Number of Conjugates is Number of Cosets of Centralizer

From ProofWiki
Jump to navigation Jump to search

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