Transport of structure, why is $\left(G=(-1,1),* : (x,y) \to \frac{x+y}{1+xy}\right )$ a group?

Question

It is closely related to this topic:

Prove that $x * y = \frac{x+y}{1+xy}$ is a stable part of $G=(-1, 1)$

As proven, $*$ is an internal operation for $G$. Plus one may observe that $\tanh$ is bijective from $R$ to $G$ and $\tanh(x+y)=\tanh(x)*\tanh(y)$

I guess I am supposed to conclude that $G,*$ is a group but I do not see precisely how ? What theorem or property should be used ?

Answer 1

$(G, *)$ is a group because it obeys the axioms of a group:

1. the operation $(G,*)$ has closure: what you call 'an internal operation'. But that is just the beginning; we also need:

2. it is associative: \begin{align} (x*y)*z &= \left(\frac{x+y}{1+xy}\right)*z\\ &= \frac{\frac{x+y}{1+xy}+z}{1 + \frac{x+y}{1 +xy}z}\\ &= \frac{x+y+z(1+xy)}{1 + xy + (x+y)z}\\ &= \frac{x+y+z + xyz}{1 + xy + xz + yz}\\ &=\frac{x(1+yz) + y + z}{1 + yz + x(y+z)}\\ &= \frac{x + \frac{y+z}{1 + yz}}{1 +x \frac{y+z}{1+yz}}\\ &= x * \left(\frac{y+z}{1 + yz}\right)\\ & = x * (y*z). \end{align}

3. there is an identity element in $G$: $\forall x\in G, x * 0 = \frac{x+0}{1+ x\cdot 0} = x$

4. every element has an inverse: for all $x \in G, \exists y \in G: x+y =0$. And $x+y=0 \implies x*y = 0$. So $\forall x \in G, \exists y \in G: x*y = 0$.

Answer 2

It is sufficient to prove this statement (thanks to @Did)

If (H,⋅) is a group, f:H→G a bijection, and ∗ the composition law on G uniquely defined by the fact that f(x)∗f(y)=f(x⋅y) for every x and y in H, then (G,∗) is a group.

Assume that just consider x,y,z this respective antecedent of a,b,c by f

1. associativity: a*(b*c)=(a*b)*c $\iff$ f(x)*f(y.z)=f(x.y)*f(z) $\iff$ f(x.(y.z))=f((x.y).z) which is obvious as (H,.) is a group.
2. neutral existance: f(z)=f(z* $0_H$ )=f(z)*f($0_H$) and f(z)=f($0_H$*z)=f($0_H$)*f(z) thus f($0_H$)=$0_G$ is neutral.
3. inverse: f(0_G)=f($x^{-1}$*x)=f($x^{-1}$)*a=$0_G$