Prove that $(G, \circ)$ is group, $G=\{x, -1

Question

If $1<x, y< 1$ then $-1<x*y<1$ and $0<1+x*y <2$, but then I can't prove that $-1<x\circ y =(x+y)/(1+x*y)<1$. Help please.

Answer 1

Knowing formula: $tanh(a + b) =\dfrac{tanh(a) + tanh(b)}{1 + tanh(a) . tanh(b)}$.

written under the form

$$a+b=tanh^{-1} \left(\dfrac{tanh(a) + tanh(b)}{1 + tanh(a) . tanh(b)}\right)$$

and setting $A=tanh(a)$ and $B=tanh(b)$, you get:

$$tanh^{-1}(A)+tanh^{-1}(B)=tanh^{-1}(A\circ B)$$

Thus you have proved that $((-1,1),\circ)$ is a group law by a mere transport of the additive law on $(\mathbb{R},+)$ by bijective function $tanh^{-1}$.

Have a look at (https://en.wikipedia.org/wiki/Formal_group) which provides different "transport" examples, "$\circ$" operation being one of them.

Answer 2

For $|x|<1,|y|<1$, to prove $$ \left|\frac{x+y}{1+xy}\right|<1\tag1 $$ Just prove $\:\:|x+y|<|1+xy|$. Square both side, we have $$ x^2+y^2+2|xy|<1+2|xy|+x^2y^2 $$ Rearrage it we get $$ (1-x^2)(1-y^2)>0 $$ This is true for $|x|<1,|y|<1$. So we conclude that $(1)$ be true.

To prove $(G, \circ)$ is group, note that $0$ is identity and $-x$ is inverse. I assume that you know how to prove for associativity and rest.