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

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 $GF(q)$.

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 $GF(q)$.

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Binzhou Xia
Binzhou Xia
  • 767
  • 4
  • 15

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 $GF(q)$.

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 $n$.

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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Binzhou Xia
Binzhou Xia
  • 767
  • 4
  • 15

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 thinkit can be easily seen that the exponent of $P$ may beis the largest power of $p$ greater than or equal to $n$, but I don't know a proof or reference.

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 the exponent of $P$ may be the largest power of $p$ greater than or equal to $n$, but I don't know a proof or reference.

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 $n$.

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Binzhou Xia
Binzhou Xia
  • 767
  • 4
  • 15