TITLE: Every group homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, \times)$ is the trivial map.
QUESTION [7 upvotes]: How do you show that every group homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, \times)$ is the trivial map? I am trying to use proof by contradiction by assuming there is an element $\frac{a}{b}$ such that some homomorphism $\phi$ has $\phi(\frac{a}{b}) = \frac{c}{d} \ne 1$. But I cannot seem to deduce any contradictions from here. Maybe using direct proof is a better approach? Any help is appreciated.

REPLY [2 votes]: Here are the key points:

$(\mathbb{Q},+)$ is a divisible group: every element is a multiple of $n$ for all $n \in \mathbb N$.
The image of $\phi$ is a divisible subgroup of $(\mathbb{Q},\times)$.
The only divisible subgroup of $(\mathbb{Q},\times)$ is the trivial subgroup: the only element of $(\mathbb{Q},\times)$ that is an $n$-th power for all $n$ is $1$.