How do I prove $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}\times\mathbb{Z}$ are not isomorphic?

Question

Here is how I proved this exercise.

Suppose $\phi:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ is a group isomorphism.

Set $\phi(1,0)=(a,b,c)$ and $\phi(0,1)=(d,e,f)$.

Then, it can be viewed as $span(\{(a,b,c),(d,e,f)\}) = \mathbb{R}^3$. This is obviously false. (Is my argument correct?)

However, is there another way to prove this not using linear algebra, just using group theory?

Answer 1

If $G= {\mathbb Z}^3$ and $H={\mathbb Z}^2$, then $|G/2G|=8$ and $|H/2H|=4$, so $G/2G \not\cong H/2H$ and hence $G \not\cong H$.

Answer 2

Hint:

If $\phi:\mathbb{Z}^2\to\mathbb{Z}^3$ is an isomorphism then so is $\phi^{-1}:\mathbb{Z}^3\to\mathbb{Z}^2$.

Show that some linear combination $\phi^{-1}(1,0,0)$, $\phi^{-1}(0,1,0)$ and $\phi^{-1}(0,0,1)$ is zero and therefore $\phi^{-1}$ is not injective.

Answer 3

Your argument is basically correct, but isn't so precise. The more precise way to say your same argument, which you probably haven't run into the language for in your course yet, is that $\phi$ cannot be an isomorphism after tensoring with $\mathbb{R}$ for dimension reasons, and thus already cannot be an isomorphism.

You may find one of the other arguments here more useful, just because it's easier to write without having to define too precisely what you call "viewing" the map in $\mathbb{R}^3$.