1
$\begingroup$

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 $25$?

$\endgroup$
4
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.