Sylow $p$-subgroups of Finite Matrix Groups Let $G$ be a subgroup of $GL_n(\mathbb{F}_p)$, with $n\le p$, and let $P$ be a Sylow $p$-subgroup of $G$.  Do all non-trivial elements of $P$ have order $p$?
I believe the answer is yes, because I think the result holds for $G=GL_n(\mathbb{F}_p)$.  I'm hoping someone can show me a quick way to see this fact, perhaps using minimal/characteristic polynomials.
 A: A Sylow $p$-subgroup is formed by the upper triangular matrices with ones on the diagonal. Let $A$ be any such matrix. We can write it as $A=I+B$, where $B$ is an upper triangular matrix with zeros on the diagonal. Then $B^n=0$. So if $p\ge n$ we have by the binomial formula (that holds, because $I$ and $B$ commute) that
$$
A^p=(I+B)^p=I+\sum_{k=1}^p{p\choose k}B^k.
$$
Here on the r.h.s. all the terms save the first are zero: Either the binomial coefficient is divisible by $p$, or $k=p$ in which case $B^k=B^p=B^{p-n}B^n=0$.
So if $B\neq0$ then $A$ is of order $p$.
We have shown that the claim holds for this particular Sylow $p$-subgroup. As all those Sylow subgroups are conjugate, and conjugation preserves the order of elements, the claim holds for all Sylow $p$-subgroups.
A: Since $1$ is the unique $p^k$-th root of unity in characteristic $p$, any $p$-subgroup of $\operatorname{GL}_n(\Bbb F_p)$ consists of unipotent elements $I+N$ with $N$ nilpotent. Also $(I+N)^{p^k}=I+N^{p^k}$ and taking $N$ a single Jordan block one sees this is forced to be $I$ if and only if $p^k\geq n$. So the exponent of a Sylow $p$-subgroup is $p^k$ for $k=\lceil \log_p n\rceil$. This remains true without change if you replace $\Bbb F_p$ by any finite field of characteristic $p$.
As Jyrki Lahtonen indicated, the Sylow $p$-subgroups are conjugates of the subgroup of upper unitriangular matrices.
A: Let $A\in P$ have order $p^m$.  Then
$$(A-I)^{p^m}=A^{p^m}-I=I-I=0$$
so that $(A-I)$ is nilpotent.  The nilpotent index of $(A-I)$ must be less than or equal to $n$, which is less than or equal to $p$, so we have:
$$0=(A-I)^p=A^p-I\quad\Longrightarrow\quad A^p=I$$
This generalizes to no conditions on $n$ as follows.  Let $m$ be such that $p^{m-1}<n\le p^m$.  Then by the same argument, we find that the exponent of $P$ is at most $p^m$.
