The structure of the unit circle in the plane $F^2$, where $F$ is a finite field with odd characteristic. In what follows $F$ denotes the finite field $GF(q)$, where $q=p^k$ with $p$ an odd prime and $k \in \mathbb N^*$. The unit circle ($UC$) on the vector space $F^2$ is defined as the set of couples $(c,s) \in F^2$ that satisfy $c^2+s^2 = 1$ (the names $c$ and $s$ have deliberately been chosen to their reference to the trigonometric functions $\sin$ and $\cos$).Using the GAP computer system I looked for the number of elements of the UC and it resulted to be $[q]_4$, where $[n]_4$ denotes the nearest multiple of $4$ to $n \in \mathbb N $ and $n$ odd (e.g. $[7]_4 = 8$ and $[9]_4 = 8$). Proving this didn't seem quite simple until I remembered the "`parametrisation"' of the $UC$ by the equations 
$$ \begin{array}{lcl} 
c(t)=\frac{t^2-1}{t^2+1}  \\  
s(t) = \frac{2t}{t^2+1} 
\end{array}$$ 


*

*If $q \bmod 4 = 1$ then for two values of $t$ the functions $c(t)$ and $s(t)$ are undefined, namely when $t^2=-1$. This accounts for at most $q-2$ points.

*If $q \bmod 4 = -1$ then the functions $c(t)$ and $s(t)$ are defined $\forall t \in F$, so the parametrization accounts for $q$ points.
The map $ \phi : F \rightarrow UC: t \mapsto (c(t),s(t))$ is injective on its image, its inverse given by:
$
\begin{cases} 
\frac{1-x}{y} 
&\text{if   } y \neq 0\
\\
0 
&\text{if  } y = 0,  x \neq 1 \
\end{cases} 
$
So for  $q \bmod 4 = 1$ the image of $\phi$ contains exactly $q-2$ points and for  $q \bmod 4 = -1$ exactly $q$ points. Since the only point of $UC$ that is missing in the image of $\phi$ is $(1,0)$ (corresponding to the value $t = \pm \infty$ for the reals) the number of points of the $UC$ is as originally stated.
Endowing the $UC$ with the composition law $(c,s)\circ (c',s') = (cc'-ss',cs'+c's)$ turns it into a finite abelian group, as can be verified by direct calculation. Futher tests in GAP shows that with this composition law the $UC$ is a cyclic group. Finding a canonical generator of this group doesn't seem to be obvious, the only thing I found is that this group is isomorphic to the matrix group formed by the matrices of the form 
$
\begin{pmatrix}
c&-s\\
s&c
\end{pmatrix}
$ having determinant $1$. Can one prove that this group is always a cyclic group?
 A: This is the story of split/non-split torus.
For $p$ odd prime (it will work the same way if you want to extend to finite fields with $q=p^k$ elements), since $\Bbb{F}_p^*$ is cyclic with $p-1$ elements


*

*If $p \equiv 3 \bmod 4$ then $-1$ has no square root in $\Bbb{F}_p$, let $i$ be a square root (in the algebraic closure) then $\Bbb{F}_p(i) = \Bbb{F}_{p^2}$ (the unique finite field with $p^2$ elements) and for $x,y\in \Bbb{F}_p$
$$x^2+y^2= (x+iy)(x-iy) = (x+iy)(x+iy)^p$$ thus $x^2+y^2=1$ iff $(x+iy)^{p+1} =1$, since $\Bbb{F}_{p^2}^*$ is cyclic with $p^2-1$ elements, there are $p+1$ elements whose order divides $p+1$ and since $\Bbb{F}_{p^2}$ is a 2 dimensional $\Bbb{F}_p$-vector space, $1,i$ is a basis and each of those $p+1$ elements is the form $x+iy$ with $x,y\in \Bbb{F}_p$ thus a solution of $x^2+y^2=1$.

In that case the group law of $SO_2(\Bbb{F}_p)$ is that of $\{ a \in \Bbb{F}_{p^2}^*,a^{p+1}=1\}$


*If $p \equiv 1 \bmod 4$ then $-1 = c^2\in \Bbb{F}_p$ and $$x^2+y^2 = (x+cy)(x-cy)$$ with the  change of variable $u=x+cy,v=x-cy, x= \frac{u+v}{2},y=\frac{u-v}{2c}$  we have $p-1$ solutions for $uv=1$ which gives $p-1$ solutions for $x^2+y^2=1$.

In that case the group law of $SO_2(\Bbb{F}_p)$  is that of $ \Bbb{F}_p^*$

