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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 ?

Thanks

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 $(12)(34)(12)=(34)(12)(12)=(34)(12)^2=(34)(1)=(34)$.

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  • $\begingroup$ Would the downvoter care to share why downvote please ? $\endgroup$ – Anjan3 Sep 9 '15 at 3:27
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    $\begingroup$ What do you mean by "$\sigma,\tau$ share no common entry"? $\endgroup$ – Quang Hoang Sep 9 '15 at 3:37
  • $\begingroup$ Fine. I will explain in the post $\endgroup$ – Anjan3 Sep 9 '15 at 3:38
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The answer is no. Consider $\sigma=\tau$ and has order $2$.

More generally, get $\sigma=\tau^n$ where $\tau^{2n}=1$, e.g. $\tau=(1\,2\,3\,4\,5\,6)$ and $\sigma=\tau^3=(1\,4)(2\,5)(3\,6)$.

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  • $\begingroup$ I think he meant two distinct ones $\endgroup$ – Learnmore Sep 9 '15 at 3:53
  • $\begingroup$ yes @learnmore you are right. If $\sigma=\tau$ both are of order 2 then $\sigma\tau\sigma=\tau$ implies $\sigma^2=1$ which is true. No contradiction is arriving here, Am i making any mistake ? $\endgroup$ – Anjan3 Sep 9 '15 at 3:54
  • $\begingroup$ No, you are not. However, it's not that hard to modify the counter-example (see edit). $\endgroup$ – Quang Hoang Sep 9 '15 at 3:55
  • $\begingroup$ it will be more clear to @Anjan3 if you could give him some specific example ;i think he needs that $\endgroup$ – Learnmore Sep 9 '15 at 3:58
  • $\begingroup$ @QuangHoang I am truely confused. Would you mind to share more explicitly please ? $\endgroup$ – Anjan3 Sep 9 '15 at 3:59
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If two permutation is disjoint, $\sigma\tau\sigma=\tau$.

In other cases, check the following. $\sigma = (a~b)$ and $\tau=(b~c)$

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  • $\begingroup$ $(ab)(bc)(ab)=(ac)\neq (ab)$ so I think your example is not a proper counter example. Please correct me if I am wrong $\endgroup$ – Anjan3 Sep 9 '15 at 4:01

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