0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $X_i$. $\endgroup$ – Dietrich Burde Dec 28 '16 at 16:02
  • $\begingroup$ @DietrichBurde Got it ! Thanks! $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 16:13
1
$\begingroup$

I found that I should no long left this question unanswered since it's answered in the comments. Therefore, I'll post it as community wiki so that it no longer appears in the unanswered queue.

  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. $$ k=\sum_{i=1}^k |\operatorname{Orb}_{X_i}(G)|=\sum_{i=1}^k \frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_{X_i}(g) \tag3 \label3$$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.