The Problem
In the following, $A$ is a subset of $\mathbb{R}$ and $G$ is a set of permutations of $A$. Show that $G$ is a subgroup of $S_A$, and write the table of $G$.
(1) $A$ is the set of all $x\in\mathbb{R}$ such that $x\neq 0,1$. $G=\{\varepsilon, f, g\},$ where $f(x)=1/(1-x)$ and $g(x)=(x-1)/x$.
What I Know
From what I understand, a nonempty subset $S$ of a group $G$ (that is equipped with an operation $*$) is a subgroup if $S$ is a group under that operation.
What I Don't Know
1.) What exactly is the operation in this problem? Addition? Subtraction?
2.) How does $0$ and $1$ not being included affect the problem?
3.) Up until now I've only dealt with small, finite groups of permutations. How does one tackle $\mathbb{R}$?
4.) Finally, what does the question mean by a "table of $G$?"
As always, thanks for taking the time!
To show a subset of a group is a subgroup, you only need to show that it contains the identity, that it is closed under products and closed under inverses.
For (1), the operation (I believe) is function composition, because $f$ and $g$ happen to be inverses of each other and would certainly be one way of forming a (sub)group.
For (2), note that $f$ isn't defined at $ x = 1$ and $g$ isn't defined at $x= 0$.
For (3), you don't need to tackle $\mathbb{R}$. You don't even need to tackle $S_A$. You are told $S_A$ is a group, and you need to show $G$ is a subgroup under the (presumably inherited) operation of composition. Show you need to show that the identity is in $G$ (which it is , assuming $\epsilon$ is the identity function), need to show it is closed under inverses and products. Indeed to show it is closed under products you need to do number 4.
For (4), I think it means to draw a Cayley table (it will be a $3 \times 3$ table) showing what the product of two elements is. If all the elements of the Cayley table are already in the set, then you have shown closure under products.