Exercise 3.1.36 (If $G/Z(G)$ is cyclic then $G$ is abelian)

Question

Prove that if $G/Z(G)$ is cyclic then $G$ is abelian. [If $G/Z(G)$ is cyclic with generator $xZ(G)$, show that every element of $G$ can be written in the form $x^a z$ for some integer $a \in \mathbb{Z}$ and some element $z \in Z(G)$.]

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Let $Z = Z(G)$ be the center of $G$. If $G/Z(G)$ is cyclic, then $G/Z(G) = \langle x Z angle$ for some element $\overline{x} = xZ \in G/Z(G)$, where $x \in G$. Since $(xZ)^k = x^k Z$ for all integers $k$, $G$ is the union of the cosets $x^k Z$, i.e.,
$$G = \bigcup_{k\in \mathbb{Z}} x^k Z.$$

If $a,b \in G$, then $a \in x^kZ$ for some integer $k \in \Z$, and similarly $b \in x^l Z$ for some $l \in \Z$. Therefore 
\begin{align*}
a &= x^k z,\
b &= x^l t,
\end{align*}
where $z \in Z$ and $t \in Z$.

Therefore
\begin{align*}
ab &= x^k z x^l t\
&=x^k x^l zt &	ext{(since $z \in Z$)}\
&=x^{k+l} zt,
\end{align*}
and similarly
\begin{align*}
ba &=  x^l t x^k z \
&= x^l x^k tz &	ext{(since $t \in Z$)}\
&= x^{k+l} zt.
\end{align*}
Therefore $ab = ba$. Since this is true for every $a \in G$ and every $b \in G$,
$$G 	ext{ is an abelian group}.$$

\end{proof}

Exercise 3.1.36 (If $G/Z(G)$ is cyclic then $G$ is abelian)

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Exercise 3.1.36 (If $G/Z(G)$ is cyclic then $G$ is abelian)

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