# Simple question on disproving a group isomorphism

1. Jul 27, 2008

### jeffreydk

I am trying to prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic. I know it is not enough to show that there are maps such as, [tex]f:\mathbb{Q}\rightarrow \mathbb{Z}[/itex] where the input of the function, some $f(x=\frac{a}{b})$, will not be in the group of integers because it's obviously coming from rationals. I just don't know how to rigorously prove this, because just because a map is not isomorphic doesn't mean that the whole thing is not isomorphic. Thanks for any help, it is much appreciated.

2. Jul 27, 2008

### morphism

Suppose we managed to find an isomorphism $f:\mathbb{Q}\rightarrow \mathbb{Z}$. Then f(1/2) would be an integer; what happens when you look at f(1/2) + f(1/2)?

3. Jul 27, 2008

### jeffreydk

It would still be in integers right? That's why I'm confused because I can't seem to show that that property disallows the isomorphism.

4. Jul 27, 2008

### n_bourbaki

Morphism is sort of on the right lines. The property that you're looking for is called divisibility. Q has it and Z doesn't. The same idea also shows that Q\{0} under multiplication is not isomorphic to R\{0} under multiplication. (One can provide an elementary counter argument based purely on set theory, of course.)

Alternatively, Z is cyclic. Can you prove that Q isn't? Actually it isn't that alternative, really. Try thinking about a map g from Z to Q. Any group hom is determined completely by where it sends 1 in Z. Can g(1)/2 be in the image?

Last edited: Jul 27, 2008
5. Jul 27, 2008

### jeffreydk

Oh ok I didn't think of showing Q isn't cyclic, that's probably the simplest way to do it now that I think of it. Thanks a bunch.

6. Jul 27, 2008

### morphism

Since the "cat is out of the bag," I might as well point out that I wanted jeffreydk to notice that f(1)=nf(1/n) for all n (and why is this bad?).

7. Jul 27, 2008

### n_bourbaki

So your proof is based on the idea that if there is an isomorphism f:Q-->Z, then it must follow that f(1) is divisible by every integer? Hmm, not thought of that one before. It struck me as more obvious that Q isn't cyclic, i.e. look at maps from Z to Q. But it's always good to be able to do one thing in two ways, rather than two things in one way.

8. Jul 27, 2008

### jeffreydk

I definitely agree, it's good to able to prove it in a number of ways. Thanks you guys for both suggestions.