## The set of all group automorphisms of a fixed group is a group

Let $G$ be a group and let $\mathsf{Aut}(G)$ be the set of all isomorphisms $G \rightarrow G$. Prove that $\mathsf{Aut}(G)$ is a group under function composition.

We need to verify that the three group axioms are satisfied: associativity, identity, and inverses.

• We know from set theory that function composition is always associative.
• Note that $\mathsf{id}_G$ is a bijection and trivially a homomorphism, so that $\mathsf{id}_G \in \mathsf{Aut}(G)$. Finally, we have $\mathsf{id}_G \circ \varphi = \varphi \circ \mathsf{id}_G = \varphi$ for all isomorphisms $\varphi : G \rightarrow G$, so that $\mathsf{id}_G$ is an identity element under composition.
• Given $\varphi \in \mathsf{Aut}(G)$, we know from set theory that an inverse $\varphi^{-1}$ exists. This inverse is a homomorphism, as we show. If $a,b \in G$, then $\varphi(\varphi^{-1}(ab)) = ab = \varphi(\varphi^{-1}(a))\varphi(\varphi^{-1}(b)) = \varphi(\varphi^{-1}(a) \varphi^{-1}(b))$. Since $\varphi$ is injective, we have $\varphi^{-1}(ab) = \varphi^{-1}(a) \varphi^{-1}(b)$. Thus $\varphi^{-1}$ is a homomorphism, and we have $\varphi \circ \varphi^{-1} = \varphi^{-1} \circ \varphi = \mathsf{id}_G$.

Thus $\mathsf{Aut}(G)$ is a group under function composition.