Frisbee Rules

Let \(A_1 = \begin{bmatrix} 1 & x_1 & y_1 \\ 0 & 1 & z_1 \\ 0 & 0 & 1 \end{bmatrix}\) and \(A_2 = \begin{bmatrix} 1 & x_2 & y_2 \\ 0 & 1 & z_2 \\ 0 & 0 & 1 \end{bmatrix}\in \mathcal{G}\).

Then \(A_1 A_2 =\begin{bmatrix} 1 & x_1+x_2 & y_1+x_1z_2+y_2 \\ 0 & 1 & z_1+z_2 \\ 0 & 0 & 1 \end{bmatrix}\in\mathcal{G}\), so we have closure.

Associativity follows from the associativity of standard matrix multiplication.

Letting \(x=y=z=0\), observe that the identity is in \(\mathcal{G}\).

Finally, if we take \(x_2 = -x_1\), \(z_2 = -z_1\), and \(y_2 = -y_1-x_1z_2\), then observe that \(A_1A_2 = I_3\), and thus inverses are of the required form! Therefore, $\mathcal{G} $ is a group.

The group is not abelian, e.g. take \(x_1=z_2=1\) and everything else to be \(0\). Then multiplying these matrices in the other order (i.e. \(x_2=z_1=1\)) gives a different answer.

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