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It is well-known that any finite $p$-group in which all its abelian subgroups are cyclic is either a cyclic group or a generalized quaternion group.

What can be said about $p$-groups in which every normal abelian subgroup is cyclic?

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2 Answers 2

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As a first-round answer and some detailed about this problem:

Let $G$ be a $p-$group of odd order such that every abelian normal subgroup has at most $k$ generators, then every subgroup of $G$ has at most $C(k+1,2)$ generators.
For your question, if we have $p>2$ and $k=1$, it is a classical result that $G$ is cyclic; see the thesis which I introduced below.

In the thesis "Abelian subgroups of $p-$groups" The thesis by Soo-Seng Siah, this problem and its generalization is studied with related to the term of "depth" and "normal-depth" of a $p-$group $G$. You can see the section 4 of this thesis for some more information.

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See Gorenstein, Finite Groups, Chapter 5, Theorem 4.10. Such groups are as follows:

  • if $p$ is odd, then $G$ is cyclic;
  • if $p=2$, then $G$ is either cyclic, or generalised quaternion of order $2^l$ for some $l\geq 3$, or dihedral of order $2^m$ for some $m\geq 4$ (note that $D_8$ does not satisfy the hypothesis), or semidihedral of order $2^n$ for some $n\geq 4$.
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