Order of image conjugacy class divides conjugacy class?

[lonewolf5999]

I have a surjective group homomorphism ψ:G → G', and I've shown that if an element x in G has conjugacy class C and its image ψ(x) has conjugacy class C', then ψ restricts to a surjective map from C to C'.

Now I'd like to show that the order of C' divides C. I know from the class equation that |C| divides |G|, that |C'| divides |G'|, and that due to surjectivity, |G'| divides |G|, so |C| divides |G|. I can't see how to piece these facts together to show that |C'| divides |C| though, or how to use what I proved in the first part, so there's definitely something I'm missing.

Any help is appreciated!

 

[Nino MMP]

Hint: Use the fact that ψ restricts to a surjective map from C to C'.Since ψ restricts to a surjective map from C to C', that means that for each element c' in C', there exists an element c in C such that ψ(c) = c'. But since ψ is a group homomorphism, that means that ψ(c^n) = (c')^n for any natural number n. Therefore, if you can show that |C| = |C'| then you will have shown that |C'| divides |C|. To do this, you can use the fact that the order of C' divides |G'|. Since ψ is a surjective group homomorphism, |G'| divides |G|, and since |C| divides |G|, it follows that |C'| divides |C|.