Use a group-theoretic proof to show that $\mathbb{Q}^*$ under multiplication is not isomorphic to $\mathbb{R}^*$ under multiplication.

Question

Use a group-theoretic proof to show that $\mathbb{Q}^*$ under multiplication is not isomorphic to $\mathbb{R}^*$ under multiplication.

I have tried this:

Suppose $$ \phi: \mathbb{Q}^*\to \mathbb{R}^* $$ where $\phi(x)=x^2$

Now for some $3 \in \mathbb{R}^*$ there is no mapping in $\mathbb{Q}^*$ since $\sqrt3$ does not belong to $\mathbb{Q}^*$.

Hence $\phi$ is not an onto function. Therefore $\mathbb{Q}^* \not\cong \mathbb{R}^*$.

But I am not sure if it is correct.

Also, What is a group-theoretic proof?

Answer 1

Your idea is in the right direction but $x\mapsto x^2$ is not onto. However, $x\mapsto x^3$ is onto and works. More precisely:

The map $\mathbb R^* \to \mathbb R^*$ given by $x\mapsto x^3$ is surjective and so every element of $\mathbb R^*$ is a cube.

The map $\mathbb Q^* \to \mathbb Q^*$ given by $x\mapsto x^3$ is not surjective and so not every element of $\mathbb Q^*$ is a cube of an element of $\mathbb Q^*$. For instance, $2 \in \mathbb Q^*$ is not a cube (of a rational number).

Answer 2

The squaring map is not onto $\mathbf R^\times$, so it does not quite work. However, it is pretty close: the image is a subgroup of index $2$. In $\mathbf Q^\times$, the index is infinite. Thus, the two groups are not isomorphic.

(In fact, this shows that they are not even elementarily equivalent.)