Having defined orbits and stabilizers, we now connect them by a precise counting principle. The orbit tells us how many points are reachable from $a$, while the stabilizer tells us how many group elements keep $a$ fixed. The orbit-stabilizer formula and orbit counting show that these two numbers are not independent: a large stabilizer forces a small orbit, and a large orbit forces a small stabilizer. This idea turns group actions into a powerful counting method for finite groups and finite sets. In this lesson we shall prove the orbit-stabilizer formula, derive the orbit decomposition formula, and use them to detect fixed points. :::definition[Orbit Size and Stabilizer Index] Let $G$ be a group, let $S$ be a $G$-set, and let $a\in S$. The orbit size of $a$ is the cardinality of $$\begin{equation*} [a]=\{g\cdot a:g\in G\}. \end{equation*}$$ The stabilizer index of $a$ is the index $$\begin{equation*} [G:G_a], \end{equation*}$$ where $G_a=\{g\in G:g\cdot a=a\}$ is the stabilizer of $a$. ::: Different elements of $G$ can send $a$ to the same point. This happens exactly when those group elements differ by an element of the stabilizer $G_a$. Thus the orbit is not usually counted by counting all of $G$ directly. Instead, it is counted by grouping elements of $G$ into left cosets of $G_a$. Students often expect $|[a]|=|G|$, but that is true only when the stabilizer is trivial. The orbit-stabilizer formula and orbit counting correct this mistake by showing that the repeated actions are precisely measured by $G_a$. :::theorem[Orbit-Stabilizer Formula] Let $G$ be a group and let $S$ be a $G$-set. For each $a\in S$, $$\begin{equation*} |[a]|=[G:G_a]. \end{equation*}$$ ::: :::proof Given that $G$ is a group, $S$ is a $G$-set, $a\in S$, $[a]=\{g\cdot a:g\in G\}$, and $G_a=\{g\in G:g\cdot a=a\}$. To prove that $|[a]|=[G:G_a]$. Let $\mathcal{L}$ be the set of all left cosets of $G_a$ in $G$. Define $f:\mathcal{L}\to [a]$ by $$\begin{equation*} f(gG_a)=g\cdot a. \end{equation*}$$ [1] To prove that $f$ is well-defined. Suppose $g_1G_a=g_2G_a$. Then $g_2^{-1}g_1\in G_a$, so $$\begin{equation*} (g_2^{-1}g_1)\cdot a=a. \end{equation*}$$ Acting by $g_2$ on both sides gives $$\begin{align*} g_2\cdot((g_2^{-1}g_1)\cdot a)&=g_2\cdot a\\ (g_2g_2^{-1}g_1)\cdot a&=g_2\cdot a \quad [\because \text{ action law}]\\ g_1\cdot a&=g_2\cdot a. \end{align*}$$ Therefore $f(g_1G_a)=f(g_2G_a)$. [2] To prove that $f$ is one-to-one. Suppose $f(g_1G_a)=f(g_2G_a)$. Then $$\begin{equation*} g_1\cdot a=g_2\cdot a. \end{equation*}$$ Acting by $g_2^{-1}$ on both sides gives $$\begin{align*} g_2^{-1}\cdot(g_1\cdot a)&=g_2^{-1}\cdot(g_2\cdot a)\\ (g_2^{-1}g_1)\cdot a&=(g_2^{-1}g_2)\cdot a \quad [\because \text{ action law}]\\ (g_2^{-1}g_1)\cdot a&=a. \end{align*}$$ Thus $g_2^{-1}g_1\in G_a$, and hence $g_1G_a=g_2G_a$. [3] To prove that $f$ is onto. Let $b\in [a]$. Then there exists $g\in G$ such that $$\begin{equation*} b=g\cdot a. \end{equation*}$$ Hence $b=f(gG_a)$. Therefore $f$ is a bijection from $\mathcal{L}$ to $[a]$. Hence $|[a]|=[G:G_a]$. $\square$ ::: :::corollary Let $G$ be a finite group and let $S$ be a $G$-set. For each $a\in S$, $$\begin{equation*} |[a]|=\frac{|G|}{|G_a|}. \end{equation*}$$ ::: :::proof Given that $G$ is a finite group, $S$ is