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 $G$.
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$\begingroup$ Is $G$ finite or not? $\endgroup$– Boris NovikovCommented Mar 29, 2013 at 9:35
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$\begingroup$ Do you want such a family for each possible prime $p$, or are you happy with an infinite family for, let's say, $p=2$? $\endgroup$– Tom De MedtsCommented Mar 29, 2013 at 9:48
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1$\begingroup$ Do you need $p$ to be fixed, because if not the wreath products $C_p\wr C_p$ works. $\endgroup$– Steve DCommented Mar 29, 2013 at 10:04
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1$\begingroup$ @Steve: I am tempted to believe that $C_{p^n} \wr C_p$ might work in general... $\endgroup$– Tom De MedtsCommented Mar 29, 2013 at 10:30
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2$\begingroup$ 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. $\endgroup$– Nick GillCommented Mar 29, 2013 at 15:29
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2 Answers
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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.
Added.
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:
elements $g \in Z(G)$. They necessarily lie in $N$.
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 $\lvert g^G \rvert = p$.
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.
answered Mar 29, 2013 at 10:00
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$\begingroup$ Yes, it is true. For other primes and $|cs(G)|>3$ what can we say? $\endgroup$ Commented Mar 29, 2013 at 10:07
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$\begingroup$ @ 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 $i$? $\endgroup$ Commented Mar 30, 2013 at 9:26
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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)
