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 $\displaystyle \sigma = (\alpha(i_1),\alpha(i_2),...,\alpha(i_r))$ and $\displaystyle \tau = \alpha (i_1,i_2,...,i_r) \alpha^{-1}$.
To show that $\displaystyle \sigma = \tau$ you need to show that $\displaystyle \sigma(n) = \tau(n)$ for all $\displaystyle n$.
If $\displaystyle n = \alpha (i_j)$ for $\displaystyle j=1,2,...,r$ then $\displaystyle \sigma(n) = \alpha (i_{j(\bmod r)+1})$ and $\displaystyle \tau (n)$ is the same thing.
If $\displaystyle n\not \in \{ \alpha (i_j)|j=1,2,...,r\}$ then $\displaystyle \sigma(n) = n$ while $\displaystyle \tau(n)=n$, the same value.
This is Mine 12,6th Post!!!