CCT groups of order $\leq 32$ 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.

*

*Is there any short conceptual proof of the proposition above?

*Does the result in the proposition above appear somewhere in the literature?


 A: 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.
A: 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.
