Proving a defined group $(G,*)$ is isomorphic to $(\mathbb{R},+)$

I am studying abstract algebra and I have this question:

Let $G=${$a\in\mathbb{R}|-1<a<1$}
Defined an operation $*$ in $G$ with $a*b=\frac{a+b}{1+ab}$ for all $a,b \in G$
Show that $(G,*)$ and $(\mathbb{R},+)$ are isomorphic.

I know that to prove $(G,*)$ and $(\mathbb{R},+)$ are isomorphic, I have to show that
1. both $(G,*)$ and $(\mathbb{R},+)$ are groups.
2. there exist a map $f$ from $G$ to $\mathbb{R}$
3. the map f is a homomorphism
4. the map f is a bijective, that is both one one and onto
So far, I have no difficulty in proving that $(G,*)$ is a group. And we all know that $(\mathbb{R},+)$ is a group as well. However, I find it very hard to find such a bijective map $f$ so that $(G,*)$ and $(\mathbb{R},+)$ are isomorphic.
I was wondering, are $(G,*)$ and $(\mathbb{R},+)$ really isomorphic? I mean, $G$ is subset of $\mathbb{R}$, right? How can there be an onto map from $G$ to $\mathbb{R}$? If there is such a map, what is it? Can anyone help me?
Thank you.

• There is no problem in G being a subset of R (in fact, they are even homeomorphic with respect to standard topology on R). For example, (R, +) is isomorphic to (R+, *) (the group of positive reals under multiplication). – lisyarus Sep 17 '14 at 12:50
• Hint: consider the hyperbolic tangent. – polmath Sep 17 '14 at 12:51
• @lisyarus and Paul, I was thinking about linear function, I never thought about trigonometric function, especially hyperbolic one. Thanks! – Monica Sendi Afa Sep 18 '14 at 12:53

Suppose there is a differentiable group isomorphism $$\phi: \mathbb{R} \to (-1, 1);$$ in particular $\phi$ must respect the group law: $$\phi(x + y) = \phi(x) \ast \phi(y),$$ which in terms of the usual operations on $\mathbb{R}$ is $$\phi(x + y) = \frac{\phi(x) + \phi(y)}{1 + \phi(x) \phi(y)}.$$
Differentiating both sides with respect to $y$ and evaluating at $y = 0$ gives $$\phi'(x) = \phi'(0) (1 - \phi(x)^2).$$ This is separable: Denoting $\lambda := \phi'(0)$ and rewriting gives $$\frac{d\phi}{1 - \phi^2} = \lambda dx,$$ and integrating gives $$\mathrm{artanh}\, \phi = \lambda x + C$$ for some constant $C$. Checking shows that the group identity for both $G$ and $\mathbb{R}$ is $0$, so substituting gives $C = 0$. Solving for $\phi$ gives $$\phi(x) = \tanh \lambda x;$$ provided $\lambda \neq 0$ this map is invertible and hence is an isomorphism by construction, but of course one can check this (or any other candidate) directly. (Less interestingly, $\lambda = 0$ gives the zero homomorphism.)
We can think of this solution just as well as a sum identity for $\tanh$: $$\tanh (x + y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}.$$ If we know this ahead of time, we can guess the isomorphism just from the formula of the group operation $\ast$; even if we don't, we might guess something like this holds given the formally similar sum identity for $\tan$.
• thanks for the answer. I don't really understand the part "Differentiating the first and third expressions with respect to $y$ and evaluating $y=0$ gives ..." but assuming it correct, O understand the rest. I would be grateful if you can explain it more detail especially about $phi '$. Thanks – Monica Sendi Afa Sep 18 '14 at 13:02
• I reworded that part of the answer for clarity. As for $\phi'$, it's just the usual derivative of $\phi$ as a real-valued function on $\mathbb{R}$. – Travis Sep 18 '14 at 13:13
• $x$ and $y$ are real numbers and thus, $\phi(x)$ and $\phi(y)$ must be real numbers as well. How can we derivate it? – Monica Sendi Afa Sep 18 '14 at 13:22
• That's true for fixed values $x, y$, but we can just as well think of $x$ as the argument of a function. Moreover, we can think of the left- and right-hand sides as functions of $(x, y)$. The next step in the solution is to differentiate both sides w.r.t. y and then evaluate the resulting expressions at $y = 0$, the result of which is the given differential equation in $\phi(x)$. – Travis Sep 18 '14 at 13:47