How to find Sylow 3-subgroups of $SL_n(F_3)$?

Question

I've enumerate all 24 elements in $SL_2(F_3)$. As $24=2^3\times 3$, I need to find groups of order 3. Sylow Theorem tells me that the number of Sylow 3-subgroups $n_3 \equiv 1\ \text{mod}\ 3$ and divides $8$, so $n_3= 1 \text{ or } 4$.

I've found 2 of them: upper/lower triangular matrices with 1 on the diagonal, i.e., $$ S_1=\left\{ \left( \begin{matrix} 1& 0\\ 0& 1\\ \end{matrix} \right) ,\left( \begin{matrix} 1& 1\\ 0& 1\\ \end{matrix} \right) ,\left( \begin{matrix} 1& 2\\ 0& 1\\ \end{matrix} \right) \right\} ,S_2=\left\{ \left( \begin{matrix} 1& 0\\ 0& 1\\ \end{matrix} \right) ,\left( \begin{matrix} 1& 0\\ 1& 1\\ \end{matrix} \right) ,\left( \begin{matrix} 1& 0\\ 2& 1\\ \end{matrix} \right) \right\} . $$ The remaining ones are given by my friend: Groups generated by $ \left( \begin{matrix} 0& 2\\ 1& 2\\ \end{matrix} \right) ,\left( \begin{matrix} 2& 2\\ 1& 0\\ \end{matrix} \right) $ respectively. However, this process is rather experimental than theoretical. I wonder if there is a logical explanation to this, or a method to figure them out more efficiently.

Any idea or help would be appreciated! Thank you!

Answer 1

The Sylow subgroups of $\mathrm{SL}_2(\mathbf{F}_3)$ are conjugate to the group of upper triangular matrices as you have noted. Another way to think about this subgroup is as the group which fixes the line generated by the vector $\mathbf{v} = (1,0)$. Any other Sylow $P$ (which is conjugate) will also fix a line generated by a vector $\mathbf{w}$, and this line is uniquely determined by $P$, since if $P$ fixes two linearly independent vectors then it fixes everything. So now you just have to enumerate the lines and write down the determinant one elements which stabilize the lines. For example, you can take the lines generated by $(1,0)$, $(0,1)$, $(1,2)$, and $(1,1)$; in that order, you get the groups you wrote down.

Unnecessary Extras: If you did the same for $\mathrm{SL}_3(\mathbf{F}_p)$, you want want to stabilize a line generated by $\mathbf{v} = \mathbf{v}_1$ but also a plane $\{\mathbf{v}_1,\mathbf{v}_2\}$ generated by two vectors. More generally, Sylow subgroups of $\mathrm{SL}_n(\mathbf{F}_p)$ or $\mathrm{GL}_n(\mathbf{F}_p)$ correspond to "flags" of subspaces. The number of choices of $\mathbf{v}_1$ is $(p^n - 1)/(p-1)$ for the first vector (any non-zero vector up to scalar), then $(p^{n-1} - p)/(p-1)$ for the second vector (any non-zero vector up to scalar in $V/\mathbf{v}_1$), then $(p^{n-2} - 1)/(p-1)$ choices of a vector up to scalar in $V/(\mathbf{v}_1,\mathbf{v}_2)$, and so on. Hence

$$n_p = \frac{1}{(p-1)^p} \prod (p^{n-i} - 1).$$

Comparing this to

$$|\mathrm{GL}_n(\mathbf{F}_p)| = \prod (p^{n-i} - 1) p^i, \qquad |\mathrm{SL}_n(\mathbf{F}_p)| = \frac{1}{(p-1)} \prod (p^{n-i} - 1) p^i,$$

you deduce by Sylow III that the normalizer $N$ of $P$ has order

$$(p-1)^{n} \prod p^i$$

for $\mathrm{GL}_n(\mathbf{F}_p)$ and one fewer factor of $(p-1)$ for $\mathrm{SL}_n$. This is consistent with the fact that the normalizer of the group of upper triangular matrices with ones on the diagonal is the group of upper triangular matrices which has these orders. (In fact, this latter approach is a quicker way to compute $n_p$, but it's also nice to have the explicit description.)

Answer 2

Sylow groups are conjugate. So once you have found one, you may try to conjugate it with suitable elements, in particular generators, of $G$, such as $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ (or both combined)