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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 $G'/G''=\mathbb{Z}/2$. –  Joe Johnson 126 May 25 '11 at 11:31
    
Or are $G',G''$ supposed to be subgroups of a larger group? –  Joe Johnson 126 May 25 '11 at 11:34
    
@Joe: your description is really right. I don't think G′,G′′ need to be subgroups of any larger one. –  Basil R May 25 '11 at 11:42
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@Basil R: you should edit your title and question. –  ogerard May 25 '11 at 12:22
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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. –  Gerry Myerson May 25 '11 at 13:03

1 Answer 1

up vote 6 down vote accepted

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.

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You may want to correct this: If $G/Z(G)$ is cyclic then $G$ is abelian –  user9413 May 25 '11 at 14:22
    
@Chandru: thanks! –  Jack Schmidt May 25 '11 at 14:23
    
Thanks for everything. –  Basil R May 25 '11 at 16:46

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