TITLE: Does $|x^G|$ divide the order $\langle x^G\rangle$?
QUESTION [7 upvotes]: Let $G$ be a finite group, $x\in G$, $x^G$ denotes the conjugacy class contained $x$, that's to say
$|x^G|=n$. My question is:
what is the relationship between $|x^G|$ and the subgroup $\langle x^G\rangle$? what's more, I want to know whether $n$ divides the order of the subgroup $\langle x^G\rangle$?

REPLY [4 votes]: We can define conjugacy as a group action on $G$.  Formally, for any $g \in G$, we define $(g,x)=gxg^{-1}$ for all $x \in G$.  The Orbit-Stabilizer Theorem gives a description of $|x^G|$ in terms of $|G|$ and the size of the stabilizer $G_x=\{g \in G:x=gxg^{-1}\}$ of $x$.  Specifically,
\[|x^G|=\frac{|G|}{|G_x|}.\]
From this, we can deduce that $|x^G|$ divides $|G|$.  Thus, we have that $|x^G|$ and $|\langle x^G \rangle|$ are both divisors of $|G|$.
From this observation (and the fact that $|x^G| \leq |\langle x^G \rangle|$), we can determine that if $|G|$ is a prime power, than $|x^G|$ divides $|\langle x^G \rangle|$.