Let $G$ be a $p$ group. If $G/G’$ is cyclic, prove $G/Z(G)$ is cyclic by induction.
I tried playing with the base cases : if $|G|=p $ or $p^2$, $G $ is abelian, so $G/Z(G)=G/G =\{G\}$ which is generated by $1G$.
If $|G|=p^3$, the center is non trivial (because it is a p-group). If $Z(G)=p^3$, $G/Z(G)=\{1G\}$ which is cyclic. If $Z(G)=p^2$, we get $G/Z(G)$ having order $p$ and being cyclic, which implies an abelian $G$. This is absurd. If $Z(G)=p$, $G/Z(G)$ is abelian so $G’\subset Z(G)$. Because $G$ is non abelian the comutator is non trivial and by Lagrange’s theorem $|G’|=p\Rightarrow Z(G)=G’$ so $G/Z(G)=G/G’$ is cyclic.
This doesn’t seem to generalize nicely. Any ideas on the inductive step?
Good work so far! Here's one way to run the inductive step:
Suppose $G/G'$ is cyclic. The canonical projection $G \to G/Z(G)$ is surjective and sends commutators to commutators, so it descends to a surjective homomorphism $G/G' \to (G/Z(G))/(G/Z(G))'$. Since $G/G'$ is cyclic, $(G/Z(G))/(G/Z(G))'$ must be cyclic as well.
Now $G/Z(G)$ is a $p$-group of strictly smaller cardinality (because $G$ has nontrivial center), so we can conclude (by inductive hypothesis) that $(G/Z(G))/Z(G/Z(G))$ is cyclic.
This implies that $G/Z(G)$ is abelian, by a standard argument.
Now this means that the canonical projection $G \to G/Z(G)$ factors through $G/G'$. Now we have a surjective homomorphism from the cyclic group $G/G'$ to $G/Z(G)$, so $G/Z(G)$ is cyclic.