1. cycle decomposition of permutations

How can I show that if alpha is any permutation then:
alpha (i1,i2,....ir) alpha -1 = (alpha(i1) alpha(i2)....alpha(ir))
Thank for any help.

2. Originally Posted by thomas_donald
How can I show that if alpha is any permutation then:
alpha (i1,i2,....ir) alpha -1 = (alpha(i1) alpha(i2)....alpha(ir))
Thank for any help.
Let $\sigma = (\alpha(i_1),\alpha(i_2),...,\alpha(i_r))$ and $\tau = \alpha (i_1,i_2,...,i_r) \alpha^{-1}$.
To show that $\sigma = \tau$ you need to show that $\sigma(n) = \tau(n)$ for all $n$.

If $n = \alpha (i_j)$ for $j=1,2,...,r$ then $\sigma(n) = \alpha (i_{j(\bmod r)+1})$ and $\tau (n)$ is the same thing.

If $n\not \in \{ \alpha (i_j)|j=1,2,...,r\}$ then $\sigma(n) = n$ while $\tau(n)=n$, the same value.

This is Mine 12,6th Post!!!

3. Thanks You! Thank You! Thank You!

4. Maybe I am asking a basic question but what is the difference between: and ?
This is a HM problem and I am not clear what the question is

5. Originally Posted by thomas_donald
Maybe I am asking a basic question but what is the difference between: and ?
This is a HM problem and I am not clear what the question is
What do you mean the difference? We proved they are the same.

6. I understand why but I am not clear why is the same thing.Could you show it in more details? Or, is it possible to get an example?

7. Originally Posted by thomas_donald
I understand why but I am not clear why is the same thing.Could you show it in more details? Or, is it possible to get an example?
If $n=\alpha(i_j)$ then $\tau (n) = \alpha (i_1,...,i_r) \alpha^{-1} (\alpha(i_j)) = \alpha(i_1,...,i_r)(i_j) = \alpha (i_{(j\bmod r)+1})$.

8. Thanks ! Now I get it slowly and surely.