0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 :

$$\tau=(2,3)(5,7)$$

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 :

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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