# Automorphism group of $\mathbb{Q}$ considered as a group under addition

I was just wondering how $Aut(\mathbb{Q})$ looks like as a group with composition, when considering $\mathbb{Q}$ as a group under addition.

I have no clue to approach it. I cannot seem to write down any non-trivial map. Clearly polynomial looking functions will not work. Stuff like $f(a/b)=b/a$ also doesn't work. Anybody have a hint for how to find or construct a non-trivial map?

• what makes you think there is a map other than multiplication by rational? Feb 22, 2016 at 22:32
• Feb 22, 2016 at 22:36
• I did not found a way to show those are the only ones, but that helps! Thnx
– Rico
Feb 22, 2016 at 22:39

The first thing to note is that, for $m\in\mathbb{Z}$, $f(m)=mf(1)$. Indeed, if $m>0$, then \begin{align*} f(m)&=f(\underbrace{1+\cdots+1}_m)\\ &=\underbrace{f(1)+\cdots+f(1)}_m\\ &=mf(1). \end{align*} As $f(-m)=-f(m)$ and $f(0)=0$, the result holds for all $m$.
Similarly, we have $f(1/n)=f(1)/n$. Indeed, $$f(1)=f(n/n)=f(\underbrace{1/n+\cdots+1/n}_n)=\underbrace{f(1/n)+\cdots+f(1/n)}_n=nf(1/n).$$ Now, you can compute $$f(m/n)=(m/n)f(1).$$ If $f(1)=r$, write $f=f_r$. Then, $f_{r}^{-1}=f_{r^{-1}}$, $f_rf_s=f_{rs}$ and the map $$\mathbb{Q}^\times\to\mathrm{Aut}(\mathbb{Q})$$ given by $r\mapsto f_r$ is an isomorphism.
• So can we say from this argument that $f:(\mathbb{Q}, +) \rightarrow (\mathbb{Q}, +)$ is an isomorphism iff there exists $r\in\mathbb{Q^*}$ such that $f(x)=rx$, $\forall x\in\mathbb{Q}$? Jan 27, 2022 at 12:50
• @MarcosPaulo Well, the map $r\to f_r$ is both injective and surjective, so yes. But I’m proving more than that. Jan 28, 2022 at 5:41
• Just to be clear $f_r(x) =xf(1)$, right? Jan 29, 2022 at 10:43