On Groups of Order 315 with a unique sylow 3-subgroup . in Dummit and Foote , an exercise asked me to prove that , if $G$ is a group of order $315$ , $G$ has a normal sylow $3$-subgroup then , $G$ is abelian .
this is exercise number $27$ , section $5$ , chapter $4$
my method on proving this is showing that $ |G| = |Z(G)|$ as follows
i showed that $|Z(G)|$ $\in$ $\{ 9 , 63 , 45 , 315 \} $
and  showed that$ |Z(G)|$ is not $63 , 45$ or $9$
so it must be $315 $
but i showed this under two conditions,
first condition:
if $X$ is the set of all elements of $G$ of order $5$ , $H$ is the subgroup generated by $X$
then $|H|$ is not $315$
second condition:
also , if $Y$ is the set of all elements of $G$ of order $7$ , $N$ is the subgroup generated by $Y$ then $|N|$ is not $315$
so , if those two conditions are true then my method of proving works .
but i don't know how to prove those conditions .
any help ?
 A: There might be a shorter/better way, still...
$315 = 3^2 \cdot 5 \cdot 7$. The $3$-Sylow subgroup $T$ is of order $3^2$, thus abelian. Such a group has no automorphisms of order $5$ or $7$, so any Sylow $5$-subgroup centralizes $T$, and so does any Sylow $7$-subgroup. 
Therefore there is an abelian subgroup $H$ of order $3^2 \cdot 7$. Let $S$ be the Sylow $7$-subgroup of $H$. 
If $N_{G}(S) = H$, then the number of $7$-Sylow subgroups of $G$ is $5 = \lvert G : H \rvert$; however $5 \not\equiv 1 \pmod{7}$, contradicting one of Sylow's theorems. So $S$ is normal in $G$, and thus unique.
There is also an abelian subgroup $K$ of order $3^2 \cdot 5$. Let $F$ be a Sylow $5$-subgroup of $K$. By a similar argument, $F$ is normal in $G$, and thus unique.
Thus $G = T \times S \times F$ is abelian.
A: There may be a direct approach wrt the hints in the book: 
$\,|G|=315=3^2\cdot 5\cdot 7\,$ , and let $\,P\triangleleft G\,\,\,,\;\;|P|=9\,$ . 
By corollary $15$ in $4.4$ (page $134$), we know that $\,G/C_G(P)\cong T\le\operatorname{Aut}{P}\,$
Since $\,|\operatorname{Aut}(P)|=6\,\,or\,\,48\,$ , and $\,|G/P|=35\,$ and $\,P\,$ is abelian, we get that $\,P\le C_G(P)\Longrightarrow |P|\mid |C_G(P)|\,$ , and from here it must be $\,\left|G/C_G(P)\right|=1\,$ (you have a very, very similar example just before the exercises begin at the end of section 4.4, in page 137) , which means
$P\le Z(G)$ , which is the hint the book has there, so
$$ G/Z(G)\cong \left(G/P\right)/\left(Z(G)/P\right)$$
But any group of order $\,35\,$ is cyclic (why?) , so the above means $\,G/Z(G)\,$ is cyclic and, thus, $\,G\,$ is abelian.
A: Although, you got your answer, I think the following may help you as well. I assume again $S$ is cyclic normal subgroup of $G$ of order $7$ and $F$ is cyclic normal subgroup of $G$ of order $5$ as @Andreas proved. We have: $$|S|=7,~ S\vartriangleleft G\to [G:S]=9\times 5\to G/S~~~~ \text{is abelian}\to G'\leq S\to|G'|\big||S|=7$$ $$|F|=5,~ F\vartriangleleft G\to [G:F]=9\times 7\to G/F~~~~ \text{is abelian}\to G'\leq F\to|G'|\big||F|=5$$ but $(5,7)=1$ so $G'=\{e\}$ and then $G$ is abelian.
