I think it wants me to show that there's no bijection between the two sets.
I first tried to show that there's simply no bijection, then I realised that it doesn't work.
If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.
Thanks!
Best Answer
In $(\mathbb{R}, +)$ there are no torsion elements (the group generated by any non-identity element is infinite cyclic). On the other hand, in $\mathbb{R}^{\times},$ the group $\langle -1 \rangle$ is finite.
That said, it is interesting to note $\mathbb{R}^{\times}/ \langle -1 \rangle \cong (\mathbb{R}_{+}, \cdot)$ which is isomorphic to $(\mathbb{R}, +)$ via the natural logarithm. In fact, $\mathbb{R}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{R}.$