Exercise 4.4.14 (If $|G| = 1575$, $H \unlhd G$ and $|H| = 9$, then $H \leq Z(G)$.)

Proof. By hypothesis,

and $|H|=9$, $H⊴G$.

Since $H⊴G$, we know that

By Corollary 15, $N_G(H)∕C_G(H)=G∕C_G(H)$ is isomorphic to a subgroup $K$ of $\mathrm{Aut}(H)$.

Moreover, by Proposition 17, $H\simeq \mathbb{Z}∕9\mathbb{Z}$ or $H\simeq {(\mathbb{Z}∕3\mathbb{Z})}^2$, then $\mathrm{Aut}(H)\simeq {(\mathbb{Z}∕9\mathbb{Z})}^{\ast }$, of order $\phi (3^2)=3^2-3=6$, or $\mathrm{Aut}(H)\simeq {\mathrm{GL}}_2({𝔽}_3)$, of order $(3^2-1)(3^2-3)=48$ (see the end of Section 4.4 p. 135), so

By Lagrange’s Theorem, in both cases,

Moreover $|G|∕|C_G(H)|$ is a divisor of $|G|=1575=3^2\cdot 5^2\cdot 7$, thus $|G|∕|C_G(H)|$ divides $g.c.d.(48,1575)=3$, therefore

But $H$ is abelian, thus $H\le C_G(H)$ and so $3^2\mid |C_G(H)|$. This gives $|C_G(H)|=3^2\cdot k$ for some positive integer $k$. Consequently,

Since $3\nmid 5^2\cdot 7$, we obtain

Therefore

In other words, every element of $H$ commutes with every element of $G$, so

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