0
$\begingroup$

Let $a_1$, $a_2$, and $a_3$ be three involutions of a finite set such that $a_1 a_2 a_3$ is a cyclic permutation. Is the group generated by $a_1, a_2, a_3$ the symmetric group?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ What do you mean exactly by "a cyclic permutation of $S$? $\endgroup$
    – abx
    Commented Sep 18, 2023 at 5:58
  • 1
    $\begingroup$ @EmilJeřábek It's been deleted, but the old question was at mathoverflow.net/questions/454759/… $\endgroup$
    – user44191
    Commented Sep 18, 2023 at 6:54
  • 1
    $\begingroup$ For what it's worth, one of the comments provided what seems to be as strong an answer to the question as possible: consider $a_1 = (12)(34), a_2 = (23)(45), a_3 = (34)(15)$; then $a_1 a_2 a_3 = (13524)$ is cyclic, but every element is in $A_5$. $\endgroup$
    – user44191
    Commented Sep 18, 2023 at 6:55
  • 5
    $\begingroup$ For the revised question, a random search in $S_6$ quickly found: $a,b,c=(1, 2)(3, 4)(5, 6),(1, 6)(2, 5)(3, 4),(1, 6)(2, 4)(3, 5)$, $abc=(1, 2, 4, 6, 5, 3)$, $\langle a,b,c \rangle=12$. $\endgroup$
    – Derek Holt
    Commented Sep 18, 2023 at 8:01
  • 1
    $\begingroup$ Maybe I am missing something but if $a_1, a_2$ and $a_3$ all leave some element $s \in S$ fixed, then so does every element of the group they generate. E.g. if we interpret the example in the comment above as acting on $S = \{1, \ldots, 7\}$ instead of $\{1, \ldots, 6\}$ then it is clear that they do not generate the whole symmetry group since they won't generate any element that sends 7 to 1 (or any other element) $\endgroup$
    – Vincent
    Commented Sep 18, 2023 at 8:08

1 Answer 1

Reset to default
9
$\begingroup$

For $n \equiv 2 \pmod 4$, let $m = n/2$ and define the following permutations in $S_n$:

$$c := (1,2,3,\ldots,n-1,n),$$ $$a_1:=(1,n)(2,n-1) \cdots (m,m+1),$$ $$a_2 := c^m = (1,m+1)(2,m+2) \cdots (m,n),$$ $$ a_3 := a_2 a_1 c.$$

Then $a_1a_2a_3 = c$, with $\langle a_1,a_2,a_3 \rangle = \langle a_1,c \rangle$ the dihedral group of order $2n$.

Since $a_1$ and $a_2$ are products of $m$ transpositions, they are odd permutations, and $a_3= a_1c^{m+1}$ with $m+1$ even is conjugate to $a_1$ in the dihedral group, so that is too.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Beautiful solution! And also a very nice presentation (pun intended) $\endgroup$
    – Vincent
    Commented Sep 18, 2023 at 15:52

You must log in to answer this question.