# Find a group $G$ and abelian subgroup $H\leq G$ such that $\langle C_G(H),H\rangle$ is proper and non-abelian.

$$C_G(H)$$ 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.

• Another obvious example is the center of $G$. Its centralizer is again $G$, which is not abelian. – Crostul Jul 22 at 23:09
• 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) – Michal Dvořák Jul 22 at 23:13
• 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. – Santana Afton Jul 23 at 0:02

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