[Math] Show that the group is not simple

Question

I want to show that:

If $G$ contains a subgroup with index at most $4$ and $G$ has not a prime order, then $G$ is not a simple group.

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Let $N\leq G$ with $[G:N]\leq 4$.

We have that $|G|=x\cdot y, \ 1<x,y<|G|$.

Could yout give me a hint how we could conclude that $G$ is not simple?

Do we maybe use Sylow subgroups?

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EDIT:

In my notes I found the following proposition:

$$H\leq G, \ [G:H]=m \text{ and } |G|\not\mid m! \text{ then } G \text{ is not simple. }$$

We have that $H\leq G$ and $[G:H]=m, \ 1\leq m\leq 4$.

Suppose that $|G|\mid m!$. Then $G$ is simple. Or isn't the above proposition an off statement?

• If $m=1$, then $G=H$ and since $|G|\mid 1\Rightarrow |G|=1$. Therefore, $G=H=1$. In that case the group is not simple.
• If $m=2$, the possible values for $|G|$ are $1$ and $2$. The case $|G|=1$ is rejected. It cannot be that $|G|=2$, since $G$ has not a prime order.

• If $m=3$, then $|G|\mid 3!=6$, then the possible values for $|G|$ are $1,2,3,6$.
  The cases $1, 2,3$ are rejected, since $G$ has not a prime order.

  If $|G|=6$ then $G$ is isomorphic to $\mathbb{Z}_6$ or to $S_3$. Both of them are not simple since $\langle 2\rangle$ is normal in $\mathbb{Z}_6$ and $A_3$ is normal in $S_3$. Therefore, $G$ is not simple, a contradiction. Is this correct?

• If $m=4$, then $|G|\mid 4!=24$, then the possible values for $|G|$ are $1,2,3,4,6,8,12,24$.

  The cases $1,2,3,6$ are rejected.

  If $|G|=4$ we have that $G$ is abelian. We have that every subgroup of an abelian group is normal. Therefore, $G$ is not simple, a contradiction.

  What can we say about the cases $|G|=8$, $|G|=12$ and $|G|=24$ ?

Answer 1

Suppose $G$ is simple and let $1<m\leq 4$ be the index of a subgroup $H$. Then the action of $G$ on the set of left cosets of $H$ yields a nontrivial homomorphism $G\to S_m$, and since $G$ is simple the homomorphism is injective, so $|G|\le 24$. The only simple groups with order this small are of prime order.