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A necessary condition for a subset $\Sigma\subseteq S_n$ to be a transitive permutation group of order $n$, is to be... transitive. Is the best algorithm to check $\Sigma$'s transitivity faster than the best algorithm to check that $\Sigma$ is a (sub)group?

Context/motivation. If the transitivity check needs no more than $O(n^2)$ runtime, I guess that the same holds for checking the non-associativity of a Latin square of order $n$. See here for further reference.

Kan't
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    Could you be more specific? 'Best' is subjective. Space and time complexities could make one 'better' than another on one scenario, and the other in another. I suspect there are fast special cases too like testing cyclic groups for example. – Simon Goater Aug 23 '25 at 09:17
  • I think I had in mind time complexity, @SimonGoater. – Kan't Aug 23 '25 at 09:33
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    Sorry, my previous comment was over optimistic. You can test for transitivity in time $O(n|\Sigma|)$. I would guess that is still quicker than the best test for a subset being a subgroup. But in practice you could find out quickly if it was not a subgroup by testing random products. – Derek Holt Aug 23 '25 at 14:23
  • I don't understand your "context". How would testing for transitivity of $\Sigma$ help with testing that $\Sigma$ is a subgroup? The set ${(1,2),(2,3),(1,3)} \subset S_3$ is transitive but is not a subgroup. – Derek Holt Aug 24 '25 at 12:42

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