# The Centralizer of a Subset A of a Group G, CG(A)

Recall from The Center of a Group G, Z(G) page that if $G$ is a group then the center of $G$ denoted $Z(G)$ is defined to be the set of all elements of $g$ that commute with every element of $G$, that is:

(1)We proved that $G$ is an abelian group if and only if $G = Z(G)$. Furthermore, we proved that $Z(G)$ is always an abelian subgroup of $G$.

We now generalize this concept.

Definition: Let $G$ be a group and let $A$ be a nonempty subset of $G$. The Centralizer (or Commutant) of $A$ in $G$ denoted $C_G(A)$ is defined to be the set of all elements of $G$ that commute with every element of $G$, that is, $C_G(A) = \{ g \in G : gag^{-1} = a \quad \: \forall g \in G \}$. |

*Note that $Z(G) = C_G(G)$. By convention, if $A$ is a singleton set, say $A = \{ a \}$ then we denote the centralizer of $A$ in $G$ by $C_G(a)$.*

Proposition 1: Let $G$ be a group. Then $Z(G) = \bigcap_{a \in G} C_G(a)$. |

**Proof:**Let $g \in Z(G)$. Then $ga = ag$ for all $a \in G$. Since $C_G(a) = \{ h \in G : ha = ah \}$ we see that $g \in C_G(a)$ for each $a \in G $}]. Thus [[$ g \in \bigcap_{a \in G} C_G(a)$. So $Z(G) \subseteq \bigcap_{a \in G} C_G(a)$.

- Let $g \in \bigcap_{a \in G} C_G(a)$. Then $g \in C_G(a)$ for each $a \in G$. So $ga = ag$ for each $a \in G$. Thus $g \in Z(G)$. So $\bigcap_{a \in G} C_G(a) \subseteq Z(G)$.

- Therefore $Z(G) = \bigcap_{a \in G} C_G(a)$. $\blacksquare$

The following proposition tells us that the centralizer of $A$ in $G$ is always an abelian subgroup of $G$.

Proposition 2: Let $G$ be a group and let $A$ be a nonempty subset of $G$. Then $C_G(A)$ is a subgroup of $G$. |

*Note that $C_G(A)$ MIGHT NOT BE ABELIAN! For example, if $G$ is a nonabelian group with identity $1$ then $C_G(1) = \{ g \in G : g1g^{-1} = 1 \} = G$, which is nonabelian.*

**Proof:**Clearly $(C_G(A), \cdot)$ is closed under the operation $\cdot$, for if $g_1, g_2 \in C_G(A)$ then $g_1ag_1^{-1} = a$ and $g_2ag_2^{-1} = a$ for all $a \in A$, so:

- So $g_1g_2 \in C_G(A)$.

- Now clearly $1 \in C_G(A)$ since by definition $1a1^{-1} = a$ for all $a \in A \subseteq G$.

- Lastly, let $g \in C_G(A)$. Then $gag^{-1} = a$. This equation can be rewritten as $g^{-1}ag = a$. So $g^{-1} \in C_G(A)$.

- So $C_G(A)$ is a subgroup of $G$. $\blacksquare$