Find a $2$-Sylow subgroup of $\mathrm{GL}_3(F_7)$

Question

We have $|\mathrm{GL}_3(F_7)| = 7^3 \cdot 2^6\cdot 3^4\cdot 19$. I can find the $3,7,19$-Sylow subgroup of it, but failed to find a $2$-Sylow subgroup. Can one help?

Answer 1

Short version: $ \newcommand{\GF}[1]{\mathbb{F}_{#1}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\Sym}[1]{\operatorname{Sym}(#1)} \newcommand{\bm}[5]{\left[\begin{array}{r|rr} #1 & 0 & 0 \\ \hline 0 & #2 & #3 \\ 0 & #4 & #5 \end{array}\right]} $A Sylow 2-subgroup is a direct product of the Sylow 2-subgroup of $\GF{7}^\times$ with the Sylow 2-subgroup of $\Gal(\GF{49}/\GF{7}) \ltimes \GF{49}^\times$.

Remember that the Galois group of $\GF{49}$ acts $\GF{7}$-linearly on $\GF{49}$. Similarly $\GF{49}^\times$ (and $\GF{7}^\times$) act $\GF{7}$-linearly on $\GF{49}$ (and $\GF{7}$ respectively). Thus those groups end up acting on $\GF{7} \times \GF{49} \cong \GF{7}^3$.

Explicit version: In terms of matrices, that is $$ \bm{-1}{1}{0}{0}{1}, \qquad \bm{1}{1}{0}{-1}{-1}, \qquad \bm{1}{0}{1}{1}{-1}$$

We split the matrices into block matrices. The top-left $1 \times 1$ block is $\GF{7}$. The bottom-right $2\times 2$ block is $\GF{49}$.

The first matrix generates the Sylow 2-subgroup of $\GF{7}^\times$, the second matrix is the Frobenius automorphism of $\GF{49}$, and the last generates the Sylow 2-subgroup of $\GF{49}^\times$ of order 16. Here I choose a basis of $\GF{49}$ of the form $\{1,\xi\}$ where $\xi$ is a primitive $16$th root of unity in $\GF{49}$. For a choice that makes the second matrix prettier, see below.

Intuitive version:

I'll repeatedly use a simple but useful fact: If $H \leq G$ are finite, and $p$ does not divide the index $[G:H]$, then every Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of $G$. I'll choose $H$ so that we can use induction. I'll try to describe all Sylow $p$-subgroups for all prime powers $q$, but I'll ignore $p$ divides $q$ (upper unitriangulars form a Sylow $p$-subgroup in this case), and I'll assume $q$ is odd to make a few statements smoother.

I put a more complete version on my website.

The order of $\GL(3,q)$ is $(q^3-1)(q^3-q)(q^3-q^2) = q^3(q-1)^3(q+1)(q^2+q+1)$ with the latter factorization driving most of the case by case analysis. The main subgroups we consider are $\GF{q}^\times \cong \GL(1,q)$, $\GF{q^2}^\times \leq \GL(2,q)$, and $\GL(1,q) \times \GL(1,q) \times \GL(1,q) \leq \GL(1,q) \times \GL(2,q) \leq \GL(3,q)$.

Assume $q$ is an odd prime power. Then $[\GL(3,q):\GL(1,q) \times \GL(2,q)] = q^2(q^2+q+1)$. For many primes $p$, a Sylow $p$-subgroup is contained in $\GL(1,q) \times \GL(2,q)$. The exceptions are those $p$ such that $q$ has order 3 mod $p$, and of course the defining characteristic where $p$ divides $q$. At any rate, if $q$ is odd, then $p=2$ does not divide $q^2(q^2+q+1) \equiv 1 \mod 2$, so a Sylow 2-subgroup is contained in $\GL(1,q) \times \GL(2,q)$.

Thus we want the direct product of a Sylow $p$-subgroup of $\GL(1,q)$ and a Sylow $p$-subgroup of $\GL(2,q)$, at least for our $p=2$.

The Sylow 2-subgroup of $\GL(1,q)\cong \GF{q}^\times$ is isomorphic to the Sylow 2-subgroup of $\GF{q}^\times$. In case $q=7$, it is generated by $\begin{bmatrix}-1\end{bmatrix}$ of order 2. In general, I'll just call $\zeta$ a generator of the Sylow $p$-subgroup of $\GF{q}^\times$.

The Sylow 2-subgroup of $\GL(2,q)$ comes in two varieties. The index $[\GL(2,q):\Sym{2} \ltimes (\GL(1,q) \times \GL(1,q))] = q(1+q)/2$, so assuming $p$ does not divide $q$ (automatic if $q$ is odd and $p=2$) and $p$ does not divide $(1+q)/2$ (for $p=2$ and $q$ odd this just means $1 \equiv q \mod 4$), then the Sylow $p$-subgroup is contained in $\Sym{2} \ltimes (\GL(1,q) \times \GL(1,q))$.

In such a case, $1 \equiv q \mod 4$, the full Sylow 2-subgroup is generated by the matrices:

$$ \bm{\zeta}{1}{0}{0}{1}, \qquad \bm{1}{0}{1}{1}{0}, \qquad \bm{1}{\zeta}{0}{0}{1} $$

In the other case, $3 \equiv q \mod 4$, view $\GF{q}^2$ as $\GF{q^2}$ by choosing a basis. I suggest choosing a basis $\{x,x^q\}$ so that the linear transformation $\GF{q^2} \to \GF{q^2}:t \mapsto t^q$ is represented by the matrix $\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$. Now choose $\xi \in \GF{q^2}$ a generator of the Sylow $p$-subgroup of $\GF{q^2}^\times$. Multiplication by $\xi$ is a $\GF{q}$-linear map of $\GF{q^2}$, so it has a matrix, denote it $A$. At any rate, $[\GL(2,q):\Gal(\GF{q^2}/\GF{q})\ltimes \GF{q^2}^\times] = q(q-1)/2$ and since $3 \equiv q \mod 4$, $p=2$ does not divide this index.

In this case, $3 \equiv q \mod 4$, the full Sylow 2-subgroup is generated by the matrices:

$$\bm{\zeta}{1}{0}{0}{1}, \qquad \bm{1}{0}{1}{1}{0}, \qquad \left[\begin{array}{r|r} 1 & 0 \\ \hline 0 & A \end{array}\right] $$

In the explicit answer above, I chose a different basis of $\GF{q^2}$, $\{1,\eta\}$, and with respect to this basis $A$ has matrix $\left[\begin{smallmatrix}0&1\\1&-1\end{smallmatrix}\right]$ because the minimum polynomial of $\eta$ is $\eta^2+\eta-1$. This is easy to calculate, but the Frobenius map is more of a pain, $\eta^7 = -1-\eta$, so its matrix is now $\left[\begin{smallmatrix}1&0\\-1&-1\end{smallmatrix}\right]$.

Answer 2

In general we can use wreath product to construct a 2-sylow subgroup, for example, see https://mathoverflow.net/questions/127917/ for some details.

But in this special case we can find a beautiful explicit 2-sylow subgroup with exactly 64 elements. Namely

$$\begin{bmatrix}\pm1 & & \\ & \pm1 & \\ & & \pm1 \end{bmatrix}.\begin{bmatrix} & & \pm1\\ & \pm1 & \\ \pm1 & & \end{bmatrix}$$