After seeing that every group action produces a permutation representation, we now use actions for a deeper counting purpose. Instead of tracking an entire orbit directly, we count how many points are fixed by each group element and then average those numbers. This principle is Burnside's lemma, one of the most useful counting results in elementary group action theory. The focus keyword fixed points and Burnside's lemma connects two ideas: a fixed point records no movement under a group element, while Burnside's lemma converts fixed-point data into the number of orbits. In this lesson we shall define fixed points, prove Burnside's lemma, and use it to count orbits in finite actions. :::definition[Fixed by an Element] Let $G$ be a group and let $S$ be a $G$-set. Let $g\in G$ and $a\in S$. Then $a$ is said to be fixed by $g$ if $$\begin{equation*} g\cdot a=a. \end{equation*}$$ The set of all elements of $S$ fixed by $g$ is denoted by $$\begin{equation*} \operatorname{Fix}(g)=\{a\in S:g\cdot a=a\}. \end{equation*}$$ ::: :::definition[Fixed by the Group] Let $G$ be a group and let $S$ be a $G$-set. An element $a\in S$ is said to be fixed by $G$ if $$\begin{equation*} g\cdot a=a \end{equation*}$$ for all $g\in G$. The set of all elements fixed by the whole group is $$\begin{equation*} S^G=\{a\in S:g\cdot a=a\text{ for all }g\in G\}. \end{equation*}$$ ::: There is an important distinction between being fixed by one group element and being fixed by every group element. A point may be fixed by a reflection but moved by a rotation, or fixed by the identity but by no other element. Burnside's lemma uses the first kind of fixing: for each $g\in G$, we count the elements fixed by that particular $g$. Students often mistakenly count only the points fixed by the whole group, but the averaging formula requires fixed points element by element. This is why fixed points and Burnside's lemma must be read together rather than as separate topics. :::definition[Fixed Point Count] Let $G$ be a finite group and let $S$ be a finite $G$-set. For each $g\in G$, define the fixed point count of $g$ by $$\begin{equation*} F(g)=|\operatorname{Fix}(g)|=|\{a\in S:g\cdot a=a\}|. \end{equation*}$$ Thus $F(g)$ is the number of elements of $S$ fixed by the action of $g$. ::: :::theorem[Burnside's Lemma] Let $G$ be a finite group and let $S$ be a finite non-empty $G$-set. If $k$ is the number of orbits of $G$ on $S$, then $$\begin{equation*} k=\frac{1}{|G|}\sum_{g\in G}F(g), \end{equation*}$$ where $F(g)=|\{a\in S:g\cdot a=a\}|$. ::: :::proof Given that $G$ is a finite group, $S$ is a finite non-empty $G$-set, and $F(g)=|\{a\in S:g\cdot a=a\}|$ for each $g\in G$. To prove that the number $k$ of orbits of $G$ on $S$ satisfies $$\begin{equation*} k=\frac{1}{|G|}\sum_{g\in G}F(g). \end{equation*}$$ Let $$\begin{equation*} T=\{(g,a)\in G\times S:g\cdot a=a\}. \end{equation*}$$ [1] Count $T$ by fixing $g\in G$. For a fixed $g\in G$, the number of elements $a\in S$ satisfying $g\cdot a=a$ is $F(g)$. Therefore $$\begin{equation*} |T|=\sum_{g\in G}F(g). \end{equation*}$$ [2] Count $T$ by fixing $a\in S$. For a fixed $a\in S$, the number of elements $g\in G$ satisfying $g\cdot a=a$ is $|G_a|$, where $G_a$ is the stabilizer of $a$. Therefore $$\begin{equation*} |T|=\sum_{a\in S}|G_a|. \end{equation*}$$ Let the distinct orbits be $[a_1],[a_2],\ldots,[a_k]$. If $b\in [a_i]$, then the stabilizers $G_b$ and $G_{a_i}$ have the same order. Hence the contribution of the orbit $[a_i]$ to the sum $\sum_{a\in S}|G_a|$ is $$\begin{equation*} |[a_i]||G_{a_i}|. \end{equation*}$$ By the orbit-stabilizer formula, $$\begin{equation*} |[a_i]|=\frac{|G|}{|G_{a_i}|}. \end{equation*}$$ Thus $$\begin{align*} |[a_i]||G_{a_i}|&=\frac{|G|}{|G_{a_i}|}|G_{a_i}|\\ &=|G|. \end{align*}$$ Each orbit contributes $|G|$, and there are $k$ orbits. Therefore $$\begin{equation*} |T|=k|G|. \end{equation*}$$ Combining the two counts of $T$, we get $$\begin{equation*} \sum_{g\in G}F(g)=k|G|. \end{equation*}$$ Dividing by $|G|$ gives $$\begin{equation*} k=\frac{1}{|G|}\sum_{g\in G}F(g). \end{equation*}$$ Hence the number of orbits is the average number of fixed points. $\square$ ::: Burnside's lemma says that the number of orbits equals the average value of $F(g)$ as $g$ runs through $G$. This is surprising at first because orbits describe the movement of points, while $F(g)$ counts points that do not move. The proof works because both quantities count the same set of pairs $(g,a)$ with $g\cdot a=a$. In practice, fixed points and Burnside's lemma are often easier to use than listing all orbits