# Infinite cyclic groups isomorphic to Z

1. Jul 31, 2008

I'm currently going through Hungerford's book "Algebra", and the first proof I found a bit confusing is the proof of the theorem which states that every infinite cyclic group is isomorphic to the group of integers (the other part of the theorem states that every finite cyclic group of order m is isomorphic to the group Zm, but I understood that part).

So, if G = <a> is some infinite cyclic group, then the mapping f : Z --> G with the formula f(k) = a^k is clearly an epimorphism. To prove that it's an isomorphism (i.e. that Z is isomorphic to G), one needs to show that f is a monomorphism, too. That's where I'm stuck. I know that f is injective iff Ker(f) = {0}, so I somehow need to show that Ker(f) = {0}.

So, by definition, Ker(f) = {k in Z | a^k = e}. Suppose Ker(f) is non-trivial. Ker(f) is a subgroup of the integers Z, and hence it is cyclic, infinite and generated with m, where m is the least positive integer in it, so Ker(f) = <m>.

2. Jul 31, 2008

### morphism

If a^k=e, what can you say about the order of a?

3. Jul 31, 2008

Well, the next theorem states that, if a has infinite order, then a^k = e iff k = 0, and hence, everything is clear. But, there is no proof for this theorem, it merely says that it is "an immediate consequence of the former theorem (the one I'm trying to prove).

4. Jul 31, 2008

### morphism

OK, how about this: if ker(f)=<m>, try to define an explicit isomorphism from Z/<m> onto G (use f).

I gotta run. Hopefully this'll be good enough!

5. Jul 31, 2008

Well, this is how the proof in the book goes on - it is shown that [k] --> a^k is an isomorphism between Zm and G, but this means that G is a finite cyclic group, doesn't it? I still don't inderstand how it's proved for infinite cyclic groups. Actually, the order of G isn't mentioned in any part of the proof. I seem to be missing something big here.

6. Aug 1, 2008

### morphism

But that's exactly it - if the kernel of f isn't trivial, then G cannot be infinite.

7. Aug 1, 2008

Forgive my (probable) stupidity, but why does this hold? Intuitive, it is somehow clear to me, but then again, why couldn't a^k = e, for some non zero k hold and G still be infinite?

8. Aug 1, 2008

### morphism

If the kernel of f is non-trivial, then:
So...

9. Aug 1, 2008

### n_bourbaki

This is so utterly trivial that I cannot believe it needs a discussion.

G is infinite and cyclic and G=<a>, i.e. G is the set of all integer powers of a if written multiplicatively. So in what way is the map

a^k ---> k

not a clear isomorphism? All powers of a are distinct.

10. Aug 2, 2008