a $G$-set, and $a\in S$. To prove that $|[a]|=\frac{|G|}{|G_a|}$. By the orbit-stabilizer formula, $$\begin{equation*} |[a]|=[G:G_a]. \end{equation*}$$ Since $G$ is finite and $G_a\leq G$, the index formula gives $$\begin{equation*} [G:G_a]=\frac{|G|}{|G_a|}. \end{equation*}$$ Hence $$\begin{equation*} |[a]|=\frac{|G|}{|G_a|}. \end{equation*}$$ $\square$ ::: When $S$ is finite, its orbits partition it into disjoint pieces. Counting $S$ then becomes counting one orbit at a time. If $A$ contains exactly one representative from each orbit, then every element of $S$ lies in exactly one orbit $[a]$ with $a\in A$. The orbit-stabilizer formula converts each orbit size into a subgroup index. This is the basic mechanism behind orbit counting in finite group actions. :::theorem[Orbit Decomposition Formula] Let $G$ be a group and let $S$ be a finite $G$-set. If $A$ is a subset of $S$ containing exactly one representative from each orbit, then $$\begin{equation*} |S|=\sum_{a\in A}[G:G_a]. \end{equation*}$$ ::: :::proof Given that $G$ is a group, $S$ is a finite $G$-set, and $A$ contains exactly one representative from each orbit. To prove that $|S|=\sum_{a\in A}[G:G_a]$. The orbit relation is an equivalence relation, so the distinct orbits partition $S$. Hence $$\begin{equation*} S=\bigcup_{a\in A}[a], \end{equation*}$$ where the union is disjoint. Therefore $$\begin{equation*} |S|=\sum_{a\in A}|[a]|. \end{equation*}$$ By the orbit-stabilizer formula, $$\begin{equation*} |[a]|=[G:G_a] \end{equation*}$$ for each $a\in A$. Hence $$\begin{equation*} |S|=\sum_{a\in A}[G:G_a]. \end{equation*}$$ $\square$ ::: Use the calculator to compare the size of a finite group with the size of a stabilizer. Enter $|G|$ and $|G_a|$ when $G_a$ is known to be a subgroup of $G$. The calculator returns the orbit size predicted by the orbit-stabilizer formula and orbit counting. Try values such as $|G|=24$ with $|G_a|=6$, or $|G|=60$ with $|G_a|=12$, and observe how a larger stabilizer produces a smaller orbit. :::calculator[Orbit-Stabilizer Size Calculator] @libraries: none ::: :::solved-problem Let $G$ be a group of order $77$ acting on a set $S$ of $20$ elements. Prove that $G$ has a fixed point in $S$. ::: :::solution Let $G$ be a group of order $77$ acting on a set $S$ such that $|S|=20$. Given that $|G|=77$. To prove that there exists $a\in S$ such that $g\cdot a=a$ for all $g\in G$. For any $a\in S$, the orbit-stabilizer formula gives $$\begin{equation*} |[a]|=[G:G_a]. \end{equation*}$$ Since $G_a\leq G$, the number $[G:G_a]$ divides $|G|=77$. Therefore the possible orbit sizes are $$\begin{equation*} 1,\quad 7,\quad 11,\quad 77. \end{equation*}$$ Since $|S|=20$, no orbit can have size $77$. If possible let $G$ have no fixed point in $S$. Then no orbit has size $1$, so every orbit has size $7$ or $11$. Thus $20$ must be expressible as a sum of $7$'s and $11$'s. The only possible sums not exceeding $20$ are $$\begin{equation*} 7,\quad 11,\quad 14,\quad 18. \end{equation*}$$ None of these equals $20$. A contradiction. Hence at least one orbit has size $1$. If $[a]=\{a\}$, then $g\cdot a=a$ for all $g\in G$. Hence $G$ has a fixed point in $S$. $\square$ ::: :::solved-problem Let $S$ be a finite $G$-set, where $|G|=p^n$ for a prime $p$. Let $$\begin{equation*} S_0=\{a\in S:g\cdot a=a\text{ for all }g\in G\}. \end{equation*}$$ Prove that $$\begin{equation*} |S|\equiv |S_0|\pmod p. \end{equation*}$$ ::: :::solution Let $S$ be a finite $G$-set with $|G|=p^n$, where $p$ is prime. Given that $S_0=\{a\in S:g\cdot a=a\text{ for all }g\in G\}$. To prove that $|S|\equiv |S_0|\pmod p$. Let $A$ contain exactly one representative from each orbit. By orbit decomposition, $$\begin{equation*} |S|=\sum_{a\in A}|[a]|. \end{equation*}$$ An element $a$ lies in $S_0$ if and only if $G_a=G$, and this is equivalent to $$\begin{equation*} |[a]|=1. \end{equation*}$$ If $a\notin S_0$, then $G_a\neq G$. Since $|G|=p^n$, the index $[G:G_a]$ is a positive power of $p$ greater than $1$. Hence $p$ divides $|[a]|$. Therefore the orbit decomposition may be written as $$\begin{equation*} |S|=|S_0|+\sum p m_i \end{equation*}$$ for suitable integers $m_i$. Hence $$\begin{equation*} |S|\equiv |S_0|\pmod p. \end{equation*}$$ $\square$ ::: :::tip For finite actions, never count group elements and set elements as though they were the same kind of object. The orbit lies inside $S$, the stabilizer lies inside $G$, and the orbit-stabilizer formula bridges them through the index $[G:G_a]$. This distinction prevents many errors in orbit counting. ::: :::exercise 1. Let $|G|=48$ and let $|G_a|=12$. Find $|[a]|$. 2. Let $S$ be a finite $G$-set with orbits of sizes $1,3,3,6$. Find $|S|$. 3. Let a group of order $35$ act on a set of $12$ elements. List the possible orbit sizes. ::: :::answer 1. By the orbit-stabilizer formula, $|[a]|=48/12=4$. 2. Since the orbits partition $S$, $|S|=1+3+3+6=13$. 3. The possible orbit sizes divide $35$, so they are $1,5,7,35$. Since the set has only $12$ elements, an actual orbit may have size only $1,5$, or $7$. ::: :::faq[Frequently Asked Questions] Q: Does the orbit-stabilizer formula require $S$ to be finite? A: The index statement $|[a]|=[G:G_a]$ is general, but the quotient formula $|[a]|=|G|/|G_a|$ is used when $G$ is finite. Q: Why does a fixed point correspond to an orbit of size one? A: If every group element fixes $a$, then no action moves $a$ away from itself, so $[a]=\{a\}$. Q: Why are orbits useful for counting? A: Orbits partition the set into disjoint pieces, so the size of the whole set is the sum of the orbit sizes. ::: The orbit-stabilizer formula says that orbit size and stabilizer size multiply back to the group size in a finite action. Use the inputs to compare a proposed orbit size and stabilizer size against $|G|$. We observe that a larger stabilizer leaves fewer distinct destinations for the point, while a smaller stabilizer allows a larger orbit. This preview makes the inverse relationship visible before using the formula in counting arguments. :::scientific-preview[Orbit Stabilizer Balance Preview] @libraries: none ::: :::call-to-action[Continue to Permutation Representations] subtitle: Learn how every group action produces a homomorphism into a permutation group. button: Study Permutation Representations | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/permutation-representations-and-extended-cayley-theorem button-ghost: Review Orbits and Stabilizers | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/orbits-and-stabilizers :::