How to prove that $(G, \cdot)$ structure is an Abelian group where $G$ is a set of matrices determined by a certain rule?

Question

I have a pretty tough algebra question for today. It is from a set of problem of a competition and I tried solving it. The problem has two parts the first is the following:

a.) Given the set $G = \{A(x)\mid x\in (-1,1)\}$, where $A(x)$ are the matrices of the form $$A(x) = \frac {1}{\sqrt {(1-x^2)}} \begin{pmatrix}1 & 0 & x \\ 0 & \sqrt{1-x^2} & 0 \\ x & 0 & 1\end{pmatrix},$$ prove that $(G, \cdot)$ is an Abelian-group where "$\cdot$" is matrix-multiplication.

b.) Prove that $f:G \rightarrow R$ $f(A(x)) = \ln \frac{1+x}{1-x}$ is an isomorphism between $(G, \cdot)$ and $(R,+)$ groups.

It is clearly easy to prove that $A(x) \cdot A(y) = A(y) \cdot A(x)$, I just calculated both $A(x) \cdot A(y)$ and $A(y) \cdot A(x)$ and compared them. The problem is first we have to prove that $(G, \cdot)$ is a group and I was stuck when I needed to prove that $\forall \space A(x), A(y) \space \space A(x) \cdot A(y) \in G$. So I failed to prove closure. I also easily calculated that it always will have an inverse matrix, and also the identity matrix is a part of it, but how could I prove that $A(x)^{-1} \in G \space\forall \space A(x) \in G $?

Isomorphism is probably not that hard to prove if I can figure out the first, but unfortunately, I could not.

I would accept some help in proving that $(G, \cdot)$ is a group, on the inverse matrix and determining the formula for $A(x) \cdot A(y) = A(\alpha)$. Thanks in advance!

Answer 1

One easily checks that $$A(x)A(y)=A\left(\frac{x+y}{1+xy}\right),$$ from which $$f(A(x)A(y))=f(A(x))+f(A(y))$$ follows. Since $f:G\to\Bbb R$ is moreover bijective, this proves both that $G$ is an abelian group and $f$ is an isomorphism.

Moreover, recognizing here above the formula for the $\tanh$ of a sum, one readily reparametrizes $G$ by letting $B(u)=A(\tanh u),$ thus obtaining $B(u)B(v)=B(u+v)$ and $f(A(x))=\operatorname{argtanh}(x).$

Answer 2

The limits $|x| < 1$ and the presence of $\sqrt{1-x^2}$ suggest a trig substitution of $x = \sin\theta$ or $x = \tanh\eta$. The latter case shows more promise, giving $$ A(\eta) = \begin{bmatrix}\cosh \eta & 0 & \sinh\eta \\ 0 & 1 & 0 \\ \sinh\eta & 0 & \cosh\eta\end{bmatrix} $$ which can be easily shown to satisfy $A(\eta)A(\nu) = A(\eta+\nu)$. This also leads nicely into part 2, since $\tanh^{-1}(x) = \ln[(1+x)/(1-x)]/2$.