If the Quotient by the Center is Cyclic, then the Group is Abelian

Problem 18

Let $Z(G)$ be the center of a group $G$.
Show that if $G/Z(G)$ is a cyclic group, then $G$ is abelian.

Contents

Steps.

1. Write $G/Z(G)=\langle \bar{g} \rangle$ for some $g \in G$.
2. Any element $x\in G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.
3. Using this expression, show that $xy=yx$ for any $x, y \in G$.

Proof.

Since the quotient group $G/Z(G)$ is cyclic, it is generated by one element.
Let $g \in G$ be an element such that $\bar{g}=gZ(G)$ is a generator of $G/Z(G)$. Namely, $\langle \bar g \rangle =G/ Z(G)$.

Then for any element $x \in G$, we have $\bar{x}\in G/Z(G)=\langle \bar g \rangle$ and hence $\bar{x}=\bar{g}^a$ for some $a \in \Z$.
It follows that any element of $G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.

Take any two elements $x, y \in G$ and write $x=g^a z$ and $y=g^b w$, where $a, b \in \Z$ and $z, w \in Z(G)$.

Then we claim that $xy=yx$. To see this we calculate as follows.
\begin{align*}
xy&= g^a z g^b w\\
&=g^a g^b z w && \text{since $z\in Z(G)$}\\
&= g^{a+b} wz && \text{since $w\in Z(G)$}\\
&=g^bg^awz\\
&=g^b w g^az&& \text{since $w\in Z(G)$}\\
&=yx.
\end{align*}

Since the product of any two elements of $G$ is commutative, we conclude that $G$ is abelian.

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3 Responses

1. 07/31/2016

[…] Use the result of the problem If the quotient by the center is cyclic, then the group is abelian. […]

2. 06/09/2017

[…] of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group. Recall that if the quotient by the center is cyclic, then the group is abelian. Thus the group $G$ is abelian, which again a […]

3. 06/18/2017

[…] If the quotient by the center is cyclic, then the group is abelian (post 2) […]

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