When $|G|=105$ and has a normal Sylow $3-$subgroup, then $G$ is abelian.

Question

$|G|=105=3\cdot 5\cdot 7$. We know that the Sylow $5-$subgroup $P_5$ must be normal and the Sylow $7-$subgroup $P_7$ must also be normal by simply counting elements. With the additional assumption of the Sylow $3-$subgroup $P_3$ being normal then we can say:

1.) $|G/C_G(P_3)|$ must divide both $|\operatorname{Aut}(P_3)|=2$ and $|G|=105$. This means that $$|G/C_G(P_3)|=1\quad\Longrightarrow\quad G=C_G(P_3)\quad\Longrightarrow\quad P_3\le Z(G).$$

2.) $|G/C_G(P_5)|$ must divide both $|\operatorname{Aut}(P_5)|=4$ and $|G|=105$. This means that $$|G/C_G(P_5)|=1\quad\Longrightarrow\quad G=C_G(P_5)\quad\Longrightarrow\quad P_5\le Z(G).$$

3.) $|G/C_G(P_7)|$ must divide both $|\operatorname{Aut}(P_7)|=6$ and $|G|=105$. This means that $$|G/C_G(P_7)|=1\quad\Longrightarrow\quad G=C_G(P_7)\quad\Longrightarrow\quad P_7\le Z(G).$$

Thus $\langle P_3,P_5,P_7\rangle\le Z(G)\le G$ and since $|\langle P_3,P_5,P_7\rangle|=105$, we must $|Z(G)|=105$ and we conclude $Z(G)=G$ and that $G$ is abelian.

Do I have the just of this right or am I making a conceptual error somewhere?

Answer 1

This all looks correct to me. It's worth checking out the hint in the comments for a quicker proof, but this technique is good.

Answer 2

If $G/Z(G)$ is cyclic, then $G$ is abelian and if $C_{G}(H) = G$, then $H \leq Z(G)$. Since $H$ is normal, let $G$ act on $H$ via conjugation. This action induces a homomorphism $\phi: G \rightarrow \text{Aut}(H) = \mathbb{Z}_{2}$ with $\ker \phi = C_{G}(H)$.

Since the order of $G$ is odd, we must have $|G/C_{G}(H)| = 1 \implies G = C_{G}(H) \implies H \leq Z(G)$. It follows that $|G/Z(G)| \in \{35, 5, 7\}$ and it follows that $G/Z(G)$ is cyclic.