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 $a\in\mathbb Q^{*}$?

  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Mar 15, 2016 at 13:04
  • $\begingroup$ 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.😢 $\endgroup$
    – Bahbi
    Mar 15, 2016 at 13:16
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Mar 15, 2016 at 13:46

1 Answer 1


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 $\phi(1)=0$.

  • $\begingroup$ Interesting, but $n\phi(m/n)=\phi(m)$ doesn't obviously follow to me. $\endgroup$ Feb 27, 2021 at 0:24
  • 1
    $\begingroup$ Ah, that makes sense. $n\phi(m/n)=\phi(m/n)+\phi(m/n)+\ldots+\phi(m/n)=\phi(m/n+m/n+\ldots+m/n)=\phi(n(m/n))$ $\endgroup$ Feb 27, 2021 at 0:38

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