2
$\begingroup$

$C_G(H)$ denotes the centralizer of $H$ in $G$. $\langle W\rangle$ denotes the subgroup generated by $W\subseteq G$

So, $G$ is necessarily non-abelian, because if It was, then $\langle H,C_G(H)\rangle$ is aswell. Taking $H=\{e\}$ to be the trivial group, we get that $C_G(H)=G$, thus $\langle \{e\},G\rangle=G$, but $G$ was not abelian.

I claim that this is it, there are no other commutative subgroups such that $\langle H,C_G(H)\rangle$ happens to be non-abelian. Suppose for the sake of contradiction that such abelian $H\leq G$ exists. Let $H$ be nontrivial and abelian, then $C_G(H)$ necessarily contains $H$, so the generated subgroup reduces to $\langle C_G(H)\rangle=C_G(H)$. Here I am quite stuck, probably my claim is not even true. Basically it ends up finding a commutative subgroup whose centralizer is not commutative. What are some examples?

Furthermore, I suggest searching for a group whose center has nontrivial proper subgroups.

$\endgroup$
  • 2
    $\begingroup$ Another obvious example is the center of $G$. Its centralizer is again $G$, which is not abelian. $\endgroup$ – Crostul Jul 22 at 23:09
  • 1
    $\begingroup$ Let me strengthen the condition, can $\langle C_G(H),H\rangle$ be proper? Maybe some nontrivial subgroup of Z(G) will work? (if Z(G) has some) $\endgroup$ – Michal Dvořák Jul 22 at 23:13
  • $\begingroup$ I would assume that this behavior is more often possible than not for arbitrary nonabelian $G$. I don’t think it is possible for free groups, but I can’t see any other such examples. $\endgroup$ – Santana Afton Jul 23 at 0:02
3
$\begingroup$

You can take $G=S_4$ and $H \cong C_2$ the centre of a Sylow $2$-subgroup. Then $\langle H,C_G(H)\rangle = C_G(H)$ is just the Sylow $2$-subgroup, which is isomorphic to $D_8$ and is thus non-abelian. Or take $G=S_3 \times S_3$ and $H$ a Sylow $2$-subgroup of either $S_3$. Then $\langle H,C_G(H)\rangle \cong C_2 \times S_3 \cong D_{12}$ is non-abelian.

$\endgroup$
  • 1
    $\begingroup$ If $H$ is abelian, $C_G(H)$ always contains $H$. $\endgroup$ – Michal Dvořák Jul 22 at 23:53
  • 1
    $\begingroup$ Yes, assuming $H$ is abelian. $\endgroup$ – the_fox Jul 22 at 23:53

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.