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I am trying to show that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q} \times \mathbb{Q}$ as additive groups. Suppose that we have an isomorphism $\phi : \mathbb{Q} \rightarrow \mathbb{Q} \times \mathbb{Q}$. Let $\frac{p}{q} \in \mathbb{Q}$. Then $q\phi(\frac{p}{q}) = \phi(p)$, and $\phi(p) = p \phi(1)$, which implies that $\phi(\frac{p}{q}) = \frac{p}{q} \phi(1)$. I saw that this implies $\phi$ is an isomorphism of vector spaces over $\mathbb{Q}$ which cannot exist by dimension arguments. Is this a correct proof? Also, how do we show that $\mathbb{Q}/\mathbb{Z} \not \cong (\mathbb{Q}/\mathbb{Z})^{2}$ as additive groups? If we have a map $\phi : \mathbb{Q}/ \mathbb{Z} \rightarrow (\mathbb{Q}/\mathbb{Z})^{2}$ that is an isomorphism. Does the existence of such a map imply that $\phi(\frac{p}{q}) = \frac{1}{q} \phi(p) = \frac{1}{q} \phi(0) = \frac{1}{q}(0,0) = (0,0)$, or is this wrong? Lastly, is it correct to argue that $\mathbb{Q} \not \cong \mathbb{Q}/\mathbb{Z}$ as additive groups because every non-identity element of $\mathbb{Q}$ has infinite order where as every non-identity element of $\mathbb{Q} / \mathbb{Z}$ has finite order, so an isomorphism cannot exist.

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You’re asking several questions at once. Let’s try to address them one by one.

For your proof about the inexistence of an isomorphism $\mathbb{Q} \rightarrow \mathbb{Q}^2$, it’s essentially fine. There is one non-obvious point, it is the “division by $q$”, which is not obvious in a general abelian group. However, in $\mathbb{Q}$ and $\mathbb{Q}^2$, $x \longmapsto qx$ is a group automorphism and it works.

The same argument about $\mathbb{Q}/\mathbb{Z}$ is wrong for precisely that reason: there is no “division by $q$” which can be transported by the function. An example of such a map: $x \longmapsto (2x,3x)$.

For the last argument about $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ not being isomorphic, your proof is correct.

Note a simpler argument for your two first questions: let $X=\mathbb{Q}$ or $X=\mathbb{Q}/\mathbb{Z}$. Let us show $X$ and $X^2$ are not isomorphic as abelian groups.

Indeed, $X$ satisfies the following property: for every elements $x,y \in X$, there is a $z \in X$ with $\mathbb{Z}z \supset \{x,y\}$.

Clearly, this property is transported by isomorphisms, but $X^2$ does not possess it (consider $x=(1/2,0),y=(0,1/2)$).

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  • $\begingroup$ When we write that $\mathbb{Z}z$, does it mean $\mathbb{Z}z = \{x \cdot z \in \mathbb{Q} : x \in \mathbb{Z} \}$? Also, why does such a property need to be fixed under isomorphism of abelian groups? $\endgroup$ Dec 10 '19 at 21:39
  • $\begingroup$ For the isomorphism: just write it out (actually, you just need the morphism to be surjective, I believe). Yes, $\mathbb{Z}z=\{n \cdot z ,\, n \in \mathbb{Z}\}$. $\endgroup$
    – Mindlack
    Dec 10 '19 at 21:54
  • $\begingroup$ Is it something like $\forall x,y \in X$, $\exists z \in X : x,y \in \mathbb{Z}z$, which implies that $\forall x,y \in X, \phi(x), \phi(y) \in \mathbb{Z}\phi(z)$, but since $\phi$ is a isomorphism and hence surjective, this is equivalent to $\forall x,y \in X^{2}, \exists z \in X^{2} : x,y \in \mathbb{Z}z$, which cannot happen by the counter example given? $\endgroup$ Dec 10 '19 at 22:00

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