# Does $|x^G|$ divide the order $\langle x^G\rangle$?

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 $\langle x^G\rangle$?

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What with the non-divisibility being resolved, below, you should realise that in an abelian group $x^G=x$ but $\langle x\rangle (=\langle x^G\rangle)$ can have, well, the order of any prime dividing the order of the group, say. So I think the answer to your question is "there is nothing obvious connecting these two things". On the other hand, if you assume that $G$ is non-abelian then you might get somewhere...(you might actually have to assume $G$ is centerless...) – user1729 Mar 29 '12 at 9:40
Since $\langle x^G\rangle$ is a $normal$ subgroup, your statement holds for any $simple$ group. (Note that a group $G$ is simple iff for every $x \in G$ it holds that $\langle x^G \rangle = G$. – Nicky Hekster Mar 30 '12 at 10:49

Take $G=S_3$, $x=(123)$, then $x^G=\{(123),(213)\}$, whereas $\langle x^G\rangle=\{id,(123),(213)\}$. So the answer to the divisibility question is no.
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 $|\langle x^G \rangle|$.