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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.

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4 Answers 4

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.

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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.

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Geoff: I am not really group theorist, so I might have used some nonstandard notations. I apologize for this. I would be thankful to you if you point out mistakes I made in my answer. –  M.B Jul 15 '12 at 18:33
    
I was not criticisng your notation. I meant that the notation in the existing literature is often ambiguous. For example, sometimes ${\rm O^{+}}(n,q)$ is used to denote the unique non-Abelian composition factor of the orthogonal matrix group, sometimes it is used to mean the group of orthogonal matrices itself. Letting ${\rm SO}(2n,q)$ denote the orthogonal group of matrices of determinant $1$ associated to the standard form (over the field of $q$ elements, $q$ odd), the point is, as you correctly note, that this group has a normal subgroup of index $2$. –  Geoff Robinson Jul 15 '12 at 21:44
    
yes, I fully agree, the notation is confusing. –  Dima Pasechnik Jul 16 '12 at 17:36

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.

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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).

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Let just consider $SO_{2n}^+(\mathbb{F}_p)$, for $n\geq 3$. Is it true now that, this group is perfect? –  M.B Jul 14 '12 at 16:59
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I find the "standard" notation for orthogonal groups confusing. When $p$ is odd, the usual orthogonal group in even dimension has more than one normal subgroup of index $2$. As well as the kernel of the determinant, there is the kernel of the spinor norm. –  Geoff Robinson Jul 14 '12 at 17:36
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you should also have a look at Kleidman and Liebeck's book The Subgroup Structure of the Finite Classical Groups". $\Omega$ is the group you are looking at and which is explained there in detail. –  Natalie Jul 14 '12 at 20:46
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@M.B. I do not think you are making a mistake. I think the Spinor norm homomorphism has different kernel from the determinant in odd characteristic. –  Geoff Robinson Jul 15 '12 at 13:18
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Geoff, thanks, I stand corrected. I've removed the wrong comment. I'll dig the stuff on spinor norm up when/if I get to teach graduate group theory :–) –  Dima Pasechnik Jul 16 '12 at 17:34

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