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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.

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1 Answer 1

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 $[G:C_G(x)]=|C|.$

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May sound like a silly question, but: |ϕ[CG(x)]| divides |CG(x)| could you clarify why this is? Is it a result of surjectivity? –  Kato Hamilton Nov 5 '12 at 5:40
    
It's just the first isomorphism theorem. Let $\phi'$ be the restriction of $\phi$ to $C_G(x)$ and $C_G(x)/\text{Ker}\,\phi'\cong \phi[C_G(x)]$. –  Alexander Gruber Nov 5 '12 at 6:13
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