# The set of ring automorphisms is an abstract group under composition

1. Mar 22, 2007

### catcherintherye

1. The problem statement, all variables and given/known data

Aut(R) denotes the set of ring automorphisms of a ring R Show formally that Aut(R) is a group under composition.

2. Relevant equations

3. The attempt at a solution

I Have a very similar question to which I have the solution viz

Aut(G) denotes the set of group automorphisms of a Group G, show that Aut(G) is a group under composition.

Proof Let a,b: G -> G be automorphisms

then $$a\circ b: G \rightarrow g is also an auto$$

$$(a\circ b)(xy) = a(b(xy))=a(b(x)b(y))= a(b(x))a(b(y))$$
$$=(a\circ b)(x)(a\circ b)(y)$$

so $$a\circb$$ is a homomorphism

it is also bijective since a,b are bijective

$$\circ: aut(G)\times aut(G) \rightarrow aut(G)$$
is automatically associative (because comp of mappings is associative)

As identity in Aut(G) $$take Id_g:G \rightarrow G$$
finally inverses

Let a: G --> G be an auto

then a^-1:G -->G is atleast a mapping and bijective

need only show

$$a^-1(xy) = a^-1(x)a^-1(y)$$

let $$x,y \in G$$

choose $$c,d \in G$$ : a(c)=x, a(d)=y

$$a^-1(xy) = a^-1(a(c)a(d))=a^-1(a(cd)) = cd= a^-1(x)a^-1(y)$$

q.e.d

.......the proof for rings is essentially the same right? The only thing that concerns me is the last part(above) since we used the fact that every element has it's inverse in a group but we don't have that in a ring....

Last edited: Mar 22, 2007
2. Mar 22, 2007

### matt grime

Where have you used that fact?

3. Mar 22, 2007

### catcherintherye

ok so i haven't and these two proofs are essentially the same?

4. Mar 22, 2007

### HallsofIvy

Staff Emeritus
You did use a-1 where a is automorphism. You say you know
Are you sure of that definition of Aut(G)? Homomorphism, in general, do not have inverses.

5. Mar 22, 2007

### catcherintherye

no i've made a mistake it's supposed to be the group of isomorphisms