# A metacyclic group $G$ and its Sylow $p$-subgroup, where $p$ is the smallest prime dividing the order of $G$

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?

• 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. – Derek Holt Jun 23 '17 at 21:22
• As always Derek, thank you, will think about this tomorrow, it looks OK! – Nicky Hekster Jun 23 '17 at 22:06
• 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. – 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.

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.

• 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 $C_3 \times C_3$. – Mikko Korhonen Jul 2 '17 at 10:52
• you are definitely right. I could not see the mistake in the proof. – mesel Jul 2 '17 at 11:22
• 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 $G = PG'$. – Mikko Korhonen Jul 2 '17 at 11:41
• @MikkoKorhonen: Thank you. Let me check that whether I can fix the proof for the cae $p$ is the smallest prime. – mesel Jul 2 '17 at 12:06
• @MikkoKorhonen: I guess I fixed the argument. – mesel Jul 2 '17 at 23:32