Let $G$ be a group of order $p_1...p_s$ where $p_1,...,p_s$ are distinct primes. If $G$ has a normal $p$-Sylow subgroup, then $G$ is solvable.
We proceed by induction on $s$. If $s = 1$, $G$ is cyclic, so $G$ is abelian and therefore it is solvable. If $H$ is a normal $p_i$-Sylow of $G$, then $G/H$ has order $p_1...p_{i-1}p_{i+1}...p_s$. Using the induction hypothesis, $G/H$ is solvable. Now $H$ is abelian, so it is solvable and so $G$ is solvable.
Is this correct? I'm not entirely sure about the induction step; shouldn't we need a normal $p$-Sylow subgroup for $G/H$?