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?
Best Answer
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.