My course says it's easily explained that $\sum_{\substack{g\in G}} |C_G(g)|=m\times|G|$ where $m$ is the number of conjugacy classes of $G$.
I don't think I see it that easily... Can you tell me about it?
I thought about using $|C_G(g)|\times|C(g)|=|G|$ or maybe trying to consider only one element of each conjugacy classes but I don't see it. I'm probably missing something quite obvious...
Thank you!
On
it is a special form of Burnside's lemma but we can do it by hand,
It is enough to show that the sum is equal to $|G|$ on one conjugacy class then reslt follows.
Let $O$ be one of the conjugacy class,
$$\sum_{x\in O} |C_G(x)|=\sum_{x\in O}\dfrac{|G|}{|O|}=|G|\sum_{x\in O}\dfrac{1}{|O|}=|G|$$
Note: notice that on one conjugacy class, $|C_G(x)|$ has fixed order and by orbit stabilizer therem $$|C_G(x)|=\dfrac{|G|}{|O|}$$
The conjugacy class of each $g \in G$ contributes $|C_G(g)| \cdot |C(g)| = |G|$ to the sum.