# If $G/Z(G)$ is cyclic, then $G$ is abelian

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"?

• 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? Sep 9, 2011 at 12:16
• @Robert: Yes, I think so. Where did the negative exponent come from? Would you want to make this comment a formal "answer"? Sep 9, 2011 at 12:22
• ok, I'll explain it better Sep 9, 2011 at 12:24
• possible duplicate of Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative Sep 9, 2011 at 15:33
• 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. Feb 11, 2016 at 11:48

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.

• Many thanks! That was easy to follow. Sep 9, 2011 at 12:31
• @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?
– user23238
Mar 15, 2013 at 1:32
• Why "by defintion $(xZ(G))^m=x^mZ(G)$? That should be proven it, shouldn't it? Mar 7, 2014 at 2:57
• Dear Twink, if $H$ is a normal subgroup of a group $G$ and $x,y\in G$, then by definition $(xH)(yH)=xyH$. Mar 7, 2014 at 12:24
• 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. 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 $G/Z(G)$.

• 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. Jul 14, 2018 at 20:47

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 $$Z(G) = G$$.