# 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.

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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?

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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