Let $H \subset G$ be finite groups. Suppose that $H$ is a $p$-group and $p$ divides $[ G : H ]$. Prove that $p$ divides $[N_G(H) : H ]$.

Question

Let $G$ be a finie group, and $H$ a subgroup of $G$. Suppose that $H$ is a $p$-group and $p$ divides $[ G : H ]$. Prove that $p$ divides $[N_G(H) : H ]$.

As this question arose from a chapter dealing with (left) $G$-actions, I am thinking that I would have to use either the conjugation action or translation action on a $G$-set, perhaps on $G/H$ by $x * aH = xaH$. I am thinking I would probably have t use the orbit decomposition theorem, together with the fact $H$ is a $p$-group. But I am not sure how to proceed. I was thinking of the conjugation action, but I didn't get anywhere - though I might have missed something.

Answer 1

Let $H$ act on its own left cosets in $G$, and let $F$ be the set of cosets fixed by this action (the fixed "points"). Then the results follows from:

• $|F|=[G:H]\pmod{p}$ (this is because $H$ is a $p$-group)
• The coset $xH$ is fixed by this action iff $xH\subset N_G(H)$

The first is true in more generality: a finite $p$-group $P$, acting on a finite set $S$. Then $|\text{Fix}|=|S|\pmod{p}$. It can be proved using the orbit-stabilizer theorem.

The second follows because $hxH=xH$ is equivalent to $x^{-1}hx\in H$.