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 $$\underbrace{a+\dots+a}_n=-\left(\underbrace{b+\dots+b}_m\right)$$

  • $\begingroup$ thanks! what is this property called? $\endgroup$ – vidyarthi May 3 at 12:45
  • $\begingroup$ @vidyarthi I've seen elements $a,b$ with this property commensurable, so you could say all elements of $\mathbb Q$ are commensurable. $\endgroup$ – Wojowu May 3 at 12:49
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    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ thanks! so the same reasoning applied here too. what a silly miss! $\endgroup$ – vidyarthi May 3 at 12:46

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