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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.

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Are there conditions on $n$? Otherwise it's false. (Take the permutation matrix which cyclically permutes $p^2$ coordinates.) –  anon Mar 19 '13 at 23:00
    
Let's require $n\le p$ and see if the result holds. –  Jared Mar 19 '13 at 23:01
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3 Answers

up vote 4 down vote accepted

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.

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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.

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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 $p^m$.

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A good answer, but I just cannot understand what "I guess this generalizes to no conditions on $n$" could mean. –  Marc van Leeuwen Mar 20 '13 at 9:02
    
I mean that if we don't require $n\le p$, we still obtain a result that bounds the order of the elements of $P$. The result I mention in the final sentence yields our initial case when $n\le p$. –  Jared Mar 20 '13 at 13:25
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