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.

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