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I have two questions regarding to $p$-groups.

  1. A $p$-group $G$ is said to be extraspecial of $G'=Z(G)$ has order $p$. Hence extraspecial groups are examples of $p$-groups with cyclic center. Of course there are many other $p$-groups with cyclic center that are not extraspecial. I would to know if there is any classification of $p$-groups with cyclic center.
  2. Is there any classification of $p$-groups of order $p^n$ and nilpotency class $k$ for suitable fixed $k$ and $n$?
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  • $\begingroup$ @1: The definition of extraspecialness is missing the condition, that G/Z(G) is elementary abelian. $\endgroup$ – Johannes Hahn May 26 '15 at 19:06
  • $\begingroup$ @ Johannes: I think the condition G'=Z(G) implies that G/Z(G) is elementary. $\endgroup$ – M.B May 26 '15 at 19:29
  • $\begingroup$ @Johannes Hahn: M.B. is right. See for example Robison, A course in the theory of groups, section "Extra-special p-groups". $\endgroup$ – tj_ May 26 '15 at 19:44
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    $\begingroup$ It was realized a while ago that it is more productive to classify finite $p$-groups by coclass rather than by class where, for $|P|=p^n$, the coclass of $P$ is defined to be $n - {\rm cl}(P)$. $\endgroup$ – Derek Holt May 26 '15 at 20:16
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    $\begingroup$ @M.B. $G'=Z(G)$ does not imply that $G/Z(G)$ is elementary abelian, but $G'=Z(G)$ of order $p$ does. I am sure that the answer to Qn 1 is no. $\endgroup$ – Derek Holt May 26 '15 at 20:27
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Every $p$-group is a homomorphic image of a $p$-group with cyclic center of order $p$, so a classification (whatever that means) of $p$-groups with cyclic center would (more-or-less) yield a construction for all $p$-groups, and I would not hold my breath waiting for that.

To see why a $p$-group $P$ is a homomorphic image of a $p$-group $G$ with center of order $p$, let $G$ be the regular wreath product of a cyclic group of order $p$ with $P$. Thus $G$ has an elementary abelian subgroup $E$ of order $p^{|P|}$, where $P$ permutes the cyclic factors of $E$ the way it permutes its own elements by right multiplication, and thus $P$ acts faithfully on $E$. Also, $G$ is the semidirect product of $E$ by $P$. It is easy to see that $E \cap {\bf Z}(G)$ has order $p$, so I need to show that every element of ${\bf Z}(G)$ lies in $E$. If $z \in {\bf Z}(G)$, write $z = au$, where $a \in E$ and $u \in P$. Since $z$ centralizes $E$ and $a$ centralizes $E$, it follows that $u$ centralizes $E$ and thus $u = 1$ by the faithfulness of the action. Thus $z = a \in E$, as required.

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    $\begingroup$ There are reasonably defined specific families of $p$-groups such that every $p$-group is a quotient of one elements of the family (for instance the family of restricted free groups of exponent some power of $p$). So given a family $F$ of $p$-groups proving that every $p$-group is a quotient of some member of $F$ doesn't mean that $F$ has no reasonable/useful description. (However in this specific case, I agree that I doubt we can expect any reasonable "parameterized" enumeration of $p$-groups of cyclic center.) $\endgroup$ – YCor Jan 26 '17 at 21:34
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This was mentioned as a brief comment by Derek Holt, but I think it deserves to be an answer. For a group of order $p^n$ and class $c$, the coclass is $n-c$. The known theory of classification by coclass is much richer than the known theory of classification by class; there is a nice account at http://www.ma.rhul.ac.uk/sepgm/Eick_Classification.pdf. Part of the story is that one can also define coclass for infinite pro-$p$-groups, which is useful for the classification. The isomorphism types of finite $p$-groups of fixed coclass can be assembled into a tree in a natural way, and the infinite paths from the root of the tree correspond to isomorphism classes of pro-$p$-groups.

