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p-Divisibility of the Number of Solutions of $x^p=1$ in a Symmetric Group
H. Ishihara1, H. Ochiai2, Y. Takegahara3, and T. Yoshida4
1Kumamoto National College of Technology, Kumamoto 861-1102, Japan
ishihara@ge.knct.ac.jp
22Department of Mathematics, Tokyo Institute of Technology, Meguro-ku 152-8551, Japan
ochiai@math.titech.ac.jp
3Muroran Institute of Technology, Muroran 050-8585, Japan
yugen@mmm.muroran-it.ac.jp
4Department of Mathematics, Hokkaido University, Sapporo 060-0810, Japan
yoshidat@math.sci.hokudai.ac.jp
Annals of Combinatorics 5 (2) p.197-210 June, 2001
AMS Subject Classification: 11B39, 11B50, 12J25, 20B30, 30G06
Abstract:
For a prime $p$ and for the number $a(n)$ of solutions of $x^p=1$ in the symmetric group on $n$ letters, ${\mathrm{ord}}_p(a(n))\geq[n/p]-[n/p^2]$, and especially, ${\mathrm{ord}}_p(a(n))=[n/p]-[n/p^2]$ provided $n\equiv 0\bmod p^2$. Let $r$ be an integer with $1\leq r\leq p^2-1$. If ${\mathrm{ord}}_p(a(r))\leq[r/p]+1$, then, for each positive integer $m$, ${\mathrm{ord}}_p(a(mp^2+r))=m(p-1)+{\mathrm{ord}}_p(a(r))$. Assume that ${\mathrm{ord}}_p(a(r))=[r/p]+2$. If $a(p^2+r)\equiv-p^{p-1}a(r)\bmod
p^{p+[r/p]+2}$, then ${\mathrm{ord}}_p(a(mp^2+r))=m(p-1)+[r/p]+2$; otherwise, there exists a $p$-adic integer $b$ such that ${\mathrm{ord}}_p(a(mp^2+r))=m(p-1)+[r/p]+2+{\mathrm{ord}}_p(m-b)$.
Keywords: Frobenius theorem, $p$-adic, symmetric group

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