= Solution
Let $H=Z(x)$. The map $G/H\to C(x)$ sending $gH$ to $gxg^{-1}$ is well defined: multiplying $g$ on the right by an element commuting with $x$ leaves its conjugate unchanged. It is onto by the <orbit> definition. If $gxg^{-1}=hxh^{-1}$, then $h^{-1}g$ commutes with $x$, so $gH=hH$, proving <injectivity>.
Every <coset> has $|H|$ elements, so $|C(x)|=|G|/|H|$. Therefore
$$
\boxed{|G|=|Z(x)|\,|C(x)|.}
$$
This is the <orbit-stabilizer theorem> specialized to <conjugation>.
Back to article page
