Non-abelian group of order 75 must have subgroup $P \cong Z_5 \times Z_5$

Question

Let $G$ be a non-abelian group of order $75$, and let $P$ be a Sylow-5 subgroup. Must we have $P \cong Z_5 \times Z_5$?

I think the answer is yes, based on this, but I am not supposed to use that. (I'm actually trying to prove a lemma to help me prove that very same result).

I see that $n_5 \equiv 1$ (mod 5) and $n_5|3$, hence $n_5=1$ so $P \unlhd G$.

I also see $n_3 \equiv 1$ (mod 3) and $n_3|5^2$, hence $n_3 \in \{1,25\}$. We can't have $n_3=1$, otherwise $G \cong P \times Q$ where $Q$ is the unique Sylow-3 subgroup, which would imply that $G$ is abelian since $P$ and $Q$ are. Therefore, $n_3=25$.

Now, by the Fundamental Theorem of Finitely Generated Abelian groups, $P \cong Z_5 \times Z_5$ or $P \cong Z_{25}$. My thought was to assume $P \cong Z_{25}$ and try to derive a contradiction. Why can't $G$ have an element of order $25$?

Answer 1

Let $H$ be any of the $3$-Sylow. $H \cong \mathbb{Z}/(3)$ (additively).

There is a natural mapping $c: H \rightarrow Aut(P)$ by conjugation (ie $c(h)$ is $p \in P \longmapsto hph^{-1} \in P$).

Assume $P \cong \mathbb{Z}/(25)$. Then $A=Aut(P) \cong (\mathbb{Z}/(25))^{\times} \cong \mathbb{Z}/(20)$. So $|H|$ and $|A|$ are coprime, thus there is no non-trivial morphism between them, hence $c$ is trivial. Thus, $P$ commutes with the subgroup generated with $H \cup P$, which is $G$ for cardinality reasons, so $P$ is central.

Thus the quotient of $G$ by its center $Z$ is either $3$ or $1$. Since $G/Z$ is cyclic iff it is trivial, $G=Z$, a contradiction.