Prove that any group $G$ with $|G|=588$ is solvable

Question

I'm stuck trying to solve this problem from my abstract algebra course:

Prove that every group of order $588$ is solvable (If you assume that all groups of certain order are solvable, you must prove it too).

First I noticed that $588=2^2\cdot 3\cdot 7^2$, so I guess the prove will use Sylow $p$-subgroups, but I don't find right what I need to do. Any help or hint will be appreciated.

Thanks in advance.

Answer 1

You have seen, using Sylow's theorems, that there is only one Sylow $7$ subgroup, calling it $P_7$.

Then $G/P_7$ has order $12$.

But any group of order $p^2q$ is solvable.

Hence $G/P_7$ is solvable.

Then since $P_7$ is solvable, $G$ is solvable.