Problem 4.4 in Isaacs (Algebra a graduate course) I'm trying to prove this (problem 4.4 in Isaac's book)
Le $\varphi:G\rightarrow H$ be a surjective homomorphism with $|G|$ finite and let $g \in G$. Show that
\begin{equation}|C_{G}(g)|\geq|C_{H}(\varphi(g))|\end{equation}
HINT: Show that the conjugacy class of $g$ in the inverse image in $G$ of $C_{H}(\varphi(g))$ has size $\leq|ker(\varphi)|$.
I'm trying to interpret the hint but I'm stuck in it. 
First, I think that it suggest to restrict the conjugacy class of $g$ (i.e., $\mathcal{O}_g$) to $C_{H}(\varphi(g))$, i.e.,
\begin{equation}\mathcal{O}^{restricted}_{g} = \{x^{-1}gx\,|\,x\in C_{H}(\varphi(g))\}\end{equation}
Second, I think that it maybe suggest that I have to take the intersection between the conjugacy class of 
$g$ and $C_{H}(\varphi(g))$, i.e.,
\begin{equation}\mathcal{O}_g\cap C_{H}(\varphi(g))\end{equation}
Either way, I'm stuck. I just want help to interpret the hint. Thank you all.
 A: The hint tells you to consider the subgroup $K = \varphi^{-1}(C_H(\varphi(g))) \subseteq G$. The map $\varphi : K \to C_H(\varphi(g))$ is surjective and we have $C_G(g) \subseteq K$. In particular $C_G(g) = C_K(g)$. Let $\mathcal{O}_{K}(g) = \{x g x^{-1} \:|\: x \in K\}$  be the conjugacy class of $g$ in $K$. We have $$|K| = |\mathcal{O}_K(g)| |C_K(g)| = |\mathcal{O}_K(g)| |C_G(g)|$$ and $$|K| = |\ker(\varphi)| |C_H(\varphi(g))|$$ so if we can show that $|\mathcal{O}_K(g)| \leq |\ker(\varphi)|$, then the claim follows. 
A: Here is my proof. Is it good?
Let $A = \varphi^{-1}(C_H(\varphi(g)))$. Note that $C_A(g) = \{a\in A\,|\,ag = ga\} = \{a\in G\,|\,\varphi(a)\in C_H(g)\,\wedge\, ag = ga\}$. But $ag = ga$ implies $\varphi(a)\varphi(g) = \varphi(g)\varphi(a)$, therefore,  $C_A(g) = C_G(g)$. Let $\mathcal{O}_A(g)$ be the orbit associated to $g$ (conjugacy class of $g$) in $A$, i.e., $\mathcal{O}_A(g) = \{a^{-1}ga\,|\,a\in A\}$. We have, by the FCP, that
$|A| = |\mathcal{O}_A(g)||C_H(g)| = |\mathcal{O}_A(g)||C_G(g)| \tag{1}$
Let $\pi$ be the restriction of $\varphi$ in $A$, i.e., $\pi:A\rightarrow\varphi(A)$. Now
$x\in ker(\varphi)\Rightarrow\varphi(g)\varphi(x) = \varphi(x)\varphi(g)\Rightarrow\varphi(x)\in C_H(\varphi(g))\Rightarrow x\in G\cap A\Rightarrow x\in ker(\pi).$
Therefore $|ker(\pi)| = |ker(\varphi)|.$ By the FCP, $|A| = |C_H(\varphi(g))||ker(\pi)| = |C_H(\varphi(g))||ker(\varphi)|$. Indeed, by $(1)$,
$|\mathcal{O}_A(g)||C_G(g)| = |C_H(\varphi(g))||ker(\varphi)| \tag{2}$
Finally, $x\in g^{-1}\mathcal{O}_A(g)\Rightarrow \exists\,a\in A$ such that $x = g^{-1}a^{-1}ga\in G\Rightarrow \varphi(x) = 1$ (because $a\in A$), which implies $x\in ker(\varphi)$. Therefore $g^{-1}\mathcal{O}_A(g)\subseteq ker(\varphi)$ and so, $|g^{-1}\mathcal{O}_A(g)|\leq |ker(\varphi)|$. Obviously $|g^{-1}\mathcal{O}_A(g)| = |\mathcal{O}_A(g)|$, the
$|\mathcal{O}_A(g)|\leq |ker(\varphi)|$. That's all; by $(2)$
$|\mathcal{O}_A(g)||C_H(\varphi(g))||\leq |C_H(\varphi(g))||ker(\varphi)| = |\mathcal{O}_A(g)||C_G(g)|\Rightarrow|C_H(\varphi(g))|\leq|C_G(g)|$. 
