# Why not $\mathbb{Q}$ and $\mathbb{Q}\times \mathbb{Q}\cong \mathbb{Q}^2$ not isomorphic as additive groups?

We know that $$\mathbb{R}$$ and $$\mathbb{R}^2$$ are isomorphic as additive groups(as they are isomorphic as $$\mathbb{Q}$$ vector spaces) under the axiom of choice. But why not $$(\mathbb{Q}, +)$$ and $$(\mathbb{Q}^2,+$$) isomorphic as additive groups?

Specifically, I think for vector space isomorphism, I think since they are the smallest field to begin with(they are field of fractions of the integers), so they do not have a common basis, hence may be they are not isomorphic as vector spaces. But, what about group isomorphism? I think there is a simple counterexample. Any hints? Thanks beforehand.

Hint: $$\mathbb Q$$ has the following property: for any two nonzero elements $$a,b$$ we can there exist natural numbers $$n,m$$ such that either $$\underbrace{a+\dots+a}_n=\underbrace{b+\dots+b}_m$$ or $$\underbrace{a+\dots+a}_n=-\left(\underbrace{b+\dots+b}_m\right)$$
• @vidyarthi I've seen elements $a,b$ with this property commensurable, so you could say all elements of $\mathbb Q$ are commensurable. – Wojowu May 3 at 12:49
• Note also that ${\mathbb Q}$ is locally cyclic, while ${\mathbb Q}^2$ is not. I think that might be equivalent in abelian groups to the property described here – Derek Holt May 3 at 12:53
If additive groups $$A$$ and $$B$$ are $$\mathbb Q$$-vector spaces, then any group homomorphism $$f : A \rightarrow B$$ is actually $$\mathbb Q$$-linear, since for $$a \in A$$ and $$n \in \mathbb N$$, we have $$n\cdot f({a\over n}) = f(a)$$ by properties of group homomorphisms, so $$f({a\over n}) = {f(a) \over n}$$ because $$nx=y$$ has a unique solution in $$\mathbb Q$$-vector spaces.
Thus, if $$A$$ and $$B$$ are isomorphic as groups, they are automatically isomorphic as $$\mathbb Q$$-vector spaces. Since $$\mathbb Q$$ and $$\mathbb Q^2$$ are not isomorphic as $$\mathbb Q$$-vector spaces (due to having different dimensions), they cannot be isomorphic as groups.