# What is the length of the shortest law of $S_n$?

What is the length of the shortest word $w\in F_2$ such that $w(x,y)$ is trivial for every $x,y\in S_n$?

There is a simple argument showing that we must have $\ell(w)\geq n$. See here for instance. It seems likely however that $\ell(w)$ must be super-polynomial in $n$.

What is the state of the art for lower bounds?

(As for upper bounds, the paper "Identical relations in symmetric groups and separating words with reversible automata" by Gimadeev and Vyalyi finds an upper bound of $\exp(\sqrt{n}\log n)$.)

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Would it be possible to be a little more explicit? What is $S_n$? The symmetric group? What is $F_2$, and what is $w$ or $\mathcal{l}(w)$? –  Christian Stump Apr 10 '13 at 11:23
$F_2$ is a free group in $x$ and $y$. Sean is asking about the shortest identity $w(x,y)$ such that $w(a, b)=1$ for any $a,b\in S_n$. –  Alexey Kvashchuk Apr 10 '13 at 11:37

As far as I know, the state of the art gives some improvements to those bounds, but they are not huge. For a better lower bound: given a nontrivial element $w \in F_2$ of word length $\ell$, using a result of Buskin (Economical separability in free groups, Sib. Math. J., 50 (2009), 603-608) there exists a subgroup, $H$, of index $\ell/2+2$ that does not contain $w$. By looking at the action of $F_2$ on $F_2/ H$ we get a representation of $F_2$ into $S_{\ell/2+2}$, that does not kill $w$. Therefore, in order for $w$ to be trivial in any representation of $F_2$ into $S_n$ we must have that $n \leq \ell /2 + 2$, or $2(n-2) \leq \ell$.
It's amazing that the lower bound $10n$ is open! –  Sean Eberhard Apr 10 '13 at 14:42