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?
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3$\begingroup$ I think you are trying to represent quadratic forms by (symmetric) matrices. This doesn't work in characteristic 2. $\endgroup$– abxCommented Aug 13, 2014 at 13:16
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1$\begingroup$ no, I am sorry, I should have explained. By even and odd I mean the Arf invariant of a quadratic form on a $\mathbb{Z}_2$-vector space. I will correct the question. $\endgroup$– IMeasyCommented Aug 13, 2014 at 13:22
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1$\begingroup$ @IMeasy: You should also make your notation $\mathbb{Z}_2$ clear, since it can mean two things. Is this what the tag 'finite-fields' is for? $\endgroup$– Jim HumphreysCommented Aug 13, 2014 at 13:23
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1 Answer
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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.
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$\begingroup$ Yes I meant the finite field of order 2, thank you! What is the precise book reference that you suggest? $\endgroup$– IMeasyCommented Aug 13, 2014 at 15:49
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1$\begingroup$ The subgroup structure of the finite classical groups by Kleidman and Liebeck. I have an e-copy and will send it to you if you want (my e-mail is on my user page). $\endgroup$ Commented Aug 13, 2014 at 16:38
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1$\begingroup$ Kleidman and Liebeck - The subgroup structure of the finite classical groups. $\endgroup$– LSpiceCommented Jul 11, 2020 at 22:19