MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I will be so thankful if someone help me about the following question. I need to know the presentation of a (if it is possible) family of finite non-abelian $p$-group $G$ with the follwing properties: 1- all non-central element have abelian centralizer. 2- $cs(G)$ has at least three integer, where by $cs(G)$ I mean the set of all conjugacy class sizes of $G$.

share|cite|improve this question
Is $G$ finite or not? – Boris Novikov Mar 29 '13 at 9:35
Do you want such a family for each possible prime $p$, or are you happy with an infinite family for, let's say, $p=2$? – Tom De Medts Mar 29 '13 at 9:48
Do you need $p$ to be fixed, because if not the wreath products $C_p\wr C_p$ works. – Steve D Mar 29 '13 at 10:04
@Steve: I am tempted to believe that $C_{p^n} \wr C_p$ might work in general... – Tom De Medts Mar 29 '13 at 10:30
The paper by David Rocke "p-groups with abelian centralizers" gives a full classification of the groups in question, although you will need to check which have $|cs(G)|>3$. Note that any $p$-group with abelian subgroup of index $p$ has all elements having abelian centralizers (prop 3.10.(a) of that paper) which confirms the comments of Tom and Steve regarding wreath products. – Nick Gill Mar 29 '13 at 15:29

The dihedral groups of order $2^n$ (with $n \geq 4$) form such a family. Indeed, for such a group, we have $$\operatorname{cs}(G) = \{ 1, 2, 2^{n-2} \} , $$ and they do have the required property that each non-central element has an abelian centralizer.


Here is another class of examples for arbitrary $p$, still with $\lvert\operatorname{cs}(G)\rvert = 3$ however.

Let $N$ be an arbitrary abelian $p$-group admitting a non-trivial action of $C_p$ (the cyclic group of order $p$), and let $G$ be the semidirect product $$ G = N \rtimes C_p .$$ Then I claim that all non-central elements of $G$ have abelian centralizer, and that $$\operatorname{cs}(G) = \{ 1, p, [N:Z(G)] \} . $$ There are three types of elements:

  1. elements $g \in Z(G)$. They necessarily lie in $N$.

  2. elements $g \in N \setminus Z(G)$. Such an element has conjugacy class of size at least $p$, but on the other hand these elements are of course centralized by $N$, so $C_G(g) = N$ and $\lvert g^G \rvert = p$.

  3. elements $g \in G \setminus N$. Notice that for such an element, $g^p \in Z(G)$. In this case, $C_G(g) = \langle g, Z(G) \rangle$, which is an abelian group of order $p \cdot \lvert Z(G) \rvert$. Hence the conjugacy class $\lvert g^G \rvert$ has size $[N: Z(G)]$ in this case.

An example of such groups is the wreath product $C_{p^n} \wr C_p$, but of course there are many more examples of this type.

share|cite|improve this answer
Yes, it is true. For other primes and $|cs(G)|>3$ what can we say? – Hamid Shahverdi Mar 29 '13 at 10:07
@ Tom, Thank you so much for your answer. Do you think that it is possible to construct a similar group $G$ that for $cs(G)$, we have $\{1,p^i,|N:Z(G)|\}$ for arbitrary $i$? – Hamid Shahverdi Mar 30 '13 at 9:26

Consider $p$-groups of maximal class with abelian subgroup of index $p$, and order at least $p^4$ (to satisfy second condition in question)

$C_p\wr C_p$ is one such group, but order of this (these) group(s) is(are) bounded by $p$, whereas, $p$-groups of maximal class, of order $p^n$, with abelian subgroup of index $p$, exists for all $p$ and all $n\geq 3$; the book of Leedham-Green and S. McKay has an interesting example (see link, Ex. 3.1.5, p. 53)

share|cite|improve this answer
Here is an infinite series of 3-groups satifying 1.and 2. Let G be a 3-group of maximal class and order $3^>3^4$ and suppose that $H=\text{C}_G(\text{Z}_2(G))$ is nonabelian; then H is minimal nonabelian. This G satifies condition 1. and 2. We have cs(G)=\{1,3,3^2,3^{n-2}\}. – yakov Jul 10 at 15:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.