Computing Sylow $p$-subgroups of classical groups

Question

Let $p>4$ be prime, and let $G=GL_2(\mathbb{F}_p)$, $H=O_3(\mathbb{F}_p)$, and $K=Sp_4(\mathbb{F}_p)$.

We know that $|G|=p(p-1)^2(p+1)$, so that a Sylow $p$-subgroup of $G$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. In fact, there are $p+1$ such subgroups. Can we write down a generator for each one?

We also know that $|H|=2p(p+1)(p-1)$, so that a Sylow $p$-subgroup of $H$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. Are there also $p+1$ such subgroups, and can we write down generators?

Finally, we know that $|K|=p^4(p-1)^2(p+1)^2(p^2+1)$. What is the isomorphism class of a Sylow $p$-subgroup of $K$, how many are there, and can we write down generators and relations?

I remember I solved the first question, involving $G$, but the solution escapes me at the moment. The other two are standard extensions of the first problem that I am also interested in.

Any reference and/or partial solution is appreciated. Thanks!

Edit: To address the comments, I would like to be able to explicitly write down matrices that generate the subgroups, one for each subgroup. As noted by Tobias, the matrix:

$$\begin{pmatrix}1&a\\0&1\end{pmatrix}$$

for $a\ne 0$, generates one Sylow $p$-subgroup of $G$, so I'd like to find $p$ more matrices of order $p$ that generate distinct subgroups. These matrices should be indexed by our field $\mathbb{F}_p$. They are certain to be conjugates of the above matrix, but two arbitrary conjugates may generate the same subgroup.

Edit 2: I've just solved the question for $G$. Here is a list of generators for the $p+1$ Sylow $p$-subgroups of $G$:

$$\begin{pmatrix}1&1\\0&1\end{pmatrix},\;\begin{pmatrix}1&0\\1&1\end{pmatrix},\;\begin{pmatrix}2&a\\-a^{-1}&0\end{pmatrix}$$

where $a=1,\ldots,p-1$. Each matrix has order $p$, and generates a distinct subgroup.

Answer 1

Symplectic group

I won't yet address identifying all maximal unipotent subgroups, but will just describe the standard ones.

The maximal unipotent subgroups of the symplectic group in four dimensions over the field $K$ have a nice description in terms of certain nice elements. $\newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}$

Let $$G=\left\{ g \in \operatorname{GL}(4,K) : gxg^T = x \right\}, \quad x=\m{0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\ -1&0&0&0}$$

be a symplectic group. It has two standard maximal unipotent subgroups, the upper and lower, which differ only by being transposes of each other. The lower on is: $$P=\left\{ \m{ 1 & 0 & 0 & 0 \\ x & 1 & 0 & 0 \\ y & w & 1 & 0 \\ z & y-xw & -x & 1 } : x,y,z,w \in K \right\}$$

Define $$ x_1(t) = \m{1&0&0&0\\t&1&0&0\\0&0&1&0\\0&0&-t&1}, \quad x_2(t) = \m{1&0&0&0\\0&1&0&0\\0&t&1&0\\0&0&0&1}, \quad x_3(t) = \m{1&0&0&0\\0&1&0&0\\t&0&1&0\\0&t&0&1}, \quad x_4(t) = \m{1&0&0&0\\0&1&0&0\\0&0&1&0\\t&0&0&1}, \quad $$

Note that $$\begin{array}{lrl} (1) & x_i(s)x_i(t) &= x_i(s+t) \\ (2) & x_2(t)x_1(s) &= x_1(s) x_2(t) x_3(st) x_4(s^2t) \\ (3) & x_3(t)x_1(s) &= x_1(s) x_3(t) x_4(2st) \\ (4) & x_4(t)x_1(s) &= x_1(s) x_4(t) \\ (5) & x_3(t)x_2(s) &= x_2(s) x_3(t) \\ (6) & x_4(t)x_2(s) &= x_2(s) x_3(t) \\ (7) & x_4(t)x_3(s) &= x_3(s) x_4(t) \\ \end{array}$$

If $K$ has characteristic not $2$, then $P$ has rank $2k$, generated by $x_1(s)$, and $x_2(t)$ where $s,t$ range over a generating set of the additive group of $K$. $[P,P]$ is isomorphic to the additive group of a two dimensional vector space $K^2$ over $K$, and is generated as a group by $x_3(s)$ and $x_4(t)$ with $s,t$ from a generating set of the additive group of $K$. $[P,P,P]$ is isomorphic to the additive group of $K$, and is generated as a group by $x_4(t)$ where $t$ ranges over a generating set of the additive group of $K$.

