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?







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


