How is it possible for this type of <G,> to be a group?
Can we assume that, since you are asking this question, that you know what a group is?
A group consists of a set of object (you given that here- the set of 2 by 2 matrices with determinant 1) and an "operation" that takes pairs of those objects to an object in the set (you are given that- usual matrix multiplication) with these properties:
The operation is "associative"- that is, (AB)C= A(BC).
(The operation is defined only for two objects. You need to show that it does not matter whether you first multiply A and B together and then C or first multiply B and C together then A.)
Take $\displaystyle A= \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, $\displaystyle B= \begin{pmatrix}w & x \\ y & z\end{pmatrix}$, and C= \begin{pmatrix}p & q \\ r & s\end{pmatrix}[/tex]
. You are told how to multiply A and B. What do you get when you multiply that by C "on the right"? What is BC? What do you get when you multiply that by A "on the left"?
There is an "identity". That is, there exist a matrix "I" such that, for any matrix, A, AI= IA= A.
You are told that $\displaystyle AB= \begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}w & x \\ y & z\end{pmatrix}= \begin{pmatrix}aw+ by & ax+ bz \\ cw+ dx & cx+ dz\end{pmatrix}$
You want that to find B (find w, x, y, and z) such that this is equal to A for any a, b, c, and d. That is, solve aw+ by= 1, ax+ bz= 0, cw+ dx= 0, and cx+ dz= 1 for w, x, y, and z. Show that you have the same solution for all a, b, c, and d.
Every member of the set has an "inverse". That is, for any A there exist B such that AB= I.
Use the same idea as above. This is where "ad- bc= 1" is important.
I can't disagree with the previous answer. Is this the first time you've seen matrix multiplication? If not, most of the verification is easy.
1. Closure: The requirement to be in G is that the determinant of the matrix is 1. So for A, B in G, you need the determinant of AB to be 1. But det(AB)=det(A)det(B), so AB has determinant 1.
2. Associativity: Matrix multiplication is associative.
3. Identity: The 2 by 2 identity matrix I (1's on the main diagonal and 0's elsewhere) is the group identity. Clearly, det(I)=1.
4. Inverses: For A in G, $A^{-1}$ exists since det(A) is not 0. $A^{-1}$ is in G since det($A^{-1}$)=1 -- proved with the determinant fact mentioned in 1.
To show G is infinite, you must give infinitely many matrices in G; also you need to find non-commuting matrices A and B in G.
I would not interpret this as asking you to prove that this group is "infinite" or "non-abelian". I would interpret "(In fact, G is an infinite Abelian group)" as an aside and not part of the problem. Of course, it is not difficult to show that.