5
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Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

Question: Is there a largest possible order of a finite group generated by $3$ class transpositions?

Remark: Given a bound $m \in \mathbb{N}$, let $n_{\rm max}(m)$ denote the largest possible order of a finite group generated by $3$ class transpositions interchanging residue classes with moduli $\leq m$. Then we have:

$n_{\rm max}(3) \ = \ 2^3 \cdot 3 \cdot 5$,

$n_{\rm max}(4) \ = \ 2^{15} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11$,

$n_{\rm max}(5) \ = \ 2^{95} \cdot 3^{47} \cdot 5^{20} \cdot 7^{14} \cdot 11^7 \cdot 13^6 \cdot 17^3 \cdot 19^2 \cdot 23^2 \cdot 29^2 \cdot 31^2$,

$n_{\rm max}(6) \ = \ 2^{200} \cdot 3^{103} \cdot 5^{48} \cdot 7^{28} \cdot 11^{16} \cdot 13^{13} \cdot 17^8 \cdot 19^6 \cdot 23^6 \cdot 29$,

$n_{\rm max}(7) \ = \ 2^{1283} \cdot 3^{673} \cdot 5^{305} \cdot 7^{193} \cdot 11^{98} \cdot 13^{84} \cdot 17^{50} \cdot 19^{41} \cdot 23^{25} \cdot 29^{13} \cdot 31^4$,

$n_{\rm max}(8) \ = \ 2^{1283} \cdot 3^{673} \cdot 5^{305} \cdot 7^{193} \cdot 11^{98} \cdot 13^{84} \cdot 17^{50} \cdot 19^{41} \cdot 23^{25} \cdot 29^{13} \cdot 31^4$,

$n_{\rm max}(9) \ = \ 2^{1283} \cdot 3^{673} \cdot 5^{305} \cdot 7^{193} \cdot 11^{98} \cdot 13^{84} \cdot 17^{50} \cdot 19^{41} \cdot 23^{25} \cdot 29^{13} \cdot 31^4$.

So maybe the sequence $(n_{\rm max}(m))$ gets constant already for $m \geq 7$, which would of course imply a positive answer to the question. In this case, $G = \langle \tau_{1(7),6(7)}, \tau_{0(5),3(5)}, \tau_{0(4),5(6)} \rangle$ would be a largest possible finite group generated by $3$ class transpositions.

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  • $\begingroup$ I have revised, updated and undeleted this question from 2012. In particular I have added data computed since then. $\endgroup$
    – Stefan Kohl ♦
    Apr 20, 2016 at 10:58
  • $\begingroup$ Do you know if, in cases 7,8 and 9, there is only one group turning up with that maximal order? Any information about, say, solvability of the group(s) in question? $\endgroup$
    – Nick Gill
    Apr 20, 2016 at 11:20
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    $\begingroup$ It's a quotient of a Coxeter group with 3 generators, in each case, right? Could you specify the order of the product of each pair of generators? $\endgroup$
    – Dima Pasechnik
    Apr 20, 2016 at 11:26
  • $\begingroup$ @NickGill: For each $m \leq 9$, up to conjugacy and choice of generators there is only one group of maximal finite order (that means in particular that the groups for $m = 7, 8, 9$ are the same), and these groups are not solvable. $\endgroup$
    – Stefan Kohl ♦
    Apr 20, 2016 at 13:04
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    $\begingroup$ @DimaPasechnik: Let $a := \tau_{1(7),6(7)}$, $b := \tau_{0(5),3(5)}$ and $c := \tau_{0(4),5(6)}$ be the generators of the group $G$ from the question. Then we have $|ab| = 4$, $|ac| = 12$ and $|bc| = 60$. $\endgroup$
    – Stefan Kohl ♦
    Apr 20, 2016 at 13:15

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