1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ A quick computer search with Maple shows that there is an example of order $32$. It is SmallGroup(32,49). $\endgroup$ – James Oct 27 '15 at 16:34
  • 2
    $\begingroup$ And the centralizer of the identity is also nonabelian! $\endgroup$ – Derek Holt Oct 27 '15 at 18:42
  • 1
    $\begingroup$ I'm not sure your definitions are quite right. A CA-group is one for which the centralizer of every non-identity element is abelian. An AC-group is one for which the centralizer of every non-central element is abelian. As it happens, John Britnell and I wrote down a result (partially) classifying the AC-groups here: arxiv.org/abs/1309.2237 (although I'm sure this was already known to experts). Our result depended on CFSG, whereas there is an important classification of CA-groups that predates CFSG. $\endgroup$ – Nick Gill Oct 28 '15 at 9:24
2
$\begingroup$

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 $$\begin{pmatrix} 1 & \ast & \ast & \ast \\ 0 & 1 & 0 & \ast \\ 0 & 0 & 1 & \ast \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.