# Automorphism of the group of rational numbers under addition

Let $\mathbb \phi$ be an automorphism of the group of rational numbers $\mathbb \ Q$ under addition. Prove that $\mathbb \phi(x)=x \phi(1), \forall x \in Q.$

• yes, thank you! – maidel b Sep 27 '15 at 23:25

The other answer is almost correct. We do have $$\phi(n) = \phi(1+1+\cdots +1) = \phi(1) + \cdots +\phi(1) = n\phi(1)\\ \phi(1) = \phi\left(\frac nn\right) = \phi\left(\frac1n+\frac1n+\cdots+\frac1n\right) = \phi\left(\frac1n\right)+\cdots + \phi\left(\frac1n\right) = n\phi\left(\frac1n\right)$$ where the last line gives $\phi\left(\frac1n\right) = \frac1n\phi(1)$. However, for a general fraction $\frac mn$, we need to do the following: $$\phi\left(\frac mn\right) = \phi\left(\frac 1n + \cdots +\frac1n\right) = \phi\left(\frac 1n\right) +\cdots + \phi\left(\frac 1n\right) = m\phi\left(\frac1n\right) = m\frac1n\phi\left(1\right)$$ which is what we want.
$\phi$ is homomorphism, therefore we have $$\phi(n) = \phi(1+1+\cdots + 1) = \phi(1) + \phi(1) + \cdots + \phi(1) = n\phi(1)$$
in the same way $$n \phi\left(\frac1n\right) = \phi(1) \to \phi\left(\frac1n\right) = \frac 1n \phi(1)$$ using the above two, we can conclude $$\phi\left(\frac mn\right) = \phi(m\frac{1}{n}) =\phi(m)\phi\left(\frac1n\right) =\frac m n \phi(1)\phi(1) = \frac m n \phi(1)$$
• This is very close to a true answer, but the transition $\phi\left(m\frac1n\right) = \phi(m)\phi\left(\frac1n\right)$ requires that we're talking about rings, and we're not. This is a group question. That's also why you get $\phi(1)\phi(1) = \phi(1)$ at the end, which is not true in general. – Arthur Sep 28 '15 at 7:32