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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?

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  • $\begingroup$ Hint: How many elements are in the basis? $\endgroup$ Feb 18, 2015 at 12:54
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    $\begingroup$ @REr I know that.. I'm just asking whether there is a direct way to prove this just using group theory $\endgroup$
    – Rubertos
    Feb 18, 2015 at 12:56
  • $\begingroup$ You mean $\mathbb Z^3$ (not $\mathbb R^3$)? $\endgroup$
    – Troy Woo
    Feb 18, 2015 at 12:59
  • $\begingroup$ @TroyWoo No I meant $\mathbb{R}^3$. This was on a graduate school entrance exam.. Should not use module argument $\endgroup$
    – Rubertos
    Feb 18, 2015 at 13:03
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    $\begingroup$ @Rubertos How, or from what, do you deduce that if $\;\phi\;$ is a hypothetical isomorphism, then Span$\,\{\phi(1,0)\,,\,\phi(0,1)\}\;$ has to be a generator set of $\;\Bbb R^3\;$ as a linear space (over itself, I presume) ? This much you should detail in your answer, imo. $\endgroup$
    – Timbuc
    Feb 18, 2015 at 13:20

3 Answers 3

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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 $G \not\cong H$.

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  • $\begingroup$ Beat me to it, though I was going to comment that any homomorphic immage of $\;\Bbb Z^2\;$ is generated by two elements at most, whereas there's an homomorphic image of $\;\Bbb Z^3\;$ which requires at least three generators. +1 $\endgroup$
    – Timbuc
    Feb 18, 2015 at 13:07
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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.

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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 $\mathbb{R}^3$.

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  • $\begingroup$ Why is it not precise? There exsits (n,m) such that $(1,0,0)=n(a,b,c)+m(d,e,f)$. And same for $(0,1,0)$ and $(0,0,1)$. Hence, considering these two tuples as vectors in $\mathbb{R}^3$, they span. $\endgroup$
    – Rubertos
    Feb 18, 2015 at 13:06
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    $\begingroup$ That new wording is perfectly precise. $\endgroup$
    – hunter
    Feb 18, 2015 at 13:10

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