# Automorphism of rational numbers

Let $\mathbb Q$ be group of rational numbers under addition. Then I want to find out the automorphism group of $\mathbb Q$. In a book it is given that Aut($\mathbb Q$) is isomorphic to $\mathbb Q^*$, the multiplicative group of rational numbers. I stuck in this.

Can you guess a function $f:\mathbb Q^*\to Aut(\mathbb Q)$? If you have a rational number $q\ne 0$, it is very natural to consider the automorphism $g_q:\mathbb Q \to \mathbb Q$ given by $x\mapsto qx$. It is easily shown that $f(q)=g_q$ is then an injective group homomorphism. The last thing to do then is show that $f$ is surjective. So, if you start with an automorphism $g:\mathbb Q \to \mathbb Q$, you really want to show that $g=g_q$ for some $q\in \mathbb q$. If that were the case, then $g(1)=g_q(1)=q\cdot 1=q$, so that tells you that the only candidate is $q=g(1)$. Now think a bit about rational numbers and figure out how to show that $g=g_{g(1)}$.