= Solution
{c}
Suppose the <quotient group> $G/A$ is a <cyclic group>, generated by $gA$. For any $x,y\in G$, there are <integers> $r,s$ and $a,b\in A$ such that
$$
x=g^ra,\qquad y=g^sb.
$$
Since $a,b$ lie in the <center of a group>, they <commute> with every element, in particular with each other and with the <integer powers of a group element> $g$. Therefore
$$
xy=g^rag^sb=g^{r+s}ab=g^{s+r}ba=g^sbg^ra=yx.
$$
As $x,y$ were arbitrary, \b[$G$ is an <abelian group>]. This is the <cyclic quotient by the center>.
Back to article page
