Show that the following set of six matrices is a group Show that the following set of six matrices is a group
  $$\begin{bmatrix}1 &amp; 0\\0 &amp; 1\end{bmatrix},\begin{bmatrix}0 &amp; 1\\-1 &amp; -1\end{bmatrix},\begin{bmatrix}-1 &amp; -1\\1 &amp; 0\end{bmatrix},\begin{bmatrix}1 &amp; 0\\-1 &amp; -1\end{bmatrix},\begin{bmatrix}-1 &amp; -1\\0 &amp; 1\end{bmatrix},\begin{bmatrix}0 &amp; 1\\1 &amp; 0\end{bmatrix}.$$


The only property that is not immediately apparent is closure. The question before this one involved the symmetric group $S_3$ and based off another post, I believe this to be necessary for this question. However, I am not really sure how to make the connection. Any help would be appreciated.
Perhaps the least tedious way is to show we have some identification (I am speaking of an isomorphism, here) with some known group. As the comments above suggest, the most likely candidate (based on your matrices' orders) is $S_3$, or (equivalently), the symmetry group $D_3$ of the equilateral triangle.

The latter group is most succinctly summed up as:

$D_3 = \langle a,b| a^3 = b^2 = e; ba = a^2b\rangle$.

With sufficient patience, one can show this determines a group of $6$ elements:

$D_3 = \{e,a,a^2,b,ab,a^2b\}$ and the rules $a^3 = b^2 = e$ and $ba = a^2b$ are sufficient to determine all $36$ entries in a multiplication table; for example:

$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = (a^3)(bab) = e(bab) = (ba)b = (a^2b)b = a^2(b^2) = a^2e = a^2.$

Then it is a matter of finding some matrix $A$ of your $6$ we can map to $a$, and another $B$ that maps to $b$, and showing that the matrix multiplication of these two matrices satisfies our $3$ rules:

$A^3 = I\\B^2 = I\\BA = A^2B.$

If that is so, then the isomorphism we would get is:

$A^kB^l \mapsto a^kb^l$, and we would get a "parallel" multiplication table for your $6$ matrices corresponding to the multiplication table for $D_3$ (thus showing closure).

As pointed out above, we have for $A = \begin{bmatrix}0&amp;1\\-1&amp;-1\end{bmatrix}$:

$A^2 = \begin{bmatrix}0&amp;1\\-1&amp;-1\end{bmatrix}\begin{bmatrix}0&amp;1\\-1&amp;-1\end{bmatrix} = \begin{bmatrix}-1&amp;-1\\1&amp;0\end{bmatrix}$, so that:

$A^3 = A^2A = \begin{bmatrix}-1&amp;-1\\1&amp;0\end{bmatrix}\begin{bmatrix}0&amp;1\\-1&amp;-1\end{bmatrix} = \begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix} = I.$

If we take $B = \begin{bmatrix}1&amp;0\\-1&amp;-1\end{bmatrix}$, we have:

$B^2 = \begin{bmatrix}1&amp;0\\-1&amp;-1\end{bmatrix}\begin{bmatrix}1&amp;0\\-1&amp;-1\end{bmatrix} = \begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix} = I$, so there's $2$ out of $3$.

The third rule is tedious to verify, but here goes:

$BA = \begin{bmatrix}1&amp;0\\-1&amp;-1\end{bmatrix}\begin{bmatrix}0&amp;1\\-1&amp;-1\end{bmatrix} = \begin{bmatrix}0&amp;1\\1&amp;0\end{bmatrix}$, while:

$A^2B = \begin{bmatrix}-1&amp;-1\\1&amp;0\end{bmatrix}\begin{bmatrix}1&amp;0\\-1&amp;-1\end{bmatrix} = \begin{bmatrix}0&amp;1\\1&amp;0\end{bmatrix}$,

so, indeed $BA = A^2B$.
Multiply every matrix with every other matrix (that are 36 multiplications) and check if the resulting matrices are in your group.
That should proof closure.