- #1

Mr Davis 97

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- 44

## Homework Statement

Prove that, for any group ##G##, the set ##\operatorname{Aut} (G)## is a group under composition of functions.

## Homework Equations

## The Attempt at a Solution

1) associativity: It is a known fact of set theory that composition of functions is an associative binary operation.

2) identity: ##\operatorname{id}_G## is in ##\operatorname{Aut} (G)## because it is a group isomorphism from ##G## to ##G##. By the properties of the identity set-theoretic map, if ##f \in \operatorname{Aut} (G)## then ##\operatorname{id}_G \circ f = f \circ \operatorname{id}_G = f##.

3) inverse: Let ##f \in \operatorname{Aut} (G)##. Then the set-theoretic inverse ##f^{-1}## is still an isomorphism from ##G## to ##G## (I have proven this fact previously), so ##f^{-1} \in \operatorname{Aut} (G)##. And since ##f^{-1} \circ f = f \circ f^{-1} = \operatorname{id}_G##, every element in ##\operatorname{Aut} (G)## has an inverse.