Confused about proving that a group of functions is a subgroup I'm studying for my Group Theory test, and one thing I keep getting stuck on is in the sections of "groups of permutations". 
For example, take $A=\{x\in\mathbb{R}:x\neq0,1\}$ and $G=\{\varepsilon, f, g\}$, where $\varepsilon(x)=x$ (i.e. the identity function) $\,f(x)=\frac{1}{1-x}$, and $\,g(x)=\frac{x-1}{x}$.
This says that $A\subseteq\mathbb{R}$ and $G$ is a set of permutations on $A$. The first is obvious (I guess you could write it as $A=\mathbb{R}-\{0,1\}$), but I'm used to seeing permutations written in the matrix-type notation, on finite sets, i.e.
$$\begin{pmatrix}1&2&3&4&5\\2&3&1&5&4\end{pmatrix}=(123)(45)$$
Given these explicit functions, which I can guess are indeed permutations of $A$ since the definition of a function $f:A\to A$ is that it maps elements of $A$ to (perhaps) other elements of $A$, how should I show that


*

*G is closed under (what I suppose to be) the operation of function composition

*G is closed under inversion


thus proving that G is a subgroup of $S_A$?
What's confusing me is the fact that each element in $G$ is not given by one explicit function, like $G=\{f\in\mathbb{R}:f(x)=x^2, \forall x\in\mathbb{R}\}$ for example, where I could plug arbitrary numbers $x_1, x_2\in\mathbb{R}$ into the function to show that any real numbers under this function are indeed also real numbers.
Do I take each function separately, and prove that $f\circ f$, $f\circ g$, $g\circ g$, etc are all closed under the operation and inversion? Or is there a way to do this more efficiently?
 A: If you're OK being pragmatic, you could note that since $G$ only consists of three functions, then if it's a group, it had better be isomorphic to the cyclic group with three elements.
So, you could try and show that $G$ is generated by, say, $f$. That is, show that $f^2 = f \circ f = g$, while $f^3 = f \circ f \circ f = \varepsilon$ so that $G = \{f^k : k = 0, 1, 2\} \cong C_3$. I haven't exactly verified this, but if $G$ is a group in which $\varepsilon$ is the identity, then this must be so. You could frame this as using the finite subgroup test and additionally exhibiting an isomorphism between your group and the cyclic group with three elements.
Just keep in mind that a permutation of $A$ is simply a bijection (that is, invertible function) $A \to A$. As you point out, there are too many real numbers to use the matrix/cycle notation, but you have definitely encountered permutations before seeing those notations (for example, any function $L(x) = mx + b$ with $m \neq 0$ is a permutation of $\Bbb R$).
A: Compute $(g(f(x))$ by substituting $f(x) = \frac 1 {1 - x}$ for $y$ in $g(y) = \frac {y - 1} y$ and simplify to obtain x:
$$
\begin{align}
g\left(\frac 1 {1-x}\right) &= \frac {{\frac 1 {1-x}} - 1} {\frac 1 {1-x}} \\
\\
&= \frac {\frac {1-(1-x)} {1-x}} {\frac 1 {1-x}} \\
&= x
\end{align}
$$
This shows that $g \circ f = \varepsilon$, the identity function, which is the two-sided identity element for function composition. Because composition is the group operation, $\varepsilon$ is the group identity element/unit/whatever term you're familiar with. 
If you now show that $f(g(x)) = x$ for any/all $x$, you will have proved similarly that $f \circ g = \varepsilon$. From these two facts, it follows that $f = g^{-1}$ and $g = f^{-1}$.
You also need to show that the group is closed under composition. Given the above, it suffices to show that $f \circ f$ and $g \circ g$ are in the group. It's enough to show that $f \circ f = g$, because then $g \circ g = f^{-1} \circ f^{-1} = (f \circ f)^{-1} = g^{-1} = f$. To show $f \circ f = g$, simplify $f(f(x))$, which is:
$$
\frac 1 {1 - \frac 1 {1-x}} 
$$
to show that it equals $\frac {x-1} x$.
