Continuing my work through Dummit & Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in this related MSE answer):

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 $x^az$ for some $a \in \mathbb{Z}$ and some element $z \in Z(G)$]

The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write:
$g, h \in G$ implies that $g = x^{a_1}z_1$ and $h = x^{a_2}z_2$, so \begin{align*}gh &= (x^{a_1}z_1)(x^{a_2}z_2)\\\ &= x^{a_1}x^{a_2}z_1z_2\\\ & = x^{a_1 + a_2}z_2z_1\\\ &= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, $G$ is abelian.
1) Is this right so far?
2) How can I prove the "hint"?

  • 7
    $\begingroup$ That is correct. To prove the hint, just think that if $G/Z(G)$ is cyclic, then we can write $G/Z(G)=\langle xZ(G)\rangle$ for some $x\in G$. This means that for every $g\in G$, $gZ(G)=x^mZ(G)$ for some $m$, and thus $x^{-m}g\in Z(G)$. Do you see how to finish the proof? $\endgroup$
    – rfauffar
    Sep 9, 2011 at 12:16
  • 1
    $\begingroup$ @Robert: Yes, I think so. Where did the negative exponent come from? Would you want to make this comment a formal "answer"? $\endgroup$
    – Altar Ego
    Sep 9, 2011 at 12:22
  • 5
    $\begingroup$ possible duplicate of Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative $\endgroup$ Sep 9, 2011 at 15:33
  • 2
    $\begingroup$ While this result isn't hard to prove and makes a decent bit of intuitive sense, it has always struck me as a strange statement. $\endgroup$ Feb 11, 2016 at 11:48
  • 1
    $\begingroup$ @Sean, See math.stackexchange.com/questions/999247/… $\endgroup$
    – lhf
    Dec 12, 2022 at 20:10

4 Answers 4


We have that $G/Z(G)$ is cyclic, and so there is an element $x\in G$ such that $G/Z(G)=\langle xZ(G)\rangle$, where $xZ(G)$ is the coset with representative $x$. Now let $g\in G$.

We know that $gZ(G)=(xZ(G))^m$ for some $m$, and by definition $(xZ(G))^m=x^mZ(G)$.

Now, in general, if $H\leq G$, we have by definition too that $aH=bH$ if and only if $b^{-1}a\in H$.

In our case, we have that $gZ(G)=x^mZ(G)$, and this happens if and only if $(x^m)^{-1}g\in Z(G)$.

Then, there's a $z\in Z(G)$ such that $(x^{m})^{-1}g=z$, and so $g=x^mz$.

The hint is then proved, and the rest is identical to the work you did.

  • 4
    $\begingroup$ Many thanks! That was easy to follow. $\endgroup$
    – Altar Ego
    Sep 9, 2011 at 12:31
  • 2
    $\begingroup$ @Robert: I am extremely confused about one thing. A group is abelian if and only its its center is the whole group. Then isn't $G/Z(G)$ the trivial group in this case? $\endgroup$
    – user23238
    Mar 15, 2013 at 1:32
  • 1
    $\begingroup$ Why "by defintion $(xZ(G))^m=x^mZ(G)$? That should be proven it, shouldn't it? $\endgroup$
    – Twnk
    Mar 7, 2014 at 2:57
  • 4
    $\begingroup$ Dear Twink, if $H$ is a normal subgroup of a group $G$ and $x,y\in G$, then by definition $(xH)(yH)=xyH$. $\endgroup$
    – rfauffar
    Mar 7, 2014 at 12:24
  • 2
    $\begingroup$ if this result holds i.e.,$G/Z(G)$ is cyclic $\implies G$ is abelian $then$ $G=Z(G) \implies \vert G/Z(G) \vert =1$.Hence, in this case $G/Z(G) $ is a trivial group. $\endgroup$
    – Styles
    Jul 30, 2016 at 9:08

The following is another way to show the hint:

We know that the left cosets of $Z(G)$ partition the group $G$. So for all $g\in G$ there exists $n\in N, \ z\in Z(G)$ such that $x^nz=g$, where $xZ(G)$ generates $G/Z(G)$.

  • 8
    $\begingroup$ Follow up question. Should this be an if and only if statement? For if $G$ is abelian, then $Z(G) = G$. So $G/G \simeq {0}$. The 0 group is trivially cyclic. $\endgroup$
    – misogrumpy
    Jul 14, 2018 at 20:47

Here's another proof of the fact that, if $G/\textbf{Z}(G)$ is cyclic, then $G$ is abelian:

Since $G/\textbf{Z}(G)\cong \text{Inn}(G)$, then $\text{Inn}(G)$ is cyclic. So there exists some $g\in G$ such that every conjugation in $G$ is a product of conjugations of $g$. Now, let $x,y\in G$. We can write $$[x,y]=x^{-1}x^y=x^{-1}x^{g^k},$$ for some integer $k$. So it suffices to show that $x$ and $g$ conmute. But this is clear, since $$[g,x]=g^{-1}g^x=g^{-1}g^{g^l}=1,$$ for some $l\in \mathbb{Z}$.


Here is another proof of the following statement:

Let $G$ be a group, $N\leq G$ a subgroup of $G$. If $N\leq Z(G)$ and $G/N$ is cyclic then $G=Z(G)$.

Let $gN$ be a generator of $G/N$. Since $N\leq Z(G)$ clearly $N\subseteq C_{G}(g)$. By definition, $g\in C_{G}(g)$ as well. Hence $C_{G}(g)/N = G/N$. From the correspondance theorem it follows that $C_{G}(g) = G$, and hence $g\in Z(G)$. So again $Z(G)/N = G/N$ and therefore by the correspondance theorem again $Z(G) = G$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .