Is there a way to classify all $p$-Sylow subgroups of $GL(n,\, \mathbb{F}_{p^k})$ where $|\mathbb{F}_{p^k}|=p^k$? Clearly the prime $p$ (that is the characteristic of the base field) is a very privileged number; for example we can say that a $p$-Sylow of $GL(n,\, \mathbb{F}_{p^k})$ has order $p^{k(1+2+\ldots+(n-1))}$. But what about the structure of these $p$-Sylows can we identify them as normalizers/centralizers of flags in the projective geometry?
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The upper triangular matrices with diagonal entries $1$ form a $p$-Sylow subgroup $P$. So the set of all $p$-Sylow subgroup is given by all the conjugates of $P$.
EDIT: In the light of the answer of Andreas Caranti, $P$ is contained in the stabilizer of the flag $$\{0\} < \langle e_n\rangle < \langle e_{n-1}, e_n\rangle < \ldots < \langle e_2, e_3,\ldots, e_n\rangle < \mathbb{F}_{p^k}^n,$$ where $e_i$ denotes the $i$-th standard vector.
EDIT 2, thanks to the comment of Jyrki Lahtonen: The full stabilizer is still a bit larger than $P$. It is the set of all invertible upper triangular matrices.
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$\begingroup$ Won't all the upper triangular matrices stabilize this flag? Irrespective of whether they have ones on the diagonal or not? IOW aren't the Borel subgroups the stabilizers of maximal flags. $\endgroup$ Commented Mar 12, 2013 at 21:55
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$\begingroup$ @JyrkiLahtonen: Thanks for pointing this out. Your are right. $\endgroup$– azimutCommented Mar 12, 2013 at 21:58
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$\begingroup$ My problem is that the above (complete) flag is not a flag of the projective geometry associated to the vector space $V$ because it contains $\{0\}$ and $V$. The elements of $PG(V)$ are only proper nonzero subspaces. $\endgroup$– DubiousCommented Mar 12, 2013 at 21:58
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$\begingroup$ This doesn't change anything. $\{0\}$ and $V$ are stabilized by any invertible linear map. So you can safely drop them. $\endgroup$– azimutCommented Mar 12, 2013 at 22:01