Why $(\mathbb{Z}, + )$ is not isomorphic to $(\mathbb{R}^+, *)$?

Question

Can someone explain to me why $(\mathbb{Z}, + )$ is not isomorphic to $(\mathbb{R}^+, *)$ where $*$ is multiplication.

My book says they aren't really isomorphic and doesn't say why. I thought that they are because of two reasons

1. All infinite groups with generator is isomorphic to $(\mathbb{Z}, + )$ and clearly $(\mathbb{R}^+, *)$ is infinite (edit, but no generator, okay)

2. The isomorphism is given by $f(x) = e^x$

Answer 1

• All infinite groups are isomorphic to $(\mathbb{Z},+)$? So, what does the classification theorem for abelian groups say about a guy like $\mathbb{Z} \oplus\mathbb{Z}/2$?
• As for your isomorphism: which would be the inverse image of $e^{1/2}$?

Answer 2

Your first claim is false. In fact, there is not even a bijection between the sets $\Bbb{Z}$ and $\Bbb{R}^+$, by Cantor's diagonalization argument.

However, the map $x \mapsto e^x$ does provide an isomorphism between the groups $(\Bbb{R}, +)$ and $(\Bbb{R}^+, \times)$.

Answer 3

The other answers are good: $(\mathbb{Z}, + )$ is cyclic and countable, while $(\mathbb{R}^+, \times)$ is neither cyclic nor countable.

Here's another, less direct answer: $(\mathbb{R}^+, \times)$ is divisible, while $(\mathbb{Z}, + )$ is not divisible.