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Let $G=SO_3(\mathbb{F}_p)$. We have $|G|=p(p^2-1)$. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Is it true that $n_p=p+1$?

We have the two conditions

  1. $n_p\equiv 1\mod p$
  2. $n_p\mid p^2-1$.

The first condition gives us $n_p=1+kp$, and then the second gives $m(1+kp)=p^2-1$. For fixed $p$, this equation at least has the two nonnegative integer solutions $(k,m)$ given by $(0,p^2-1)$ and $(1,p-1)$ which leads to the two possibilities $n_p=1$ and $n_p=p+1$. Are there more solutions to this equation?

Next, I'd like to rule out the possibility that $n_p=1$. Is $SO_3(\mathbb{F}_p)$ simple?

Any help in solving my initial question is appreciated, even if it doesn't follow my thinking.

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You can easily rule out that $kp+1|p^2-1$ for any $k\gt 1$; any such factorization needs to have $m\equiv -1\pmod p$ and $m\gt 0$, so $m\geq p-1$ and $m(kp+1)\gt m(p+1)\geq (p-1)(p+1) = p^2-1$. –  Steven Stadnicki Mar 20 '13 at 3:12
    
Great, thanks a ton. Now to rule out $n_p=1$ which should be the easy part. I need someone who is familiar with the classification of finite simple groups. –  Jared Mar 20 '13 at 4:09
2  
You might be better off simply exhibiting two distinct $p$-subgroups to conclude that $n_p\gt 1$ - or even just finding a single $p$-subgroup and then finding an element in $SO_3(\mathbb{F}_p)$ that doesn't conjugate it back to itself (which is, of course, essentially equivalent). You shouldn't need the full machinery of finite simple groups for this. –  Steven Stadnicki Mar 20 '13 at 4:48
    
You're right. This I think I can do. I'll look into it. –  Jared Mar 20 '13 at 5:16
    
Why the diophantine-equations tag? –  Álvaro Lozano-Robledo Mar 20 '13 at 11:40

1 Answer 1

up vote 4 down vote accepted

(compiling my comments so the question will have an answer...)

The first question has an easy answer: you can easily rule out any factors $kp+1$ of $p^2-1$ with $k\gt 1$. Suppose $m\cdot (kp+1) = p^2-1$; then since $p^2-1\equiv -1$ and $kp+1\equiv 1\pmod p$, it must be the case that $m\equiv -1$. Since $m\gt 0$, then $m\geq p-1$. But then $m\cdot (kp+1) \gt m\cdot (p+1)\geq (p-1)\cdot(p+1)=p^2-1$.

As for the second question, rather than breaking out the heavy machinery you're probably better off just trying to exhibit two distinct $p$-Sylow subgroups: try finding a particular subgroup and then find an element in $SO_3(\mathcal{F}_p)$ that doesn't conjugate it back to itself. If you find the right subgroup, you should be able to just conjugate it by a permutation matrix to find another distinct one.

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