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  • $\begingroup$ I am very fond of the coclass conjectures. However, one should be careful and emphasize that they say very little about most $p$-groups. In other words, the constant $f$ in conjecture A is so big that in most cases the normal subgroup that we get is just the trivial subgroup. $\endgroup$ – Yiftach Barnea May 27 '15 at 8:29
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    $\begingroup$ To Strickland. Please, let us know for which $k$ the groups of coclass are classified. Also, for $k>1$ the classification of $p$-groups of class $k$ is absent (it is known that the most $p$-groups have class $2$). $\endgroup$ – yakov Jul 10 '16 at 17:36
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To echo Marty's sentiment, a classification of $p$-groups based on nilpotency class and size seems far off. As an example of where this problem can get complicated, see

http://www.degruyter.com/view/j/jgth.2011.14.issue-6/jgt.2010.081/jgt.2010.081.xml.

Basically, Halasi and Palfy construct a collection of $p$-groups (allowing $p$ to vary) of nilpotency class 2 where the number of conjugacy classes is not polynomial in the prime $p$. This is related to the (unsolved) question of whether the number of conjugacy classes of $UT_n(\mathbb{F}_q)$, the group of unipotent upper-triangular matrices over the field with $q$ elements, is polynomial in $q$.

To summarize, $p$-groups, even those of nilpotency class 2, get complicated.

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I think that a ``proper" question which must be proposed instead of Question 1, is the following:

Is there a classification of finite $p$-groups $G$ such that both $G'$ and $Z(G)$ are cyclic?

The latter question is inspired by the example of extra-special groups as in this case both center and commutator subgroups are cyclic with further conditions.

The answer of the above question is yes: To start finding a series of papers answering the question you may see:

A. A. Finogenov, Finite $p$-Groups With Cyclic Commutator Subgroup and Cyclic Center, Matematicheskie Zametki, Vol. 63, No. 6, pp. 911-922, June, 1998.

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Classification of $p$-groups by nilpotency class is hard in general. As pointed above, coclass seems to be a better invariant. On the other hand, Ahmad, Magidin and Morse recently finished a classification of 2-generator $p$-groups of class 2:

Ahmad, Azhana(MAL-USM); Magidin, Arturo(1-LA); Morse, Robert Fitzgerald(1-EVAN-ECS) Two generator p-groups of nilpotency class 2 and their conjugacy classes. (English summary) Publ. Math. Debrecen 81 (2012), no. 1-2, 145–166.

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This is not quite what you were asking, since it involves the lower central series instead of the upper one, but Miech classified the $2$-generator $p$-groups with cyclic commutator subgroup and $p$ odd in:

Miech, R.J. On $p$-groups with a cyclic commutator subgroup J. Austral. Math. Soc. 20 no. 2 (1975), 178-198, MR0404441 (53 #8243).

Their nilpotency class can be arbitrarily large, though of course they are all metabelian. As far as I know, no similar classification exists for the $p=2$ case.

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(Addition to Isaacs' example) Let $G$ be a $p$-group and $d$ the minimal degree of representation of $G$ by permutations. Let $d<=p^n$. where $n$ is as small as possible. Then $G$ is a subgroup of $\Sigma_{p^n}$, a Sylow $p$-subgroup of the symmetric group of degree $p^n$. Since the center of $\Sigma_{p^n}$ has order $p$, we conclude that $G$ is a subgroup of a $p$-group with center of order $p$.

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  • $\begingroup$ Why -1? I gave a correct example. $\endgroup$ – yakov Aug 9 '16 at 18:22
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It is easy to prove that if a nonabelian $p$-group $G$ of exponent $>p$ contains $<p$ maximal abelian subgroups of exponent $>p$, then the Hughes subgroup of $G$ is abelian of index $p$.

PROBLEM. Classify the nonabelian $p$-groups of exponent $>p$ containing exactly $p$ maximal abelian subgroups of exponent $>p$.

If $G$ satisfies the problem, then $\exp(G)=p^2$ (Z. Janko). Moreover, all such $G$ are classified for $p=2$.

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    $\begingroup$ Is this an answer to the question? or is it a new question? $\endgroup$ – Gerry Myerson Jun 20 '16 at 22:54

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