group of order 135 solvable?

Question

I need to prove that a group G of order 135 is solvable.

$|G| = 135 = 5\cdot 3^3$

i found that the Sylow-subgroups are unique, so they are both normal. Let H be the Sylow-5-subgroup and F be the Sylow-3-subgroup.

Because they are normal, we can say that:

$G\cong H\times F$

Here is where i get stuck... I know that H is abelian and hence solvable. But that is it.

Is there a way to prove that F is abelian or solvable? or is this a dead end and do i need to prove this in a whole other way?

Answer 1

Let $G$ be a group with $H\lhd G$. Then $G$ is solvable if and only if $H$ and $G/H$ are solvable. Also all $p$-groups are solveable. So, then $F$ is solvable, and since $(G:F)=5$, $G/F$ is solvable.

Answer 2

Note that finite $p$-groups are always solvable, since the center is always nontrivial.