[proofplan] We verify the group axioms for $\operatorname{Aut}(G)$ with operation given by composition of maps. The main points are that the composite of two bijective homomorphisms is again a bijective homomorphism, the identity map on $G$ is an automorphism, and the inverse function of an automorphism is again a homomorphism. Associativity is inherited from associativity of ordinary function composition. [/proofplan] [step:Define the composition operation on automorphisms] Let $\operatorname{Aut}(G)$ be the set of all bijective group homomorphisms $\varphi: G \to G$. Define the binary operation \begin{align*} \circ: \operatorname{Aut}(G) \times \operatorname{Aut}(G) &\to \operatorname{Aut}(G) \\ (\varphi,\psi) &\mapsto \varphi \circ \psi \end{align*} provided that $\varphi \circ \psi$ is again an automorphism. For elements $g \in G$, the composite map is defined by \begin{align*} (\varphi \circ \psi)(g)=\varphi(\psi(g)). \end{align*} We now verify that this operation is well-defined and satisfies the group axioms. [/step] [step:Show that the composite of two automorphisms is an automorphism] Let \begin{align*} \varphi: G &\to G,\\ \psi: G &\to G \end{align*} be elements of $\operatorname{Aut}(G)$. Since $\varphi$ and $\psi$ are bijections, their composite $\varphi \circ \psi: G \to G$ is a bijection by the standard property of [bijective functions](/page/Bijection). It remains to verify the homomorphism property. Let $a,b \in G$. Since $\psi$ is a group homomorphism, $\psi(a \cdot b)=\psi(a)\cdot \psi(b)$. Since $\varphi$ is a group homomorphism, applying $\varphi$ to this product gives \begin{align*} (\varphi \circ \psi)(a \cdot b) &= \varphi(\psi(a \cdot b))\\ &= \varphi(\psi(a)\cdot \psi(b))\\ &= \varphi(\psi(a))\cdot \varphi(\psi(b))\\ &= (\varphi \circ \psi)(a)\cdot(\varphi \circ \psi)(b). \end{align*} Thus $\varphi \circ \psi$ is a group homomorphism. Therefore $\varphi \circ \psi \in \operatorname{Aut}(G)$, so composition is closed on $\operatorname{Aut}(G)$. [guided] Take two arbitrary automorphisms \begin{align*} \varphi: G &\to G,\\ \psi: G &\to G. \end{align*} To prove that their composite belongs to $\operatorname{Aut}(G)$, we must check exactly the two defining conditions of an automorphism: bijectivity and preservation of the group operation. First, $\varphi$ and $\psi$ are bijections by definition of $\operatorname{Aut}(G)$. The composite of bijections is a bijection, so $\varphi \circ \psi: G \to G$ is bijective. Second, we verify the homomorphism property. Let $a,b \in G$. The map $\psi$ preserves multiplication, so \begin{align*} \psi(a \cdot b)=\psi(a)\cdot \psi(b). \end{align*} The map $\varphi$ also preserves multiplication, so applying $\varphi$ to the product on the right gives \begin{align*} (\varphi \circ \psi)(a \cdot b) &= \varphi(\psi(a \cdot b))\\ &= \varphi(\psi(a)\cdot \psi(b))\\ &= \varphi(\psi(a))\cdot \varphi(\psi(b))\\ &= (\varphi \circ \psi)(a)\cdot(\varphi \circ \psi)(b). \end{align*} This proves that $\varphi \circ \psi$ is a group homomorphism. Since it is both bijective and a homomorphism from $G$ to $G$, it is an automorphism. Hence composition is closed on $\operatorname{Aut}(G)$. [/guided] [/step] [step:Use associativity of function composition] Let \begin{align*} \varphi: G &\to G,\\ \psi: G &\to G,\\ \theta: G &\to G \end{align*} be elements of $\operatorname{Aut}(G)$. For every $g \in G$, \begin{align*} ((\varphi \circ \psi)\circ \theta)(g) &= (\varphi \circ \psi)(\theta(g))\\ &= \varphi(\psi(\theta(g)))\\ &= \varphi((\psi \circ \theta)(g))\\ &= (\varphi \circ (\psi \circ \theta))(g). \end{align*} Since the two maps agree on every element $g \in G$, we have \begin{align*} (\varphi \circ \psi)\circ \theta=\varphi \circ (\psi \circ \theta). \end{align*} Thus composition is associative on $\operatorname{Aut}(G)$. [/step] [step:Identify the identity automorphism] Define the identity map \begin{align*} \operatorname{id}_G: G &\to G\\ g &\mapsto