Question: Prove that the group $G$ consisting of the set $\mathbb{R}\setminus\{-1\}$ with multiplication defined by $a\sim b=a+b+ab$ is isomorphic to the multiplicative group of nonzero real numbers, $(\mathbb{R}\setminus\{0\}, \times)$.
I'm trying to find a function that equates the two groups. Let's call the isomorphism Y. I've tricked around with $Y(a\sim b)=(ab+a+b)-(a+b)$, but this is taking in two input variables, so when it comes to trying to show $Y(a\sim b)=Y(a)(b)$, the function loses all of its meaning. So then I tried $Y(a)=-a$, but then $Y(a\sim b)=-a-b-ab \neq Y(a)y(b) = ab$.
A little help here? It's probably something obvious, but I'm having difficulties at forming a legitimate map.
Best Answer
Hint:
Consider the isomorphism function
where $e$ is the multiplicative identity. Then we have
Taking $e=1$ for the set $\Bbb R\setminus\{-1\}$, we see that we would have $Y(-1)=-1+1=0$, and $Y(-1\text~b)=Y(a\text~-1)=0$ if we allowed $-1$ to be part of the set.