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I know that one of the Sylow $p$-subgrouops of $GL_n(\mathbb F_q)(q=p^a)$ is the subgroup $P$ of upper triangular matrices with $1$ on diagonal, but is there a simple and direct way to see that every Sylow $p$-subgroup is obtained by change of basis isomorphism $T^{-1}PT=\{T^{-1}UT: U\in P\}, T\in GL_n(\mathbb F_q)$ from this one(i.e. conjugate)? (This is Sylow's second theorem, but I'm seeking a more direct proof in this particular case. In fact, I also want to know if we can use this case to prove Sylow's second theorem by imbedding any group into $GL_n(\mathbb F_q)$?)

Eric Ley
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    You could regard $P$ as the stabilizer in ${\rm GL}_n(q)$ of a maximal chain of subspaces of ${\mathbb F}_q^n$, and show that any Sylow $p$-subgroup (in fact any $p$-subgroup) stabilizes such a chain. Serre had a proof of the existence of Sylow $p$-subgroups of finite groups that involved embedding into a linear group, but I am not sure about the conjugacy part. – Derek Holt Jan 30 '24 at 08:22
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    @Derek Holt: I think the stablizer (the subgroup of all upper triangular matrices)is bigger than $P$? How to see any Sylow $p$-subgroup stablizes some maximal filtration directly? – Eric Ley Jan 30 '24 at 09:44
  • Yes, by the stabilizer, I really meant the subgroup that fixes each subspace in the chain and induces the trivial representation on each quotient in the chain. That subgroup really is $P$. The fact that it stabilizes a maximal filtration (in that sense) follows from the fact that the only irreducible representation of a finite $p$-group in characteristic $p$ is the trivial one of dimension $1$. For finite fields you can prove that by a simple counting argument (it's essentially the same as the proof that $p$-groups have nontrivial centres). – Derek Holt Jan 30 '24 at 09:56
  • “ The fact that it stabilizes a maximal filtration (in that sense) follows from the fact that the only irreducible representation of a finite p-group in characteristic p is the trivial one of dimension 1.” Is this invoking Jordan-Holder’s theorem? – Eric Ley Jan 30 '24 at 10:24
  • Not really. See this post – Derek Holt Jan 30 '24 at 10:26
  • I mean why all irreducible representations trivial implies the existence of a such maximal filtration? – Eric Ley Jan 30 '24 at 10:36
  • That's a straightforward induction on $n$. – Derek Holt Jan 30 '24 at 10:37

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