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This question has already been asked here. However, the answer there isn't particularly helpful.

Consider $G = GL_{n}(\mathbb{F}_{q})$ where $q = p^{r}$ where $p$ is prime. Show that the upper unipotent matrices (say $P$) are a p-Sylow subgroup of $G$.

Furthermore, find conditions so that every element of $P$ has order dividing $p$.

Attempt at a solution -

I recognize that this is true. However, is there an easy way to determine that it is? I know that the cardinality of $P$ is $q^{\frac{n(n-1)}{2}}$ and that the cardinality of $G$ is $(q^{n} - 1)(q^{n} - q) ... (q^{n} - q^{n-1})$ but there doesn't seem to be an easy way to factor the above equation.

For the second part, you can write any upper nilpotent matrix $A$ as $I + B$ where $B$ is an upper triangular matrix with $0's$ along the diagonal. It seems as though by the binomial theorem that $p \geq n$ and $r \geq p - n$ guarantees this. However, is this the exact characterization of when this property is true?

Edit - this is homework by the way.

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    $\begingroup$ For the first part, you just need to extract the highest power of $q$ form the factored expression. i.e $q^n-q=q(q^{n-1}-1)$, $q^n-q^2=q^2(q^{n-2}-1)$, etc. $\endgroup$
    – Derek Holt
    Nov 16, 2014 at 21:45
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    $\begingroup$ For the first part: $|G| = (q^n-1)(q^n-q)\cdots(q^n-q^{n-1}) = q^{0+1+\cdots+(n-1)}(q^n-1)(q^{n-1}-1)\cdots(q-1)$. Here, the exponent of the power of $q$ on the left is $\frac{n(n-1)}{2}$ and all the other factors in brackets are relatively prime to $q$ and thus also to $p$. $\endgroup$
    – jflipp
    Nov 16, 2014 at 21:46

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