Let $G$ be a group and let $G'$ denotes its commutator subgroup, that is the group generated by all elements of the form $g^{-1}h^{-1}gh$. Is the following true:

If $G/G'$ is cyclic, then G is abelian.

Recall that the claim is true for the $G$ modulo the center of $G$. See If $G/Z(G)$ is cyclic, then $G$ is abelian

  • $\begingroup$ but my problem is if G/G′ is cyclic then G is abelian. so my prove is incorrect?, $\endgroup$ – Rachel Oct 9 '13 at 18:51
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    $\begingroup$ No. The statement is not correct, so your proof is also not correct. $\endgroup$ – Tobias Kildetoft Oct 9 '13 at 18:54
  • $\begingroup$ but the exercise is not correct ? $\endgroup$ – Rachel Oct 9 '13 at 18:59
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    $\begingroup$ If the exercise asks you to prove this, then no, it is not correct. Take the smallest non-abelian group to get an example of why. $\endgroup$ – Tobias Kildetoft Oct 9 '13 at 19:00
  • $\begingroup$ (For why your proof fails, you change the order of $z$ and $x^b$ at some point with no mention of why that would be allowed, and then you do the same with $z$ and $w$). $\endgroup$ – Tobias Kildetoft Oct 10 '13 at 6:29

The fact is that in general the question is false, but if one add the hypothesis, that the group $G$ is nilpotent (e.g. a $p$-group for some prime $p$), then the result become true:

Indeed, we have the chain $\gamma_{3}(G)\subset \gamma_{2}(G)\subset G$, where $\gamma_{2}(G)= G'$ and $\gamma_{3}(G)=[\gamma_{2}(G), G]$ (the commutator). Now from the last relation we have $\frac{\gamma_{2}(G)}{\gamma_{3}(G)}=Z( \frac{G}{\gamma_{3}(G)} ) $ (the center); so the quotient $\frac{\frac{G}{\gamma_{3}(G)}}{Z( \frac{G}{\gamma_{3}(G)} ) }=\frac{G}{\gamma_{2}(G)}$ and this is cyclic, and as someone have remarked above, this implies that $\frac{G}{\gamma_{3}(G)}$ is abelian and then $G'=\gamma_{2}(G)\subset \gamma_{3}(G)$ and finally $\gamma_{2}(G)= \gamma_{3}(G)$. Since $\gamma_{3}(G)=[\gamma_{2}(G),G]=[\gamma_{3}(G),G]=\gamma_{4}(G)$ and so on... But in last we know that $G$ is nilpotent and so there is a $c \in \mathbb{N}^{+}$ such that $\gamma_{c}(G)=1$ and so $G'=1$ that means $G$ is abelian.

For the general case one can see to the alternating group $A_{4}$: indeed one can prove that $|A_{4}|=12$, $|A'_{4}|=4$ so that the quotient $\frac{A_{4}}{A'_{4}}$ has 3 element and it is cyclic; but $A_{4}$ is not abelian.

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    $\begingroup$ You could add in your answer a counterexample for the general case, so that one doesn't need to go back to the comments for it. Almost you might consider removing the first paragraph of your post: in 2-3 years, who will care that your answer was posted 3 years after the question was asked? $\endgroup$ – Najib Idrissi Aug 22 '16 at 15:53
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    $\begingroup$ @Najib Idrissi; I do it! thanks! $\endgroup$ – The Number Theorist Sep 15 '16 at 13:06
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    $\begingroup$ Summarizing: generally, the argument shows that if $G$'s abelianization is cyclic, then the lower central series is stationary after the commutator. If it is nilpotent, this forces $G$ to be abelian (and therefore cyclic). $\endgroup$ – Ben Blum-Smith Sep 15 '16 at 13:18

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