Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

$(1)l>1$;

$(2)i_0=1$;

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

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

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.

share|improve this question
2  
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
1  
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

1 Answer 1

up vote 3 down vote accepted
+50

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 $\{2,3,n \}$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.