How to define a group for a complex rational function

Question

I'm trying to show that the set of mappings $z\mapsto \frac{az+b}{cz+d}$ with the requirement, $ad − bc \neq 0$ satisfies the properties of a group.

I know that a group is a set with an operation that sends ordered pairs of elements to elements of the set (group multiplication) with the added requirements that the multiplication operation is associative, there is an identity element, and every element has an inverse.

I have no idea how to attempt to show this as I've never encountered proving properties for a group before.

Answer 1

1.Show that calculation is closed. 2.Show that identity element exist. 3.Show that for any elements, the inverse element exist. I do this proof with honesty but complicated way ( You can do it. Just calculation).

I'm moved by Peetrius way.That's very beautiful.I explain why this beautiful relation exists.Peetrius separete denominater and numerator to the elements of vector and substitue the operation to matrix operation. You can show $z↦(az+b)/(cz+d)$ is equal to $z/z'↦(az+bz')/(cz+dz')$ Oh,the former mapping is equal to$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} z\\ 1 \end{bmatrix}$ and latter one is $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} z\\ z' \end{bmatrix}$.This mean you can separate denominater and numerator freely (You don't need to make denominater to 1). Now the mapping operation is completely equal to the matrix operation.

Answer 2

As it turns out we can use the natural matrix structure for the rational mapping $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

to help show that the rational mapping is a group. The identity of the matrix is $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ The inverse of the matrix when the determinant doesn't equal zero is $$\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$

Finally, we know that matrix multiplication is associative.