58
$\begingroup$

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"?

$\endgroup$
  • 2
    $\begingroup$ 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 '11 at 12:16
  • $\begingroup$ @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 '11 at 12:22
  • $\begingroup$ ok, I'll explain it better $\endgroup$ – rfauffar Sep 9 '11 at 12:24
  • 5
    $\begingroup$ possible duplicate of Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative $\endgroup$ – Jonas Meyer Sep 9 '11 at 15:33
  • $\begingroup$ 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. $\endgroup$ – Sean English Feb 11 '16 at 11:48
66
$\begingroup$

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 $(x^m)^{-1}g\in Z(G)$. There then exists 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.

$\endgroup$
  • 2
    $\begingroup$ Many thanks! That was easy to follow. $\endgroup$ – Altar Ego Sep 9 '11 at 12:31
  • $\begingroup$ @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 '13 at 1:32
  • $\begingroup$ Yes, the proof shows then that $G/Z(G)$ is actually trivial. $\endgroup$ – rfauffar Mar 20 '13 at 19:07
  • 2
    $\begingroup$ 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 '14 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 '16 at 9:08
4
$\begingroup$

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 $G/Z(G)$.

$\endgroup$
  • 1
    $\begingroup$ 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 '18 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.