Proving Equality of Orders in Group Isomorphisms

[SNOOTCHIEBOOCHEE]

Homework Statement

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

The Attempt at a Solution

I don't 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?

 

[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.

 

[HallsofIvy]

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'.

 

[SNOOTCHIEBOOCHEE]

Ok I am still lost on this problem.

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

but i don't know how to do this.

 

[HallsofIvy]

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

 

[SNOOTCHIEBOOCHEE]

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

because phi is an isom

phi(xⁿ)= phi(x)ⁿ

and phi(x)ⁿ=e_g'

thus both x and x' have order n

//

That good?