Orders of Elements in GL(2,R) Let A =
$$\begin{pmatrix}
0&1\\
-1&0\\
\end{pmatrix}$$
and B =
$$\begin{pmatrix}
0&-1\\
1&-1\\
\end{pmatrix}$$
be elements in $GL(2, R)$. Show that $A$ and $B$ have finite orders but AB does not.
I know that $AB$ =
$$\begin{pmatrix}
1&-1\\
0&1\\
\end{pmatrix}$$
and that $GL(2,R)$ is a group of 2x2 invertible matrices over $R$ with the matrix multiplication operation.
Firstly, I am confused as to how an element of this group can have an order. Is it that A and B are the products of invertible matrices in this group? Given that this is over $R$ the orders should be infinite. My book did not define the general linear group very well. Secondly, I would like some guidance on how to proceed following this problem. Any help is much appreciated.

To have an order n means that n is the smallest positive number such that $A^n = I$. In this case, $I$ is of course the identity matrix.
$A^2$ = $\begin{pmatrix}
0&1\\
-1&0\\
\end{pmatrix}\begin{pmatrix}
0&1\\
-1&0\\
\end{pmatrix} = \begin{pmatrix}
-1&0\\
0&-1\\
\end{pmatrix} = -I$
and so $A^4 = I$, order of A is 4
$B^3$=
$\begin{pmatrix}
0&-1\\
1&-1\\
\end{pmatrix}^3=I$ so the order of B is 3
Given $AB$ =
$\begin{pmatrix}
1&-1\\
0&1\\
\end{pmatrix}$, I'll let you check that $(AB)^n \neq I$ for any $n$ , thus proving $AB$ does not have finite order
Now to rigorously prove that $AB^n \neq I$ for any n, one way is to show that 
$AB^n$ = $\begin{pmatrix}
1&-n\\
0&1\\
\end{pmatrix}$
by induction.
