# Computing Sylow $p$-subgroups of classical groups

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.

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For $G$, you can choose a $p$-Sylow to be the group of upper triangular unipotent matrices, so it is easy to find a generator for that (and then you need to take conjugates to get generators of the others). – Tobias Kildetoft Jul 30 '13 at 7:07
@Tobias: well, I think that's a key point: if you have an explicit generator $x$ for one Sylow $p$-subgroup, does the OP regard $\{ \langle gxg^{-1} \rangle \mid g \in G\}$ as being an explicit description of all the Sylow $p$-subgroups? It will certainly make things easier if the answer is "yes"...Note that in the first case, if the answer is "no" then the OP is essentially asking for all unipotent matrices without allowing a Jordan form description. – Pete L. Clark Jul 30 '13 at 7:16
I've got the beginnings of a Sp(4,K) answer (each non-standard maximal unipotent has a 2-element generating set with a certain zero pattern). However, the calculations are pretty bad, so I'm (a) waiting for a computer to make some progress on them, and (b) trying to get a more theoretical understanding. Unfortunately I suspect describing how to find these special generators is probably the same as having a fairly complete understanding of the quillen complex. For GL2 and GO3, the Sylows had trivial intersection, so it was easy, but for Sp4 things are more complicated. – Jack Schmidt Jul 31 '13 at 21:40
@JackSchmidt: What is the isomorphism class of a Sylow $p$-subgroup of $Sp_4$? – Jared Jul 31 '13 at 22:06
By the way, I'm working on an explanation of how to find all maximal unipotent subgroups of the classical groups (that is, Sylow p-subgroups of GL(n,p^k), Sp(n,p^k), O±(n,p^k), and U(n,p^k)). I've got the explanation for GL(n,p^k), but I'm still working on the details for the others. For GL(2,p^k) parameterize the Sylows by their fixed vector, and you get a nice expression for your "2" matrix, but also for all of its powers (which at least to me was not at all obvious from trying to raise that matrix to a power). For GO(3,p^k) this also works, but you have to restrict to a special kind. – Jack Schmidt Aug 6 '13 at 15:11

## 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 $(|K|^2+1)(|K|+1)^2$.

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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):

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

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Wonderful. Thank you! I see you've define the orthogonal group as the invertible matrices preserving the symmetric bilinear form given by your $x$. Is there any advantage to using $x$ instead of $I$, so that instead we work with the usual orthogonal matrices? – Jared Jul 31 '13 at 22:12
In even dimensions you have to use weird $x$s, but in odd one should be able to use the identity for $x$. However, perhaps due to my lack of theoretical understanding, I was not able to give good descriptions of the Sylow $p$-subgroups. In the "it'd be nice" paragraph, an answer there would also almost certainly give an answer for $x=I$ as well. – Jack Schmidt Jul 31 '13 at 23:40
I've now shown every maximal unipotent contains one of those guys and parameterised the maximal unipotents over arbitrary fields (of characteristic not 2, for a specific bilinear form, the antidiagonal). – Jack Schmidt Aug 1 '13 at 20:39