Cyclic over central implies abelian

Statement

Suppose $H\le K\le G$ are groups, such that $H$ is a central subgroup of $G$ (in other words, $H$ is contained in the center of $G$), and $K/H$ is cyclic. Then $K$ is an abelian subgroup of $G$, i.e., it is Abelian as a group.

Related facts

Applications

• A nontrivial cyclic group is not a capable group: it cannot be realized as the quotient of a group by its center. For full proof, refer: Cyclic and capable implies trivial
• Characteristically metacyclic and commutator-realizable implies abelian

Proof

Given: A group $G$, subgroups $H\le K\le G$. $H$ is in the center of $G$, and $K/H$ is cyclic.

To prove: $K$ is abelian.

Proof: Suppose $\overline{a}$ is a generator of $K/H$ and $a$ is an element of $K$ whose image mod $H$ is $\overline{a}$. Then, we have $⟨H,a⟩$ contains $H$ and intersects every coset of $H$ in $K$. Hence, $⟨H,a⟩=K$.

1. $H$ is in the center of $K$: This follows from the fact that $H$ is in the center of $G$.
2. $a$ is in the center of $K$: The centralizer of $a$ in $G$ contains $a$, and also contains $H$, since $H$ is in the center of $G$. Hence, the centralizer of $a$ contains $⟨H,a⟩=K$, so $a$ is in the center of $K$.
3. The center of $K$ is $K$, and hence $K$ is abelian: From the previous two steps, $⟨H,a⟩$ is in the center of $K$, which in turn is contained in $K$. But $⟨H,a⟩=K$, so $K$ equals its own center.