# Abstract Algebra: isomorphism proof

1. Apr 17, 2012

### tiger4

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

Let G be an abelian group of order n. Define phi: G --> G by phi(a) = a^m, where a is in G. Prove that if gcd(m,n) = 1 then phi is an isomorphism

2. Relevant equations

phi(a) = a^m, where a is in G
gcd(m,n) = 1

3. The attempt at a solution

I know since G is an ableian group it is a commutative group (so ab=ba). Also since we have the special converse we know there exists a r, s \in G such that mr + ns = 1. They only way i know if proving an isomorphism is proving that it is one-to-one and onto and I'm not sure what to do with these puzzle pieces. Is there another way to prove isomorphisms?

2. Apr 17, 2012

### Dick

To start, try to show that if phi(a)=e (where e is the group identity) and a is not equal to e, then m and n have common divisor that is not 1. Consider the set {e,a,a^2,...,a^(m-1)}. Is it a subgroup of G?

Last edited: Apr 17, 2012
3. Apr 19, 2012

### tiger4

I think I proved the contrapositive,

Let a be a non-zero nilpotent element in R such that a^m = 0. If psi(a) is 0, then a is in the kernel and we're done. Otherwise, psi(a) is non-zero, and psi(a)^m = psi(a^m) = psi(0) = 0, which means that psi(a) is a non-zero nilpotent element of the image.

4. Apr 19, 2012

### Dick

Proving psi can't have a nontrivial kernel is the whole point. The rest of that is gibberish. The proof has to involve gcd(n,m)=1.

5. Apr 20, 2012

### tiger4

Does this work???

Since H is abelian then for all a,b in H we have that (ab)^m = a^m b^m. Thus phi(ab)=(ab)^m = a^m b^m = phi(a)phi(b) so phi is a group homomorphism.

Since H is finite it suffices to show phi is surjective. Let x be in H , we need to find a in H such that a^m = x.

By assumption (m,n)=1 so we can find integers u,v such that um + vn = 1.

But then x = x^1 = x^(um+vn) = x^(um) x^(vn). Since |H|=n then x^n = e (the identity) so x = x^(um).

Thus taking a = x^u yields phi(a) = a^m = (x^u)^m = x^(um) = x so phi is surjective. Since phi is a surjective endomorphism of a finite group then phi is injective as well, thus an isomorphism.

6. Apr 20, 2012

### Citan Uzuki

Yes, that proof is correct. Well done. Note that you can actually go further here: you have proved that (x^u)^m = x for all x, so that tells you that the inverse homomorphism is the function $x \mapsto x^u$. In general, if you are given a group homomorphism $\varphi : G \rightarrow H$ and you can find a function $\psi: H \rightarrow G$ such that $\varphi \circ \psi = \mathrm{id}_{H}$ and $\psi \circ \varphi = \mathrm{id}_{G}$, then you have proved that $\varphi$ is an isomorphism and $\varphi^{-1} = \psi$. This gives you another technique that you can use to prove that a function is an isomorphism.