Let $n>3$ be a positive integer. We denote the symmetric group of $n$ elements by $S\_n$ and the identity mapping by $\mathrm{id}$. 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:


$$f\_k:(1,\ldots,k,k+1,k+2,\ldots,n-1,n)\mapsto (1,\ldots,k,n,n-1,\ldots,k+2,k+1).$$


Then I conjecture that if $i\_0,i\_1,\ldots,i\_l\in \{1,2,\ldots,n-2\}$ satisfy:


(1) $l>1$;


(2) $i\_0=1$;


(3) for any $1\leq u\leq v\leq l$ such that $\{u,u+1,...,v\}$ is a proper subset of $\{0,1,...,l\}$, one has $f\_{i\_v}\circ ...\circ f\_{i\_{u+1}}\circ f\_{i\_u}\neq \mathrm{id}$;


(4) $f\_{i\_l}\circ \ldots\circ f\_{i\_1} \circ f\_{i\_0}=\mathrm{id}$,


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.



A conjecture about a specific subset of the symmetric group $S_n$
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 $\{2,3,n \}$.


