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.


The order of an element $g$ (in a group $G$) is the smallest
  positive integer $n$ such that $g^n=e$, where $e$ is the identity of
  the group. If such an $n$ exists, we say the the element $g$ has finite order.

With this in mind, you should look for some positives integers $n,m$ such that
$$\begin{pmatrix}
0&1\\
-1&0\\
\end{pmatrix}^n=\begin{pmatrix}
1&0\\
0&1\\
\end{pmatrix}$$
and
$$\begin{pmatrix}
0&-1\\
1&-1\\
\end{pmatrix}^m=\begin{pmatrix}
1&0\\
0&1\\
\end{pmatrix}$$
but, you must show that such integer does not exist for the product. This will be because, once you take successive powers, you reach a point where you get back to the beginning without reaching the identity. Try it by yourself!
