After learning how homomorphisms compare two groups through structure-preserving maps, we now ask a different but equally powerful question: how can one group move the elements of a set? This is the starting point of group actions and G-Sets. A group action translates an abstract group into visible transformations of a set, so that algebraic multiplication becomes movement, symmetry, or rearrangement. Students often think that a group action is only a formula, but the essential point is compatibility: multiplying first in the group must give the same result as acting step by step. In this lesson we shall define left actions, introduce $G$-sets, and verify the standard actions coming from permutations, conjugation, and matrices. :::definition[Left Action of a Group on a Set] Let $G$ be a group with identity element $e$, and let $S$ be a non-empty set. A left action of $G$ on $S$ is a function $$\begin{equation*} G\times S\to S \end{equation*}$$ written as $$\begin{equation*} (g,x)\mapsto g\cdot x \end{equation*}$$ such that the following conditions hold for all $g_1,g_2\in G$ and all $x\in S$: (i) $(g_1g_2)\cdot x=g_1\cdot(g_2\cdot x)$, (ii) $e\cdot x=x$. ::: The first action law says that the group operation and the action are compatible. If $g_2$ acts first and $g_1$ acts next, the final result must be the same as letting the product $g_1g_2$ act once. The second action law says that the identity element of the group does not move any element of the set. A common mistake here is to check only that $g\cdot x$ belongs to $S$; that proves the formula is closed in $S$, but it does not prove the two action laws. The focus keyword group actions and G-Sets should always remind us that the group structure and the set movement must work together. :::definition[G-Set] Let $G$ be a group and let $S$ be a non-empty set. If a left action of $G$ on $S$ is given, then $S$ is called a $G$-set. In that case we say that $G$ acts on $S$ on the left. ::: When the action is understood, it is common to write $gx$ instead of $g\cdot x$. With this convention, the two defining laws become $$\begin{equation*} (g_1g_2)x=g_1(g_2x) \end{equation*}$$ and $$\begin{equation*} ex=x. \end{equation*}$$ This notation is convenient, but it must be read carefully. The expression $gx$ is not necessarily a product inside the group $G$; it means that the group element $g$ acts on the set element $x$. :::theorem[Elements Act as Bijective Transformations] Let $G$ be a group and let $S$ be a $G$-set. For each $g\in G$, define $\tau_g:S\to S$ by $\tau_g(x)=g\cdot x$. Then $\tau_g$ is a bijection from $S$ to $S$. ::: :::proof Given that $G$ is a group, $S$ is a $G$-set, and $\tau_g:S\to S$ is defined by $\tau_g(x)=g\cdot x$ for a fixed $g\in G$. To prove that $\tau_g$ is a bijection. [1] To prove that $\tau_g$ is one-to-one. Let $x,y\in S$ and suppose $\tau_g(x)=\tau_g(y)$. Then $$\begin{equation*} g\cdot x=g\cdot y. \end{equation*}$$ Acting by $g^{-1}$ on both sides gives $$\begin{align*} g^{-1}\cdot(g\cdot x)&=g^{-1}\cdot(g\cdot y)\\ (g^{-1}g)\cdot x&=(g^{-1}g)\cdot y \quad [\because \text{ action law}]\\ e\cdot x&=e\cdot y\\ x&=y. \end{align*}$$ Therefore $\tau_g$ is one-to-one. [2] To prove that $\tau_g$ is onto. Let $y\in S$. Put $x=g^{-1}\cdot y$. Then $x\in S$, and $$\begin{align*} \tau_g(x)&=g\cdot x\\ &=g\cdot(g^{-1}\cdot y)\\ &=(gg^{-1})\cdot y \quad [\because \text{ action law}]\\ &=e\cdot y\\ &=y. \end{align*}$$ Therefore $\tau_g$ is onto. Hence $\tau_g$ is a bijection from $S$ to $S$. $\square$ ::: :::example[Natural Action of a Permutation Group] Let $G$ be a permutation group on a non-empty set $S$. For $\sigma\in G$ and $x\in S$, define $$\begin{equation*} \sigma\cdot x=\sigma(x). \end{equation*}$$ [1] The identity permutation $e$ satisfies $$\begin{equation*} e\cdot x=e(x)=x \end{equation*}$$ for every $x\in S$. [2] For $\sigma_1,\sigma_2\in G$ and $x\in S$, composition of functions gives $$\begin{align*} (\sigma_1\sigma_2)\cdot x&=(\sigma_1\sigma_2)(x)\\ &=\sigma_1(\sigma_2(x))\\ &=\sigma_1\cdot(\sigma_2\cdot x). \end{align*}$$ Hence, $S$ is a $G$-set under the natural permutation action. $\square$ ::: :::example[Conjugation Action on a Normal Subgroup] Let $G$ be a group and let $H\trianglelefteq G$. Define a function $G\times H\to H$ by $$\begin{equation*} g\cdot h=ghg^{-1}. \end{equation*}$$ [1] Since $H\trianglelefteq G$, we have $ghg^{-1}\in H$ for all $g\in G$ and all $h\in H$. Therefore the function is well-defined. [2] For the identity element $e\in G$, $$\begin{equation*} e\cdot h=ehe^{-1}=h \end{equation*}$$ for every $h\in H$. [3] For $g_1,g_2\in G$ and $h\in H$, $$\begin{align*} (g_1g_2)\cdot h&=(g_1g_2)h(g_1g_2)^{-1}\\ &=g_1g_2hg_2^{-1}g_1^{-1}\\ &=g_1(g_2hg_2^{-1})g_1^{-1}\\ &=g_1\cdot(g_2\cdot h). \end{align*}$$ Hence, $H$ is a $G$-set under conjugation. $\square$ ::: Use the inputs to test a familiar finite action: $\mathbb{Z}_n$ acting on itself by addition modulo $n$. Enter a modulus $n$, two group elements $r,s$, and a set element $x$. Compare $(r+s)\cdot x$ with $r\cdot(s\cdot x)$, and also compare $0\cdot x$ with $x$. This is a small numerical model of group actions and G-Sets, because it shows the action laws without hiding them behind notation. :::scientific-preview[Cyclic Action Law Explorer] @libraries: none ::: :::solved-problem Let $G=GL(2,\mathbb{R})$ and let $S=\mathbb{R}^2$. For $$\begin{equation*} A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in GL(2,\mathbb{R}) \end{equation*}$$ define $$\begin{equation*} A\cdot(x,y)=(ax+by,cx+dy). \end{equation*}$$ Show that $S$ is a $G$-set. ::: :::solution Let $G=GL(2,\mathbb{R})$ and $S=\mathbb{R}^2$. Given that $A\cdot(x,y)=(ax+by,cx+dy)$ for $A\in GL(2,\mathbb{R})$ and $(x,y)\in\mathbb{R}^2$. To prove that $S$ is a $G$-set. The formula is ordinary multiplication of the matrix $A$ with the column vector determined by $(x,y)$, so $A\cdot(x,y)\in\mathbb{R}^2$. [1] The identity matrix $$\begin{equation*} I=\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{equation*}$$ satisfies $$\begin{align*} I\cdot(x,y)&=(1x+0y,0x+1y)\\ &=(x,y). \end{align*}$$ [2] Let $A,B\in GL(2,\mathbb{R})$ and $(x,y)\in\mathbb{R}^2$. Matrix multiplication gives $$\begin{equation*} (AB)\cdot(x,y)=A\cdot(B\cdot(x,y)). \end{equation*}$$ Therefore the identity law and compatibility law both hold. Hence $\mathbb{R}^2$ is a $GL(2,\mathbb{R})$-set under the given action. $\square$ ::: :::tip To verify a proposed group action, first check that the formula really lands inside the set being acted upon. Then check the identity law. Only after that should you verify the compatibility law, because most incorrect examples fail at this third step. For group actions and G-Sets, the order of multiplication matters in non-abelian groups. ::: :::exercise 1. Let $G$ act on itself by left multiplication, $g\cdot x=gx$. Prove that this is a left action of $G$ on $G$. 2. Let $G$ act on itself by conjugation, $g\cdot x=gxg^{-1}$. Prove that this is a left action of $G$ on $G$. 3. Let $\mathbb{Z}$ act on $\mathbb{R}$ by $n\cdot x=x+n$. Prove that $\mathbb{R}$ is a $\mathbb{Z}$-set. ::: :::answer 1. Since $e\cdot x=ex=x$ and $(g_1g_2)\cdot x=(g_1g_2)x=g_1(g_2x)=g_1\cdot(g_2\cdot x)$, this is a left action. 2. Since $e\cdot x=exe^{-1}=x$ and $(g_1g_2)\cdot x=g_1g_2xg_2^{-1}g_1^{-1}=g_1\cdot(g_2\cdot x)$, this is a left action. 3. Since $0\cdot x=x$ and $(m+n)\cdot x=x+(m+n)=(x+n)+m=m\cdot(n\cdot x)$, this is a left action. ::: :::faq[Frequently Asked Questions] Q: Is a group action the same thing as a group operation? A: No. A group operation combines two elements of $G$, while a group action lets an element of $G$ move an element of a set $S$. Q: Why must $S$ be non-empty? A: In the usual undergraduate definition, $S$ is taken to be non-empty so that the action describes movement of actual set elements. Q: Does every group action give permutations of the set? A: Yes. For each fixed $g\in G$, the function $x\mapsto g\cdot x$ is a bijection of $S$. ::: A group action calculator helps separate the group element from the set element. Enter a modulus $n$, a group element $a$, and a point $x$ to evaluate the action $a\cdot x=a+x\pmod n$. Notice that changing $a$ changes the motion, while changing $x$ changes the point being moved. This numerical check reinforces the definition of a $G$-set before more abstract examples appear. :::calculator[Modular Action Calculator] @libraries: none ::: :::call-to-action[Continue to Orbits and Stabilizers] subtitle: Study how a group action partitions a set into reachable parts and fixes elements through stabilizer subgroups. button: Study Orbits and Stabilizers | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/group-actions/orbits-and-stabilizers button-ghost: Review Permutation Groups | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/permutation-groups :::