Orthogonal Groups over finite fields Hello
Let $\mathbb{F}_p$ be the finite field with $p$ elements. One can show that over finite fields, there are just two non-degenerate quadratic forms. 
So here I want to pick 
any non-degenerate symmetric matrix $B$, and then look at the special orthogonal group defined by
$$
SO_{n}(\mathbb{F}_p,B):=\{ A\in SL_n(\mathbb{F}_p): ABA^T=B \}
$$
Is it true that the commutator subgroup of $SO_{n}$ is the whole group? In the other words I would like to know if $SO_{n}$ over finite fields is a perfect group.
 A: For those who have the same question like me I should say that, based on Prof. Robinson's answer, the group is not perfect. I would like to expand Prof. Robinson.  
Indeed let $q$ be an isotropic quadratic form over $\mathbb{F}_p$ for $p>2$. for any non-zero anisotropic vector (i.e $q(v)\neq 0$) consider the symmetries $$
\sigma_v(x):=x-\frac{x.v}{v.v}v,
$$
then one can show that $O(q)$, the orthogonal group of the quadratic form, is generated by the symmetries. Therefore for any $\sigma\in O(q)$ we have
$$
\sigma=\sigma_{v_1}\cdots\sigma_{v_n}.
$$
$v_i$'s are not uniquely determined, but the following map is independent of choosing of $v_i$'s. 
$$
\theta(\sigma):=q(v_1)\cdots q(v_n)(\mathbb{F}_p^*)^2.
$$ 
Hence we have a group homomorphism known as "Spinor Norm" defined by
\begin{equation}
\begin{split}
\theta: SO(q) & \to \mathbb{F}_p^*/(\mathbb{F}_p^*)^2\\\\
 \sigma &\to q(v_1)\cdots q(v_n)(\mathbb{F}_p^*)^2
\end{split}
\end{equation}
Notice that, since $q$ is an isotropic then $q$ takes any value in $\mathbb{F}_p$, so the Spinor norm is surjective, and hence $SO(q)$ is not perfect. 
A: This is, roughly (sometimes you might have to take the commutator subgroup), true if $n$ and/or $q$ are big enough. Wikipedia and sufficiently advanced textbooks on finite group theory spell it out in all the detail.
EDIT: E.g., from the discussion in comments, $SO^+_{2n}(2^k)$ for $n\geq 3$ is not perfect, for it has a simple index 2 subgroup (specified by Dickson invariant). 
A: Since nobody has attempted to say what happens in even characteristic, let me do that briefly.
As someone observed earlier in a comment that seems to have been deleted, in even characteristic and even dimension, alternating bilinear forms are symmetric. There is a unique isometry class of such forms - the matrix of one has 1's on the "antidiagonal" and 0's elsewhere. The group that preserves such a form is the symplectic group ${\rm Sp}(2n,2^e)$.
This group, however, has two subgroups, which preserve the two types of quadratic form in characteristic 2 (the convenient one-one correspondence between symmetric bilinear and quadratic forms does not work in characteristic 2), and these are the orthogonal groups ${\rm GO}^+(2n,2^e)$ and ${\rm GO}^-(2n,2^e)$. All of their elements have determinant 1, so they are equal to ${\rm SO}^+(2n,2^e)$ and ${\rm SO}^-(2n,2^e)$.
These groups are not perfect and have subgroups of index 2, often denoted by ${\Omega}^+(2n,2^e)$ and ${\Omega}^-(2n,2^e)$, which consist of those elements that are a product of an even number of reflections. (The homormorphism from the special orthogonal group to the cyclic group of order 2 is still usually called the spinor norm homomorphism, although its definition is not identical to the one in odd characteristic.)  These have trivial centres (in dimension at least 4) and so are isomorphic to their projective versions, and the $\Omega$ groups are simple in dimension at least 6.
A: I think it's worth adding that there is a very detailed analysis of the orthogonal groups over arbitrary fields (not just finite fields, and including characteristic 2) in Dieudonné's "La Géométrie des Groupes Classiques". (The first edition of this book already goes back to 1954, but I'm referring to the 3rd edition from 1971.)
In particular, Chapter II, Section 8, seems to answer the question you're interested in when the characteristic is not 2, and Section 10 (in particular item (8) on p.67) answers your question when the characteristic is 2.
