# A doubt in the proof of Burnside's formula

I read a proof of Burnside's formula and I got stuck in a step.

## Burnside's formula

Let $X$ be a finite set, and $G$ be a finite group acting on $X$.
Denote the $G$-orbits of $X$ as $X_1,\dots,X_k$, and
$\operatorname{Fix}_X(g) = \{x \in X \mid g \cdot x = x\}$.
Then $$\bbox[yellow,5px,border:1px solid red]{k=\frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_X(g)\rvert.}$$

Proof in the link: Since $\operatorname{Fix}_X(g) = \bigcup\limits_{i=1}^k \operatorname{Fix}_{X_i}(g)$, one has $\lvert\operatorname{Fix}_X(g)\rvert = \sum\limits_{i=1}^k \lvert\operatorname{Fix}_{X_i}(g) \rvert$. \begin{align} k &= \sum\limits_{i=1}^k 1 \\ & \stackrel{?}= \sum\limits_{i=1}^k \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_{X_i}(g) \rvert \tag{?}\label{?} \\ &= \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \sum\limits_{i=1}^k \lvert\operatorname{Fix}_{X_i}(g) \rvert \\ &= \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_{X}(g) \rvert \end{align}

I don't understand step \eqref{?}. For each fixed $G$-orbit $X_i$, why is $\sum\limits_{g \in G} \lvert \operatorname{Fix}_{X_i}(g) \rvert = \lvert G \rvert$?

Thanks!

• This is explained in the text: "Ainsi $1=Orb_X(G)|=\frac{1}{|G|}\sum_{g\in G}Fix_X(g)"$. Now apply it for each $X_i$. – Dietrich Burde Dec 28 '16 at 16:02
• @DietrichBurde Got it ! Thanks! – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 16:13

1. In the linked text, the case for $G$ acting transitively on $X$ is first proved, so that $$\bbox[5px, border:1px solid black]{1=|\operatorname{Orb}_X(G)|=\frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_X(g).} \tag1 \label1$$
2. Apply \eqref{1} to all $G$-orbits $X_i$. $$1=|\operatorname{Orb}_{X_i}(G)|=\frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_{X_i}(g) \tag2 \label2$$
3. Sum \eqref{2} over $i = 1,\dots,k$ on both sides. $$k=\sum_{i=1}^k |\operatorname{Orb}_{X_i}(G)|=\sum_{i=1}^k \frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_{X_i}(g) \tag3 \label3$$