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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.

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

(1) We know from set theory that function composition is always associative.

(2) 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.

(3) 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.


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This Post Has One Comment

  1. Do we need to show that function composition is "closed"? In other words, that the composition of two arbitrary functions in Aut(G) produces another isomorphism?

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