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

This is from Michael Artin's algebra chapter 7 section 3. I'm quite unsure as to how I should start.

My first thought was to prove that $G/Z$ is isomorphic to some subgroup of $G$. Then we would have that $G \cong G/Z \times Z$, which is Abelian, assuming that $G/Z \cap Z = \{0\}$. However, I don't see a way to prove that such a subgroup of $G$ exists.

A second idea was to note that $G/Z = \{Z, xZ, (xZ)^2, \cdots, (xZ)^{n-1} \}$ for some $x\not \in Z$. Since $Z$ is the center, which is a normal group, we can say that $(xZ)^k = xZxZxZ\cdots xZ = x^kZ$ for all $k$. Sadly, this only guarantees an $x$ such that $x^n$ is in $Z$, rather than guaranteeing a subgroup, which would allow the first idea to follow.

Some help would be appreciated.

Tae Hyung Kim
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1 Answers1

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HINT: Use your idea of letting $xZ$ generate $G/Z$. Show that for any $a,b\in G$ there are $y,z\in Z$ and $m,n\in\Bbb Z$ such that $a=x^my$ and $b=x^nz$. Now calculate $ab$ and $ba$. (By the way, $G/Z$ need not be finite.)

Brian M. Scott
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