I will prove this by the contrapositive:

If $G/Z(G)$ is cyclic then $G$ is abelian.

Proof: We assume that $G/Z(G)$ is cyclic. This means it is generated by a left coset $(aZ(G))^n=e$ for some integer $n$. By defined operation $a^n Z(G)=e$.
Let $x,y \in G$ and $z_0, z_1 \in Z(G)$, so this leads to $$a^n z_0=x,\ a^m z_1=y.$$.
This implies: $$xy=a^n z_0,\ a^m z_1 = a^{nm}(z_0 z_1)=a^m z_1,\ a^n z_0=yx.$$

This what I've been told is the write answer but I am wondering however, why is it necessary to let $x,y \in G$ and set our two elements $a^n z_0$ and $a^m z_1$ to elements $x, y \in G$? Also is the fact that $a^n Z(G)=e$ wrong?

Thanks in advance.

  • 2
    $\begingroup$ The statement that $(aZ(G))^n=e$ for some $n$ is not true when $G/Z(G)$ is infinite, and it is not used in the proof anyway, so you can just delete that sentence, but you need to say that $G/Z(G)$ is generated by a coset $aZ(G)$. At the moment, you are using $n$ twice, with different meanings. $\endgroup$
    – Derek Holt
    Oct 31, 2014 at 1:13

2 Answers 2


Your proof is correct (except, as the comment points out, for the statement that $aZ(G)$ has finite order).

You need to start with two arbitrary elements $x, y\in G$ because you want to show that $G$ is abelian --- that is, that for any two elements of $G$, that $xy = yx$. So you start the proof by picking two arbitrary elements.

Next, the only way to make progress is to recognize that $x$ and $y$ are each in some $Z(G)$-coset; since the quotient is cyclic, that means that they are in the cosets $a^mZ(G)$ and $a^nZ(G)$. You aren't really setting $x$ and $y$ to those values; rather, you are recognizing the fact that they are in these cosets and using that to manipulate the values to prove what you want.


If $G/Z(G)$ is cyclic, then there is $g\in G$ such that, for every $a\in G$, there is an integer $n$ such that: $$aZ(G)=(gZ(G))^n=g^nZ(G)\iff g^{-n}a\in Z(G)$$ Therefore, for every $a,b\in G$, there are integers $n$ and $m$, and central elements $z$ and $z'$, such that: $$ab=g^nzg^mz'=g^ng^mzz'=g^mg^nz'z=g^mz'g^nz=ba$$ So, $G$ is Abelian (and hence the only cyclic quotient $G/Z(G)$ is the trivial one).

  • $\begingroup$ There is no contradiction. The trivial group is cyclic. $\endgroup$ Jun 8, 2022 at 5:38
  • 2
    $\begingroup$ This is a proof by fake contradiction. You have a direct proof that if $G/Z(G)$ is cyclic, then $G$ is abelian. Then you add a useless assumption that $G$ is non-abelian at the top, and add an unnecessary line claiming a contradiction with the assumption you just added at the bottom. The only time you use the assumption that $G$ is non-abelian is to say $g\notin Z(G)$, and to claim a contradiction at the end, which is why I call it "useless": it is literally never used in the actual argument. You do not need $g\notin Z(G)$. This is never used either. $\endgroup$ Jun 8, 2022 at 5:44

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