 Solution

The identity belongs to the centralizer $C_G(H)$. If $x,y\in C_G(H)$, then for every $h\in H$,Also $xh=hx$ implies $h{x}^{-1}=x^{-1}h$ by multiplying on both sides by $x^{-1}$. Hence $xy$ and $x^{-1}$ belong to $C_G(H)$, proving by the subgroup test that $\boxed{C_G(H)\le G}$. This argument works even when $H$ is any subset, giving the centralizer of a subset.