Please help me on the following. I got stuck.
We consider the symmetric group $S_n$ of order $n!$. Suppose that $\sigma, \tau$ be two permutation in it satisfying the condition $\sigma\tau\sigma=\tau$.
We are willing to prove/disprove that $\sigma, \tau$ share no common entry. Mean to say $\sigma$ and $\tau$ are totally disjoint.
Is it true ? I believe so. Although no counter example I have been able to found or to prove the statement.
What to do ?
P.S. By the phrase "$\sigma, \tau$ share no common entry" i meant to say if $\sigma, \tau$ be two permutation, they are formed by product of disjoint cycles. No matter whatever the cycles are, the complete expression $\sigma\tau$ is product of disjoint cycle.
For example, in $S_8$ if $\sigma=(12), \tau=(34)$ then note that they share no common entry and also satisfy $\sigma\tau\sigma=\tau$ viz $(12)(34)(12)=(34)$ because $(12)(34)(12)=(34)(12)(12)=(34)(12)^2=(34)(1)=(34)$.