= Solution
The identity commutes with every $h\in H$, so the <centralizer> $C_G(H)$ is nonempty. If $x,y\in C_G(H)$, then for every $h\in H$,
$$
(xy)h=x(yh)=x(hy)=(xh)y=(hx)y=h(xy).
$$
Thus it is closed under multiplication. From $xh=hx$, multiplication by $x^{-1}$ on both sides gives $hx^{-1}=x^{-1}h$, so $x^{-1}\in C_G(H)$. These are exactly the <subgroup> conditions; hence
$$
\boxed{C_G(H)\leq G.}
$$
The proof works for the <centralizer of a subset> as well; it does not require that $H$ itself be a <subgroup>.
Back to article page
