Let \(G\) be a finite group. Show that the size of every conjugacy class in \(G\) divides \(|G|\).
Solution
Let \(x\in G\).
1. Under the conjugation action of \(G\) on itself, the orbit of \(x\) is its conjugacy class and its stabilizer is the centralizer \(C_G(x)\).
The conjugation orbit is \[ G\cdot x=\{gxg^{-1}:g\in G\}=\operatorname{Cl}_G(x). \] Moreover, \[ g\cdot x=x \quad\Longleftrightarrow\quad gxg^{-1}=x \quad\Longleftrightarrow\quad gx=xg, \] so \[ \operatorname{Stab}_G(x)=C_G(x). \]
2. Hence\[ |\operatorname{Cl}_G(x)|=[G:C_G(x)]. \]
This is the orbit-stabilizer theorem applied to the conjugation action, using 1.
3. Therefore the size of the conjugacy class divides \(|G|\).
Since \(C_G(x)\le G\) and \(G\) is finite, Lagrange’s theorem gives \[ |G|=|C_G(x)|[G:C_G(x)]. \] By 2, the second factor is \(|\operatorname{Cl}_G(x)|\), so \[ |\operatorname{Cl}_G(x)|\mid |G|. \]