Do nontrivial semisimple elements of $q$-bad order exist in $\operatorname{PSL}_2(q)$ for odd $q>3$?
问题内容
Definition 1: An element of $\operatorname{PSL}_2(q)$ is semisimple if it is diagonalisable in $\operatorname{PSL}_2(\overline{\Bbb F_q})$, where $\overline{\Bbb F_q}$ is the algebraic closure of $\Bbb F_q$.
Definition 2: Let $q$ be a power of a prime. Then we say $n\in \Bbb N$ is $q$-good if:
- $n$ is odd and ($n\mid q-1$ or $n\mid q+1$), or
- $n$ is even and ($4n\mid q-1$ or $4n\mid q+1$).
We say $n$ is $q$-bad if it is not $q$-good.
The Question:
Do nontrivial semisimple elements of $q$-bad order exist in $\operatorname{PSL}_2(q)$ for odd $q>3$?
Context:
This seems to be a matter of fitting together a few pieces:
- For each odd $q>3$, there always exist $q$-bad numbers less than $q$.
- For each finite cyclic group $G$, there is an element $g$ of $G$ with order $m$ iff $m\mid |G|$.
- The order of $\operatorname{PSL}_2(q)$ is $(q-1)q(q+1)/2$.
- There are cyclic subgroups of $\operatorname{PSL}_2(q)$ of orders $(q-1)/2$ and $(q+1)/2$.
- There exist nontrivial semisimple elements in each of those cyclic subgroups.
I am unsure of 4. and 5., but they appear to do with maximal tori of $\operatorname{PSL}_2(q)$, and, for what it is worth (i.e., not much), Google's AI asserts them (but upon probing, it contradicts itself). I haven't proven 1., but it's almost by inspection, right? Some help there would be appreciated too.
回答 (1)
If $q$ is odd, then an irreducible polynomial in $\mathbb{F}_q[x]$ cannot have repeated roots. Therefore, all you're asking is, what is the order of $\left[ {\begin{array}{cc} 1 & 0 \\ 0 & a \\ \end{array} } \right]$ for $a\in\mathbb{F}_q-\{0,1\}$, or that of $\left[ {\begin{array}{cc} b & 0 \\ 0 & c \\ \end{array} } \right]$ where $(x-b)(x-c)\in\mathbb{F}_q[x]$ is irreducible.
The order of the former divides $q-1$.
As for the order of the latter: let $u\in\mathbb{F}_{q^2}$ have order $q^2-1$. We can choose, for $b$, any power of $u$ which is not a multiple of $q+1$. We know that $c=b^q$, because $\sigma(x)=x^q$ is the only non-trivial automorphism of $\mathbb{F}_{q^2 }$ fixing $\mathbb{F}_q$. Let $b=u^n$. We seek the smallest positive integer $m$ such that $mn\equiv mqn\mod q^2-1$. In other words, we want the smallest positive integer $m$ such that $q+1|mn$. This means that $m=\dfrac{q+1}{\gcd(q+1,n)}$.