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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.

TfwBear
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2 Answers2

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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.

More water plz
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  • Of course the matrix stated for $AB$ is not in $GL(2,\Bbb R)$ anyway (because the OP calculated it incorrectly). – David Oct 28 '15 at 04:27
  • haha I fixed it. Thanks for pointing that out I didn't even check the value of AB – More water plz Oct 28 '15 at 04:29
  • Ok. So I manually did this A and B just for my reference. I got back to order 4 for A and order 3 for B like you said. I've done a few successive steps for AB and clearly I'm not going to get back to I. But I=is there a way to show generally that there exists no such $n$? So how do I know that it would't take say $n=100000$ for example? I suspect it would have to do with divisibility of the orders. – TfwBear Oct 28 '15 at 04:41
  • Oh wait never mind. This brings me back to AB. Ignore my question. – TfwBear Oct 28 '15 at 04:43
  • No it doesnt bring you back to AB. I edited my answer and added one more hint for proving that the order of AB is not finite – More water plz Oct 28 '15 at 04:44
  • My mistake. I was looking A again. Ok so then by induction AB has infinite order. Thank You all so much. – TfwBear Oct 28 '15 at 04:53
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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!

Daniel
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