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

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 $f(x) = e^x$

• Your map is a homomorphism. Why is it not an isomorphism? (This isn't enough, but it is of dire importance that you check your logic). In response to 1., not all infinities are created equal. Any real analysis class should tell you this. – PVAL-inactive Jan 5 '14 at 8:57
• Not all infinite groups are isomorphic to $\Bbb{Z}$. For one that would mean that all infinite groups are abelian, which is definitely not true. – Nick D. Jan 5 '14 at 8:58
• All infinite cyclic groups are isomorphic to $(\Bbb Z,+)$. Not every infinite group is cyclic, however. – Stahl Jan 5 '14 at 8:58
• @NickD, I misquoted the theorem. There must be a generator for the group to be isomorphic to $\mathbb{Z}$. Clearly, the one I have has no generator. – Hawk Jan 5 '14 at 9:01
• You can't even embedded a $(\mathbb{Z},+)$ inside an arbitrary infinite group. For example, $C_{2}^{\mathbb{N}}$ – Gina Jan 5 '14 at 9:01

• 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 $e^{1/2}$?
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 $(\Bbb{R}^+, \times)$.
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.