TITLE: If $f\colon G\to H$ is a surjective homomorphism, then $|C_G(g)| \geq |C_H(f(g))|$
QUESTION [7 upvotes]: Let $G$ be finite, $f\colon G\to H$  be a surjective homomorphism (hence $H$ is finite) and $g \in G$. Prove the order of center of $g$ in $G$ is greater than or equal to the order of the center of $f(g)$ in $H$, i.e. $$|C_G(g)| \ge |C_H(f(g))|.$$

Attempts at solution. 
If $f$ is an isomorphism this is clear. So if $\ker f=K$ we can consider $G \to G/K \xrightarrow{f'} H$, where $f'$ is the induced map. So we can reduce this problem to the case where $f$ is a projection map. 
Also if $f(x)f(g) = f(g)f(x)$ then $f(x^{-1}g^{-1}xg) = e$.
So if $h = f(y)$, $h \in C_H(f(g))$ iff $[h,g] \in K$. 
The center of $g$ is the set of fixed points of the conjugation map (by $g$). And the center of $h$ is the set of fixed points under conjugation by $h$. So maybe this can be done by considering how the orbits collapse mod a normal subgroup?

REPLY [2 votes]: An alternative way by character theory:
Lifting all the irreducible characters of $H$ to a possibly proper subset of irreducible characters of $G$ and using column orthogonality relations for $g$ and $f(g)$, we have the result.
So, the equality holds if and only if the other irreducible characters of $G$ has value $0$ at $g$.