$C_G(H)$ 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.
Best Answer
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.