Is There a Nontrivial Normal Subgroup in a Finite Group with Index Not Divisible by the Group's Order?

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Here is this week's POTW:

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Let $G$ be a finite group, and let $H$ be a subgroup of $G$ of index $n$. Prove that if the order of $G$ does not divide $n!$, then $H$ contains a nontrivial normal subgroup of $G$.

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $G$ act on the set of left cosets of $H$ by left multiplication, and let $K$ be the kernel of the induced permutation representation. Then $K$ is the intersection of all conjugates of $H$, so $K$ is a normal subgroup of $G$ contained in $H$. Since $H$ has index $n$, $K$ has index dividing $n!$. Using the fact that $n!$ is not divisible by $|G|$, $K$ must be nontrivial.