If permutation p=(148)(25)(396)(7) how to find p^123 ?


closed as off-topic by Hans Lundmark, José Carlos Santos, Shailesh, Ak19, Xander Henderson Aug 15 at 13:06

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  • $\begingroup$ OK, what have you tried? $\endgroup$ – Parcly Taxel Aug 15 at 12:14
  • $\begingroup$ Hint: Is the exponent divisible by 2? 3? Think about each cycle independently. $\endgroup$ – Don Thousand Aug 15 at 12:16


Your permutation is very conveniently written in disjoint cyclic form. A very nice property of permutations written as disjoint cycles is that raising the permutation to a power as a whole is equivalent to raising each cycle to that power individually (seen easily by the fact that disjoint cycles commute with one another and induction).

$p^{123} = (148)^{123}(25)^{123}(396)^{123}(7)^{123}$. Now, consider simplifying each of these cycles individually.


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