$(G,\circ)$ is an abelian group, where $x\circ y=\frac{x+y+a(1+xy)}{1+xy+a(x+y)}$

Let $G=(-1,1)$ and $a\in G$ be fixed. Prove that $(G,\circ)$ is an abelian group, where $$x\circ y=\frac{x+y+a(1+xy)}{1+xy+a(x+y)}, \forall x,y\in G.$$

To me, it seems extremely tedious to prove the axioms of the group in this case. Proving associativity is horrendous and I don't believe that any other of the axioms (apart from commutativity) is provable without extremely long computations.

In order to avoid this, I tried to use the so-called structure transport i.e. finding a bijective function from $G$ to some well-known group. I couldn't come up with any function, so I don't know how to actually solve this question. I doubt that it can be solved by proving each of the group axioms, but if anyone finds a way to do this I would be both amazed and grateful.

Solution 1:

the structure transport is the composition of two functions, first is hyperbolic tangent, as $$ \tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}. $$ Second is the Mobius transformation $$ m(z) = \frac{z+a}{az+1} $$

Note that $$ \frac{p+aq}{q+ap} = m\left(\frac{p}{q} \right) $$

Works very nicely. Note that the group identity is $(-a).$ All we are doing is mapping the interval to itself by Mobius transformation, in such a way that $-a$ maps to $0.$ Then we use the known operation $ \frac{u+v}{1+uv} \; , \; $ then back by the inverse of the Mobius transformation.