MathWiki
|
November 21, 2024, at 11:22 AM | MathWiki / MathWiki / CircularFerreroPairs |
|
CircularFerreroPairsLet $k\geq 4$ be a positive integer. Modisett showed that there is a finite set of prime numbers $\mathcal P_k$ such that if $F=F_q$ is a finite field where $q=p^s$ for some prime $p$ such that $k|(q-1)$ and $\Phi$ a subgroup of $F^*=F\setminus \{0\}$ with $|\Phi|=k $, then $(F,\Phi)$ is a circular Ferrero pair if and only if $p\not\in\mathcal P_k$. Using an algorithm of Beidar, Fong and Ke, $\mathcal P_k$ can be computed fairly quickly for $k\leq 60$. Here, a GAP program is provided for computing the set of primes. On Modisett primes page one finds the list of $\mathcal P_k$ for $k$ up to $32$, and the lists compressed for downloads. On the Pk Max page, one finds the information on the number of primes in a $\mathcal P_k$ and the maximal element in it. Let $\Phi$ be a subgroup of $\mathrm{GL}(2,q)$ with $|\Phi|=q+1$ where $q=p^\alpha$ and $p$ a prime. Denote $N=\mathrm{GF}(q)\times\mathrm{GF}(q)$. Suppose that $\Phi$ is a fixed point free group. Then all Sylow subgroups of $\Phi$ are cyclic or generalized quaternion.
Circular Group Without Fixed Points A metacyclic group has presentation $\langle x,y\mid x^m=e=y^n,y^{-1}xy=x^r\rangle$ for some $m$, $n$ and $r$ such that $\gcd(m,n)=\gcd(m,r-1)=1$ and $r^n\equiv 1\pmod m$. We say that a metacyclic group $\Phi$ is circular if there is some group $N$ such that $\Phi$ acts fixed point freely and $(N,\Phi)$ is circular. It holds that if $\Phi$ is circular, then there is a finite subset $\mathcal P_\Phi$ of primes such that for every group $N$ on which $\Phi$ acts fixed point freely, the Ferrero pair $(N,\Phi)$ is circular if and only if for every prime $p$ dividing $|N|$, $p$ is not in $\mathcal P_\Phi$. |
Validate the XHTML and CSS of this page. | Page last modified on October 19, 2018, at 02:26 PM | Edit History Print Recent Changes |
Powered by PmWiki |