[proofplan] A group action of $G$ on $X$ is a map $G \times X \to X$, $(g, x) \mapsto gx$, satisfying $1x = x$ and $(g_1 g_2) x = g_1 (g_2 x)$. From the action we build, for each $g \in G$, the function $\theta_g: X \to X$, $x \mapsto gx$. The proof has two parts: first we show each $\theta_g$ is a bijection by exhibiting $\theta_{g^{-1}}$ as its two-sided inverse (the action axioms force $\theta_{g^{-1}} \circ \theta_g = \operatorname{id}_X = \theta_g \circ \theta_{g^{-1}}$); then we show that the map $\theta: G \to \operatorname{Sym}(X)$, $g \mapsto \theta_g$, is a homomorphism, again directly from the associativity axiom of the action. [/proofplan] [step:Define the candidate map $\theta_g: X \to X$ for each $g \in G$] Fix $g \in G$. Define \begin{align*} \theta_g: X &\to X \\ x &\mapsto gx, \end{align*} where $gx$ denotes the result of the given group action $G \times X \to X$ applied to the pair $(g, x)$. The map $\theta_g$ is well-defined as a function $X \to X$ because the action takes values in $X$. [/step] [step:Verify $\theta_g$ is a bijection by exhibiting $\theta_{g^{-1}}$ as a two-sided inverse] Let $g \in G$ and consider the maps $\theta_g$ and $\theta_{g^{-1}}$ defined as in the previous step. We compute the two compositions. For any $x \in X$: \begin{align*} (\theta_{g^{-1}} \circ \theta_g)(x) &= \theta_{g^{-1}}(\theta_g(x)) = \theta_{g^{-1}}(gx) = g^{-1}(gx) = (g^{-1} g) x = 1 \cdot x = x, \\ (\theta_g \circ \theta_{g^{-1}})(x) &= \theta_g(\theta_{g^{-1}}(x)) = \theta_g(g^{-1} x) = g(g^{-1} x) = (g g^{-1}) x = 1 \cdot x = x, \end{align*} where the third equality on each line uses the associativity axiom $(ab)x = a(bx)$ of the action, and the fourth uses the identity axiom $1 \cdot x = x$. Therefore $\theta_{g^{-1}} \circ \theta_g = \operatorname{id}_X = \theta_g \circ \theta_{g^{-1}}$, so $\theta_g$ is a bijection $X \to X$ with inverse $\theta_{g^{-1}}$. In particular $\theta_g \in \operatorname{Sym}(X)$. [guided] We need to show that the map $\theta_g$, defined by $\theta_g(x) = gx$, is a bijection — i.e. it lies in $\operatorname{Sym}(X)$, the group of all bijections $X \to X$. The slickest way to prove that a function is bijective is to exhibit its inverse: if there is a function $\psi: X \to X$ with $\psi \circ \theta_g = \operatorname{id}_X$ and $\theta_g \circ \psi = \operatorname{id}_X$, then $\theta_g$ is automatically bijective and $\psi = \theta_g^{-1}$. The natural candidate is $\theta_{g^{-1}}$: morally, "if $g$ moves $x$, then $g^{-1}$ should move it back." Let us verify this rigorously using the action axioms. For any $x \in X$: \begin{align*} (\theta_{g^{-1}} \circ \theta_g)(x) &= \theta_{g^{-1}}(\theta_g(x)) && \text{(definition of composition)}\\ &= \theta_{g^{-1}}(gx) && \text{(definition of } \theta_g\text{)} \\ &= g^{-1}(gx) && \text{(definition of } \theta_{g^{-1}}\text{)} \\ &= (g^{-1} g) x && \text{(associativity axiom of the action)} \\ &= 1 \cdot x && \text{(group inverse: } g^{-1} g = 1\text{)} \\ &= x && \text{(identity axiom of the action).