If $G$ is a group of order $1575$ with normal Sylow $3$ subgroup, then show that Sylow $5$ and $7$ subgroups are both normal. This is Dummit & Foote's Exercise 4.5.28 of "Abstract Algebra".

If $G$ is a group of order $1575$ with normal Sylow $3$ subgroup, then show that Sylow $5$ and $7$ subgroups are both normal.

This question is answered here but I don't understand the answer. Why the preimage of Sylow $5$ subgroup of $G/P_3$ is normal in $G$?
 A: Some explanations are given by commenters so here's a simpler approach that involves semidirect products. First of all notice that $|G| = 3^2\cdot5^2\cdot 7$. Let $P \triangleleft G$ be the (unique) Sylow $3$-subgroup. Since $|P| = 3^2$ it has to be abelian so $P\simeq \mathbb{Z}_9$ or $P \simeq \mathbb{Z}_3 \times \mathbb{Z}_3$. Now let $P_5 \in Syl_5(G)$, $P_7 \in Syl_7(G)$ and $Q := P_5P_7$. Now observe that $|Q| = |P_5||P_7| = 5^2\cdot 7$ and that since $|PQ|=|P||Q| = |G|$ it must be the case that $G=P \rtimes Q$. So the possibilities for $G$ are determined by the homomorphisms $\varphi: Q \to \text{Aut}(P)$. Now consider the cases:

*

*$P\simeq \mathbb{Z}_9$: In this case, $\text{Aut}(P) \simeq (\mathbb{Z}_9)^*$ so $|\text{Aut}(P)|=\varphi(9) =6$ which is comprime to $|Q|$ so $\varphi$ has to be trivial and $G$ abelian.

*$P \simeq \mathbb{Z}_3 \times \mathbb{Z}_3$: Just like before, $\text{Aut}(P) \simeq GL_2(\mathbb{Z}_3)$ which has order $(3^2-1)(3^2-3)=8\cdot 6$ which is again coprime to $|Q|$ so again $\varphi$ has to be trivial and $G$ abelian.

In any case, $G$ has to be abelian so $G$ is abelian, and since every subgroup of an abelian group is normal both $P_5$ and $P_7$ have to be normal. Moreover, $G = P \times P_5 \times P_7$ so you can go one step further and find all the possibilities for $G$ (up to isomorphism).
A: Let $G$ have order $3^2\cdot 5^2\cdot 7$, and let $P$ be the normal Sylow $3$-subgroup. Let $Q$ be a Sylow $5$-subgroup and let $R$ be a Sylow $7$-subgroup. Write $n_p$ for the number of Sylow $p$-subgroups in $G$, and note that $n_5=|G:N_G(Q)|$ and $n_7=|G:N_G(R)|$.
Since $P$ is normal, $PQ$ and $PR$ are subgroups of $G$, of order $3^2\cdot 5^2$ and $3^2\cdot 7$ respectively. They have a unique Sylow $5$- and $7$-subgroup respectively as well, since $n_p\equiv 1\bmod p$ and divides the order of the group (so is $1$, $3$ or $9$, but only $1$ has the right congruence). Thus $P\leq N_G(Q)$ (and hence $PQ\leq N_G(Q)$) and $P\leq N_G(R)$ (and hence $PR\leq N_G(R)$). Then $n_5\mid |G:PQ|=7$ so is $1$, and the Sylow $5$-subgroup is normal. Similarly, $n_7\mid |G:PR|=25$ so is also $1$, and the Sylow $7$-subgroup is normal.
