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

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 $$25$$?

• If you can use semidirect products it is easy, as there must be a non-trivial automorphism $\;Q\to\text{ Aut}\,(P)\;$ , but if $\;P=\Bbb Z_5\times\Bbb Z_5\;$ the the order of its automorphism group is $\;5\times 4=20\;$ ...Now, if you can't use semidirect products then I've no idea how to do this. – DonAntonio Nov 15 '19 at 21:48
• @DonAntonio I can't imagine OP isn't allowed to use semidirect products, or at least the semidirect product morphism. They are literally built for this. – Don Thousand Nov 15 '19 at 21:49
• @DonAntonio I wouldn't mind using semidirect products, so long as we are starting from scratch (i.e. not taking the linked result as an axiom) – Pascal's Wager Nov 15 '19 at 21:50
• Use $x\mid y$ for $x\mid y$. For comparison: $x|y$ renders as $x|y$. – Shaun Nov 15 '19 at 22:05

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.
• Ok, I see why $\operatorname{ker} (c)$ is trivial. Would you mind elaborating on why $P$ commutes with the subgroup generated with $H \cap P$? – Pascal's Wager Nov 15 '19 at 22:32
• $P$ is abelian because it is a group of cardinality $p^2$. Since $c$ is trivial (not its kernel — usually trivial kernels refer to injective morphisms), $H$ commutes with $P$. So let $C$ the subgroup made with all $x \in G$ that commute with $P$. $H < C$ so $3=|H|$ divides $|C|$. Similarly, $P < C$ so $25=|P|$ divides $|C|$. Finally, $75=|G|$ divide $|C| \leq |G|$, so $|C|=|G|$, so $C=G$ ie $P$ is central. – Mindlack Nov 15 '19 at 23:13