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Fix a prime $p$.

How many Sylow $p$-groups are there in $GL_3(\Bbb F_p)$?

Let $s_p$ be the number of Sylow $p$-groups in $GL_3(\Bbb F_p)$. By the Sylow theorems, $s_p\equiv 1\mod p$ and if $p^e$ is the maximal prime power of the order of $GL_3(\Bbb F_p)$, then $s_p$ divides $\frac{\left|GL_3(\Bbb F_p) \right|}{p^e}$. Unfortunately I don't even really know how to figure out the cardinality of that group.

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    $\begingroup$ To compute the cardinality of the group, note that a general element of $GL_3(\mathbb F_p)$ is an invertible $3\times 3$ matrix over $\mathbb F_p$. The first column can be anything except $0$, so there are $p^3 - 1$ possibilities. The second column can be anything except a scalar multiple of the first column, so there are $p^3 - p$ possibilities. The third column can be anything except a linear combination of the first two columns, so there are $p^3 - p^2$ possibilities. So $|GL_3(\mathbb F_p)| = (p^3 - 1)(p^3 - p)(p^3 - p^2)$. $\endgroup$ – Bungo Jul 22 '16 at 18:09
  • $\begingroup$ The general formula for $|GL(n,q)|$ is computed the same way: see en.wikipedia.org/wiki/General_linear_group#Over_finite_fields $\endgroup$ – Bungo Jul 22 '16 at 18:09
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I assume you use column vectors.

One particular Sylow $p$-subgroup is $$ L = \left\{ \begin{bmatrix} 1 & 0 & 0\\ a & 1 & 0\\ c & b & 1\\ \end{bmatrix} : a, b, c \in \Bbb F_p \right\}. $$ You may try and compute its normalizer. The index will give you the number of Sylow $p$-subgroups.

Alternatively, you may note that $x \in L$ iff $x - 1$ maps the underlying vector space to $\langle e_{2}, e_{3} \rangle$, $\langle e_{2}, e_{3} \rangle$ to $\langle e_{3} \rangle$ and $\langle e_{3} \rangle$ to $\{ 0 \}$, where $e_{1}, e_{2}, e_{3}$ is the standard basis. So there are as many Sylow $p$-subgroups as the number of pairs $U_{2} \supseteq U_{1}$ of subspaces, with $\dim(U_{i}) = i$. You will obtain the same number as with the previous method.

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