Having defined group actions and G-Sets, the next natural question is what happens to a single element when every group element is allowed to act on it. The collection of all positions reachable from that element is called its orbit, while the group elements that leave it unchanged form its stabilizer. These two ideas are the first structural tools in group actions because they separate movement from symmetry. The focus keyword for this lesson is orbits and stabilizers. In this class note, we shall prove that orbits form equivalence classes, define stabilizers as subgroups, and compute them in the natural action of $S_3$. :::definition[Orbit Relation] Let $G$ be a group and let $S$ be a $G$-set. Define a relation $\sim$ on $S$ by $$\begin{equation*} a\sim b \end{equation*}$$ if and only if there exists $g\in G$ such that $$\begin{equation*} g\cdot a=b. \end{equation*}$$ This relation says that $b$ is reachable from $a$ by the action of some element of $G$. ::: :::theorem[Orbit Relation Is an Equivalence Relation] Let $G$ be a group and let $S$ be a $G$-set. The relation $\sim$ on $S$ defined by $a\sim b$ if and only if $g\cdot a=b$ for some $g\in G$ is an equivalence relation on $S$. ::: :::proof Given that $G$ is a group, $S$ is a $G$-set, and $a\sim b$ if and only if $g\cdot a=b$ for some $g\in G$. To prove that $\sim$ is an equivalence relation on $S$. [1] To prove reflexivity. Let $a\in S$. Since $e$ is the identity element of $G$, the identity action law gives $$\begin{equation*} e\cdot a=a. \end{equation*}$$ Therefore $a\sim a$. [2] To prove symmetry. Let $a,b\in S$ and suppose $a\sim b$. Then there exists $g\in G$ such that $$\begin{equation*} g\cdot a=b. \end{equation*}$$ Acting by $g^{-1}$ on both sides gives $$\begin{align*} g^{-1}\cdot b&=g^{-1}\cdot(g\cdot a)\\ &=(g^{-1}g)\cdot a \quad [\because \text{ action law}]\\ &=e\cdot a\\ &=a. \end{align*}$$ Therefore $b\sim a$. [3] To prove transitivity. Let $a,b,c\in S$ and suppose $a\sim b$ and $b\sim c$. Then there exist $g_1,g_2\in G$ such that $$\begin{equation*} g_1\cdot a=b \end{equation*}$$ and $$\begin{equation*} g_2\cdot b=c. \end{equation*}$$ Using $b=g_1\cdot a$, we get $$\begin{align*} c&=g_2\cdot b\\ &=g_2\cdot(g_1\cdot a)\\ &=(g_2g_1)\cdot a \quad [\because \text{ action law}]. \end{align*}$$ Since $g_2g_1\in G$, we get $a\sim c$. Hence $\sim$ is an equivalence relation on $S$. $\square$ ::: :::definition[Orbit of an Element] Let $G$ be a group and let $S$ be a $G$-set. For $a\in S$, the orbit of $a$ is the equivalence class of $a$ under the orbit relation. It is denoted by $[a]$ or $G\cdot a$, and it is given by $$\begin{equation*} [a]=G\cdot a=\{g\cdot a:g\in G\}. \end{equation*}$$ ::: An orbit measures movement. Starting with one element $a\in S$, we let every element of $G$ act on it and collect all results. If two elements lie in the same orbit, then the group action can move one into the other. Since the orbit relation is an equivalence relation, the orbits partition $S$ into disjoint reachable regions. Students often confuse the orbit of $a$ with the whole set $S$; that is true only when the action is transitive. :::definition[Stabilizer of an Element] Let $G$ be a group and let $S$ be a $G$-set. For $a\in S$, the stabilizer of $a$ is the set $$\begin{equation*} G_a=\{g\in G:g\cdot a=a\}. \end{equation*}$$ It is also called the isotropy group of $a$. ::: The stabilizer measures symmetry at a single point. While the orbit asks where $a$ can go, the stabilizer asks which group elements keep $a$ fixed. In geometry, these are the symmetries that preserve a chosen point; in permutation groups, they are the permutations that do not move a chosen symbol. This contrast between movement and fixing is the reason orbits and stabilizers appear together throughout finite group theory. Later, the orbit-stabilizer formula will connect the size of the orbit to the size of the stabilizer. :::theorem[Stabilizer Is a Subgroup] Let $G$ be a group and let $S$ be a $G$-set. For $a\in S$, the stabilizer $G_a=\{g\in G:g\cdot a=a\}$ is a subgroup of $G$. ::: :::proof Given that $G$ is a group, $S$ is a $G$-set, $a\in S$, and $G_a=\{g\in G:g\cdot a=a\}$. To prove that $G_a$ is a subgroup of $G$. [1] To prove that $G_a$ is non-empty. Since $$\begin{equation*} e\cdot a=a, \end{equation*}$$ we get $e\in G_a$. Therefore $G_a$ is non-empty. [2] To prove closure under products. Let $g,h\in G_a$. Then $$\begin{equation*} g\cdot a=a \end{equation*}$$ and $$\begin{equation*} h\cdot a=a. \end{equation*}$$ Now $$\begin{align*} (gh)\cdot a&=g\cdot(h\cdot a) \quad [\because \text{ action law}]\\ &=g\cdot a\\ &=a. \end{align*}$$ Therefore $gh\in G_a$. [3] To prove closure under inverses. Let $h\in G_a$. Then $$\begin{equation*} h\cdot