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 $\langle x^G\rangle$?
Does $|x^G|$ divide the order $\langle x^G\rangle$?
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group-theory
finite-groups
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0What 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...) – 2012-03-29
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1Since $\langle x^G\rangle$ 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 $\langle x^G \rangle = G$. – 2012-03-30