Let \(G = \operatorname{GL}_3(\mathbb{F}_p)\) for \(p\) prime. (1) What is the order of \(G\), and what is the order of a Sylow \(p\)-subgroup? (2) Give the standard matrix form for the canonical Sylow \(p\)-subgroup \(P \le G\). (3) Give the matrix form for the normalizer \(N_G(P)\) and compute the number of conjugates (Sylow \(p\)-subgroups) \(n_p\). (4) Explain conjugacy of Sylow \(p\)-subgroups in terms of complete flags / eigenspaces.
Solution
We have \[ |\operatorname{GL}_3(\mathbb F_p)|=(p^3-1)(p^3-p)(p^3-p^2) =p^3(p-1)^3(p+1)(p^2+p+1). \] Thus a Sylow \(p\)-subgroup has order \(p^3\). The standard one is \[ P=\left\{\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}:a,b,c\in\mathbb F_p\right\}. \] Its normalizer is the upper-triangular Borel subgroup \[ B=\left\{\begin{pmatrix}d_1&a&b\\0&d_2&c\\0&0&d_3\end{pmatrix}:d_i\in\mathbb F_p^\times\right\}, \] so \[ |B|=p^3(p-1)^3,\qquad n_p=[G:B]=(p+1)(p^2+p+1). \]
The subgroup \(P\) is the unipotent radical of the stabilizer of the standard complete flag \[ 0<\langle e_1\rangle<\langle e_1,e_2\rangle<\mathbb F_p^3. \] Conjugating \(P\) corresponds to conjugating this flag. Hence Sylow \(p\)-subgroups are in bijection with complete flags. Their number is \[ (p^2+p+1)(p+1), \] matching the index calculation.