} \end{align*} The reverse composition is computed symmetrically, swapping $g$ and $g^{-1}$: \begin{align*} (\theta_g \circ \theta_{g^{-1}})(x) = g(g^{-1} x) = (g g^{-1}) x = 1 \cdot x = x. \end{align*} Both compositions equal $\operatorname{id}_X$, so $\theta_g$ is a bijection with inverse $\theta_{g^{-1}}$, i.e. $\theta_g \in \operatorname{Sym}(X)$. Note where each axiom enters: associativity is what allows us to "move" $g$ and $g^{-1}$ next to each other inside the action, and the identity axiom lets us conclude that $1$ acts as the identity. Both axioms are needed; neither one alone suffices. [/guided] [/step] [step:Verify $\theta: G \to \operatorname{Sym}(X)$ is a group homomorphism] Define \begin{align*} \theta: G &\to \operatorname{Sym}(X) \\ g &\mapsto \theta_g. \end{align*} This is well-defined by the previous step, which produced $\theta_g \in \operatorname{Sym}(X)$ for each $g \in G$. We must verify that $\theta$ respects the group operations: for all $g_1, g_2 \in G$, \begin{align*} \theta(g_1 g_2) = \theta(g_1) \circ \theta(g_2), \end{align*} i.e. $\theta_{g_1 g_2} = \theta_{g_1} \circ \theta_{g_2}$ as elements of $\operatorname{Sym}(X)$. Two functions on $X$ are equal if and only if they agree pointwise. For any $x \in X$: \begin{align*} \theta_{g_1 g_2}(x) = (g_1 g_2) x = g_1 (g_2 x) = g_1 \cdot \theta_{g_2}(x) = \theta_{g_1}(\theta_{g_2}(x)) = (\theta_{g_1} \circ \theta_{g_2})(x), \end{align*} where the second equality is the associativity axiom of the action and the remaining equalities are the definitions of $\theta_{g_1}$, $\theta_{g_2}$ and composition. This proves $\theta_{g_1 g_2} = \theta_{g_1} \circ \theta_{g_2}$. Therefore $\theta: G \to \operatorname{Sym}(X)$ is a group homomorphism, completing the proof. [guided] We have built a candidate map $\theta: G \to \operatorname{Sym}(X)$ taking $g$ to the bijection $\theta_g$. To conclude that this is a **group homomorphism**, we must verify the homomorphism axiom: for all $g_1, g_2 \in G$, \begin{align*} \theta(g_1 g_2) = \theta(g_1) \circ \theta(g_2). \end{align*} The product on the left is the group product in $G$; the operation on the right is composition in $\operatorname{Sym}(X)$. Unpacked, this asks that $\theta_{g_1 g_2}$ and $\theta_{g_1} \circ \theta_{g_2}$ are the same function $X \to X$. Two functions on $X$ are equal iff they agree at every point. Take an arbitrary $x \in X$ and compute both sides: \begin{align*} \theta_{g_1 g_2}(x) &= (g_1 g_2) x && \text{(definition of } \theta_{g_1 g_2}\text{)} \\ (\theta_{g_1} \circ \theta_{g_2})(x) &= \theta_{g_1}(\theta_{g_2}(x)) = \theta_{g_1}(g_2 x) = g_1 (g_2 x). \end{align*} The two are equal precisely when $(g_1 g_2) x = g_1(g_2 x)$ — which is **exactly** the associativity axiom of the action. So the homomorphism property of $\theta$ is, in disguise, the associativity axiom. (The identity axiom $1 \cdot x = x$ tells us further that $\theta_1 = \operatorname{id}_X$, the identity element of $\operatorname{Sym}(X)$, but for an arbitrary group homomorphism this follows automatically from preservation of products.) Therefore $\theta$ is a group homomorphism $G \to \operatorname{Sym}(X)$. This completes the proof. What this construction shows conceptually: a group action of $G$ on $X$ and a homomorphism $G \to \operatorname{Sym}(X)$ are **the same data in two different forms**. The proof above goes in one direction; the converse — given a homomorphism $\theta$, define $gx := \theta(g)(x)$ — is also straightforward, and the two constructions are mutual inverses. [/guided] [/step]