise 4.5.51. It looks like a very interesting result.

My idea is that $|G:M||M:N_G(P)|=|G:N_G(P)|=n_p\equiv1$ ($\mathrm{mod}$ $p$). But I couldn't find a way to proceed. Any help will be appreciated.

Answer 1

By Sylow's third theorem, $|M:N_M(P)|=1\mod p$. Since $N_G(P)\leq M$, $N_M(P)=N_G(P)\cap M=N_G(P)$. So $|M:N_G(P)|=1\mod p$. Hence $|G:M|=1\mod p$.

Answer 2

Let $X$ denote the set of cosets $\{gM:g \in G\}$, and let $P$ act on $X$ by left multiplication. Then $X$ is partitioned into $P$-orbits, and by the orbit-stabilizer theorem the length of any orbit is a divisor of $|P|$, hence is either a proper power of $p$ or $1$. It is easily to see that $[M]$ is a single-element orbit, because $PM=M$. It suffices to prove $[M]$ is the only one, since then any other orbit has length divided by $p$, and the total number of elements in $X$ is the foem $pm+1$.

Suppose $gM \in X$ forms another orbit of a single element. Let $a\in P$, by definition $PgM=gM$, and thus $g^{-1}Pg \in M$, which is another Sylow $p$-subgroup of $M$. By the second Sylow theorem, there exists $m \in M$ such that $g^{-1}Pg = m^{-1}Pm$. Henceforth $gm^{-1} \in N_G(P)$, and it follows trivially that $gm^{-1} \in M$. So $g \in M$, $gM = M$, $[M]$ is the only single orbit under the action of $P$.