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 $25$?

  • $\begingroup$ 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. $\endgroup$ – DonAntonio Nov 15 '19 at 21:48
  • $\begingroup$ @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. $\endgroup$ – Don Thousand Nov 15 '19 at 21:49
  • $\begingroup$ @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) $\endgroup$ – Pascal's Wager Nov 15 '19 at 21:50
  • $\begingroup$ Use $x\mid y$ for $x\mid y$. For comparison: $x|y$ renders as $x|y$. $\endgroup$ – 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.

  • $\begingroup$ Ok, I see why $\operatorname{ker} (c)$ is trivial. Would you mind elaborating on why $P$ commutes with the subgroup generated with $H \cap P$? $\endgroup$ – Pascal's Wager Nov 15 '19 at 22:32
  • $\begingroup$ $P$ 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. $\endgroup$ – Mindlack Nov 15 '19 at 23:13

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