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.
Best Answer
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.