0
$\begingroup$

According to Wikipedia (current revision) the cardinality of $O(n,q)$ depends on the properties of the field we're working over. These are the results:

We have the following formulas for the order of $\operatorname{O}(n, q)$, when the characteristic is not two: $$\left|\operatorname{O}(2n + 1, q)\right| = 2q^n\prod_{i=0}^{n-1}\left(q^{2n} - q^{2i}\right).$$ If $-1$ is a square in $\mathbf F_q$ $$\left|\operatorname{O}(2n, q)\right| = 2\left(q^n - 1\right)\prod_{i=1}^{n-1}\left(q^{2n} - q^{2i}\right).$$ If $-1$ is a non-square in $\mathbf F_q$ $$\left|\operatorname{O}(2n, q)\right| = 2\left(q^n + (-1)^{n+1}\right)\prod_{i=1}^{n-1}\left(q^{2n} - q^{2i}\right).$$

I'm considering both $O^+$ and $O^−$. The definitions: Suppose $V$ is a vector space on which the orthogonal group $G$ acts, then $V=L_1\oplus L_2\oplus\dots\oplus L_m\oplus W$, with $L_i$ hyperbolic lines and $W\le V$ contains no singular vectors. If $W=0$, then $G$ is of plus type. If $\dim(W)=2$, then $G$ is of minus type. If $W$ is one-dimensional then $G$ has odd dimension.

As for the question: I'm studying the Sylow subgroups of these groups. The cardinalities of a Sylow $q$-subgroup of $O(2n+1,q)$ (see picture) would be $q^{n^2}$ and for the last two $q^{n(n-1)}$. I'm now wondering how the Sylow $q$-subgroups of these orthogonal groups look like, i.e. what is their structure? What matrices generate such a Sylow subgroup? In addition, I would like to know how the normalizers look like.

Note: I don't need lengthy proofs (or even any proofs), results only suffice.

Thanks in advance.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ An expert may correct me, but I think p-Sylow in a group defined in characteristic p is usually the unipotent subgroup (upper triangular matrices). $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 11, 2019 at 5:44
  • 2
    $\begingroup$ Your wiki reference doesn't do justice to the finite orthogonal groups. R Carter's "Simple Groups of Lie Type" and D Taylor's "The Geometry of the Classical Groups" are two fine books that answer your questions. E.g. in even dimensions $n$, $O^+(n,q)$ and $O^-(n,q)$ are different groups, with different orders & different Sylow structures. Given a vector space $V$ and a nondegenerate quadratic form $Q$ on $V$, the orthogonal group $O(V,Q)$ is the subgroup of $GL(V)$ preserving $Q$. If $dim(V)$ is even there are two inequivalent quadratic forms on $V$. The wiki page only considers one of them. $\endgroup$
    – Richard Lyons
    Commented Nov 11, 2019 at 14:28
  • 1
    $\begingroup$ @TheoJohnson-Freyd is correct. See, for example, Corollary 24.11 in Linear algebraic groups and finite groups of Lie-type, by Testerman and Malle, and use the fact that your group acts transitively by conjugation on its $p$-Sylow subgroups and on radical unipotents. $\endgroup$
    – kneidell
    Commented Nov 11, 2019 at 19:27
  • $\begingroup$ A sketch proof: $G/B$ is paved by affines (Bruhat decomposition), one of which is 0-dimensional, so has prime-to-$p$ order, so $B$ contains a $p$-Sylow. $B/U \cong T$ consists of semisimple elements, so has prime-to-$p$ order. $\endgroup$
    – LSpice
    Commented Dec 1, 2019 at 15:40

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .