12
$\begingroup$

In the symmetric group $S_n$ what is the shortest sequence $c_1,\ldots,c_k\in S_n$ such that, for all $x\in S_n$ the following product of conjugates of $x$: $$x^{c_1}x^{c_2}\ldots x^{c_k}$$ equals the identity?


I've previously asked, on M.SE, whether the degree of the minimal monomial equals the group's exponent and received $S_3$ as a counterexample. The counterexample leverages the fact that $S_3$ has a normal subgroup $G$ (in this case $A_3$) and an element $r$ such that, for $g\in G$ we have $rgrg=e$. Unfortunately, $S_n$ for $n>3$ does not satisfy the above property, so the approach does not generalize, and lacking further insight into the problem, even $S_4$ seems too large to computationally search the space of such monomials.

$\endgroup$
5
  • 3
    $\begingroup$ Note that $k$ must be even. $\endgroup$ Jan 25, 2015 at 20:13
  • 2
    $\begingroup$ The question looks quite nontrivial, but is there some specific motivation for it? $\endgroup$ Jan 25, 2015 at 20:43
  • $\begingroup$ @Jim It has no particular motivation, beyond the thought that a proof that $k<\text{lcm}(1,\ldots,n)$ for infinitely many $n$ would likely reveal something interesting about some "additional structure" allowing that to happen; in the other case, if $k=\text{lcm}(1,\ldots,n)$, it might be interesting to see why the exponent of a group is "irreducible"; I ask about the symmetric group in particular, because it has exactly one normal subgroup - so any approach applicable to it is likely to extend well to simple groups, but an approach could also leverage that $S_n$ is not itself simple. $\endgroup$ Jan 25, 2015 at 20:56
  • 1
    $\begingroup$ Another small remark is that you might as well suppose that $c_{1} = 1.$ $\endgroup$ Jan 25, 2015 at 22:03
  • 1
    $\begingroup$ You can prove $k\geq n$ by adapting the argument here: mathoverflow.net/questions/20471/… $\endgroup$ Jan 30, 2015 at 13:56

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.