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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 largest least power of $p$ greater than or equal to $n$. This gives a upper bound for exponents of classical groups of dimension $n$ over $GF(q)$.

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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 largest power of $p$ greater than or equal to $n$. This gives a upper bound for exponents of classical groups of dimension $n$ over $GF(q)$.

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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 $G=GL(n,q)$, I think it can be easily seen that the exponent of $P$ may be is the largest power of $p$ greater than or equal to $n$, but I don't know a proof or reference.n\$.

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