TITLE: Proof of $|C_{G/N}(gN)| \leq |C_G(g)|$ without character theory
QUESTION [9 upvotes]: In the book Character Theory of Finite Groups by Isaacs the following is proven (Corollary 2.24) using character theory:

Proposition: Let $G$ be a finite group and $N \trianglelefteq G$. Then $|C_{G/N}(gN)| \leq |C_G(g)|$ for all $g \in G$.

Isaacs mentions that "The following could be proved without characters but it is somewhat tricky to do so".  
So what is a character-free proof for the above proposition? 
I know how to do this in a special case. If $g$ has order coprime to $N$, then it is possible to prove (the proof I know is a bit long) that $C_{G/N}(gN) = C_G(g)N/N$. Then $$|C_{G/N}(gN)| = |C_G(g)N/N| = |C_G(g) / N \cap C_G(g)| \leq |C_G(g)|$$

REPLY [10 votes]: $\newcommand{\Size}[1]{\lvert #1 \rvert}\newcommand{\Set}[1]{\{ #1 \}}$Consider the actions of $G$ and $G/N$ by conjugacy on themselves, and write $x^{y} = y^{-1} x y$, and $x^{G} = \{ x^{y} : y \in G \}$.
We have 
$$
\Size{G} = \Size{g^{G}} \cdot \Size{C_{G}(g)}
$$
and
$$
\frac{\Size{G}}{\Size{N}} = \Size{G/N} = \Size{(gN)^{G/N}} \cdot \Size{C_{G/N}(gN)}.
$$
Now note that 
$$(gN)^{G/N} = \Set{gN^{yN} : y \in G} = \Set{g^{y} N : y \in G} = g^{G} N,$$
 so that we get the key inequality $$\Size{g^{G}} \le \Size{N} \cdot \Size{(gN)^{G/N}},\tag{key}$$ because $g^{G}$ is contained in the union of the cosets that make up $(gN)^{G/N}$.
In other words (since this is the key point), the map $f : g^{G} \to (gN)^{G/N}$ that sends $g^{y} \to g^{y} N = (gN)^{yN}$ is surjective, and we have for the pre-image of a point $f^{-1} (g^y N) \subseteq g^{y} N = \Set{g^{y} n : n \in N}$, so that we get (key).
It follows that
\begin{align}
\Size{gN^{G/N}} \cdot \Size{C_{G/N}(gN)} &= \frac{\Size{G}}{\Size{N}} 
= \frac{\Size{g^{G}} \cdot \Size{C_{G}(g)}}{\Size{N}} 
\\&\le
\frac{\Size{N} \cdot \Size{(gN)^{G/N}} \cdot  \Size{C_{G}(g)}}{\Size{N}}
\\&\le \Size{(gN)^{G/N}} \cdot  \Size{C_{G}(g)},
\end{align}
form which one gets
$$
\Size{C_{G/N}(gN)} \le \Size{C_{G}(g)}.
$$