a=a. \end{equation*}$$ Acting by $h^{-1}$ on both sides gives $$\begin{align*} h^{-1}\cdot(h\cdot a)&=h^{-1}\cdot a\\ (h^{-1}h)\cdot a&=h^{-1}\cdot a \quad [\because \text{ action law}]\\ e\cdot a&=h^{-1}\cdot a\\ a&=h^{-1}\cdot a. \end{align*}$$ Thus $h^{-1}\cdot a=a$, so $h^{-1}\in G_a$. Hence $G_a$ is a subgroup of $G$. $\square$ ::: :::example[Natural Action of S Three] Let $I_3=\{1,2,3\}$ and let $S_3$ act on $I_3$ by $$\begin{equation*} \sigma\cdot a=\sigma(a) \end{equation*}$$ for $\sigma\in S_3$ and $a\in I_3$. The identity permutation fixes every element, and composition of permutations satisfies the action law. Therefore $I_3$ is an $S_3$-set. Since a permutation in $S_3$ can send any element of $I_3$ to any other element of $I_3$, all three elements lie in one orbit: $$\begin{equation*} [1]=[2]=[3]=\{1,2,3\}. \end{equation*}$$ Hence the natural action is transitive. ::: :::example[Stabilizers in S Three] Let $S_3$ act naturally on $I_3=\{1,2,3\}$. The stabilizer of $1$ is $$\begin{equation*} G_1=\{\sigma\in S_3:\sigma(1)=1\}. \end{equation*}$$ The permutations fixing $1$ are $e$ and $(23)$, so $$\begin{equation*} G_1=\{e,(23)\}. \end{equation*}$$ Similarly, $$\begin{equation*} G_2=\{e,(13)\} \end{equation*}$$ and $$\begin{equation*} G_3=\{e,(12)\}. \end{equation*}$$ Thus each stabilizer has two elements, although the fixed point is different in each case. ::: Use the inputs to explore a simple finite action. The group $\mathbb{Z}_n$ acts on itself by addition modulo $n$, and the orbit of a point $x$ under a chosen subgroup generated by $r$ is the set of all values $x+kr\pmod n$. Change $n$, $r$, and $x$ to see when the orbit is small and when it becomes the whole set. This numerical experiment prepares the intuition for orbits and stabilizers before the counting formula appears. :::scientific-preview[Modular Orbit Explorer] @libraries: none ::: :::solved-problem Let $G=\mathbb{Z}_6$ act on $S=\mathbb{Z}_6$ by addition modulo $6$, so that $$\begin{equation*} g\cdot x=g+x\pmod 6. \end{equation*}$$ Find the orbit and stabilizer of $0$. ::: :::solution Let $G=\mathbb{Z}_6$ and $S=\mathbb{Z}_6$. Given that $g\cdot x=g+x\pmod 6$. To find the orbit and stabilizer of $0$. The orbit of $0$ is $$\begin{align*} [0]&=\{g\cdot 0:g\in\mathbb{Z}_6\}\\ &=\{g+0\pmod 6:g\in\mathbb{Z}_6\}\\ &=\{0,1,2,3,4,5\}. \end{align*}$$ The stabilizer of $0$ is $$\begin{align*} G_0&=\{g\in\mathbb{Z}_6:g\cdot 0=0\}\\ &=\{g\in\mathbb{Z}_6:g+0\equiv 0\pmod 6\}\\ &=\{0\}. \end{align*}$$ Hence $[0]=\mathbb{Z}_6$ and $G_0=\{0\}$. $\square$ ::: :::tip The orbit is a subset of the acted-upon set $S$, while the stabilizer is a subset of the group $G$. This distinction is essential. In the natural action of $S_3$ on $\{1,2,3\}$, an orbit contains numbers such as $1,2,3$, but a stabilizer contains permutations such as $e$ and $(23)$. ::: :::exercise 1. Let $S_4$ act naturally on $\{1,2,3,4\}$. Find the orbit of $1$. 2. For the same action, find the stabilizer of $1$. 3. Let $G$ act on itself by left multiplication. Find the stabilizer of $e$. ::: :::answer 1. The orbit of $1$ is $\{1,2,3,4\}$ because a permutation in $S_4$ can send $1$ to any element of the set. 2. The stabilizer of $1$ is $\{\sigma\in S_4:\sigma(1)=1\}$, which is naturally identified with the permutations of $\{2,3,4\}$. 3. If $g\cdot e=ge=e$, then $g=e$. Hence the stabilizer of $e$ is $\{e\}$. ::: :::faq[Frequently Asked Questions] Q: Are two different orbits allowed to overlap? A: No. Since orbits are equivalence classes, two orbits are either equal or disjoint. Q: Is the stabilizer always a subgroup? A: Yes. The identity fixes the point, products of fixing elements fix the point, and inverses of fixing elements also fix the point. Q: Can an orbit have only one element? A: Yes. That happens exactly when every group element fixes that element. ::: The natural action of $S_n$ on $\{1,2,\ldots,n\}$ gives a clean model for stabilizers. Choose $n$ and a point, and the calculator gives the size of the orbit and the size of the stabilizer of that point. We observe that every point can move to $n$ possible places, while the permutations fixing one chosen point may freely permute the remaining $n-1$ symbols. This prepares the numerical form of the orbit-stabilizer formula. :::calculator[Stabilizer Size Calculator] @libraries: none ::: :::call-to-action[Continue to the Orbit-Stabilizer Formula] subtitle: Connect the number of reachable elements with the number of group elements that fix a point. button: Study Orbit-Stabilizer Formula | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/orbit-stabilizer-formula-and-orbit-counting button-ghost: Review Group Actions and G-Sets | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/group-actions-and-g-sets :::