Exercise 4.5.27 (If $|G| = 315$ and if $G$ has a normal Sylow $3$-subgroup, then $G$ is abelian)

Proof. Here $|G|=3^2\cdot 5\cdot 7$.

By hypothesis, $G$ has a unique Sylow $3$-subgroup $P$, of order $9$, which is normal in $G$.

Since $P⊴G$, $G$ acts on $P$ by conjugation. This action affords an homomorphism

with kernel $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=C_G(P)$, thus $G∕C_G(H)$ is isomorphic to a subgroup $T$ of $\mathrm{Aut}(G)$:

(see Proposition 13 and Corollary 15 p. 133, 134.)

Moreover $|P|=3^2$, thus $P$ is abelian, isomorphic to $Z_9$ or $Z_3\times Z_3$, and by Section 4.3,

In both cases

Since $P$ is abelian, $P\le C_G(P)$, thus $9\mid |C_G(P)|$, i.e., $|C_G(P)|=9k$ for some integer $k$, so

so

Since $|T|$ is a divisor of $|\mathrm{Aut}(P)|=48$ and $35$, where $g.c.d.(48,35)=1$, then $|T|=1$. Therefore $|G∕C_G(P)|=1$, thus $G=C_G(P)$. This shows that

Put $\bar{G}=G∕P$. Then $|\bar{G}|=5\cdot 7=35$.

By the first example (“groups of order pq”, p. 143), since $p=5\nmid q-1=6$, $\bar{G}$ is cyclic, so

By the Third Isomorphism Theorem, since $P\le Z(G)$,

Therefore $G∕Z(G)$ is a quotient group of the cyclic group $G∕P$, so is cyclic. By Exercise 3.1.36, this proves that $G$ is abelian. □