----- > [!proposition] Proposition. ([[general linear group over a prime field]]) > Let $\mathbb{F}_{p}$ be a [[prime field]] and consider the [[general linear group]] $GL_{n}(\mathbb{F}_{p})$. We have that $|GL_{n}(\mathbb{F}_{p})|=\prod_{i=0}^{n-1}(p^{n}-p^{i}).$ > A [[p-Sylow subgroup]] of $GL_{n}(\mathbb{F}_{p})$ consists of [[upper-triangular matrix|upper-triangular matrices]] with [[diagonal]] fixed to $1$: $ \begin{bmatrix} 1 & * & * & \cdots & * \\ & 1 & * & \cdots & * \\ & & 1 & \cdots & * \\ & & & \ddots & \vdots \\ & & & & 1 \end{bmatrix}$ the number of such subgroups is $n_{p}=\frac{1}{p-1}\prod_{i=1}^{p-1} (p^{n-i}-1).$ > ^92af21 > [!proof]- Proof. ([[general linear group over a prime field]]) > Recall that a [[matrix]] $A \in \mathbb{F}_{p}^{n \times n}$ [[inverse matrix|invertible]] iff it [[rank]] $A=n$; i.e., iff $A$ has [[linearly independent]] [[row rank|rows]]. We want to count the number of [[matrix|matrices]] in $\mathbb{F}_{p}^{n \times n}$ satisfying this property. The idea is to proceed row-by-row, counting the number of ways in which row $r_{j}$ can avoid being written as a linear combination its preceding rows $r_{1},\dots,r_{j-1} < r_{j}$. \ To make the set $\{r_{1}\}$ [[linearly independent]], we just need to make sure $r_{1} \neq \b 0$ (so, there is a single invalid choice). Each 'slot' in $r_{1}$ has one of $p$ entries; thus there are $p^{n}-1$ valid choices. \ To make the set $\{ r_{1},r_{2} \}$ [[linearly independent]], we need to make sure that $r_{2}$ is not a [[scalar|scalar multiple]] of $r_{1}$. Since there are $p$ scalar multiples of $r_{1}$, there are are $p^{n}-p$ valid choices of $r_{2}$ per valid choice of $r_{1}$; $(p^{n}-1)(p^{n}-p)$ total choices for the pair $r_{1},r_{2}$. \ To make the set $\{ r_{1},r_{2},r_{3} \}$ [[linearly independent]], we need to make sure that $r_{3}$ is not a [[linear combination]] of $r_{1}$ and $r_{2}$. Since there exist $p^{2}$ [[linear combination]]s of $r_{1}$ and $r_{2}$, there are $p^{n}-p^{2}$ ways to do this per choice of $r_{1}$ and $r_{2}$, hence $(p^{n}-1)(p^{n}-p)(p^{n}-p^{2})$ choices total. Induction ultimately yields $\prod_{i=0}^{n-1}(p^{n}-p^{i})=|GL_{n}(\mathbb{F}_{p})|. (*)$ \ The first step to finding a [[p-Sylow subgroup]] of $GL_{n}(\mathbb{F}_{p})$ is to find $r,m$ such that $|GL_{n}(\mathbb{F}_{p})|=p^{r}m$. We can write the expression $(*)$ has $\begin{align} \prod_{i=0}^{n-1} (p^{n}-p^{i})= & (p^{n}-1)(p^{n}-p)(p^{n}-p^{2}) \cdots (p^{n}-p^{n-1}) \\ = & (p^{n}-1)p(p^{n-1}-1)p^{2} (p^{n-2}-1)\cdots p^{n-1}(p-1) \\ = & p^{n \choose 2} \textcolor{Thistle}{\prod_{i=1}^{p-1} (p^{n-i}-1)}. \\ = & p^{n \choose 2} \textcolor{Thistle}{m}. \end{align}$ By [[Euclid's Lemma]], if $p$ does not [[divides|divide]] $\textcolor{Thistle}{m}$, then there because there does not exist $i \in [p-1]$ for which $p | (p^{n-i}-1)$. Hence we conclude that $r={n \choose 2}$ in the statement of [[the Sylow theorems]]. Thus, to find a [[p-Sylow subgroup]] of $GL_{n}(\mathbb{F}_{p})$ is to find a [[subgroup]] whose [[order of a group|order]] is $p^{{n \choose 2}}=p^{\frac{n(n-1)}{2}}$. We claim that the [[upper-triangular matrix|upper triangular matrices]] with $1$s along the [[diagonal]] form such a [[subgroup]]: $ K := \{ \text{matrices of the form} \begin{bmatrix} 1 & * & * & \cdots & * \\ & 1 & * & \cdots & * \\ & & 1 & \cdots & * \\ & & & \ddots & \vdots \\ & & & & 1 \end{bmatrix}\}.$ We see that $K$ has order $p^\frac{n(n-1)}{2}$ because the [[matrix]] above is determined by the $\frac{n(n-1)}{2}$ entries above the [[diagonal]], each of which can take on $p$ different values. We also see that $K$ is [[inverse matrix|invertible]] because its entries have [[determinant of a matrix|determinant]] $1$. Clearly $I_{n} \in K$. Moreover, it is clear that the product of [[upper-triangular matrix|upper-triangular matrices]] is [[upper-triangular matrix|upper-triangular]] with $1$ on the [[diagonal]]: if $A$ and $B$ are two [[matrix|matrices]] satisfying $a_{ij}=0=b_{ij}$ when $i>j$ and $a_{ii}=b_{ii}=1$, then fixing $i>j$ we see that \ $(AB)_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}=\sum_{k=1}^{i-1}\cancel{a_{ik}}^{0}b_{kj} + \sum_{k=i}^{n}a_{ik}\cancel{b_{kj}}^{0}=0$ and fixing $i=j$ we see that $(AB)_{ii}=\sum_{k=1}^{n} a_{ik}b_{ki}=\sum_{k=1}^{i-1}\cancel{a_{ik}}^{0}b_{ki}+ a_{ii}b_{ii} + \sum_{k=i+1}^{n} a_{ik }\cancel{b_{ki}}^{0}=a_{ii}b_{ii}=1.$ Finally, we know from linear algebra that $K$ is closed under inverses. \ Next we compute the number of [[p-Sylow subgroup]]s. We start by computing the [[order of a group|order]] of the [[normalizer of a subgroup|normalizer]] of $K$ in $GL_{n}(\mathbb{F}_{p})$. The claim is that the [[normalizer of a subgroup|normalizer]] is the set of invertible [[upper-triangular matrix|upper-triangular matrices]] with entries in $\mathbb{F}_{p}$. \ Now, the order of the normalizer is counted as follows: Each entry on the diagonal can by anything except $0$; there are $(p-1)^{ n}$ ways to accomplish this. Then, the $\frac{n(n-1)}{2}$ above-diagonal entries can be whatever they like. There are $p^{{n \choose 2}}$ ways to accomplish this. Thus the order of the normalizer is $(p-1)^{n}p^{{n \choose 2}}$. We conclude that the index of the normalizer (and therefore the number of $p$-Sylows) is $p^{n \choose 2} \frac{\textcolor{Thistle}{\prod_{i=1}^{p-1} (p^{n-i}-1)}}{(p-1)^{n}p^{n \choose 2}}=\frac{1}{p-1}\prod_{i=1}^{p-1} (p^{n-i}-1).$ ^3091a3 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```