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Let $G$ be a group and $g \in G$ and $C_G(g)=\{x \in G|x^{-1}gx=g\}$ and $Cl_g=\{x^{-1}gx|x \in G \}$.Prove that $[G:C_G(g)]=|Cl_g|$.

Can someone help me with this by giving me a hint?

Thank you in advance.

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    Given an action of a group $G$ on a set $X,$ the cardinality of the orbit $|G \cdot x|,$ where $x \in X$ is equal to the index $|G:G_x|$ of the stabilizer $G_x$ of $x$ in $G.$ – 2017-02-19
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    Sorry i will edit my question.i meant centralizers – 2017-02-19

2 Answers 2

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This is an application of the Orbit-Stabiliser Theorem where $G$ acts on itself by conjugation. Specifically for $g,x\in G$ $g^x=x^{-1}gx$. $Cl_g$ is the orbit of $g$ and $C_G(g)$ the stabiliser. Thus $|Cl_g|=|G|/|C_G(g)|=[G:C_G(g)]$.

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There's a map $G\rightarrow{\rm Cl}_g$ given by $x\mapsto x^{-1}gx$. Clearly the map is constant on cosets $xC_G(g)$ and so descends to a map $$ G/C_G(g)\rightarrow{\rm Cl}_g $$ which is obviously surjective and injective.