Restudying Marty Isaacs' book Finite Group Theory, Chapter 5 - Transfer, I thought of the following by working through some easy examples and I am wondering if it is true.

Suppose $G$ is finite and metacyclic in the sense that $G'$ and $G/G'$ are both cyclic. Let $P \in Syl_p(G)$, where $p$ is the smallest prime divisor of $|G|$. Does it follow that $P$ is cyclic?

I tried to prove that $P/P'$ is cyclic, since then we are done ($P' \subseteq \Phi(P)$). Tried to work through $P \cap G' \unlhd G$ and one shows that in fact $G=PC_G(P \cap G')$ and $G/C_G(P \cap G')$ is a cyclic $p$-group. But my analysis does not lead to anything useful. Any thoughts?

  • 1
    $\begingroup$ It's getting late, so I might not get this right, but will the following work? I think we can assume that $G'$ is a $p$-group, because we can factor out the $p'$-part. So $P \cap G' = G'$. Now let $Q$ be a Sylow $p$-complement of $C_G(G')$. Then $Q$ centralizes both $P/G'$ and $G'$ and so, since $|Q|$ is coprime to $p$, $Q$ centralizes $P$ and hence $G' = (QP)' = P'$ and $P/P'$ is cyclic. $\endgroup$ – Derek Holt Jun 23 '17 at 21:22
  • $\begingroup$ As always Derek, thank you, will think about this tomorrow, it looks OK! $\endgroup$ – Nicky Hekster Jun 23 '17 at 22:06
  • 1
    $\begingroup$ I think you can even shorten/skip your argument of the automorphism of a $p$-group: note that $C_G(G')$ is nilpotent (this is a general truth, since $[C_G(G')',C_G(G')] \subseteq [G',C_G(G')]=1$, so even of class $2$). Hence $Q \ char \ C_G(G') \unlhd G$, and $Q$ is normal. Because $Q \cap G'=1$, $Q$ must be central and the rest follows. $\endgroup$ – Nicky Hekster Jun 24 '17 at 9:11

Here is my comment expanded slightly. By factoring out Sylow $q$-subgroups of $G'$ for $q \ne p$, we can assume that $G'$ is a $p$-group, so $G' \cap P = G'$. Hence you have shown (using the fact that $p$ is the smallest prime dividing $|G|$) that $G = PC_G(G')$.

Let $Q$ be a Sylow $p$-complement of $C_G(G')$. So $Q$ is also a Sylow $p$-complement of $G$, and $G=PQ$. Since $Q \cap G' = 1$, $Q$ must be abelian (in fact cyclic).

Since $G/G'$ is abelian, $Q$ centralizes $P/G'$ and $Q$ centralizes $P'$ from its definition. Now any automorphism of a finite $p$-group $P$ that centralizes a normal subgroup $N$ of $P$ and induces the identity on $P/N$ must have order a power of $p$. This is a standard result, and is not hard to prove. So in fact $Q$ centralizes $P$, and hence $Q \le Z(G)$.

So $G' = (PQ)' = P'$, and hence $P/P'$ is cyclic, which imples that $P$ is cyclic.

| cite | improve this answer | |

I would like to give another solution;

$G/G'=<xG'>$ for $x\in G$. It is easy to see that $G'=[x,G']$.

Now the map $\phi:G'\to G'$ by $g\mapsto [g,x]$ is an homomorphism as $G'$ is abelain. More clealrly,($[gh,x]=[g,x]^h[h,x]=[g,x][h,x]$)

Since $\phi(G')=G'$, $Ker(\phi)=1=C_{G'}(x)$. Now, let $p$ the smallest prime dividing the order of $G$.

Let $Q\in Syl_p(G')$ then we have $C_Q(x)=1$. Note that $p'$ part of $x$ act trivially on $Q$ as $|Aut(Q)|=p^{n-1}(p-1)$ and $p$ is the smallest prime. Moreover $p$-part of $x$ definitly fixes something on $Q$ if $Q\neq 1$. Thus, we have $Q=1$. Let $P\in Syl_p(G)$ then $G'\cap P=1\implies$ $P$ is cyclic and $G$ is $p$-nilpotent.

| cite | improve this answer | |
  • 1
    $\begingroup$ But I think the result is not true if $p$ is not the smallest prime. For example for $G = C_3 \times S_3$, we have $G'$ cyclic of order $3$ and $G/G'$ cyclic of order $6$; however a $3$-Sylow of $G$ is $C_3 \times C_3$. $\endgroup$ – Mikko Korhonen Jul 2 '17 at 10:52
  • $\begingroup$ you are definitely right. I could not see the mistake in the proof. $\endgroup$ – mesel Jul 2 '17 at 11:22
  • $\begingroup$ In the example I gave, for a $3$-Sylow $P$ the subgroup $PG' = P$ does not have $P/P'$ cyclic, so you cannot reduce to the case $G = PG'$. $\endgroup$ – Mikko Korhonen Jul 2 '17 at 11:41
  • $\begingroup$ @MikkoKorhonen: Thank you. Let me check that whether I can fix the proof for the cae $p$ is the smallest prime. $\endgroup$ – mesel Jul 2 '17 at 12:06
  • $\begingroup$ @MikkoKorhonen: I guess I fixed the argument. $\endgroup$ – mesel Jul 2 '17 at 23:32

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.