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Prove that $G$ is solvable given its order is $3^k5^l$ while $k,l \in \mathbb{N} , k \le 3$. we are not allowed to use burnside's theorem and Feit–Thompson.

I tried to use sylow's theorems to prove all sylow subgroups are normal. If i'm not mistaken its equivalent to $G$ being nil-potent (is it ?) and all nil-potent groups are solvable .

So using the sylow theorems i get $n_5|3^k$ and $n_5 \equiv1\mod5$ given $k\le3$ we get $n_5 = 1$. so sylow-5 subgroup is normal.

Now for sylow-3 subgroups we get $n_3|5^l$ and $n_3 \equiv1\mod3$ but since $l$ is not bounded i'm not sure how to countinue

Boris Morozov
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    Use the fact: If $G$ is a group, $H\leq G$ its normal subgroup and $H$ and $G/H$ are solvable, then $G$ is also solvable. Take $H$ to be Sylow-5 subgroup, then $H$ and $G/H$ are $p$-groups, hence solvable. – SMM Jan 26 '15 at 06:47

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