Prove: A group of order 315 with a normal 3-Sylow group is Abelian.

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Prove: A group of order 315 with a normal 3-Sylow group is Abelian. I know that $315 = 3^2\times5\times7$ I've tried using Sylow's theorems on 5, 7 but haven't got anywhere meaningful. I would appreciate any help.

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Let $G$ be such a group. Let $P_k$ be Sylow k-subgroup for $3,5,7$ of $G$, respectively. First, $P_3$ is a normal abelian subgroup. Then $P_3P_5$ and $P_3 P_7$ are subgroups of $G$ since $P_3$ is normal. Furthermore, they are abelian subgroups by the Sylow theorem. We may conclude that the centralizer $C_G(P_3)$ of $P_3$ has oreder at least $3^2\cdot 5 \cdot 7$. Thus $C_G(P_3) = G$.