## Centralizer is inclusion-reversing

Let $G$ be a group. Prove that if $A$ and $B$ are subsets of $G$ with $A \subseteq B$, then $C_G(B)$ is a subgroup of $C_G(A)$.

Let $x \in C_G(B)$. Then for all $b \in B$, $xbx^{-1} = b$. Since $A \subseteq B$, for all $a \in A$ we have $xax^{-1} = a$, so that $x \in C_G(A)$. Thus $C_G(B) \subseteq C_G(A)$, and hence $C_G(B) \leq C_G(A)$.