- #1

RJLiberator

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## Homework Statement

Let G be a group. An Isomorphism Φ: G --> G is called an automorphism of G. Let Aut(G) denote the set of all automorphisms of G.

Prove that Aut(G) is a subgroup of Bij(G).

## Homework Equations

For it to be a subgroup we need to show:

i) e ∈ Aut(G)

ii) For all x,y ∈ Aut(G), xy∈ Bij(G)

iii) For all x ∈Aut(G), x^-1 ∈ Aut(G)

## The Attempt at a Solution

i) Clearly the identity, e: G--> G is an automorphism. This is trivial.

ii) Φ(xy) = Φ(x) Φ(y) ??

Let Φ = Φ2⋅Φ1

Then Φ(xy) = Φ2( Φ1(xy)) = Φ2( Φ1(x) Φ1(y)) = Φ2( Φ1(x)) Φ2( Φ1(y)) = Φ(x) Φ(y)

iii) Let u,v ∈ G, want to show that Φ^-1(uv) = Φ^1(u)⋅ Φ^-1(v)

Let x = Φ^-1(u) and y = Φ^-1(v) so that Φ(x) = u and Φ(y) = v.

Φ^-1(uv) = Φ^-1( Φ(x) Φ(y)) = Φ^-1( Φ(xy)) = xy = Φ^-1(u) Φ^-1(v)

Thus, Φ^-1 is an automorphism.

Therefore, by i, ii, iii, Aut(G) is a subgroup of Bij(G).

Correct? :D