Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers

Theorem
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.

Let $\circ: G \times G \to G$ be the binary operation defined as:
 * $\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$

It has been established in Group Examples: $\dfrac {x + y} {1 + x y}$ that $\struct {G, \circ}$ forms a group.

$\struct {G, \circ}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.

Proof
To prove $G$ is isomorphic to $\struct {\R, +}$, we need to find a bijective homorphism $\phi: (-1, 1) \to \R$ that preserves the group structure, specifically $\phi(x \circ y) = \phi(x) + \phi(y)$ for all $x,y \in G$. From some analysis of the groups, we can surmise that the identity element of $G$ is 0, and an inverse of an element $x$ is its negative ($x^{-1}=-x$). This is true for both groups. This hints at the structure of a possible isomorphism, namely that it is an odd function, that it passes through 0, and is defined on the $(-1, 1)$ open interval. One such function is the inverse hyperbolic tangent function $tanh^{-1}$, which is defined on the $(-1, 1)$ interval.

Using $tanh^{-1}(x)=\frac{1}{2}ln(\frac{x+1}{x-1})$:

$tanh^{-1}(x \circ y)=\frac{1}{2}ln(\frac{\frac{x+y}{1+xy}+1}{\frac{x+y}{1+xy}-1})=\frac{1}{2}ln(\frac{x+y+1+xy}{x+y-1-xy})=\frac{1}{2}ln(\frac{(x+1)(y+1)}{(x-1)(y-1)})=\frac{1}{2}ln(\frac{x+1}{x-1}) + \frac{1}{2}ln(\frac{y+1}{y-1}) = tanh^{-1}(x) + tanh^{-1}(y)$

And so $tanh^{-1}$ is a homomorphism between $\struct {G, \circ}$ and $\struct {\R, +}$. It is bijective because its inverse is just $tanh$ defined over $\R$, therefore $(G, \circ) \cong \struct {\R, +}$