# Showing that Aut(G) is a group

• Mr Davis 97
In summary, we have proven that the set ##\operatorname{Aut} (G)## is a group under composition of functions by showing associativity, the existence of an identity element, the existence of inverses, and closure. This is because composition of automorphisms results in an automorphism, making it a closed binary operation.
Mr Davis 97

## 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.

Mr Davis 97 said:

## 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
fresh_42 said:
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')##.

Mr Davis 97 said:
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

## 1. What is Aut(G)?

Aut(G) is the group of automorphisms of a group G, which are bijective homomorphisms from G to itself. In simpler terms, it is the set of all ways in which a group can be rearranged without changing its structure or properties.

## 2. How do you show that Aut(G) is a group?

To show that Aut(G) is a group, we need to prove that it satisfies the four group axioms: closure, associativity, identity, and inverse. This means that the composition of two automorphisms must be an automorphism, the composition is associative, the identity function is an automorphism, and every automorphism has an inverse.

## 3. Why is it important to prove that Aut(G) is a group?

Proving that Aut(G) is a group is important because it helps us understand the structure of the group G better. It also allows us to use group theory techniques to study the properties and behaviors of automorphisms, which can be applied to various fields such as algebra, geometry, and physics.

## 4. How do you prove closure for Aut(G)?

To prove closure for Aut(G), we need to show that the composition of two automorphisms is also an automorphism. This can be done by showing that the composition preserves the group structure and is bijective. In other words, if f and g are automorphisms of G, then (f ∘ g)(x) = f(g(x)) is also an automorphism.

## 5. Can Aut(G) be isomorphic to G?

No, Aut(G) cannot be isomorphic to G. This is because Aut(G) is a group of functions, while G is a set of elements. Isomorphism can only exist between groups with the same algebraic structure, and a group and its automorphism group have different structures.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Differential Geometry
Replies
0
Views
530
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
24
Views
3K
• Calculus and Beyond Homework Help
Replies
2
Views
1K