Let $G$ be a group of order
$$72=2^3\cdot 3^2$$
Without using Burnside's Theorem, how to show that $G$ is solvable?
Atempt:
If we can show that $G$ has at least one non-trivial normal subgroup $N$, then it would be easy to show it is solvable. Indeed,
$$1\longrightarrow N\longrightarrow G\longrightarrow G/N\longrightarrow 1$$
would be a short exact sequence with $N$ and $N/G$ of order $2^i\cdot3^j$ for some $i,j\in\{0,1,2\}$ and it is not too hard to show that such groups are always solvable. However, I can't find a way to show that $G$ is not simple.
Added: If $G$ is not simple, then Sylow's Theorem implies that there are $4$ subgroups of order 9 and 3 or 9 subgroups of order 8. Then, I don't see how to use that to show that $G$ is not simple.
Best Answer
If $G$ has 4 Sylow-3 subgroups, $G$ acts on those subgroups via conjugation, inducing a homomorphism $G\to S_4$. Since $|S_4|=24<72=|G|$, this map must have a non-trivial kernel. If the morphism is not the trivial map, you are done. What can you say if the kernel is all of $G$?