g. \end{align*} The map $\operatorname{id}_G$ is bijective, with inverse function itself. For $a,b \in G$, \begin{align*} \operatorname{id}_G(a\cdot b) &= a\cdot b\\ &= \operatorname{id}_G(a)\cdot \operatorname{id}_G(b), \end{align*} so $\operatorname{id}_G$ is a group homomorphism. Hence $\operatorname{id}_G \in \operatorname{Aut}(G)$. For every $\varphi \in \operatorname{Aut}(G)$ and every $g \in G$, \begin{align*} (\operatorname{id}_G \circ \varphi)(g) &= \operatorname{id}_G(\varphi(g))\\ &= \varphi(g), \end{align*} and \begin{align*} (\varphi \circ \operatorname{id}_G)(g) &= \varphi(\operatorname{id}_G(g))\\ &= \varphi(g). \end{align*} Therefore $\operatorname{id}_G \circ \varphi=\varphi$ and $\varphi \circ \operatorname{id}_G=\varphi$, so $\operatorname{id}_G$ is the identity element for composition on $\operatorname{Aut}(G)$. [/step] [step:Show that the inverse function of an automorphism is an automorphism] Let $\varphi \in \operatorname{Aut}(G)$. Since $\varphi: G \to G$ is bijective, it has an inverse function \begin{align*} \varphi^{-1}: G &\to G. \end{align*} The inverse function $\varphi^{-1}$ is bijective. It remains to prove that $\varphi^{-1}$ is a group homomorphism. Let $x,y \in G$. Since $\varphi$ is surjective, there exist $a,b \in G$ such that \begin{align*} x=\varphi(a), \qquad y=\varphi(b). \end{align*} By definition of $\varphi^{-1}$, this gives $a=\varphi^{-1}(x)$ and $b=\varphi^{-1}(y)$. Since $\varphi$ is a homomorphism, \begin{align*} xy &= \varphi(a)\cdot \varphi(b)\\ &= \varphi(a\cdot b). \end{align*} Applying $\varphi^{-1}$ to both sides yields \begin{align*} \varphi^{-1}(xy) &= a\cdot b\\ &= \varphi^{-1}(x)\cdot \varphi^{-1}(y). \end{align*} Thus $\varphi^{-1}$ is a group homomorphism, and hence $\varphi^{-1}\in \operatorname{Aut}(G)$. Moreover, \begin{align*} \varphi^{-1}\circ \varphi=\operatorname{id}_G, \qquad \varphi\circ \varphi^{-1}=\operatorname{id}_G, \end{align*} by the defining property of the inverse function. Therefore every element of $\operatorname{Aut}(G)$ has an inverse in $\operatorname{Aut}(G)$ under composition. [guided] Let $\varphi \in \operatorname{Aut}(G)$. We need to find its inverse with respect to composition. Because $\varphi$ is an automorphism, it is a bijective map $\varphi: G \to G$, so it has an inverse function \begin{align*} \varphi^{-1}: G &\to G. \end{align*} This inverse function is automatically bijective. The only non-formal point is that $\varphi^{-1}$ must preserve the group operation. Let $x,y \in G$. Since $\varphi$ is surjective, we may write $x=\varphi(a)$ and $y=\varphi(b)$ for some elements $a,b \in G$. These elements satisfy $a=\varphi^{-1}(x)$ and $b=\varphi^{-1}(y)$ by the definition of inverse function. Now use the homomorphism property of $\varphi$: \begin{align*} xy &= \varphi(a)\cdot \varphi(b)\\ &= \varphi(a\cdot b). \end{align*} Applying $\varphi^{-1}$ to this equality gives \begin{align*} \varphi^{-1}(xy) &= \varphi^{-1}(\varphi(a\cdot b))\\ &= a\cdot b\\ &= \varphi^{-1}(x)\cdot \varphi^{-1}(y). \end{align*} Thus $\varphi^{-1}$ is a group homomorphism. Since it is also bijective, $\varphi^{-1}$ is an automorphism of $G$. Finally, the equations \begin{align*} \varphi^{-1}\circ \varphi=\operatorname{id}_G, \qquad \varphi\circ \varphi^{-1}=\operatorname{id}_G \end{align*} are exactly the defining equations for the inverse function of a bijection. Therefore $\varphi^{-1}$ is the inverse of $\varphi$ inside $\operatorname{Aut}(G)$ under composition. [/guided] [/step] [step:Conclude that the automorphisms form a group] We have proved that composition is a closed binary operation on $\operatorname{Aut}(G)$, that composition is associative, that $\operatorname{id}_G$ is an identity element, and that every $\varphi \in \operatorname{Aut}(G)$ has inverse $\varphi^{-1}\in \operatorname{Aut}(G)$. These are precisely the group axioms for the operation $\circ$. Hence $\operatorname{Aut}(G)$ is a group under composition. [/step]