Cyclic over central implies abelian

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., cyclic group) must also satisfy the second group property (i.e., epabelian group)
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Statement

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.

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.