Showing that group of orientation preserving isometries of Icosahedron is a simple group

Question

Let $G$ denote the group of orientation preserving isometries of Icosahedron. To prove the claim, I have shown that

1. $\nexists \ N \ \triangleleft \ G$ such that $|N|=5.$
2. $\nexists \ N \ \triangleleft \ G$ such that $|N|=3.$
3. $\nexists \ N \ \triangleleft \ G$ such that $|N|=2.$

I know the following:

1. Rotation around a vertex gives a subgroup of order 5.
2. Rotation around the center of a face gives a subgroup of order 3.
3. Rotation around the midpoint of an edge gives a subgroup of order 2.

We have $|G|=60=2^2\times 3\times 5.$ What more do I need to show to conclude that $G$ is simple? Is it that I have to show there are no normal subgroup of order $4,6,10,12,15,20,30$? Is there a general theorem that can be useful in my case? I would be glad to know in that case.

Answer 1

Let $G$ be a group of order $60$ with no normal subgroups of order $2$, $3$, or $5$. In each of the following cases, suppose that $H \unlhd G$ were a normal subgroup of order $n$:

1) $n = 10, 15, 20, 30$: $H$ has a normal Sylow $5$-subgroup $P_5$, so $P_5 \text{ char } H \unlhd G \implies P_5 \unlhd G$.

2) $n = 6$: Same argument as in (1), considering a (normal) Sylow $3$-subgroup of $H$.

3) $n = 12$: Either $H$ has a normal Sylow $3$-subgroup, which would be normal in $G$ (contradiction), or $H$ has a normal Sylow $2$-subgroup, which is then normal in $G$. Thus it suffices to show the last case:

4) $n = 4$: Then $H = [G,G]$ (since $G/H$ has order $15$, hence is cyclic $\implies [G,G] \le H$, and $|[G,G]| \ne 1, 2$). If $P_5$ is a Sylow $5$-subgroup of $G$, then $HP_5$ is a subgroup of $G$ of order $20$. But $[G,G] = H \le HP_5$, so $HP_5$ is normal in $G$, impossible by (1).