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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.

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    $\begingroup$ Note that $k$ must be even. $\endgroup$
    – Geoff Robinson
    Commented Jan 25, 2015 at 20:13
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    $\begingroup$ The question looks quite nontrivial, but is there some specific motivation for it? $\endgroup$
    – Jim Humphreys
    Commented 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. $\endgroup$
    – Milo Brandt
    Commented Jan 25, 2015 at 20:56
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    $\begingroup$ Another small remark is that you might as well suppose that $c_{1} = 1.$ $\endgroup$
    – Geoff Robinson
    Commented Jan 25, 2015 at 22:03
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    $\begingroup$ You can prove $k\geq n$ by adapting the argument here: mathoverflow.net/questions/20471/… $\endgroup$
    – Sean Eberhard
    Commented Jan 30, 2015 at 13:56

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