Respected All.

Today I post following problem regarding clearing my doubt. I have got the answer.

Now after going through the counter example provided by Quang Hoang, now I am willing to establish the following.

Let $\sigma, \tau\in S_n$ where $S_n$ be symmetric group of order $n!$. If $\sigma\tau\sigma=\tau$ holds then either $\sigma$ and $\tau$ are disjoint permutation or $\sigma=\tau^k$ for some positive integer $k$.

Well, for the first part, I am able to do that. For the second part, this is what I have done. Of course credit goes to Quang Hoang first.

Let $\sigma=\tau^m$. Then $\sigma\tau\sigma=\tau$ gives $\tau^{2m+1}=\tau$ i.e. $\tau^{2m}=1$. hence $k=m$ and we are done.

My question/doubt: (1) Are there any other optional cases that are left ?

(2) Would you mind telling me what can other possibilities should be under consideration?

(3) Did I make any mistake above ?

Thanks in advance

  • $\begingroup$ What should be reason for downvote? Seems like the same downvoter who did the same thing in my previous post. Whats the issue ? $\endgroup$ – Anjan3 Sep 9 '15 at 5:02
  • $\begingroup$ Can someone please help me out ?? $\endgroup$ – Anjan3 Sep 9 '15 at 5:49

I don't think this hold, basically your equation is equivalent to :

$$\sigma=\tau\sigma^{-1}\tau^{-1} $$

That is $\tau$ conjugates $\sigma$ and its inverse. In some sense you say that your equation holds if $\sigma$ and $\tau$ commutes, we actually see that in this case $\sigma=\sigma^{-1}$. But we could construct examples where this does not hold. For instance take :

$$\sigma=(1,2,3)(4,5,6,7)(8,9) $$

We know that :

$$\sigma^{-1}=(1,3,2)(4,7,6,5)(8,9) $$

Furthermore if :

$$\tau=\begin{pmatrix}1&2&3&4&5&6&7&8&9\\1&3&2&4&7&6&5&8&9\end{pmatrix} $$

That is :


Then we see that :

$$\sigma=\tau\sigma^{-1}\tau^{-1} $$

Clearly $\tau$ is not disjoint from $\sigma$ nor a power of $\sigma$.

1)There are other cases left.

2) Find non-commutative examples.

3) I would like you to confirm that you are looking for solutions of the equation $\sigma\tau\sigma=\tau$ and not $\sigma\tau\sigma^{-1}=\tau$.

Edit : In general there is a formula to understand the conjugate of a cycle. That is if $c$ is a cycle and $\tau$ any permutation.

$$\text{ If } c=(c_1,...,c_n)\text{ then } \tau c\tau^{-1}=(\tau(c_1),...,\tau(c_n)) $$

Now if you have two explicit permutations $\sigma_1$ and $\sigma_2$ with the same decomposition into disjoint cycles, it is easy to use this to find a $\tau$ such that :

$$\sigma_1=\tau\sigma_2\tau^{-1} $$

  • $\begingroup$ Dear Sir. Thank you. Let me confirm it, yes I am indeed in search of solution to the equation $\sigma\tau\sigma=\tau$. Also your example truely amazing. would you mind to share with me how did you manage to find such example please? $\endgroup$ – Anjan3 Sep 9 '15 at 6:51
  • 1
    $\begingroup$ Let me edit my answer. $\endgroup$ – Clément Guérin Sep 9 '15 at 6:53

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