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> [!proposition] Proposition. ([[size of conjugacy class divides order of finite group]])
> If $G$ is a finite [[group]], the size of any [[conjugate|conjugacy class]] in $G$ [[divides]] the [[order of a group|order]] of $G$.
> [!basicexample]
> Take $S_{3}$. The [[conjugate|conjugacy class]] $C_{\tau}$ of $\tau$ is $\{ \tau, \tau^{2} \}$. Its [[centralizer of a subgroup|centralizer]] $Z_{S_{3}}(\tau)$ is $\{ e, \tau, \tau^{2} \}$. It is of course a left [[coset]] of itself; by [[Lagrange's Theorem]] the other left [[coset]] must be $\{ \sigma, \sigma \tau, \sigma \tau^{2} \}$. So $[S_{3}: Z_{S_{3}}(\tau)]=2=|C_{\tau}|$. Since $|G|=[S_{3}: Z_{S_{3}}(x)] |Z_{S_{3}}(\tau)|$, we see the result.
> [!proof]- Proof. ([[size of conjugacy class divides order of finite group]])
> Denote the [[conjugate|conjugacy class]] of $x \in G$ by $C_{x}$. We will show that $|C_{x}|=[G:Z_{G}(x)]=\frac{|G|}{|Z_{G}(x)|}$, since then we may write $|G|=|C_{x}| |Z_{G}(x)|$ and the result is clear.
\
Begin by remarking that $(*)$ $g_{1}xg_{1} ^{-1}=g_{2} x g_{2}^{-1} \iff g_{2}^{-1}g_{1}x=xg_{2}^{-1}g_{1} \iff g_{2}^{-1}g_{1} \in Z_{G}(x),$
for this will be of use later.
\
Let's construct a [[bijection]] between $C_{x}$ and $G / Z_{G}(x)$.
>
Define $f:G \to C_{x}$ to map $g \mapsto gxg^{-1}$, and note that $f$ is a [[surjection]] by construction. And $\begin{align}
f(g_{1})=f(g_{2}) \iff & g_{1}xg_{1}^{-1} = g_{2}x g^{-1} \\
\iff & g_{2}^{-1}g_{1} \in Z_{G}(x) \\
\iff & g_{1} \in g_{2} Z_{G}(x),
\end{align}$
where the second equivalence is from $(*)$. In words: $f(g_{1})=f(g_{2})$ if and only if $g_{1}$ and $g_{2}$ represent the same left [[coset]] of $Z_{G}(x)$. It follows that to form a [[bijection]] $C_{x} \leftrightarrow G / Z_{G}(x)$ we should map the [[coset]] $gZ_{G}(x)$ to $gxg^{-1} \in C_{x}$. This is the [[bijection]] desired.
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#### References
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