The multiplicative groups $\mathbb{Q}^\ast$ and $\mathbb{R}^\ast$ are not isomorphic [duplicate]
Solution 1:
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.
Solution 2:
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.