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$C_G(H)$ 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.

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    $\begingroup$ Another obvious example is the center of $G$. Its centralizer is again $G$, which is not abelian. $\endgroup$
    – Crostul
    Commented Jul 22, 2019 at 23:09
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    $\begingroup$ 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) $\endgroup$ Commented Jul 22, 2019 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. $\endgroup$ Commented Jul 23, 2019 at 0:02

1 Answer 1

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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.

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    $\begingroup$ If $H$ is abelian, $C_G(H)$ always contains $H$. $\endgroup$ Commented Jul 22, 2019 at 23:53
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    $\begingroup$ Yes, assuming $H$ is abelian. $\endgroup$
    – the_fox
    Commented Jul 22, 2019 at 23:53

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