When $K$ has characteristic 2, things make less sense to me. The nilpotency class is 2. When $|K|>2$, $P/[P,P] \cong [P,P] \cong (K^+)^2$, but when $|K|=2$, $P\cong C_2 \times D_8$ behaves differently.

The normalizer of $P$ is the semidirect product of $P$ and a maximally split maximal torus $$H=\left\{ \m{ s & 0 & 0 & 0 \\ 0 & t & 0 & 0 \\ 0 & 0 & t^{-1} & 0 \\ 0 & 0 & 0 & s^{-1} } : s,t \in K^\times \right\}$$

In particular, the collection of maximal unipotent subgroups is in bijection with $(|K|^2+1)(|K|+1)^2$.

Answer 2

Here is a similar answer for the three dimensional general orthogonal group $$G=O_3(K) = \left\{ g \in \operatorname{GL}(3,K) : g \cdot x \cdot g^T = x \right\}, \quad x = \left[\begin{smallmatrix}0&1&0\\1&0&0\\0&0&-1\end{smallmatrix}\right]$$ over a field $K$ of characteristic not 2. In the following, “unipotent element” means $p$-element, “unipotent subgroup” means $p$-subgroup, and “maximal unipotent subgroup” means Sylow $p$-subgroup. However, saying unipotent lets the statements remain true over non-prime fields and fields of characteristic zero.

First we classify a special type of element of $G$: suppose $(g-1)^3=0$, $gxg^T=x$, $g_{2,2}=0$. Then a calculation shows $$g=g(t) =\begin{bmatrix} 4 & t^2/8 & t \\ 8/t^2 & 0 & 0 \\ -8/t & 0 & -1 \end{bmatrix}$$

In particular, these elements are parameterized by $K^\times$ where $K$ is the field. Obviously $g(t)$ is unipotent, so if $K$ has characteristic $p$, $g(t)$ lies in some Sylow $p$-subgroup.

If a subgroup contains both $g(s)$ and $g(t)$, then it must also contain $g(s) \cdot g(t)$ and $g(s) \cdot g(t) \cdot g(s)$. However, a (long and tedious) calculation shows that if both $g(s)\cdot g(t)$ and $g(s) \cdot g(t) \cdot g(s)$ are unipotent, then $s=t$.

Hence a unipotent subgroup contains at most one of $g(s)$ or $g(t)$.

I haven't shown in general that every maximal unipotent subgroup other than the two standards (positive and negative root subgroups) contains a $g(t)$, but since you already know the count, we are done: there are the two standard maximal unipotents, and then one for each $t \in K^\times$, namely, the one containing $g(t)$.

The standard maximal unipotent subgroups are defined by $g_{12}=0$ (the lower standard) and $g_{21}=0$ (the upper standard).

Define $$ g^+(t) = \begin{bmatrix} 1 & 0 & 0 \\ t^2/2 & 1 & t \\ t & 0 & 1 \end{bmatrix} \quad g^-(t) = \begin{bmatrix} 1 & t^2/2 & t \\ 0 & 1 & 0 \\ 0 & t & 1 \end{bmatrix} $$

Then one easily checks $g^+(s) \cdot g^+(t) = g^+(s+t)$ and $\{ g^+(t) : t \in K \}$ is a unipotent subgroup, which is maximal by considering the rank of $O_3(K)$. Similarly $\{ g^-(t) : t \in K \}$ is a maximal unipotent subgroup.

It'd be nice to have a similar parameterization of the elements of the maximal unipotent subgroups containing $g(t)$. This would give a nice description of all unipotent elements in the group.

At any rate, the description is that every maximal unipotent subgroup of $O_3(K)$ contains exactly one of the following elements (which are generators if $K$ is a field of prime order):

$$ g^+(1) = \begin{bmatrix} 1 & 0 & 0 \\ 1/2 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}, \quad g^-(1) = \begin{bmatrix} 1 & 1/2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}, \quad\text{ or } g(t), t \in K^\times $$