Cyclic over central implies abelian

Straightforward formulation
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.

In terms of cyclic and epabelian groups
Any cyclic group is an epabelian group.

Applications

 * A nontrivial cyclic group is not a capable group: it cannot be realized as the quotient of a group by its center.
 * 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 $$\langle H, a \rangle$$ contains $$H$$ and intersects every coset of $$H$$ in $$K$$. Hence, $$\langle H, a \rangle = K$$.


 * 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 $$\langle H, a \rangle = K$$, 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, $$\langle H, a \rangle$$ is in the center of $$K$$, which in turn is contained in $$K$$. But $$\langle H, a \rangle = K$$, so $$K$$ equals its own center.