I am working through an example in Hungerford. In this example, he asserts that it's not difficult to show that the only homomorphism from $\Bbb{Q}$, regarded as an additive group, to $S_3$ is the trivial homomorphism. I am having trouble seeing this. Treat $\Bbb{Q}$ as an additive group, and let $G$ be some group (written multiplicatively). If $f : \Bbb{Q} \to G$ a homomorphism, I believe that $f(p/q) = f(1)^{-pq}$, so that the homomorphism is determined by how it maps $1$ (at least is what I proved on my own; it may be wrong). In our case $G = S_3$. If $f(1)= e$, then $f$ is the trivial homomorphism. If $f(1) \neq e$, then...
Not sure what to do at this point. I could use a hint.
If $G$ is a finite group and $\phi:\mathbb Q\to G$ then for any $r$:
$$\phi(r)=\phi\left(\frac{r}{|G|}\right)^{|G|}=e$$
because for each $g\in G$, $g^{|G|}=e$.