Show that the set of these functions form a group $\mathit G$, where the group operation is composition of functions. Let $\mathit X$ = $\mathbb R$ \ {0,1}. Consider the follow six functions:
$x$, $1-x$, $\frac 1x$, $\frac{1}{1-x}$, $\frac{x-1}{x}$, $\frac{x}{x-1}$
which are all functions from $\mathit X$ to $\mathit X$. Show that the set of these functions form a group $\mathit G$, where the group operation is composition of functions.
So, in order to show that it is a group, I need to show that the closure, associative, identity exist and inverse exist.
For identity element, the identity element is 1 because all functions multiply by 1 is equal to itself. So, identity exist.
For inverse, $(x)(\frac 1x)=1$, $(1-x)(\frac {1}{1-x})=1$, ...etc. So, inverse exist.
Is that how to show identity and inverse exist? However, I don't know how to show the closure and associativity.
After proving that it is a group. How to find a group among cyclic groups, dihedral groups and symmetric groups which is isomorphism to $\mathit G$?
Does it mean that I need to think of a group which is cyclic and think of a function such that they are bijective and homomorphism?
 A: You already know the composition of functions is associative. As such, the group operation is associative as well!
The identity, however, is not 1. if the group operation were multiplication, you would be correct, but as you note, the group operation is composition of functions, NOT multiplication. Also, note that 1 is not in the group. If 1 were indeed the identity, then the group would not contain the identity.
To show that inverses exist, you must once again use function composition, not multiplication.
As a gentle reminder, function composition works as such. Take $f(x)=1-x$ and $g(x)=\frac{1}{x}$. Then, we have:
$$f(g(x))=f\left(\frac{1}{x}\right)=1-\frac{1}{x}=\frac{x-1}{x}$$
To show that closure holds, it might be useful to construct a multiplication table to show that the result of the composition of any two functions of the group remains in the group.
Once you have constructed this multiplication table, the properties of closure and the existence of inverses should follow.
As an aside, it should be noted that in mathematics, once must be careful in proving that a definition is met. Also, if you are lost and unsure of how to proceed, restating the definition is always helpful!
