Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 1D (Groups)

Suppose $G/Z(G)$ is generated by $gZ(G)$. Every element of $G$ can then be written $g^{a}z$ with $z\in Z(G)$. HenceThus the cyclic quotient by the center condition forces $G$ to be abelian.

If $|H/Z(H)|=7$, the quotient has prime order and is therefore cyclic. The result just proved would make $H$ abelian, so $Z(H)=H$ and the quotient would have order one, a contradiction. Therefore

Such a group does exist. Take $H=S_3$. Its center is trivial, sohas order six. Therefore