Let $k(G)$ be the number of conjugation classes of the finite group $G$. Defining $C:=\{(x,y) \in G \times G: xy=yx\}$. Prove that $|C|=k(G)|G|$.

Question

(I need help to understand some steps in this solution.)

P.S. My question is in $\color{red}{\text{Color Red}}$

I'm going to type down the right answer found in the manual solutions, which is:

SOLUTION: Let $c_x$ be the number of conjugates of $x$ in $G$, we have $c_x=|G:C_{G}(x)|$ then $|C_{G}(x)|=|G|/c_{x}$, where it follows that $|C|=|G|\displaystyle\sum_{x\in G} 1/c_{x}$. We can join all elements that are in the same class: renaming by $C_1, \cdots, C_k$ the conjugation classes of $G$, where $k=k(G)$, if $x \in C_i$, then $c_x=|C_i|$, thus $$\begin{align} |C|&=|G|\displaystyle\sum_{i=1}^{k}\sum_{x \in C_i} \frac{1}{c_x}\\ &=|G|\displaystyle\sum_{i=1}^{k}\sum_{x \in C_i}\frac{1}{|C_i|}\\ &=|G|\displaystyle\sum_{i=1}^{k}\underbrace{1}_{\color{red}{\text{why is equal to 1?}}}\\ &=|G|k. \end{align}$$

Answer 1

Conjugacy is an equivalence relation, so the conjugacy classes partition the set, which implies that $\sum_i |C_i| = |G|$. But that fact is not needed here.

More generally, for any nonempty finite set $A$, $$\sum_{j\in A} \frac{1}{|A|}=\frac{1}{|A|}\sum_{j\in A} 1=\frac{1}{|A|}|A|=1$$

Answer 2

This is a particular case of Burnside's lemma. The action of $G$ on $X=G$ is $(x,y) \mapsto x\star y \colon =x y x^{-1}$. We have $ x\star y = y \Leftrightarrow x y = y x$.