Let $O(2,\mathbb{Z}\_2)$ be the orthogonal group of order two matrices. On $\mathbb{Z}\_2$ there should exist just one odd quadratic form, hence the stabilizer subgroup $O^-$ of an odd quadratic for should be the whole group. Is it really so or I am confused? On the other hand the even stabilizer $O^+$ should have index 3, so - since the order of $O$ is 6 - I guess $O^+$ is just $\pm Id$. Can someone confirm this?



orthogonal group in characteristic 2
Assuming that, by $\mathbb{Z}\_2$, you mean the finite field of order $2$, then 
$$ O\_2^{\pm}(q) \cong D\_{2(q\mp 1)}$$
where $D\_{2(q\mp 1)}$ is the dihedral group of order $2(q\mp 1)$. Taking $q=2$, one obtains that $O\_2^+(2)$ is cyclic of order $2$ (although not equal to $\pm I$ you suggest, because $-I=I$), while $O\_2^-(2)$ is dihedral of order $6$.


One can calculate all this directly, or you can refer to Proposition 2.9.1 of Kleidman and Liebeck's book.


