Show the only homomorphism $F:\Bbb Q\rightarrow \Bbb Q\setminus\{0\}$ is the trivial map

Question

I've been working on this problem for a long time and am stuck.

I was thinking I could use Lagrange's theorem and the fact that the image of any homomorphism must have an order that divides both the order of the domain group and the codomain group.

Since they differ by 1, their orders must be coprime meaning that the order of the image must be 1, namely the trivial map.

I'm not exactly sure if this is right, if I could get some feedback that would be great.

Thanks!!!

Answer 1

I assume $F:(\mathbb{Q},+)\rightarrow (\mathbb{Q}-\{0\},\times)$. If $F$ is not trivial there exists $u$ such that $F(u)\neq 1$ you must have $F({u\over n})^n=F(u)$ implies that the $n$-root of $F(u)$ exist in $\mathbb{Q}$ for every $n$ contradiction.

Answer 2

Let $f$ be a homomorphism and let $f(1/2) = x$. Then $f(1) =f(\frac12 + \frac12)= f(1/2)\times f(1/2)=x^2$. This means $f(1)$ is a rational number that is a square of $f(1/2)$. Now with 1/2 playing the role of 1 above we get $f(1/2) = f(\frac14)\times f(\frac14)$. So $f(1/2)$ is a square, and so $f(1)$ is a 4th power. Digging deeply we will see that $f(1)$ is an 8th power, 16th power and so on. But now focussing on the denominators (which are positive integers) we get that the denom of $x$ is square, 4th power etc. This contrdicts Unique Factorization Theorem on integers.