Abstract Algebra 31: How do you write a product of permutations in disjoint cycle notation?

[@wallflower6942]

Thank you!

[@HenryAdamsMath]

You're welcome!

[@japindersingh3848]

Wonderful

[@HenryAdamsMath]

Thank you! Cheers!

[@sayanjitb]

Can you provide the proof this theorem?

[@HenryAdamsMath]

Unfortunately this video is too short for a proof. But most textbooks in abstract algebra will contain a proof that any permutation can be written as a product of disjoint cycles! The main idea is that the process for doing so is not so different than the process for multiplying permutations together...

[@miftahulanambinahmadyani4647]

how about (1 2)(1 2)(3 4)?

[@HenryAdamsMath]

Great question! The answer is (1 2)(1 2)(3 4) = (3 4), which you could also write as (1)(2)(3 4) if you wanted to, but it's better to write this as simply (3 4). Does that make sense? Note that 1 -> 2 -> 1, that 2 -> 1 -> 2, that 3-> 4, and that 4 -> 3, leaving us with the permutation (3 4). Another way to explain this would be to note that doing (1 2)(1 2) is the same as doing the identity, and so we are left with (3 4).