Vtyjkyuk

Of course, since the cardinality of $mathbbR$ exceeds the cardinality of $mathbbQ$, there does not exist a bijection between $mathbbQ^ast$ and $mathbbR^ast$, 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.

In $mathbbR^ast$ every element has a cuberoot; that is, for any $a$ in $mathbbR^ast$ there is an element $b$ such that $b^3=a$. In $mathbbQ^ast$ there are elements that do not.

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 $xin R^*, f(x)=2$, if $x>0, f(sqrt x)^2=2$ impossible. If $x<0, f(sqrt -xsqrt-x)=f(-x)=f(-1)f(x)=-2=f(sqrt-x)^2$ Impossible.

Suppose $phi: mathbbQ^ast to mathbbR^ast$ is an isomorphism. Let $a in mathbbQ^ast$ satisfy $phi(a)=2$. For each $n$, there is an element $b in mathbbQ^ast$ 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.

Suppose that $f:BbbR^*toBbbQ^*$ 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 $mathbbR^*$. 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 $mathbbQ^*$, so this is a contradiction.