15.3: Burnside's Lemma
Burnside's lemma relates the number of equivalence classes of the action of a group on a finite set to the number of elements of the set fixed by the elements of the group. Before stating and proving it, we need some notation and a proposition. If a group \(G\) acts on a finite set \(\mathcal{C}\), let ~ be the equivalence relation induced by this action. (As before, the action of \(\pi∈G\) on \(\mathcal{C}\) will be denoted \(\pi^∗\).) Denote the equivalence class containing \(C∈\mathcal{C}\) by ⟨\(C\)⟩. For \(\pi∈G\), let \(fix_C(\pi)=\{C∈ \mathcal{C}:\pi^∗(C)=C\}\), the set of colorings fixed by \(\pi\). For \(C∈ \mathcal{C}\), let \(stab_G(C)=\{\pi∈G: \pi (C)=C\}\) be the stabilizer of \(C\) in \(G\), the permutations in \(G\) that fix \(C\).
To illustrate these concepts before applying them, refer back to Figure 15.2 . Using that information, we can determine that \(fix_C(r_2)=\{C_1,C_{10},C_{11},C_{16}\}\). Determining the stabilizer of a coloring requires finding the rows of the table in which it appears. Thus, \(stab_{D8}(C_7)=\{ι,h\}\) and \(stab_{D8}(C_{11})=\{ι,r_2,p,n\}\).
Let a group \(G\) act on a finite set \(\mathcal{C}\) . Then for all \(C \in \mathcal{C}\) ,
\(\displaystyle \sum_{C' \in ⟨C⟩} |stab_G(C')| = |G|\).
- Proof
-
Let \(stab_G(C)=\{\pi_1,…,\pi_k\}\) and \(T(C,C′)=\{\pi∈G:\pi^∗(C)=C′\}\). (Note that \(T(C,C)=stab_G(C)\).) Take \(\pi∈T(C,C′)\). Then \(\pi \circ \pi_1 ∈T(C,C′)\) for \(1≤i≤k\). Furthermore, if \(\pi \circ \pi_i = \pi \circ \pi_j\), then \(\pi^{-1} \circ \pi \circ \pi_i = \pi^{-1} \circ \pi \circ \pi_j\). Thus \(\pi_i = \pi_j\) and \(i=j\). If \(\pi'∈T(C,C′)\), then \(\pi^{−1} \circ \pi′∈T(C,C)\). Thus, \(\pi−1 \circ \pi′=\pi_i\) for some \(i\), and hence \(\pi′= \pi \circ \pi_i\). Therefore \(T(C,C′)=\{\pi \circ \pi_1,…,\pi \circ \pi_k\}\). Additionally, we observe that \(T(C′,C)=\{\pi^{-1}:\pi∈T(C,C′)\}\). Now for all \(C′∈⟨C⟩\),
\(|stab_G(C′)|=|T(C′,C′)|=|T(C′,C)|=|T(C,C′)|=|T(C,C)|=|stab_G(C)|\).
Therefore,
\(\displaystyle \sum_{C' \in ⟨C⟩} |stab_G(C')| = \sum_{C' \in ⟨C⟩} |T(C,C')|\).
Now notice that each element of \(G\) appears in \(T(C,C′)\) for precisely one \(C′∈⟨C⟩\), and the proposition follows.
With Proposition 15.8 established, we are now prepared for Burnside's lemma.
Let a group \(G\) act on a finite set \(\mathcal{C}\) . If \(N\) is the number of equivalence classes of \(\mathcal{C}\) induced by this action, then
\(N = \dfrac{1}{|G|} \displaystyle \sum_{\pi \in G} |fix_{\mathcal{C}}(\pi)|\).
Before we proceed to the proof, note that the calculation in Burnside's lemma for the example of 2-coloring the vertices of a square is exactly the calculation we performed at the end of Section 15.1.
- Proof
-
Let \(X=\{(\pi,C) \in G \times \mathcal{C}: \pi (C)=C\}\). Notice that \(\sum_{\pi \in G} |fix_C(\pi )|=|X|\), since each term in the sum counts how many ordered pairs of \(X\) have \(\pi\) in their first coordinate. Similarly, \(\sum_{C \in \mathcal{C}}|stab_G(C)|=|X|\), with each term of this sum counting how many ordered pairs of \(X\) have \(C\) as their second coordinate. Thus, \(\sum_{C \in \mathcal{C}} |fix_C(\pi)|=\sum_{C \in \mathcal{C}} |stab_G(C)|\). Now note that the latter sum may be rewritten as
\(\displaystyle \sum_{equivalence \\ classes ⟨C⟩} (\sum_{C' \in ⟨C⟩} |stab_G(C')|)\).
By Proposition 15.8 , the inner sum is \(|G|\). Therefore, the total sum is \(N \cdot |G|\), so solving for \(N\) gives the desired equation.
Burnside's lemma helpfully validates the computations we did in the previous section. However, what if instead of a square we were working with a hexagon and instead of two colors we allowed four? Then there would be \(4^6=4096\) different colorings and the dihedral group of the hexagon has 12 elements. Assembling the analogue of Figure 15.2 in this situation would be a nightmare! This is where the genius of Pólya's approach comes into play, as we see in the next section.