$|G|=112$, prove that $G$ is solvable.

I already proved that $G$ is not simple so I know that it has normal subgroup but I don't know from what order. Another solution that I tried was looking at the sylow subgroups. In that case, if 2-sylow subgroup is normal or 7-sylow subgroup is normal then the solution is obvious, but in case that there are 7 2-sylow subgroups and 8 7-sylow subgroups, I don't know how to continue.

## Best Answer

The number $112=2^4\cdot 7$ has prime factors $2$ and $7$. By Burnside's $p^aq^b$-theorem, all groups of such order, having only two different prime divisors, are solvable. For order $p^aq$ the proof is not difficult, see here.

In case you do not want to use Burnside's result, we already know from the other question,

Proving that a group of order $112$ is not simple

about the Sylow subgroups. For example, one possibility is that the group has a normal Sylow $7$-subgroup $N$ of order $7$. Hence the quotient has order $2^4$ and is nilpotent (since it is a $p$-group), hence solvable. Now, if $N$ and $G/N$ are solvable, so is $G$, and we are done.