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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.

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    $\begingroup$ $(\mathbb{Q}^*,\times)$. $\endgroup$ – C.Ding Nov 8 '18 at 5:38
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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.

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  • $\begingroup$ You can show that $a/b$ (with $a$, $b$ coprime) has rational square root if and only if $a$ and $b$ are both perfect squares. Now, continue taking square roots indefinitely - only possible if $a = b = 1$. $\endgroup$ – Ennar Nov 8 '18 at 5:11
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    $\begingroup$ Suppose $0\ne \phi(r)=x/y\ne \pm 1$ with $x,y\in \Bbb Z$ and $\gcd(x,y)=1 $. Take $n\in \Bbb Z^+$ such that $2^n>\max ( |x|,|y|).$ If $a,b\in \Bbb Z$ with $\gcd (a,b)=1$ and $(a/b)^n=x/y$ then $a^ny=b^nx,$ and $ a,b$ cannot both belong to $\{\pm 1\}.$ But if $p$ is a prime divisor of $a$ then $ p^n\geq 2^n>| x|$, so let $p^k$ be the largest power of $p$ that divides $x.$ Then $n-k>0$ and $p^{n-k} $ divides $b^n,$ so $ p$ divides $b,$ contrary to $\gcd (a,b)=1.$ Similarly we get a contradiction if $q$ is a prime divisor of $b.$ $\endgroup$ – DanielWainfleet Nov 8 '18 at 5:17
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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.

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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 $1$.

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