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I will be so thankful if someone helps me with the following question. There exists finite non-abelian p-groups G (except non-abelian groups of order $p^3$) with the following properties:

  1. all non-central elements have abelian centralizer.

  2. cs(G) has exactly two integers, where cs(G) is the set of all sizes of conjugacy classes of G.

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    $\begingroup$ Could you please give some background to this problem, such as where it originates, and why you need presentations for these groups? $\endgroup$
    – Yemon Choi
    Commented Aug 12, 2013 at 23:24
  • $\begingroup$ Have you seen this? plms.oxfordjournals.org.libproxy.cc.stonybrook.edu/content/… $\endgroup$
    – Steve D
    Commented Aug 13, 2013 at 2:56
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    $\begingroup$ @Steve: Your link attempts to proxy through the Stony Brook University libraries... I think the link you want is plms.oxfordjournals.org/content/s3-30/1/55.full.pdf $\endgroup$
    – Ricardo Andrade
    Commented Aug 13, 2013 at 10:48
  • $\begingroup$ @RicardoAndrade: And your link goes to a page which offers 1-day access to the article for US$39.00. $\endgroup$
    – Stefan Kohl ♦
    Commented Aug 13, 2013 at 13:05
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    $\begingroup$ @Stefan: Unfortunately, you are absolutely correct... :( Perhaps it would have been better to simply give the reference for the article: p-Groups with Abelian Centralizers, Proc. London Math. Soc. (1975) s3-30 (1): 55-75. $\endgroup$
    – Ricardo Andrade
    Commented Aug 13, 2013 at 21:06

2 Answers 2

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Let be $G=F/F^p[F,F,F]$, with $F$ denotes the free group on $n$ generators. Then $G$ satisfies $Z(G) = \Phi (G)=G'$.

If $x \in G-Z(G)$ then $C_G(x)= \langle x, Z(G) \rangle$ which is abelian as $C_G(x)/Z(G)$ is cyclic. Now we have your condition (1). For (2) observe that $C_G(x)$ has index $p^{n-1}$ in $G$, whenever $x \notin Z(G)$.

You can also consider any non-abelian $p$-groups (of order $>p^3$) with a center of index $p^2$. Clearly, for such a group the centralizer of any non-central element $x$ is a maximal subgroup, so has index $p$.

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  • $\begingroup$ I think yes. Ok I was thinking about a group with 2 non-central conjugacy classe sizes. $\endgroup$
    – Yassine Guerboussa
    Commented Aug 14, 2013 at 21:55
  • $\begingroup$ I modified the answer to deal with the case of two conjugacy class sizes. $\endgroup$
    – Yassine Guerboussa
    Commented Aug 14, 2013 at 22:33
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    $\begingroup$ Thank you Yassine. Every $2$-generator finite nonabelian p-group of class 2 such that $exp(G')=p$, satisfy above conditions. We know that G is of class at most 3. Is there a finite non-abelian p-group of class 3 such that satisfy above conditions? $\endgroup$
    – A.Mohammadian
    Commented Aug 16, 2013 at 19:41
  • $\begingroup$ You are right, in general if $M(G)$ is the subgroup generated by the elmts having a minimal (non-central) conj class size, then $M(G)$ has class at most $3$. If $p=2$ the bound can be decreased by $1$, so for your question you can assume that $p>2$. Also if the centralizers are normal in $G$ then $G$ must have class $\leq 2$ (this would be the case if you have {1,p} as a set of conj class sizes). But in general I have no idea about what the answer would be. $\endgroup$
    – Yassine Guerboussa
    Commented Aug 16, 2013 at 21:46
  • $\begingroup$ I think the following question is still open: having a set of powers of $p$ different from {1,p}, find a $p$ group of class >2 with this set of conj class sizes. Good luck. $\endgroup$
    – Yassine Guerboussa
    Commented Aug 16, 2013 at 21:53
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Any nonabelian $p$-group with center of index $p^2$ satisfies conditions 1 and 2. Such groups were studied by Isaacs-Passman.

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    $\begingroup$ (1) Groups whose center has index $p^2$ where already mentioned by Yassine Guerboussa's answer. What's the point of repeating others answers ? In my opinion this is very bad practice. (2) Extraspecial groups don't satisfy 1. in general: Take the extraspecial group of exponent $p$ (odd) and order $p^5$: $G=\langle x_1, y_1,x_2,y_2, z\mid x_i^p=y_j^p=1, [x_i,y_i]=z, [x_1,x_2]=[x_1,y_2]=[y_1,x_2]=[y_1,y_2]=1\rangle$. Hence $C_G(x_1) \supseteq \langle x_2,y_2\rangle$ contains the non-abelian extraspecial group of order $p^3$ and exponent $p$. $\endgroup$
    – tj_
    Commented Jul 10, 2016 at 17:40

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