Subgroup of a metacyclic group is metacyclic

Question

Here I am referring to the older definition of metacyclic, from Rose, where

$G$ is metacyclic if its commutator subgroup $G'$ is cyclic and $G/G'$ is cyclic.

This is stronger than the typical definition of metacyclic but not equivalent. I am struggling to prove that subgroups of metacyclic finite groups (using the above definition) are themselves metacyclic in the same sense. Certainly given $H \subseteq G$ the commutator of $H$ is a subgroup of $G'$ and therefore cyclic, but is $H/H'$ cyclic? I don't see a way to show why.

Answer 1

Here is a counterexample. Let $G$ be the group of order $54$ defined by the presentation $$\langle x,y \mid x^9=y^6=1,y^{-1}xy=x^2 \rangle.$$

(This is $\mathtt{SmallGroup}(54,6)$ in GAP and Magma.)

Then $G' = \langle x \rangle$, and $G/G' \cong \langle y \rangle$ are both cyclic.

But $P \in {\rm }Syl_p(G)$ is a nonabelian group of order $27$ with $|P'|=3$ and $P/P'$ elementary abelian of order $9$. So $P/P'$ is not cyclic.

Answer 2

Let us call the property quasimetacyclic, to make a difference with metacyclicness. Partial answer: if $H \cap G'=H'$, then $H$ is again quasimetacyclic: observe that $G'H/G' \cong H/(H \cap G')$ and $G'H/G'$ as a subgroup of $G/G'$ is cyclic. But I am sure that there must be a counterexample for the general case.