Number of Conjugates is Number of Cosets of Centralizer
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Theorem
Let $G$ be a group.
Let $\map {C_G} a$ be the centralizer of $a$ in $G$.
Then the number of different conjugates of $a$ in $G$ equals the number of different (left) cosets of $\map {C_G} a$:
- $\card {\conjclass a} = \index G {\map {C_G} a}$
where:
- $\conjclass a$ is the conjugacy class of $a$ in $G$
- $\index G {\map {C_G} a}$ is the index of $\map {C_G} a$ in $G$.
Consequently:
- $\card {\conjclass a} \divides \order G$
Proof
Let $x, y \in \conjclass a$.
By definition of $\conjclass a$:
- $x a x^{-1} = y a y^{-1}$
By Conjugates of Elements in Centralizer, this is the case if and only if $x$ and $y$ belong to the same left coset of $\map {C_G} a$.
Hence:
- $\card {\conjclass a} = \index G {\map {C_G} a}$
It follows from Lagrange's Theorem that:
- $\card {\conjclass a} \divides \order G$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 48.2$ Conjugacy