The existence of group isomorphism between Euclidean space. Is there any group isomorphism for addition $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n\neq m$? I could prove that there exists any vector space isomorphism or smooth map, but I still could not know that if we consider only abelian group structure for the addition of them, then the group isomorphism between them exits or not.
 A: Hint: For vector spaces over $\mathbb{Q}$, you can see the underlying group homomorphism as a $\mathbb{Q}$-linear map. This reduces the problem to seeing if dimensions of the vector spaces over $\mathbb{Q}$ are equal as this is equivalent to the underlying groups being isomorphic. In turn, it is known (facts section) that the dimension of an infinite dimensional vector space over  $\mathbb{Q}$ is the cardinality of the vector set.
A: Partial answer:
Assume $n \neq m$ and that $f : \mathbb R^n \rightarrow \mathbb R^m$ is a group isomorphism for addition, i.e. for all $u,v \in \mathbb R^n$, $f(u+v) = f(u) + f(v)$. Then $f$ cannot be continuous.
Indeed, you can easily successively show that for all $u \in \mathbb R^n$, $f(nu) = nf(u)$ for all $n \in \mathbb N$, then for all $n \in \mathbb Z$, then $f(qu) = qf(u)$ for all $q \in \mathbb Q$. Now if $f$ is continuous you have $f(\lambda u) = \lambda f(u)$ for all $\lambda \in \mathbb R$. So $f$ is linear and therefore $n = m$, which is a contradiction.
Here we have also shown that any group isomorphism for addition from $\mathbb R^n$ to $\mathbb R^m$ is a $\mathbb Q$-linear map.
I didn't manage to disprove the statement for non-continuous applications though but I will look into it!
