Answer to Question #288540 in Abstract Algebra for Siva

Question

Find aba^-1 where (I) a= 5,7,9 b=1,2,3 (ii) a= (1,2,5)(3,4) b=(1,4,5)

Answer 1

Let us find $aba^{-1}$ where

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

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

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

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

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

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