Let $\displaystyle G$ be a group of order 315 and $\displaystyle G$ contains a normal Sylow subgroup. Show that $\displaystyle G$ must be abelian.
I cannot finish this . Maybe someone can finish this.
Let $\displaystyle C = G'$, the commutator subgroup. To show $\displaystyle G$ is abelian it is sufficient to show $\displaystyle |C| = 1$. Let $\displaystyle H$ be the unique Sylow $\displaystyle 3$-subgroup, then it is normal. Construct the group $\displaystyle G/H$ and note $\displaystyle |G/H| = 5\cdot 7$. Since $\displaystyle 7\not \equiv 1 (\bmod 5)$ it means $\displaystyle G/H$ is an abelian group. This implies that $\displaystyle C$ is contained in $\displaystyle H$. But since $\displaystyle |H| = 3^2$ it means by Lagrange's theorem $\displaystyle |C| = 1,3,9$.
Lemma: If $\displaystyle G$ is a group with $\displaystyle |G| = 3^2\cdot 7=63$ then $\displaystyle G$ is abelian.
Proof: There is exactly one Sylow $\displaystyle 3$-subgroup and exactly one Sylow $\displaystyle 7$-subgroup. Call them $\displaystyle M_1,M_2$. Then $\displaystyle |M_1\cap M_2| = 1$ therefore $\displaystyle |M_1M_2| = 63$ i.e. $\displaystyle M_1M_2 = G$. Furthermore, both $\displaystyle M_1,M_2$ are normal subgroups. It follows by isomorphism theorems that $\displaystyle G \simeq M_1 \times M_2 \simeq \mathbb{Z}_3 \times \mathbb{Z}_7$.
Consider the Sylow $\displaystyle 7$-subgroups. By Sylow theorems there are either $\displaystyle 1$ or $\displaystyle 15$ of them. If there is just one, call it, $\displaystyle N$ then form $\displaystyle G/N$. It follows that $\displaystyle |G/N| = 63$. This is abelian by lemma. Thus, $\displaystyle N$ contains $\displaystyle C$. This means by Lagrange's theorem $\displaystyle |C| = 1,7$. But by the first paragraph it would force $\displaystyle |C|=1$. Therefore, $\displaystyle G$ abelian. Thus, it is safe to assume that there are $\displaystyle 15$ Sylow $\displaystyle 7$-subgroups.
But I do not know how to show that having 15 is impossible.