Example of a group that centralizers of every non-identity elements is not abelian I am study $AC$-groups that centralizer of every non-identity elements is abelian . Now i need an example of group that centralizer of every non-identity elements is not abelian.
Where can i find this?
 A: Let $k$ be any field, let $V$ be a $k$ vector space of dimension $4$ equipped a nondegenerate skew symmetric form $\langle  \ , \ \rangle$. (If $k$ has characteristic $2$, you'll need to modify this slightly.) Let $G$ be the group whose ground set is $k \times V$ with multiplication
$$(k_1, v_1) \ast (k_2, v_2) = (k_1+k_2+\langle v_1, v_2 \rangle, v_1+v_2).$$
A group element of the form $(k,0)$ is central, so has centralizer $G$, which is not abelian.
A group element of the form $(k, v)$ with $v \neq 0$ has centralizer those elements of the form $(k', w)$ with $\langle v,w \rangle = 0$. The set of such $w$ forms a $3$-dimensional subspace $W$ of $V$, and $\langle \ , \rangle$ does not restrict to zero on $W$. So we can find $(0,w_1)$ and $(0,w_2)$ centralizing $(k,v)$ with $\langle w_1, w_2 \rangle$ nonzero, and then $(0, w_1) \ast (0, w_2) \neq (0, w_2) \ast (0, w_1)$.
A concrete realization of this group is the group of matrices of the form
$$\begin{pmatrix} 
1 & \ast & \ast & \ast \\
0 & 1 & 0 & \ast \\
0 & 0 & 1 & \ast \\
0 & 0 & 0 & 1 \\
\end{pmatrix}$$
