Group of automorphisms of $(\mathbb{R}, +)$

Question

I understand that the group of automorphisms of $\mathbb{Q}$ (as a group) is isomorphic to $\mathbb{Q}^{\times}$.

I am wondering what the group of automorphisms of $\mathbb{R}$ (as an additive group) is.

I think this is isomorphic to $\mathbb{R}^\times$, but I don't know how to prove it. My idea is that (just as with $\mathbb{Q}$) an isomorphism $\phi: \mathbb{R} \to \mathbb{R}$ must satisfy $$ \phi\left( \frac{n}{m}\right) = \frac{n}{m}\phi(1). $$ But what should I do about the non-rational real numbers?

Answer 1

No, it's much, much larger. Assuming the axiom of choice, $\mathbb{R}$ has a basis as a $\mathbb{Q}$-vector space, so it's isomorphic to an (uncountable) direct sum $\mathbb{R} \cong \bigoplus_I \mathbb{Q}$. Its automorphism group is $\text{GL}_I(\mathbb{Q})$, which is very big, and in particular nonabelian.

Answer 2

The group of continuous automorphisms of $(\mathbb{R}, +)$ is indeed isomorphic with $\mathbb{R}^{\times}$ (use your argument, and the density of rationals).

But the group of arbitrary automorphisms of $(\mathbb{R}, +)$ is way bigger. If $\phi$ is any automorphisms of $(\mathbb{R}, +)$, we can indeed show that $\phi(qx) = q .\phi(x)$ for all $x \in \mathbb{R}$ and $q \in \mathbb{Q}$. This means $\phi$ is a $\mathbb{Q}$-linear automorphism of $\mathbb{R}$ (seen as a $\mathbb{Q}$-vector space), but that's all you can say. So if you pick two arbitrary basis $(e_i)_{i \in I}$ and $(f_i)_{i \in I}$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, the linear map that sends $e_i$ to $f_i$ defines such an automorphism of $(\mathbb{R}, +)$.