# group of order 135 solvable?

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?

• i think not, we haven't seen it in class so i'm no familiar with that theorem... – Robbe Motmans Apr 6 '15 at 16:01
• @Robbe: The person above you was really making a joke. The joke is that, that theorem requires 255 pages to prove and is complete overkill. (The theorem is that, if a group is odd order, then it is solvable). – Nicolas Bourbaki Apr 6 '15 at 16:04

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.
• Even better: all finite $\;p$-groups are nilpotent (and thus also solvable) – Timbuc Apr 6 '15 at 16:22
• @Timbuc You need to be careful here, because if you allow infinite $p$-groups, then it is not true that all $p$-groups are solvable. So the word "finite" needs adding to this answer! – Derek Holt Apr 6 '15 at 16:56
• @DerekHolt I can't see how after I clearly remarked finite you still can see any suggestion otherwise, honest...not to mention the question was intended to a group of order $\;135\;$ ...finite. – Timbuc Apr 6 '15 at 19:58
Note that finite $p$-groups are always solvable, since the center is always nontrivial.
• Again, you mean finite $p$-groups are always solvable. (Too many people assume that all groups are finite!) – Derek Holt Apr 6 '15 at 16:58