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A finite, non-abelian group $G$ is said to be a center commutative-transitive group $($or a CCT-group, for short$)$ if commutativity is a transitive relation on the set on non-central elements. In other words, if $x, \, y, \, z \in G-Z(G)$ and $[x, \, y]=[y, \, z]=1$, then $[x, \, z]=1$.

A quick search with GAP4 allows to prove the following

Proposition. Let $G$ be a non-abelian finite group with $|G| <32$; then $G$ is a CCT-group, unless it is isomorphic to $S_4$. In the case $|G|=32$, there are precisely seven groups that are not CCT: the two extra-special groups (whose nilpotency class is $2$), and five further groups having nilpotency class $3$.

I tried to give a computer-independent proof, but it turned very soon into a lenghty and messy case-by case analysis. So let me ask the following

Question.

  1. Is there any short conceptual proof of the proposition above?
  2. Does the result in the proposition above appear somewhere in the literature?
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    $\begingroup$ It's quite easy to check for groups whose order is a product of at most 3 primes (equal or not). Below $32$ this only excludes $16$ and $24$. $\endgroup$
    – YCor
    Sep 8, 2020 at 12:55
  • $\begingroup$ May I ask you your argument? $\endgroup$ Sep 8, 2020 at 14:38
  • $\begingroup$ @YCor: oh, right, thanks. In fact, a characterization of CCT-groups is that the centralizer of any non-central element is abelian, and this is true (by your argument) for groups whose order is the product of at most three primes. $\endgroup$ Sep 8, 2020 at 14:54
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    $\begingroup$ Eventually I posted an answer, also covering order $p^4$. $\endgroup$
    – YCor
    Sep 8, 2020 at 14:57
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    $\begingroup$ The paper D. M. Rocke, p-groups with abelian centralizers, Proc. London Math. Soc. 30 (1975), 55-75, describes, to a certain extent, finite $p$-groups with your property. Abdollahi et al. (Non-commuting graph of a group, J. Algebra 298 (2006), 468-492) call these groups AC-groups, and give some information about them. Otherwise, CCT-groups are precisely groups with the property that all centralizers of non-central elements are abelian. This can be used together with GAP to efficiently scan through the database of small groups and look for the CCT ones. $\endgroup$
    – Primoz
    Sep 9, 2020 at 8:21

2 Answers 2

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Proposition.

(a) Let $G$ be a group of order a product of $\le 3$ primes (possibly equal). Then $G$ is CCT (the smallest numbers not of this form are $16$, $24$, $32$, $36$, $48$, $54$, $60$).

(b) Let $G$ be a group of order $p^4$, $p$ prime. Then $G$ is CCT.

Proof. (a) Suppose by contradiction that $G$, of order a product of $\le 3$ (possibly equal) primes is not CCT. So there are $x,y,z$ with $y$ non-central, such that $x,z$ don't commute and both commute with $y$. Let $C$ be the centralizer of $y$: its order is a product of at most two primes, and it has a nontrivial center (as $1\neq y\in C$), so $C$ is abelian. Contradiction as $x,z\in C$.

(b) Let $x,y,z$ be as in (a). Then they generate a non-abelian subgroup $N$, which is not all of $G$, since $y$ is not central in $G$. Hence $N$ has order $p^3$, and hence its center $Z$ has order $p$. As a subgroup of index $p$ in a finite $p$-group, it is normal. The $G$-action by conjugation on $N$ preserves $Z$, whose automorphism group has order $p-1$. Hence the $G$-action on $Z$ is trivial, so $y$ is central, contradiction.

Corollary. Every group of order $<32$ and $\neq 24$ is CCT.

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Well, you are looking at small groups, so you might expect to have to do some case-by case analysis. But here are a few general remarks. A direct product $A \times B$ is a CCT group if and only either if $A,B$ both are, or if one of $A,B$ is Abelian and the other is CCT. So, a non-Abelian nilpotent group is a CCT group if each if its Sylow subgroups is CCT or Abelian (but not all are Abelian).

A dihedral group $D$ is a CCT group. This is clear if $|D| = 2n$ with $n$ odd. If $D = \langle t \rangle N$ with $[D:N] = 2$ and $t$ inverting each element of $N$, then $Z(D)$ has order $2$, and $Z(D) \leq N$. Each element $x \in N \setminus Z(D)$ has centralizer $N$, while each element $y \in D \setminus N$ has centralizer $\langle y \rangle Z(D)$.

Any Frobenius group with Abelian kernel and Abelian complement is a CCT group (recall that a Frobenius group is a group of the form $G= KH$ with $K \lhd G$ and $H \cap H^k = 1$ for each $k \in K \setminus \{1\}$.

Any non-Abelian $p$-group whose maximal subgroups are all Abelian is a CCT group.

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