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The order of an element $g$ is the smallest positive number $k$ such that $g^k = e$.

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

Claim, $S$ is a group.

\begin{enumerate}
\item (Closure) The product of two $2\times2$ matrices is always a $2\times 2$ matrix so we only need to check the determinant. Suppose $\det(\mathbf{A}) = \det (\mathbf{B}) = 1$, then $\det(AB) = \det(A)\det(B) = 1$, so our operation is closed
\item (Associativity) Inherited from matrix multiplication
\item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\
	1 & 1
	\end{pmatrix}$ has determinant $1$.
\item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer.
\end{itemize}

If $\mathbf{A}^-1 = \mathbf{A}$ then assuming $\mathbf{A} = \begin{pmatrix}a & b\\
	c & d
	\end{pmatrix}$ then $\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\
	-c & a
	\end{pmatrix}$ so we must have $a=d, -b=b, -c=c$, so $b = c = 0$ and $a = d$. For the determinant to be $1$ we must have $ad = a^2 = 1$, ie $a = \pm 1$. Therefore we must have $\mathbf{A} = \begin{pmatrix}1 & 0\\
	0 & 1
	\end{pmatrix}$ or $\mathbf{A} = \begin{pmatrix}-1 & 0\\
	0 & -1
	\end{pmatrix}$.

For an element to have order $2$ then $\mathbf{A}^2 = \mathbf{I}$ ie, $\mathbf{A} = \mathbf{A}^{-1}$ and $\mathbf{A} \neq \mathbf{I}$ therefore the only element of order $2$ is $\begin{pmatrix}-1 & 0\\
	0 & -1
	\end{pmatrix}$.

For an element to have order $3$ we must have $\mathbf{A}^2 = \mathbf{A}^{-1}$, ie

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

Therefore $w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w$.

The second and third equations are satisfied iff $w+z+1 = 0$ or $x = 0$ and $y = 0$, but if $x = 0$ and $y = 0$ then we aren't order $3$, so we just need to check this is sufficient for the first and last equations. Since $\det(\mathbf{A}) = 1$ we have $wz =xy +1$, so the first and last equations are equivalent to $w^2 + wz - 1 = z$ and $x^2 + wz-1 = w$ which are equivalent to $w(w+z) = z+1$ or $w + z+ 1 = 0$ as required

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