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

# 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)
\begin{align} \quad Z(G) = \{ a \in G : ag = ga\quad \: \forall g \in G \} \end{align}

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:
(2)
\begin{align} \quad g_1g_2a(g_1g_2)^{-1} = g_1(g_2ag_2^{-1})g_1^{-1} = g_1ag_1^{-1} = a \end{align}
• 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 $G$. $\blacksquare$