Are all group homorphisms of the reals also linear transformations?

Question

Is $Hom_\mathbb{R}(\mathbb{R},\mathbb{R})$ (linear transformations) = $Hom(\mathbb{R},\mathbb{R})$ (morphisms of abelian groups)? No hypothesis of continuity is assumed.

Obviously $Hom_\mathbb{R}(\mathbb{R},\mathbb{R}) \subseteq Hom(\mathbb{R},\mathbb{R})$.

Answer 1

Because $(\mathbb{R},+)$ is a divisible group a map $f \colon \mathbb{R} \to \mathbb{R}$ is a group homomorphism if and only if it is $\mathbb{Q}$-linear. So your question is if any $\mathbb{Q}$-linear map $\mathbb{R} \to \mathbb{R}$ is already $\mathbb{R}$ linear, i.e. given by multiplication with a real scalar.

To see that this doesn’t hold let $(x_i)_{i \in I}$ is a $\mathbb{Q}$-basis of $\mathbb{R}$. Then then any permutation of two basis $x_i$ and $x_j$ with $i \neq j$ induces a $\mathbb{Q}$-linear map $f_{ij} \colon \mathbb{R} \to \mathbb{R}$ which is not given by multiplication with a real scalar, as $f_{ij}$ fixes $x_k$ for $k \notin \{i,j\}$. (Notice that we used the axiom of choice to construct these counterexamples.)

Answer 2

$\newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \DeclareMathOperator{Hom}{Hom}$ No. If we consider $\RR$ as a vector space over $\QQ$, we can get some very nasty group homomorphisms.

First, note that $\Hom_\QQ(\RR, \RR)$ is the same as $\Hom(\RR, \RR)$. If $f$ is a $\QQ$-linear map, then $f(a + b) = f(a) + f(b)$, and so $f$ is a group homomorphism. If $f$ is a group homomorphism, we have additivity down, so we just need to show that $f(\frac{p}{q} x) = \frac{p}{q} f(x)$. By induction, we know that for integer $n$, $f(nx) = n \cdot f(x)$. $$ q \cdot f(\frac{p}{q} x) = pq \cdot f(\frac{1}{q} x) = p \cdot f(x) \implies f(\frac{p}{q} x) = \frac{p}{q} f(x) $$

So let's show that $\Hom_\QQ(\RR, \RR)$ is ludicrously bigger than $\Hom_\RR(\RR, \RR)$. We know that $\Hom_\RR(\RR, \RR) = \{ x \mapsto \alpha x \mid \alpha \in \RR \}$, so it's got the same size as $\RR$.

Let $B = \{ b_i \}_{i \in I}$ be a basis for $\RR$ over $\QQ$. Since $B$ is a basis, any map from $B$ to $\RR$ generates a $\QQ$-linear map from $\RR$ to $\RR$. Additionally, any $\QQ$-linear map from $\RR$ to itself gives a map from $B \to \RR$. So how many maps are there from $B \to \RR$? There's $|\RR|^{|B|}$, and since $B$ is uncountable, this is a lot.