7
$\begingroup$

Let $p$ be a prime, and let $G$ be any of the finite classical groups $SL_n(\mathbb{F}_p)$, $O_n(\mathbb{F}_p)$, or $SP_n(\mathbb{F}_p)$. Let $P$ be a Sylow $p$-subgroup of $G$. What is $P$ as a subgroup of $G$, that is, how can we describe the matrices in $P$? Also, how many Sylow $p$-subgroups are in $G$?

Answers and/or references are welcome. Thanks!

$\endgroup$
  • 1
    $\begingroup$ The Sylow p-subgroup of $SL_n(F_p)$ is the subgroup of upper triangular matrices (this should be well-known). $\endgroup$ – Ralph Mar 25 '13 at 0:03
  • $\begingroup$ You're right; I am aware of this result. Do you know how many conjugates there are? $\endgroup$ – Jared Mar 25 '13 at 0:06
6
$\begingroup$

As indicated in the comments, a Sylow subgroup is the group $U$ of unipotent upper triangular matrices in each case. I'll indicate how to compute the normalizer of $U$ in the $SL_n$ case (the others could be handled similarly, I bet). Let $e_1,\dots,e_n$ be the standard basis of $F^n$, where $F$ is a field. We will actually compute the normalizer of the group of unipotent upper triangular matrices in this generality, showing that it is simply the group $B$ of upper triangular matrices of determinant one.

It is obvious that this group normalizes $U$. So now we prove that the normalizer $N$ of $U$ is contained in $B$. The vector $e_1$ is the unique eigenvector with eigenvalue $1$ for all of $U$ (i.e., the intersection of the kernels of $1-u$ for all $u \in U$). It follows that $Ne_1 \subseteq F e_1$. Likewise, the intersection of the kernels of $(1-u)^2$ for $u \in U$ is the span $F \{e_1,e_2 \}$ of $e_1$ and $e_2$, and hence this span is $N$ stable as well. Evidently one can continue in this way to prove that $N \subseteq B$, done. Now you know how many Sylow subgroups there are.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.