According to [Wikipedia](https://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields) ([current revision](https://en.wikipedia.org/w/index.php?title=Orthogonal_group&oldid=925232711#Over_finite_fields)) the cardinality of $O(n,q)$ depends on the properties of the field we're working over.  These are the results:

> We have the following formulas for the order of $\operatorname{O}(n, q)$, when the characteristic is not two:
$$\left|\operatorname{O}(2n + 1, q)\right| = 2q^n\prod_{i=0}^{n-1}\left(q^{2n} - q^{2i}\right).$$
If $-1$ is a [square in $\mathbf F_q$](https://en.wikipedia.org/wiki/Quadratic_residue)
$$\left|\operatorname{O}(2n, q)\right| = 2\left(q^n - 1\right)\prod_{i=1}^{n-1}\left(q^{2n} - q^{2i}\right).$$
If $-1$ is a non-square in $\mathbf F_q$
$$\left|\operatorname{O}(2n, q)\right| = 2\left(q^n + (-1)^{n+1}\right)\prod_{i=1}^{n-1}\left(q^{2n} - q^{2i}\right).$$ 

  [1]: https://i.sstatic.net/69y1o.png

 I'm considering both $O^+$ and $O^−$. The definitions: Suppose $V$ is a vector space on which the orthogonal group $G$ acts, then $V=L_1\oplus L_2\oplus\dots\oplus L_m\oplus W$, with $L_i$ hyperbolic lines and $W\le V$ contains no singular vectors. If $W=0$, then $G$ is of plus type. If $\dim(W)=2$, then $G$ is of minus type. If $W$ is one-dimensional then $G$ has odd dimension. 

As for the question: I'm studying the Sylow subgroups of these groups. The cardinalities of a Sylow $q$-subgroup of $O(2n+1,q)$ (see picture) would be $q^{n^2}$ and for the last two $q^{n(n-1)}$.
I'm now wondering how the Sylow $q$-subgroups of these orthogonal groups *look* like, i.e. what is their structure? What matrices generate such a Sylow subgroup? In addition, I would like to know how the normalizers look like.

Note: I don't need lengthy proofs (or even any proofs), results only suffice. 

Thanks in advance.