📂Abstract Algebra

Centralizer in Abstract Algebra

Definition¹

For an element $a \in G$ of a group $G$, the set of elements that commute with $a$ is called the centralizer of $a$ and is denoted by $C(a)$.

$$ C(a) := \left\{ g \in G : ga = ag \right\} $$

Properties

• For any $a \in G$, we have $Z(G) \subset C(a)$. Here $Z(G)$ is the center of $G$.
• If $G$ is an Abelian group, then $C(a) = G$ for all $a$. The converse also holds.

Theorem

For any $a \in G$, the centralizer of $a$ is a subgroup of $G$.

$$ C(a) \le G $$

Proof

Subgroup Test

For a nonempty subset $H$ of a group $G$, if the following two conditions are satisfied, then $H$ is a subgroup of $G$.

1. $a$, $b \in H \implies ab \in H$
2. $a \in H \implies a^{-1} \in H$

For a given group $G$, fix an arbitrary $a \in G$.

Step 1: $C(a) \ne \emptyset$

Since it is trivial that the identity element is contained in $C(a)$, it is nonempty.

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Step 2: $b, d \in C(a) \implies bd \in C(a)$

Suppose $b, d \in C(a)$. Then the following holds.

$$ (bd)a = b(da) = b(ad) = (ba)d = (ab)d = a(bd) $$

Therefore $bd \in C(a)$.

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Step 3: $b \in C(a) \implies b^{-1} \in C(a)$

Suppose $b \in C(a)$. Then $ab = ba$ holds. Multiplying both sides by $b^{-1}$ on the front and back, we obtain

$$ \begin{align*} && b^{-1}(ab)b^{-1} &= b^{-1}(ba)b^{-1} \\ \implies && b^{-1}a(bb^{-1}) &= (b^{-1}b)ab^{-1} \\ \implies && b^{-1}a &= ab^{-1} \\ \end{align*} $$

Therefore $b^{-1} \in C(a)$.

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