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.


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 $x=(1/2,0),y=(0,1/2)$).

  • $\begingroup$ 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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