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There is a claim saying that if both $G'/G''$ and $G''$ are cyclic groups, then $G''=1$, where $G'$ is the derived subgroup of the group $G$. I have been thinking of this by focusing the N/C Lemma to clear the problem for myself. I need a useful igniting hint(s). Furthermore, may I ask: are these kinds of groups well known? Of course, any group satisfying the above conditions will be metabelian and obviously is soluble.

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    If $G'=\mathbb{Z}/4$ and $G''=\mathbb{Z}/2$ then $G'/G''=\mathbb{Z}/2$.2011-05-25
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    Or are $G',G''$ supposed to be subgroups of a larger group?2011-05-25
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    @Joe: your description is really right. I don't think G′,G′′ need to be subgroups of any larger one.2011-05-25
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    $G''=Z_p$, $G'=Z_p\times Z_q$, $G'/G'' = Z_q$. ;-)2011-05-25
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    Are you sure $G'$ and $G''$ are not supposed to be $[G,G]$ and $[G',G']$? Otherwise, it is clearly not true.2011-05-25
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    What is $G'$ and what is a calim?2011-05-25
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    @Tobias: @Qiaochu: You are right. G′ and G′′ are [G,G] and [G′,G'] reletively.2011-05-25
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    @Basil R: you should edit your title and question.2011-05-25
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    $G''=1$ is the same thing as saying $G'$ is abelian, so you're trying to prove that if $G'/G''$ and $G''$ are both cyclic then $G'$ is abelian.2011-05-25
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    @Gerry: You said what I exactly meant. :)2011-05-25

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This is theorem 9.4.2, page 146, in M. Hall's textbook on the Theory of Groups.

You are going in the right direction. It uses the N/C theorem, as in, the normalizer modulo the centralizer is a subgroup of the automorphism group.

Another hint: It is very similar to showing "If G/Z(G) is cyclic, then G is abelian.".


These groups were known as "metacyclic groups" for a few decades, though the name is now used slightly differently.

A special case where G′ and G/G′ have coprime order is very special: these are exactly the groups in which all Sylows are cyclic. They are also known as "Z-groups", though again the name means different things to different people.