Isomorphism between vector spaces and abelian groups 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.
 A: 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 $x=(1/2,0),y=(0,1/2)$). 
