D+F 4.5.27 (modified)

Let $G$ be a group with order 315 and suppose $G$ has a normal subgroup $H$ that is cyclic of order 9. Prove that $G$ is cyclic.

$\textit{Proof}$: By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ In this case, since $H \lhd G$, $H$ normalizes all of $G$, i.e. $N_G(H)=G$. Also, $H$ being cyclic and $|H|=9 \Longrightarrow H \cong \mathbb{Z}_9$. By Proposition 16 p. 135, $$Aut(H) \cong \mathbb{Z}_9^* \cong \mathbb{Z}_6 $$ Thus, $(\star)$ becomes: $$G/C_G(H) \lesssim \mathbb{Z}_6$$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 6$$ This means that $$|G/C_G(H)| = 1,2,3,6 \Longrightarrow |C_G(H)|=315, \frac{315}{2}, \frac{315}{3}, \frac{315}{6}$$ But the order of a subgroup must be an integer, so $$|C_G(H)|=315, 105 \quad (\star \star)$$ By problem 2.2.6(b), $H \leq C_G(H) \Longleftrightarrow H$ is Abelian. Since $H \cong \mathbb{Z}_9$, it is cyclic, and thus Abelian. So, $H \leq C_G(H)$ and $|H| \mid |C_G(H)|$ by Lagrange.

$9 \mid |C_G(H)|$ but also $C_G(H)$ is always a subgroup of $G$. So, the possible values for the order of $C_G(H)$ are multiples of 9 that divide 315. $$|C_G(H)|=9,45,63,315$$ The only value above that is also consistent with $(\star \star)$ is 315. So, $|C_G(H)|=315$. So, every element of $G$ commutes with elements of $H$ i.e. $H \leq Z(G)$.

We now establish that $G$ is Abelian:
Since $H \cong \mathbb{Z}_9 \leq Z(G)$ and $Z(G) \leq G$, then by Lagrange, $$9 \mid |Z(G)| \mid 315$$ Thus, $|Z(G)|=9,45,63,315$ (The multiples of 9 that divide 315) So then, $$|G/Z(G)|=\frac{315}{315}, \frac{315}{63}, \frac{315}{45}, \frac{315}{9} = 1,5,7,35$$ Groups of order 1,5,7, and 35 are all cyclic. (We can show the 35 case by the method on p.143) By exercise 3.2.36, We get that $G$ is Abelian. Finally, by the Fundamental Theorem of Finitely Generated Abelian Groups, $$G \cong \mathbb{Z}_9 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_7$$ $gcd(9,5,7)=1$. So, by Proposition 6(1) P. 163, $G$ is cyclic. $\blacksquare$