Prove in two different ways-“If the centre of $G$ is of index $n$,prove that every conjugacy class has atmost $n$ elements .”

Solution:By Orbit-Stabilizer theorem,$\vert G:G_g\vert$.$\vert G_g:Z(G) \vert$=$\vert G:Z(G)\vert =n \implies \vert G:G_g\vert$ divides $n$.

When the action is by conjugation,the no. of conjugates of $g$ is $\vert G:C_G(g)\vert$.

Thus,$\vert G:C_G(g)\vert$ divides $n$,which implies that every conjugacy class has atmost $n$ elements.

I'm not sure whether this is correct,please tell me if i'm missing some detail.

Also,i don't know how to prove this in another way.

• It is not really a different way: If $C$ is a conjugacy class, then $G/Z(G)$ acts on $C$ (by conjugation) and the action is transitive. – Hagen von Eitzen Jul 30 '16 at 12:39