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Let $n>3$ be a positive integer.We denote the symmetric group of $n$ elements by $S_n$ and the identity mapping by $id$. For every $f\in S_n$, $f(1,2,\ldots,n)=(a_1,a_2,\ldots,a_n)$, denote $a_n$ by $m(f)$.

For any positive integer $1\leq k\leq n-2$, define $f_k\in S_n$ as follow:

$f_k:(1,\ldots,k,k+1,k+2,\ldots,n-1,n)\to (1,\ldots,k,n,n-1,\ldots,k+2,k+1)$.

Then I conjecture that if $i_0,i_1,\ldots,i_l\in \{1,2,\ldots,n-2\}$ satisfy:



$(3)$For any $1\leq u\leq v\leq l$ such that $\{u,u+1,...,v\}$ is a proper subset of $\{0,1,...,l\}$,$f_{i_v}\circ ...\circ f_{i_{u+1}}\circ f_{i_u}\neq id$;

$(4)f_{i_l}\circ \ldots\circ f_{i_1} \circ f_{i_0}=id$,

we must have $\{m(f_{i_0}),m(f_{i_1}\circ f_{i_0}),\ldots,m(f_{i_l}\circ \ldots\circ f_{i_1}\circ f_{i_0})\}=\{2,3,\ldots,n\}$.

Is it true? If not, please give a counterexample.

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Why are you looking at these particular permutations? What made you come up with this conjecture (how large examples and how many have you checked?) – Tobias Kildetoft Oct 31 '13 at 8:23
What is $l$?... – Boris Novikov Oct 31 '13 at 8:39
$l$ is some positive integer. – user40096 Oct 31 '13 at 9:09
If your conjecture is true then one must be $l+1\ge |\{2,\ldots,n\}|=n-1$. Yes? – Boris Novikov Oct 31 '13 at 10:57
Yes,I think $l$ should not be too small if the conditions hold. – user40096 Oct 31 '13 at 11:24
up vote 4 down vote accepted

The conjecture is true for $n=3$ or $4$, and false for $n>4$.

It is trivially true for $n=3$. For $n=4$, we can only use $f_1$ and $f_2$, so we simply check that $(f_2\circ f_1)^3 = id$, and that $m(f_1)=2$ and $m((f_2\circ f_1)^2)=3$.

Assume now that $n>4$. We have that $f_2\circ f_3 \circ f_2 \circ f_1$ is the transposition exchanging $2$ and $n$. Hence $(f_2\circ f_3 \circ f_2 \circ f_1)^2 = id$. We also see by direct computation that it satisfies condition (3) of the statement, and that the set $\{m(f_{i_0}), \ldots, m(f_{i_l}\circ \cdots \circ f_{i_0}) \}$ is just $\{2,3,n \}$.

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Thank you very much!Pierre-Guy Plamondon,I also noticed it and I found I am so careless that I forgot a very important condition.Please look at my modified question. – user40096 Oct 31 '13 at 23:55
The answer to the modified question seems to be the same. I have edited the proof. – Pierre-Guy Plamondon Nov 1 '13 at 0:52
Pierre-Guy Plamondon:Yes,you are right!Thank you very much! – user40096 Nov 1 '13 at 2:13

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