Group or not?

We first calculate all possible operations between the elements of the set $G$ . The element $e$ is the identity $e (x) = x$ , so that $e \circ h = h \circ e = h$ for any function $h (x)$ . We must calculate only the operations between the elements $f$ and $g$ : $\begin{aligned} f(f(x)) &= -\frac{1}{1 - \frac{1}{1+x}} = -\frac{1}{1 - \frac{1}{1+x}} \frac{1+x}{1+x} = -\frac{1 + x}{1 + x - 1} = - \frac{1 + x}{x} = - \frac{1}{x} - 1 = g(x)\\ f(g(x)) &= -\frac{1}{1 - (1 + \frac{1}{x})} = - \frac{1}{- \frac{1}{x}} = \frac{1}{\frac{1}{x}} \frac{x}{x} = x = e(x) \\ g(f(x)) &= f(f(f(x))) = f(g(x)) = e(x) \\ g(g(x)) &= f(f(g(x))) = f(e(x)) = f(x) \end{aligned}$ Here we can extend the fractions with the terms $x$ and $x + 1$ , since these are non-zero for $x \in \mathbb{R} \! \setminus \! \{- 1,0 \}$ . In lines 3 and 4 we took advantage of the associativity of the composition $\circ$ . The results can be summarized in the following Cayley table So we can summarize the results as follows:

• $G$ is closed under the operation $\circ$ .
• $e$ is the neutral element for the operation $\circ$ .
• For the other two elements $f$ and $g$ there is an inverse element $f ^{- 1} = g$ and $g ^{- 1} = f$ , so that $f ^{- 1} \circ f = g ^{- 1} \circ g = e$ .

Since the operation $\circ$ is also associative, all group axioms are satisfied. The group $(G, \circ)$ is isomorphic to the group $D_3$ .