Find all group morphisms $f:(Q,+)\rightarrow(Q^∗,⋅)$

Question

this was a task from a previous homework that I couldn't do. I am now trying to understand it. We were provided with a solution but I think it could be wrong. This is what it was:

$$f(1) = f\left(\frac bb\right) = f\left(\frac1b + ... + \frac 1b \right) = f\left(\frac 1b\right)^b \quad \Rightarrow \quad \forall b \in \mathbb Z, \ f \left(\frac 1b\right) = f(1)^{\frac1b} \in \mathbb Q$$

this next line is the part I don't understand:

$$f \left(\frac 1b\right) = f(1) = 1$$

$$f(2) = f \left(\frac ab\right) = f \left(\frac 1b\right)^a = 1^a = 1 \quad \forall 2 \in Q , a,b \in \mathbb Z$$

so $f$ is the trivial group morphism.

I don't understand how f(1) = 1. I think maybe the person that provided the solution thought 1 is the neutral element of (Q, +) but that is 0. But I guess it doesn't have to be injective so f(1) could still also be 1. But I don't understand how you would get there from the previous steps.

Sorry for not quite knowing how to write this neatly. Thanks in advance!

Answer 1

Assume $f$ is not trival. Then there is some $a\ne 0$ with $f(a)\ne 1$. Note that for any $n\in\Bbb N$ we can find $a'=\frac1na\in \Bbb Q$ with $$a=\underbrace{a'+a'+\ldots +a'}.$$ It follows that $$f(a)=\underbrace{f(a')f(a')\cdots f(a')}_n=f(a')^n.$$ But that means that you can take $n$th roots of $f(a)$ for arbitrary $n$. This is not possibly because $f(a)\ne 1$: If $f(a)<0$ this fails already for $n=2$; and if $f(a)=\frac uv$ with $u,v\in\Bbb N$ coprime and not both $=1$, pick a prime $p$ that divides $uv$. Now if $n$ is large enough and $p^n\nmid uv$, there is no rational $n$th root of $\frac uv$.