TITLE: $H$ is a subgroup of a finite group $G$ such that $|H|$ and $\big([G:H]-1\big)!$ are relatively prime. Prove that $H$ is normal in $G$.
QUESTION [5 upvotes]: Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and  $\big([G:H]-1\big)!$ are relatively prime. Prove that $H$ is normal in G

Let $[G:H]=m$
Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have 
$$\phi : G \to S_A\simeq S_m$$
where  $$K= \ker\phi \subseteq H$$
$$\frac{G}{K} \simeq \phi(G)\leq S_m$$
So $$[G:K] \mid m!$$
$$\Rightarrow [G:H][H:K]\mid m!$$
$$\Rightarrow [H:K]|(m-1)!$$
But $\gcd\big(|H|,(m-1)!\big) = 1 \Rightarrow |K|>1 $
How do I proceed to prove $K=H \lhd G$?
Or is this approach wrong

REPLY [4 votes]: Note that $[H:K]$ divides both $|H|$ and $\big([G:H]-1\big)!=(m-1)!$.  Therefore, it divides the greatest common divisor of  $|H|$ and $\big([G:H]-1\big)!$, which is $1$.  Hence, $[H:K]=1$, making $K=H$.