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I am interested in determining a decomposition of the identity of a symmetric group $S_n$ like

$$\sigma_1\sigma_2...\sigma_m=1$$

for distinct $\sigma_i\in S_n$ (distinct to remove the case $\sigma_i=1\forall i$). Some questions I would like answered:

Such a decomposition is clearly unique for $m=1$, but not unique for $m=2$ (since $\sigma_1\sigma_1^{-1}=\sigma_2\sigma_2^{-1}=...=1$). In fact, for any even $m$ we can always find another decomposition with products of $\sigma_i\sigma_i^{-1}$ for any $i$.

Q1: Can someone determine uniqueness in the odd case?

I guess if we have one such case like $\sigma_1\sigma_2\sigma_3=1$, then we know this is true if $\sigma_1\sigma_2=\sigma^{-1}_3$ or $\sigma_2\sigma_3=\sigma_1^{-1}$, but how do we know those aren't the same?

Q2: What if we restrict to a single conjugacy class? Can we say anything about the uniqueness of the decomposition then?

Q3: How can we find such a decomposition for a given $m$?

levitopher
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    Do you require the $\sigma_i$ to be distinct? If not, there is always the trivial decomposition where $\sigma_i=1$ for all $i$. Also, for $m<n$, let $\sigma_i$ be any $m$-cycle for all $i$. More generally, if there exists an element $\tau$ of order $m$ in $S_n$, then let $\sigma_i=\tau$ for all $i$. – Jared May 17 '13 at 00:56
  • Oh great point yeah. I'll edit. – levitopher May 17 '13 at 00:58
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    I don't understand: for instance if $m=3$ and $a,b,c$ are a solution, that is they are distinct and their product has $abc=1$, then we also have $abc=bca=cab=1$. – Olivier Bégassat May 17 '13 at 02:30
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    If we restrict to just transpositions, then this is something that has been researched to some extend, inspired by episodes of Futurama and Stargate. See for example http://arxiv.org/abs/1204.6086 or http://arxiv.org/abs/1209.4991 – Tobias Kildetoft May 17 '13 at 15:04

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