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On my mid-term exam tonight, I had a problem which went something like this:

Which algebraic structure is $(M,*)$, where $M= \left \{ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \right., \begin{bmatrix} 0 & 1\\ -1 &0 \end{bmatrix},\begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix},\begin{bmatrix} 0 &-1 \\ 1 & 0 \end{bmatrix}\left. \right \} $ and * is matrix multiplication.

There must be some smart way of doing it, but I couldn't figure it out. In the end I made a table with all elements of M and relationship between them and then just read structure's properties. To me this just feels like excessive application of brute force.

When I was trying to prove that operation is closed, I tried with something like $x=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, y= \begin{bmatrix} p & q\\ s& t \end{bmatrix}$ and multiply that and then prove that result is part of the set, but I run out of ideas on how to prove that result is part of the set.

I had basically same problem for inverse element and commutativity too.

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    @AndrejaKo: Sorry the second part of my previous comment does not apply to your problem. I misread it.2010-11-28

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Max has got the right idea in the comments. If you look at what these matrices do to points in the plane the answer should be clear.

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Another option is to take one of the matrices and apply it several times. Say you take the second matrix and apply it to $(x,y)$. Magically, you will immediately recognize the other matrices (in order), so you've got a cyclic group of order $4$.

As Max and Qiaochu have commented, there's a geometric interpretation (think what geometric transformation keeps the origin and has order $4$).

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    H$e$ m$e$ans multiply th$e$ v$e$ctor by the matrix.2010-11-28