Exercise 4.5.28 (A group of order 1575 with a normal Sylow 3-subgroup is abelian)

Proof. Here $|G|=1575=3^2\cdot 5^2\cdot 7$. We assume that $G$ contains a normal Sylow $3$-subgroup $P$, of order $3^2$.

Put $\bar{G}=G∕P$. Then $|\bar{G}|=|G|∕|P|=5^2\cdot 7$. By Sylow’s Theorem,

Therefore

hence there is a unique Sylow $5$-subgroup $\bar{H}$ of $\bar{G}$, and there is a unique Sylow $7$-subgroup $\bar{K}$ of $\bar{G}$.

By the Lattice Isomorphism Theorem (Correspondence Theorem), there is a unique subgroup $H$ of $G$ which contains $P$, of index $|G:H|=7$ (of order $3^2\cdot 5^2$), and similarly, there is a unique subgroup $K$ of $G$ which contains $P$, of index $5^2$ (of order $3^2\cdot 7$).

We want to show that there is a unique Sylow $5$-subgroup $Q$ of $G$, and that $Q\le H$.

Let $Q,Q^{\prime }$ be Sylow $5$-subgroups of $G$. Since $P⊴G$, $\mathit{𝑃𝑄}$ and $P{Q}^{\prime }$ are subgroups of $G$ which contain $P$. Moreover, since $|P|=3^2$ and $|Q|=|Q^{\prime }|=5^2$, $P\cap Q=P\cap Q^{\prime }=\{1\}$, thus $|\mathit{𝑃𝑄}|=|P{Q}^{\prime }|=3^2\cdot 5^2$. The unicity of $H$ as subgroup of index $7$ among the subgroups of $G$ which contain $P$ shows that

This shows that $Q\le H$ and $Q^{\prime }\le H$, so $Q$ and $Q^{\prime }$ are Sylow $5$-subgroups of $H$.

We count the Sylow $5$-subgroups of $H$, where $|H|=3^2\cdot 5^2$:

therefore

Since $Q$ and $Q^{\prime }$ are Sylow $5$-subgroups of $H$, this proves $Q=Q^{\prime }$, so there is only one Sylow $5$-subgroup of $G$: $n_5(G)=1$, so $Q⊴G$.

Similarly, let $R$ and $R^{\prime }$ be Sylow $7$-subgroup of $G$. Then $K=\mathit{𝑃𝑅}=P{R}^{\prime }$, so $R\le K$ and $R^{\prime }\le K$: $R$ and $R^{\prime }$ are Sylow $7$-subgroups of $K$, where $|K|=3^2\cdot 7$. Moreover

therefore

This proves $R=R^{\prime }$, thus $G$ has a unique Sylow $7$-subgroup $R$, and $R⊴G$. We know that

(Here $S=\mathit{𝑄𝑅}$.)

Now we prove that $G$ is abelian. By Theorem 9 of Section 5.4 (characterization of direct products), since $P⊴G$, $Q⊴G$ and $P\cap Q=\{1\}$,

Moreover $H⊴G$: if $h\in H$, then $h=\mathit{𝑝𝑞}$, where $p\in P$ and $q\in Q$. For all $g\in G$, $\mathit{𝑔h}{g}^{-1}=(\mathit{𝑔𝑝}{g}^{-1})(\mathit{𝑔𝑞}{g}^{-1})\in \mathit{𝑃𝑄}=H$. Moreover $R⊴G$ and $H\cap R=\{1\}$. Therefore $\mathit{𝑃𝑄𝑅}=(\mathit{𝑃𝑄})R=\mathit{𝐻𝑅}\simeq H\times R\simeq P\times Q\times R$. Thus $|\mathit{𝑃𝑄𝑅}|=|P|\cdot |Q|\cdot |R|=3^2\cdot 5^2\cdot 7=|G|$, so $\mathit{𝑃𝑄𝑅}=G$. We obtain

where $|R|$ is a prime, and $|P|$ and $|Q|$ are squares of primes, so $P,Q$ and $R$ are abelian. This proves that $G\simeq P\times Q\times R$ is abelian. □