TITLE: Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$ 
QUESTION [7 upvotes]: Let $G$ be a classical group of dimension $n$ over $GF(q)$ where $q=p^f$ is a prime power, and $P$ be a Sylow $p$-subgroup of $G$. What is the maximal order of elements, i.e. the exponent, of $P$? 
For $G=GL(n,q)$, it can be easily seen that the exponent of $P$ is the least power of $p$ greater than or equal to $n$. This gives a upper bound for exponents of classical groups of dimension $n$ over $GF(q)$.

REPLY [3 votes]: For some classical groups (arising from simple algebraic groups), the answer is given in Proposition 0.5 in a paper by Donna Testerman: $A_1$-type overgroups of elements of order $p$ in semisimple algebraic groups and the associated finite groups, J. Algebra, 177 (1995), no. 1, 34--76. I quote this proposition: 
Let $G$ be a simple algebraic group defined over an algebraic closed field of arbitrary characteristic $p>0$ and $\sigma$ be a surjective endomorphism of $G$ such that $G_\sigma$ is finite. Then the exponent of a Sylow $p$-subgroup of $G_\sigma$ is the least power of $p$ greater than the height of the highest root in the root system of $G$. 
The height of the highest root is equal to $h-1$, where $h$ is the mentioned Coxeter number, and so is $n$, $2n-1$, $2n-1$ and $2n-3$ in $A_n$, $B_n$, $C_n$ and $D_n$ respectively (the Coxeter number in type $A_n$ is $n+1$ not $n$).