# Orders of Isomorphisms

1. Oct 21, 2008

### SNOOTCHIEBOOCHEE

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

let phi : G --->G' be an isomorphism of groups. let x element of G and let x'=phi(x)

Prove that the orders of x and x' are equal

3. The attempt at a solution

I dont even know what the order of a isomorphism means. As far as i know, an isomorphism is just a bijective map from G to G'. How does this have order?

Last edited: Oct 21, 2008
2. Oct 21, 2008

### morphism

The question isn't asking you to find the order of an isomorphism. It's asking you to look at the orders of x and x'.

Also, generally speaking, an isomorphism can have an order: there are groups whose elements are maps.

3. Oct 21, 2008

### HallsofIvy

Staff Emeritus
An isomophism is NOT just a "bijective map from G to G'". It is a bijective map that preserves the operation: phi(x*y)= phi(x).phi(y) where * is the operation in G and . is the operation in G'.

As morphism told you, the question does not ask anything about "order of an isomophism"- it asks about the orders of x and phi(x), a member of G and a member of G'.

4. Oct 21, 2008

### SNOOTCHIEBOOCHEE

Ok im still lost on this problem.

I know we want to show that xn=eg and phi(x)n = eg' for some integer n.

but i dont know how to do this.

5. Oct 21, 2008

### HallsofIvy

Staff Emeritus
No, you want to show that IF xn= eG, then (phi(x)n= eG'. try applying phi to both sides of the first equation.

Last edited: Oct 22, 2008
6. Oct 21, 2008

### SNOOTCHIEBOOCHEE

ok so phi(xn)= phi(eg)
==> phi(xn)= eg'

because phi is an isom

phi(xn)= phi(x)n

and phi(x)n=eg'

thus both x and x' have order n

//

That good?