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

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 HendersonAug 15 at 13:06

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

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