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Of course, since the cardinality of $\mathbb{R}$ exceeds the cardinality of $\mathbb{Q}$, there does not exist a bijection between $\mathbb{Q}^\ast$ and $\mathbb{R}^\ast$, let alone a group isomorphism.

My question is whether it is also possible to prove that these multiplicative groups are not isomorphic without using a cardinality argument.

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In $\mathbb{R}^\ast$ every element has a cuberoot; that is, for any $a$ in $\mathbb{R}^\ast$ there is an element $b$ such that $b^3=a$. In $\mathbb{Q}^\ast$ there are elements that do not.

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Let $f:R^*\rightarrow Q^*$ be an isomorphism, remark that if $f(x)=-1, f(x^2)=1$, since $f$ is an isomorphism, it implies that $x^2=1$ and $x=-1$ since $f(1)=1$.

There exists $x\in R^*, f(x)=2$, if $x>0, f(\sqrt x)^2=2$ impossible. If $x<0, f(\sqrt{ -x}\sqrt{-x})=f(-x)=f(-1)f(x)=-2=f(\sqrt{-x})^2$ Impossible.

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Suppose $\phi: \mathbb{Q}^\ast \to \mathbb{R}^\ast$ is an isomorphism. Let $a \in \mathbb{Q}^\ast$ satisfy $\phi(a)=2$. For each $n$, there is an element $b \in \mathbb{Q}^\ast$ with $\phi(b)=2^{1/n}$. Now $\phi(b^n) = \phi(b)^n = (2^{1/n})^n = 2 = \phi(a)$, so $b^n = a$. It follows that $a$ is an $n$-th power for all $n$. But then it follows that $a=1$, contradiction.

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  • $\begingroup$ Similarly, one could show that $a'=1$ where $\phi(a')=3$. So then $\phi(1)$ is equal to both $2$ and $3$. $\endgroup$ – user133281 Aug 23 '16 at 21:58
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Suppose that $f:\Bbb{R}^*\to\Bbb{Q}^*$ is a such homomorphism. Then we have some $x$ such that $f(x)=2$. Now take a cube root $\sqrt[3]{x}$ of $x$, which always exists in $\mathbb{R}^*$. Then $(f(\sqrt[3]{x}))^3 = f(x) = 2$, i.e. $f(\sqrt[3]{x})$ is a cube root of $2$. But $2$ has no cube root in $\mathbb{Q}^*$, so this is a contradiction.

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    $\begingroup$ This is exactly the argument in the accepted answer. $\endgroup$ – Tobias Kildetoft Aug 16 '18 at 7:51
  • $\begingroup$ sorry@TobiasKildetoft $\endgroup$ – jasmine Aug 16 '18 at 7:53

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