Is the group of all group homomorphisms from $\mathbb{R}$ to $\mathbb{R}^n$ isomorphic to $\mathbb{R}^n$? That is, with the usual additive structures, do we have ${Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R}^n)\cong \mathbb{R}^n$?
 A: This is not true. Consider the case $n = 1$. Any group homomorphism $\varphi \colon (\mathbb{R},+) \rightarrow (\mathbb{R},+)$ is automatically a homomorphism of $\mathbb{Q}$-vector spaces but the converse doesn't hold (see here for more information). Assuming the axiom of choice, $\mathbb{R}$ as a vector space over $\mathbb{Q}$ has a basis of cardinality $|\mathbb{R}|$. Then
$$ \operatorname{Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R}) = \operatorname{Hom_{\mathbb{Q}}}(\mathbb{R},\mathbb{R}) = \operatorname{Hom_{\mathbb{Q}}}\left( \bigoplus_{\mathbb{R}} \mathbb{Q}, \mathbb{R} \right) = \prod_{\mathbb{R}} \operatorname{Hom}_{\mathbb{Q}}(\mathbb{Q},\mathbb{R}) = \prod_{\mathbb{R}} \mathbb{R} = \mathbb{R}^{\mathbb{R}}$$
and clearly $|\mathbb{R}^{\mathbb{R}}| > |\mathbb{R}|$ so $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R})$ is much larger than $\mathbb{R}$.
A: As levap said, this is false for $\mathbb{Z}$-linear maps (by RAPL nonsense + ac). But, it is true for $\mathbb{R}$-linear maps. For $\varphi : \mathbb{R} \to \mathbb{R}^n$ with $\varphi(1)= x \in \mathbb{R}^n$ we have $$\varphi(r ) = \varphi(r*1) = r\varphi(1) = rx$$
by $\mathbb{R}$-linearity. So every $\varphi \in \text{Hom}(\mathbb{R},\mathbb{R}^n)$ is uniquely determined by the image of $1$.
