Showing that Aut(G) is a group

[Mr Davis 97]

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.

 

[fresh_42]

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.

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?

 

[Mr Davis 97]

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]

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.
$h(ab)=f(g(ab))=f(g(a)g(b))= f(g(a))f(g(b))=h(a)h(b)$