Image of conjugacy class under surjective homomorphism There is a surjective homomorphism from $G$ to $G'$. Let $C$ denote the conjugacy class of element $x$ in $G$, $C'$ the conjugacy class of the image of $x$ in $G'$. Prove the order of $C'$ divides the order of $C$.
So far, using the class equation I can observe that $|C|$ divides $|G|$ and $|C'|$ divides $|G'|$, and it's also obvious that the homomorphism maps $C$ surjectively to $C'$. But I can't quite piece it all together.
Any help appreciated.
 A: Call the homomorphism $\phi$.  By the Orbit Stabilizer theorem, $[G:C_G(x)]=|C|$ for any $x\in C$.  Now observe that the image of $C_G(x)$ under $\phi$ centralizes the $\phi(x)$ in $G'$, whence $\phi[C_G(x)]\leqslant C_{G'}(\phi(x))$.  
$|\phi[C_G(x)]|$ divides $|C_G(x)|$ so $|C'|=[G':C_{G'}(\phi(x))]$ divides $[G':\phi[C_G(x)]]$ divides $[G:C_G(x)]=|C|.$
A: Call $\phi$ the homomorphism. By applying the First Homomorphism Theorem to $\phi_{|C_G(x)}$, we get:
$$C_G(x)/\operatorname{ker}\phi\cap C_G(x)\cong \phi(C_G(x))\tag 1$$
Besides:
\begin{alignat}{2}
g'\in\phi(C_G(x)) &\Rightarrow \exists g\in C_G(x)\mid g' &&=\phi(g) \\
& &&=\phi(xgx^{-1}) \\
& &&=\phi(x)\phi(g)\phi(x)^{-1} \\
& &&=\phi(x)g'\phi(x)^{-1} \\
&\Rightarrow g'\in C_{G'}(\phi(x)) \\
\end{alignat}
whence:
$$\phi(C_G(x))\le C_{G'}(\phi(x)) \tag 2$$
Therefore, by calling $l:=[C_{G'}(\phi(x)):\phi(C_G(x))]\in \Bbb Z$, we get:
\begin{alignat}{1}
|C'| &= \frac{|G'|}{|C_{G'}(\phi(x))|} \\
&\stackrel{(2)}{=}\frac{|G'|}{l|\phi(C_G(x))|} \\
&\stackrel{(1)}{=}\frac{|G'||\operatorname{ker}\phi\cap C_G(x)|}{l|C_G(x)|} \\
&\stackrel{}{=}\frac{|\operatorname{ker}\phi\cap C_G(x)|}{l}\cdot\frac{|G'|}{|G|}\cdot\frac{|G|}{|C_G(x)|} \\
&\stackrel{}{=}\frac{|\operatorname{ker}\phi\cap C_G(x)|}{l}\cdot\frac{|G'|}{|G|}\cdot |C| \\
&\stackrel{}{=}\frac{|\operatorname{ker}\phi\cap C_G(x)|}{l|\operatorname{ker}\phi|}\cdot |C| \\
&\stackrel{}{=:}\frac{|C|}{\kappa} \\
\end{alignat}
where
$$\kappa:=l\frac{|\operatorname{ker}\phi|}{|\operatorname{ker}\phi\cap C_G(x)|}\in \Bbb Z$$
