# Looking for a smart way to find out what algebraic structure this is

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.

• For example, the product of the second and third matrix is the fourth matrix and the first matrix is identity. To prove closure under matrix multiplication, you just need to show that multiplying any two matrices in the set you get another one in the set. Also, remember to get the inverse of a 2x2 matrix, just swap diagonal entries, negate non-diagonal entries and divide each by the determinant. Commented Nov 28, 2010 at 0:40
• @Timothy Wagner Yeah, I got that. But it boils down to what I did. Multiply every matrix by every matrix and prove that they are all in the set. The identity and negated identity matrix matrix made it a bit easier. Commented Nov 28, 2010 at 0:44
• @AndrejaKo: OK. Also, observe that second and fourth matrices are inverses of each other and first and third matrices are self-inverses. That should simplify your calculations a bit. You can either do this kind of elementwise computation or alternately establish an isomorphism to a well known algebraic structure. Commented Nov 28, 2010 at 0:48
• @Timothy Wagner I did figure out that too and it did make calculations simpler. How would I start on the isomorphism part. Commented Nov 28, 2010 at 0:51
• Every matrix corresponds to a linear map. What maps of the plane do the 4 matrices you have represent?
– Max
Commented Nov 28, 2010 at 0:55

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 $4$).