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Let $n,k \in \mathbb{N}$ and $x_1, \dots, x_k \in S_n$, symmetric group. Is there an efficient algorithm to determine the $$ \text{order of }\langle x_1, \dots, x_k\rangle\text{?} $$

If necessary, you can assume that $\langle x_1\rangle, \dots, \langle x_k \rangle$ have pairwise trivial intersection. Indeed, you can assume whatever you think is needed, but of course the fewer assumptions, the better.

Subquestion
GAP can calculate the order of this group: $$\begin{align*} \langle &(1\ 2\ 3\ 4)(5\ 6\ 7\ 8)(12\ 25\ 42\ 19)(15\ 32\ 45\ 22)(11\ 28\ 41\ 18),\\ &(9\ 10\ 11\ 12)(13\ 14\ 15\ 16)(34\ 26\ 2\ 18)(37\ 29\ 5\ 21)(33\ 25\ 1\ 17),\\ &(17\ 18\ 19\ 20)(21\ 22\ 23\ 24)(9\ 1\ 41\ 35)(16\ 8\ 48\ 38)(12\ 4\ 44\ 34),\\ &(25\ 26\ 27\ 28)(29\ 30\ 31\ 32)(11\ 33\ 43\ 3)(14\ 40\ 46\ 6)(10\ 36\ 42\ 2),\\ &(33\ 34\ 35\ 36)(37\ 38\ 39\ 40)(10\ 17\ 44\ 27)(13\ 24\ 47\ 30)(9\ 20\ 43\ 26),\\ &(41\ 42\ 43\ 44)(45\ 46\ 47\ 48)(4\ 28\ 36\ 20)(7\ 31\ 39\ 23)(3\ 27\ 35\ 19) \rangle \end{align*}$$ It is $43\,252\,003\,274\,489\,856\,000$. How does it do that?
Any resemblance to real cubes, living or dead, is purely coincidental ;)

Olexandr Konovalov
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Τίμων
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  • Yes, but this is not at all a trivial problem, you need to know the generator relations to do it, and even then, afaik, there is no one algorithm that will always succeed thanks to the general undecidability of the word problem for groups (though this is not entirely applicable to the finite case you are dealing with). – Justin Benfield Mar 14 '16 at 15:18
  • As a side note: Nice Rubik's Cube group. – Justin Benfield Mar 14 '16 at 15:20
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    The fundamental algorithm for solving this problem is known as the Schreier-Sims Algorithm, so I would suggest that you google that as a first step! It has various modifications and improvements for large values of $n$. For a small worked example, see http://math.stackexchange.com/questions/1662723 – Derek Holt Mar 14 '16 at 16:11
  • @DerekHolt. Thank you! I will read about it. It seems to be what I was looking for. – Τίμων Mar 14 '16 at 20:39

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