Question #300236

Find aba -1


where (i) a = (5 , 7 , 9) , b = (1 , 2, 3) (ii) a = (1 , 2,5)(3 , 4) , b = (1 , 4 , 5) .

1
Expert's answer
2022-02-22T03:20:39-0500

Let us find aba1aba^{-1} where

(i) a=(5,7,9), b=(1,2,3)a=(5,7,9),\ b=(1,2,3)


It follows that a1=(5,9,7).a^{-1}=(5,9,7). Taking into account that the circles aa and bb are independent, and hence they commute, we conclude that

aba1=(5,7,9)(1,2,3)(5,9,7)=(1,2,3)(5,7,9)(5,9,7)=(1,2,3)=b.aba^{-1}=(5,7,9) (1,2,3)(5,9,7)=(1,2,3)(5,7,9)(5,9,7)=(1,2,3)=b.



(ii) a=(1,2,5)(3,4), b=(1,4,5)a=(1,2,5)(3,4),\ b =(1,4,5)


It follows that a1=(3,4)(1,5,2).a^{-1}=(3,4)(1,5,2). Therefore,

aba1=((1,2,5)(3,4))(1,4,5)((3,4)(1,5,2))=((134)(2,5))((3,4)(1,5,2))=(123).aba^{-1}=((1,2,5)(3,4))(1,4,5)((3,4)(1,5,2))=((134)(2,5))((3,4)(1,5,2)) =(123).



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