Generalization of index 2 subgroups are normal

Let $G$ be a finite group and $H$ a subgroup of index $p$, where $p$ is a prime. If $\operatorname{gcd}(|H|, p-1)=1$, then $H$ must be normal. Does somebody have a quick proof of this?

By induction on $|G|$, we can assume that there is no non-trivial normal subgroup $N$ of $G$ contained in $H$. Thus the action of $G$ on the conjugates of $H$ gives an embedding of $G$ into $S_p$, of order $p!$. Thus $|H|$ divides $(p-1)!$, and so $p$ does not divide $|H|$. Thus the Sylow p-groups of $G$ are cyclic of order $p$.

Now if any $h\in H$ normalized a Sylow p-group $P$, then $h$ would map into $Aut(P)\cong C_{p-1}$, and by hypothesis the image would be trivial. That is, any $h\in H$ normalizing $P$ also centralizes $P$. It follows that $N_G(P)=C_G(P)$, and by the Burnside Transfer theorem, $G$ has a normal subgroup $M$ of index $p$. Of course, any Sylow q-group of $H$ is then contained in $M$, and so $H$ is contained in $M$; that is, $H=M$ is normal.

Note: As pointed out in the comments this answer is missing a crucial detail, and unfortunately I have no idea if it can be salvaged.

This is a slightly more elementary way to show this:

Assume $H$ is not normal. Then clearly $N_G(H) = H$ and $G$ acts on the $|G:H| = p$ conjugates of $H$ by conjugation. The stabilizer of $H$ under this action is just $H$, which then acts on the other $p-1$ conjugates.

The orbit of a $gHg^{-1}$ under this action consists of all $g'Hg'^{-1}$ with $g'\in Hg$ and thus all the orbits have the same size. On the other hand, their sizes all divide the order of $H$, which is a contradiction since $gcd(|H|,p-1) = 1$