Show that A is a field which has $9$ elements For $A=\left\{\left( {\begin{array}{*{20}{c}}
a&b\\
{ - b}&a
\end{array}} \right) | a,b\in\mathbb{Z/3Z}\right\}$ . Show that A is a field which has $9$ elements .  $(A^*, .)$ is a cyclic group $G$ and $|G|=8$. 
 A: Hint $\ $ Use the following to show $\rm\: A\,\cong\, \Bbb F_3{\bf [}\,{\it i}\:{\bf ]}\, \cong\, \Bbb F_3{\bf [}x{\bf ]}/(x^2\!+1)$
$${\bf i}\, = \left[\begin{array}{rr} 0 & 1\\ -1 & 0\end{array}\right]\ \Rightarrow\,\ {\bf i}^2 =\, -1 ,\qquad \left[\begin{array}{rr} \rm a &\rm b\\ \rm -b &\rm a\end{array}\right]\, =\rm \,\ a + b\ {\bf i}$$
A: Hints:
$$(1)\;\;\;\;\;\;\;\;\;|A|=3^2=9$$
$$(2)\;\;\;\;\;\;\;\;\;\forall \;0\neq\begin{pmatrix}a&b\\\!\!-b&a\end{pmatrix}\in A\;,\;\;\text{ also}\;\;\frac{1}{a^2+b^2}\begin{pmatrix}a&\!\!-b\\b&a\end{pmatrix}\in A$$
$$(3)\;\;\;\;\;\;\;\forall\;X,Y\in A\;,\;\;\;XY=YX\ldots$$
A: Hint. 
There is only one isomorphism class of finite field for every prime power $p^e$, so if you don't want to make a multiplication table or check the axioms one by one, you may instead construct a ring homomorphism from your $A$ to the usual presentation of $\displaystyle\mathbb{F}_9=\frac{\mathbb{F}_3[x]}{\langle x^2+x+2\rangle}$.  Everything in there looks like $a_0+a_1x$ and the multiplication is done reduced modulo $ x^2+x+2$, so all you need to do is figure out which elements you'll send to the element $1$ (i.e. $a_0=1,a_1=0$) and which to the element $x$ ($a_0=0,a_1=1$).  The way the matrices multiply is suggestive of how you could make this isomorphism.
