Exercise 4.5.51 (If $N_G(P) \leq M$, where $P$ is a Sylow $p$-subgroup of $G$, then $|G : M| \equiv 1 \pmod p$.)

Proof. Let $G$ be a group of order $p^{\alpha }m$, where $p\nmid m$, and let $P$ be a Sylow $p$-subgroup of $G$

Consider the following tower of subgroups of $G$:

Note that $|M|=|M:P||P:1|=p^{\alpha }{m}^{\prime }$, where $m^{\prime }=|M:P|$ divides $m=|G:P|$, thus $p\nmid m^{\prime }$. This shows that $P$ is a Sylow $p$-subgroup of $M$.

We prove that

• If $g\in N_M(P)$, then $g\in M$ and $\mathit{𝑔𝑃}{g}^{-1}=P$. Since $M\le G$, $g\in N_G(P)$, so $N_M(P)\le N_G(P)$.
• If $g\in N_G(P)$, then $g\in G$ and $\mathit{𝑔𝑃}{g}^{-1}=P$. Therefore $g\in N_G(P)$, where $N_G(P)\le M$, thus $g\in M$ and so $g\in N_M(P)$. This shows that $N_G(P)\le N_M(P)$.

By the third part of Sylow’s Theorem,

Since

we obtain

If $N_G(P)\le M$, where $P$ is a Sylow $p$-subgroup of $G$, then $|G:M|\equiv 1(\mathit{𝑚𝑜𝑑}p)$. □