Let $G$ be a group of order 315 with a normal 3-Sylow subgroup. Prove $G$ is abelian. I know this it a prevalent question, I really do. It's just that every proof requires using Automorphisms groups about which we were barely taught. I can't start learning everything about Automorphisms by myself (Nor can I tell how far to learn about it) for I don't know how necessary it is if we weren't taught about it. I would appreciate that if you could help me circumvent those proofs that use the autumorphisms. Thank you. 
 A: Let $\{P\}=Syl_3(G)$. Since $|P|=9=3^2$ (the square of a prime), $P$ must be abelian. Hints:
(1) Study the N/C Theorem. Observe that $P \subseteq C_G(P) \subseteq G=N_G(P)$. Hence $G/C_G(P)$ embeds homomorphically into $Aut(P)$.
(2) Note that $Aut(P) \cong C_6$ or $\cong GL(2,3)$. Hence $|Aut(P)|$ divides $6$ or $48$, so $|G/C_G(P)|$ divides these numbers, as well as $5 \cdot 7$.
(3) Conclude that $G=C_G(P)$, that is $P \subseteq Z(G)$, the center of $G$.
(4) Use the fact that there is only one group of order $5 \cdot 7$, and that has to be $\cong C_{35}$.
(5) Use the general fact that if $G/Z(G)$ is cyclic, then $G$ must be abelian.
A: OK, without automorphisms: put $\{P\}=Syl_3(G)$, $Q \in Syl_5(G)$ and $R \in Syl_7(G)$. Observe that $G/P \cong C_{35}$ (I leave it to you to prove this with again Sylow Theory). Hence $QP \unlhd G$, since $QP/P \in Syl_5(G/P)$. Further it is easy to see that $Q$ is the only Sylow $5$-subgroup of $QP$. Hence $Q \text{ char } QP \unlhd G$, so $Q$ is normal in $G$. Same reasoning applies to $R$. Then $G=PQR$, three normal subgroups that mutually intersect trivially, since their orders are relatively prime. Hence $G \equiv P \times Q \times R$ and is abelian.
