# Automorphisms of rationals as a group [duplicate]

Given to me was the following assignment:

Prove that for each $a\in\mathbb Q^{*}$, the mapping from $\mathbb Q$ to $\mathbb Q$ which sends $x$ to $ax$ is an automorphism of the additive group $(\mathbb{Q},+)$. Vice versa, every automotphism of $(\mathbb{Q},+)$ is of this form. Conclude that $\mathrm{Aut}(\mathbb{Q},+)$ is isomorphic to $\mathbb Q^{*}$.

My first problem was that I didn't really 'understand' this problem (still don't). I do understand the definitions. And my idea was that I was asked to basically prove that each homomorphism from $(\mathbb{Q},+)$ to itself can be assigned to a unique $a\in\mathbb Q^{*}$?

• It can be assigned a unique $a$, and yes, I believe that that's basically what you're asked. $\Bbb Q^+$ is not cyclic, but any group homomorphism $h:\Bbb Q^+\to G$ into some group $G$ is still uniquely determined by $h(1)$ (side question: does anyone know whether such groups have a name?). Also note that the group of automorphisms does not consist of homomorphisms. It consists of isomorphisms, so you may limit your work to those of you like. Mar 15, 2016 at 13:04
• Why is it uniquely determined through $h(1)$? I don't see how $h(1)$ determines $h(2/3)$ for example. I mean my my case $G=Q^{+}$, so I don't see how we can flip scalars in here.ðŸ˜¢ Mar 15, 2016 at 13:16
• Note that $h(2/3)+h(2/3)+h(2/3)=h(1)+h(1)$, so if $h(1)$ is decided, then there is only one thing that $h(2/3)$ can be. Or, at least there is only one thing in $\Bbb Q^+$ it can be. I was wrong to claim this for general $G$, so please forget that. Mar 15, 2016 at 13:46

Let $\phi$ be an automorphism of the additive group $(\mathbb Q,+)$.
Then $\phi(n) = \phi(1+1+\cdots+1)=\phi(1)+\phi(1)+\cdots+\phi(1)=n\phi(1)$.
Thus, $n\phi(m/n)=\phi(m)=m\phi(1)$ and so $\phi(m/n)= (m/n)\phi(1)$ and $\phi$ is determined by $\phi(1)$.
Since $\phi$ is injective and $\phi(0)=0$, we cannot have $\phi(1)=0$.
• Interesting, but $n\phi(m/n)=\phi(m)$ doesn't obviously follow to me. Feb 27, 2021 at 0:24
• Ah, that makes sense. $n\phi(m/n)=\phi(m/n)+\phi(m/n)+\ldots+\phi(m/n)=\phi(m/n+m/n+\ldots+m/n)=\phi(n(m/n))$ Feb 27, 2021 at 0:38