Given $f(x)=\frac{1+x}{1-x},\,x\neq -1,0,1,$ find:
(a) Compute the composition $f^2, f^3, f^4.$ On the basis of your findings, what are $f^{10}, f^{100}?$
(b) Let $S$ be the set of distinct functions found in part (a). List the elements of $S,$ show $S$ is closed under composition, make a Cayley table for $S.$
(c) Show that composition is a commutative operation on $S,$ there is an identity element in $S,$ and every element of $S$ has an inverse in $S.$
(a) $f^2= \frac{1+\frac{1+x}{1-x}}{1-\frac{1+x}{1-x}}=-1/x.$ $f^3= -\frac1{\frac{1+x}{1-x}} = \frac{x-1}{1+x}.$ $f^4= \frac{\frac{1+x}{1-x} -1}{1+\frac{1+x}{1-x}} = \frac{1+x-1+x}{1+x+1-x}= \frac{2x}{2}=x.$ $f^5=\frac{1+x}{1-x}.$
So, the operation of function composition yields a cycle of order $4$:
$f(x)=\frac{1+x}{1-x}, f^2(x)=-1/x,
f^3(x)= \frac{x-1}{1+x}, f^4(x)=x, \cdots$
Hence, $f^{10}= f^2= -1/x, f^{100}=x.$
(b) Let the composition be denoted by $\circ.$ $$\begin{array}{ccccc} \hline \circ & f^1= \frac{1+x}{1-x} & f^2 = -1/x & f^3 = \frac{x-1}{1+x} & f^4= x\\ \hline f^1 & f^2 & f^3 &f^4& f^1\\ \hline f^2 &f^3&f^4&f^1& f^2\\ \hline f^3 &f^4&f^1&f^2&f^3\\ \hline f^4 &f^1&f^2&f^3&f^4\\ \hline \end{array}$$
Not clear how to show, except by taking all possible compositions $x\circ y$ of any two elements: $x, y\in S,$ that $S$ is closed under composition.
(c) As stated among the methods here, the easiest method is to show if the group is cyclic. Here, all elements generate the group.
Only by examination of the table, can state $f_4$ is the identity element of $S.$
Although, found while constructing the Cayley table, but is there an easy way to show that every element has an inverse in the given set, under the operation of function composition?
I found that $(f^1)^{-1}= f^3,$ and vice-versa. Inverse of $f^2, f^1$ is itself.
Have a related doubt: why associative property check was not asked?