Are $([0,1),+_{\text{mod}1})$ and $(\mathbb{R}_{>0},\cdot)$ isomorphic?

Question

I have the following groups:

(1) $([0,1),+_{\text{mod}1})$ where:

$x+_{\text{mod}1}y= \begin{cases} x+y & \mbox{if }x+y<1 \\ x+y-1 & \mbox{if }x+y \geq1 \\ \end{cases}$

(2) $(\mathbb{R}_{>0},\cdot)$ which is the usual multiplication of positive real numbers.

The question is to check if these groups are isomorphic.

I started by assuming that there is a bijection from $[0,1)$ to $\mathbb{R}_{>0}$ where:

$f(x,y)= \begin{cases} x+y & \mbox{if }x+y<1 \\ x+y-1 & \mbox{if }x+y \geq1 \\ \end{cases}$

I got stuck here and I'm not sure what to do next

Answer 1

In $([0,1),+_{\text{mod}})$, $0$ is the identity element $e$, and there are two elements ($0$ and $1/2$) that satisfies $x^2=e$.

In $(\mathbb{R}_{>0}, \cdot)$, $1$ is the identity element $e$. However, there is only one element ($1$) that satisfies $x^2=e$.

So they are not isomorphic.