Every group homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, \times)$ is the trivial map. 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.
 A: Hint: For any $r\in \mathbb Q$ and any $n\in \mathbb Z$, 
$$ \phi(r/n)^n = \phi(n * r/n) = \phi(r),$$
that is $\phi(r)$ has rational $n$-th root for any $n$. Now, it is left to show $1$ is the only such element in $(\mathbb Q, \times)$, which, honestly, is not that trivial to me.
A: Suppose $\phi(1)=a$, then $\phi(\frac{1}{k})^k=\phi(1)=a$,i.e.,$\phi(\frac{1}{k})=a^{\frac{1}{k}}$($k\in \mathbb{N}^*$).
If $a=-1$, take $k=2$, then $\phi(\frac{1}{k})\notin\mathbb{Q}$.
 If $a\in \mathbb{Q}\backslash\{-1,0,1\}$, suppose $a=\frac{m}{n}$($m,n$ are relative prime integers and $m>0$) and $m=p_1^{\alpha_1}\cdots p_n^{\alpha_n}(n\geq 1)$ be the standard decomposition. Take $k>\alpha_1$, then $\phi(\frac{1}{k})\notin\mathbb{Q}$. Otherwise, suppose $a^{\frac{1}{k}}=\frac{p}{q}$($p,q$ are relative prime integers),i.e.,$mq^k=np^k$, so, $p_1|p^k\Rightarrow p_1^k|p^k \Rightarrow p_1^k|m\Rightarrow p_1^k|p_1^{\alpha_1}$, a contradiction. Hence $a=1$.
$\phi(1)=1\Rightarrow \phi(\frac{1}{k})=1^{\frac{1}{k}}=1\Rightarrow \phi(\frac{l}{k})=(\phi(\frac{1}{k}))^l=1$, so $\phi$ is trivial.
A: 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 $1$.
