Problem source

Explain what is meant by the order of an element $g$ of a group $G$. 

	The set $S$ consists of all $2\times2$ matrices whose determinant is $1$. Find the inverse of the element $\mathbf{A}$ of $S$, where
	\[
	\mathbf{A}=\begin{pmatrix}w & x\\
	y & z
	\end{pmatrix}.
	\]
	Show that $S$ is a group under matrix multiplication (you may assume
	that matrix multiplication is associative). For which elements $\mathbf{A}$
	is $\mathbf{A}^{-1}=\mathbf{A}$? Which element or elements have order
	2? Show that the element $\mathbf{A}$ of $S$ has order 3 if, and
	only if, $w+z+1=0.$ Write down one such element.

Solution source

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