The condition wz −xy = 1 is the same as |A| = 1; it follows that the set contains an identity element (with w = z = 1 and x = y = 0). Moreover, each matrix in S has an inverse and, since |{A}^{−1}| |A| = | I | = 1 implies that |{A}^{−1}| = 1, the inverses also belong to the set.

If A and B belong to S then, since |AB| = |A| |B| = 1× 1 = 1, their product also belongs to S, i.e. the set is closed.

These observations, together with the associativity of matrix multiplication establish that the set S is, in fact, a group under this operation.

If A is to have order 2 then

\begin{pmatrix} w & x \\ y & z \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.

i.e. {w}^{2} + xy = 1, x(w + z) = 0, y(w + z) = 0, xy+ {z}^{2} = 1.

These imply that {w}^{2} = {z}^{2} and that either z = −w or x = y = 0. If z = −w, then both

{w}^{2} + xy = 1, from the above condition,

and − {w}^{2} − xy = 1, from wz − xy = 1.

This is not possible and so we must have x = y = 0, implying that w and z are either both +1 or both −1. The former gives the identity (of order 1), and so the matrix given by the latter, A = −I, is the only element in S of order 2.

If w + z + 1 = 0 (as well as xy = wz − 1), {A}^{2} can be written as

{ A }^{ 2 }=\begin{pmatrix} { w }^{ 2 }+xy & x(w+z) \\ y(w+z) & xy+{ z }^{ 2 } \end{pmatrix}

=\begin{pmatrix} { w }^{ 2 }+wz-1 & -x \\ -y & wz-1+{z}^{2} \end{pmatrix}

=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}.

Multiplying again by A gives

{ A }^{ 3 }=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}

=-\begin{pmatrix} w(w+1)+xy & (w+1)x+xz \\ wy+y(z+1) & xy+z(z+1) \end{pmatrix}

=-\begin{pmatrix} w(w+1)+wz-1 & x\times 0 \\ y\times 0 & wz-1+z(z+1) \end{pmatrix}

=-\begin{pmatrix} (w\times 0)-1 & 0 \\ 0 & (z\times 0)-1 \end{pmatrix}

=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Thus A has order 3.