# Simple question on disproving a group isomorphism

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.

Related Linear and Abstract Algebra News on Phys.org
morphism
Homework Helper
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)?

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.

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:
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.

morphism
Homework Helper
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?).

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.

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