If $f\colon G\to H$ is a surjective homomorphism, then $|C_G(g)| \geq |C_H(f(g))|$ 
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?
 A: Let us denote by $[g]$ the conjugacy class of $g$ in $G$. The set
$$
[g]\cap Kg=\{g=g_1,g_2,\ldots,g_m\}
$$
is obviously finite, and $m\le |K|$.
On the other hand $f(x)$ commutes $f(g)$ in $H$, iff $xgx^{-1}g^{-1}\in K$, or iff
$$
xgx^{-1}\in Kg.
$$
All the elements $g_i,i=1,2,\ldots,m,$ are conjugates in $G$ so the equation
$$xgx^{-1}=g_i$$ holds for exactly $|C_G(g)|$ different elements $x\in G$. Irrespective of the choice of the index $i$.
Denote the subgroup $f^{-1}(C_H(f(g)))$ by $M$.
Putting all this together we have calculated that
$$
|M|=m\cdot |C_G(g)|.
$$
On the other hand $C_H(f(g))=f(M)\cong M/K$, so
$$
|C_H(f(g))|=\frac{|M|}{|K|}=\frac{m\cdot |C_G(g)|}{|K|}\le |C_G(g)|,
$$
as $m\le |K|$.
A: 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 $g$.