directly, especially when the set $S$ is large. Enter fixed point counts for the group elements, separated by commas. The calculator adds the counts and divides by the number of entries, which represents $|G|$. Use it after you have already found the values $F(g)$ from the action. For example, counts $4,2,2,0$ give an average of $2$, so the action has $2$ orbits. :::calculator[Burnside Average Calculator] @libraries: none ::: :::example[A Two-Element Group Acting on Two Points] Let $G=\{e,r\}$ act on $S=\{1,2\}$, where $e$ fixes both points and $r$ interchanges them. Then $$\begin{equation*} F(e)=2 \end{equation*}$$ and $$\begin{equation*} F(r)=0. \end{equation*}$$ Burnside's lemma gives $$\begin{align*} k&=\frac{1}{|G|}\sum_{g\in G}F(g)\\ &=\frac{1}{2}(2+0)\\ &=1. \end{align*}$$ Thus there is one orbit, namely $\{1,2\}$. ::: :::solved-problem Let $G=\{e,r\}$ act on $S=\{1,2,3,4\}$. Suppose $e$ fixes all four elements and $r$ acts as the permutation $(12)(34)$. Find the number of orbits. ::: :::solution Let $G=\{e,r\}$ and $S=\{1,2,3,4\}$. Given that $e$ fixes all elements of $S$ and $r=(12)(34)$. To find the number of orbits of $G$ on $S$. The identity element fixes every element, so $$\begin{equation*} F(e)=4. \end{equation*}$$ The permutation $r=(12)(34)$ fixes no element of $S$, so $$\begin{equation*} F(r)=0. \end{equation*}$$ By Burnside's lemma, $$\begin{align*} k&=\frac{1}{|G|}\sum_{g\in G}F(g)\\ &=\frac{1}{2}\big(F(e)+F(r)\big)\\ &=\frac{1}{2}(4+0)\\ &=2. \end{align*}$$ Hence the action has $2$ orbits. $\square$ ::: :::solved-problem Let a group $G=\{e,r\}$ act on a set $S$ of $5$ elements. Suppose $F(e)=5$ and $F(r)=1$. Find the number of orbits. ::: :::solution Let $G=\{e,r\}$ act on a set $S$ with $|S|=5$. Given that $F(e)=5$ and $F(r)=1$. To find the number of orbits. By Burnside's lemma, $$\begin{align*} k&=\frac{1}{|G|}\sum_{g\in G}F(g)\\ &=\frac{1}{2}\big(F(e)+F(r)\big)\\ &=\frac{1}{2}(5+1)\\ &=3. \end{align*}$$ Hence the action has $3$ orbits. $\square$ ::: :::remark The value obtained from Burnside's lemma must be an integer. If the average of the proposed fixed point counts is not an integer, then the counts cannot come from a genuine action of a finite group on a finite set. This is a useful way to detect arithmetic errors while applying fixed points and Burnside's lemma. ::: :::tip When applying Burnside's lemma, make a table with one row for each group element or each conjugacy type when elements of the same type have the same number of fixed points. First compute $F(g)$ carefully, then add the values, and only then divide by $|G|$. Do not divide each fixed point count separately before summing. ::: :::exercise 1. Let $G=\{e,r\}$ act on $S=\{1,2,3,4,5,6\}$, where $e$ fixes all elements and $r=(12)(34)(56)$. Find the number of orbits. 2. Let $G=\{e,r\}$ act on a set $S$ of $6$ elements with $F(e)=6$ and $F(r)=2$. Find the number of orbits. 3. Suppose a group of order $3$ acts on a finite set and the fixed point counts are $7,1,1$. Find the number of orbits. ::: :::answer 1. Since $F(e)=6$ and $F(r)=0$, Burnside's lemma gives $k=(6+0)/2=3$. 2. Burnside's lemma gives $k=(6+2)/2=4$. 3. Burnside's lemma gives $k=(7+1+1)/3=3$. ::: :::faq[Frequently Asked Questions] Q: What does $F(g)$ count? A: It counts the elements $a\in S$ satisfying $g\cdot a=a$ for that fixed group element $g$. Q: Does Burnside's lemma count fixed points or orbits? A: It counts orbits by averaging fixed point counts over all group elements. Q: Why is the identity element important in Burnside's lemma? A: The identity fixes every element of $S$, so $F(e)=|S|$, and this term is always part of the average. ::: Fixed point data can be organized before applying Burnside's lemma. Enter the size of the acted-upon set and a candidate permutation cycle form to see how many points stay fixed. We observe that fixed points depend on the particular group element, not only on the size of the set. Try changing the number of fixed symbols and notice how the preview feeds directly into the average used in Burnside's lemma. :::scientific-preview[Fixed Point Data Preview] @libraries: none ::: :::call-to-action[Review Group Action Counting] subtitle: Revisit the orbit-stabilizer formula before using Burnside's lemma in larger symmetry-counting problems. button: Review Orbit-Stabilizer Formula | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/orbit-stabilizer-formula-and-orbit-counting button-ghost: Review Permutation Representations | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/permutation-representations-and-extended-cayley-theorem :::