proof of a useful counting result in group theory

Question

Let G be a finite group, H a subgroup of G satisfying |G| |̸| [G : H]!. Prove there exists a normal subgroup N of G satisfying 1 < N ⊂ H.

maybe the General Cayley's Theorem works. I am not sure. It seems that it requires a complicated proof. Any idea?

Answer 1

Consider the action of $G$ on the left cosets of $H$. This action affords a homomorphism from $G$ to the symmetric group of order $[G:H]!$. The kernel of this homomorphism is called the core of $H$ and is the largest normal subgroup contained by $H$.

Notice the order of the image of this homomorphism is a divisor of $[G:H]!$ Call this order $m$. So by the first isomorphism theorem $\frac{|G|}{|N|}=m$. We cannot have $|N|=1$ since otherwise $|G|=m$ and then the order of $G$ divides $[G:H]!$