# Showing that Aut(G) is a group

## Homework Statement

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

## 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.

fresh_42
Mentor

## Homework Statement

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

## 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.
Correct, but you have forgotten what most students tend to forget. It's as easy as the rest, but for the sake of completion, it is very important: closure! Why is a composition of automorphisms an automorphism again?

member 587159
Correct, but you have forgotten what most students tend to forget. It's as easy as the rest, but for the sake of completion, it is very important: closure! Why is a composition of automorphisms an automorphism again?
4) closure: Let ##f,g \in \operatorname{Aut} (G)##. Then ##h = f \circ g \in \operatorname{Aut} (G)## becuase ##h## is still a bijection, and ##h## is a homomorphism because composition of homomorphisms results in a homomorphism: ##h (gg') = f(g(hh')) = f(g(h)g(h')) = f(h(g))f(h(g')) = h(g)h(g')##.

fresh_42
Mentor
4) closure: Let ##f,g \in \operatorname{Aut} (G)##. Then ##h = f \circ g \in \operatorname{Aut} (G)## becuase ##h## is still a bijection, and ##h## is a homomorphism because composition of homomorphisms results in a homomorphism: ##h (gg') = f(g(hh')) = f(g(h)g(h')) = f(h(g))f(h(g')) = h(g)h(g')##.
I just type this to control your line. There are a bit too many ##g,h## meaning different things.
##h(ab)=f(g(ab))=f(g(a)g(b))= f(g(a))f(g(b))=h(a)h(b)##

Mr Davis 97