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Given $\alpha = (1, 2, 3, ... , n)$ and $\tau = (1, 2)$

Show that $\alpha^k\tau\alpha^{-k}=(k+1, k+2)$

I'm having trouble figuring out what $\alpha^k$ is. I know that I can split $\alpha$ up into transpositions but I think overall I'm just stuck and need a hint. Is there any exponentiation rule that I may be missing?

Ook
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  • $\alpha^k$ is the permutation that maps $i$ to $i +k$ if $i +k \le n$ and to $i + k - n$ otherwise (think of adjusting a digital clock by pressing the $+$ or $\uparrow$ button $k$ times). – Rob Arthan Nov 03 '21 at 23:44
  • See here https://math.stackexchange.com/q/2526127/437127 – Trevor Gunn Nov 03 '21 at 23:44
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    If you want to understand intuitively what $\alpha^k$ does, consider what the number $1, 2, \ldots, n$ as modulo $n$ classes, and what $\alpha$ does to one of these classes. Then, think about what happens when you do it $k$ times. But, if you want a nice formal way of doing this, I would recommend induction. You just need to compute $\alpha^{k+1}\tau\alpha^{-(k+1)} = \alpha(\alpha^k \tau \alpha^{-k})\alpha^{-1}$. – Theo Bendit Nov 03 '21 at 23:45

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