Let f:G -> H be a surjective homomorphism. |C_G(g)| >= |C_H(f(g)|

[Esran]

Let f:G --> H be a surjective homomorphism. |C_G(g)| >= |C_H(f(g)|

Homework Statement

Suppose G is a finite group and H is a group, where θ:G→H is a surjective homomorphism. Let g be in G. Show that |C_G(g)| ≥ |C_H(θ(g))|.

Homework Equations

This problem has been bugging me for a day now. I'm studying for my qualifying exam and doing very well otherwise. I sure could use some peace of mind though concerning this problem. I tend to obsess over things I can't figure out.

The Attempt at a Solution

Obviously, θ[C_G(g)] ≤ C_H(θ(g)), so C_G(g) is contained in the pullback of C_H(θ(g)). Beyond that, I'm stuck and I would greatly appreciate assistance.

 

[deluks917]

This question was asked by someone (you?) recently on math stack exchange. Here are the answers.

http://math.stackexchange.com/quest...a-surjective-homomorphism-then-c-gg-geq-c-hfg

 

[Esran]

It wasn't me, but thanks for the link!