# 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.

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

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.

• 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 $a = b = 1$. – Ennar Nov 8 '18 at 5:11
• 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 $b.$ – DanielWainfleet Nov 8 '18 at 5:17

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.

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 $$1$$.