suppose that $k$ is a finite field of characteristic 2 and $(V,q)$ a quadratic space, i.e., $V$ is a $k$-vector space and $q:V\to k$ quadratic form. Suppose that $\dim(V)\geq 4$ and that $q$ is non-defective, i.e., the $k$-bilinear $B(x,y)=q(x+y)-q(x)-q(y)$ is non-degenerated:


The question is:


Given a non-singular vector $0\neq v\in V$, (that is $q(v)\neq 0$), What is the isomorphism type of the stabilizer of $v$ in $O'(V,q)$?


Note: I know the answer in the case that $k=\mathbb F\_2$. In this case the stabilizer is $O(2n-1,2)\cong Sp(2n-2,2)$, where $2n=\dim(V)$ (but I do not have a reference). I suppose that the same is true in general, that is the stabilizer of a non-singular vector is isomorphic to $Sp(2n-2,k)$.


Thanks. 



Stabilizer of a nonsingular vector in a quadratic space (char (k)=2)
A precise reference for this result is the book "The Subgroup Structure of the Finite Classical Groups" by P. Kleidman and M. Liebeck, Proposition 4.1.7.


The notation used for the various orthogonal groups varies a lot from book to book, so I am not completely sure what you mean by $O(n,q)$, but presumably $O'(n,q)$ is intended to denote the simple subgroup, which is often denoted $\Omega(n,q)$ or $\Omega\_n(q)$.


The stabilizer of a non-singular $1$-space in the simple group $\Omega^{\pm}\_{2n}(q)$ is isomorphic to ${\rm Sp}\_{2n-2}(q)$. There is a unique conjugacy class of subgroups of this type, so it is normalized by the diagonal outer automorphism of $\Omega^{\pm}\_{2n}(q)$, and the stabilizer in ${\rm SO}^{\pm}\_{2n}(q) = {\rm GO}^{\pm}\_{2n}(q)$ is isomorphic to ${\rm Sp}\_{2n-2}(q) \times C\_2$.


You asked about the stabilizer of a vector rather than a $1$-space, but that will be the same, because there are no scalars of order $2$ when $q$ is even.


