# Order of conjugacy class divides order of group

https://proofwiki.org/wiki/Number_of_Conjugates_is_Number_of_Cosets_of_Centralizer

and I don't understand the very last part, why the order of the conjugacy class divides the order of the group. It says that it follows by Lagrange's but the conjugacy class is not necessarily a subgroup? I'm thinking that perhaps the index of the centralizer is exactly equal to the order of some subgroup of $G$, but I'm not sure. Can someone explain?

• No; in general if $d$ is a divisor of $n$, then $\tfrac{n}{d}$ is also a divisor of $n$, because $d\cdot\tfrac{n}{d}=n$. – Servaes Nov 7 '16 at 0:14
Since $|C_a|=[G:C_G(a)]$, if $G$ is a finite group, we have $$|C_a|=\frac{|G|}{|C_G(a)|}$$ $$|G|=|C_a||C_G(a)|$$ So obviously, $|C_a|$ divides $|G|$
Maybe the proof there wants to mean that the index of a subgroup $[G:H]$ divides $|G|$.