Let α=(1 7 3)(5 4 2 9)andβ=(2 3)(7 4)(5 1 8) be elements of S10. Then α2β−1 is: a) (1 2 7 9 4)(3 8 5) b) (1 2 7 4 9)(3 5 8) c) (1 2 7 9 4)(3 5 8) d) (1 2 9 7 4)(3 5 8)

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Let α=(1 7 3)(5 4 2 9)andβ=(2 3)(7 4)(5 1 8) be elements of S10. Then α2β−1 is: a) (1 2 7 9 4)(3 8 5) b) (1 2 7 4 9)(3 5 8) c) (1 2 7 9 4)(3 5 8) d) (1 2 9 7 4)(3 5 8)

mauricio0815sh · Accepted Answer

Step 1  α=[123456789107972473951]  β=[1234567891083271745101]  [123456789107912473951]  α2⇒[123456789103579231547]  β−1⇒[1234567891083271745101]  β−1=[123456789105234847118]  α2β−1⇒[123456789103579231547]  ×[123456789105234847118]  ⇒[1234579827813945]  α2β−1⇒(1 2 7 9 4)(3 8 5)

limacarp4 · Accepted Answer

Step 1  α=(1 7 3)(5 4 2 9)   (α)2=(1 7 3)(5 4 2 9)×(1 7 3)(5 4 2 9)   that is  (1 3 7)(5 2)(4 9)  and beeta inverse will be  (3 2)(4 7)(8 1 5)   that is  (α)2×(β)−1=(1 3 7)(5 2)(4 9)×(3 2)(4 7)(8 1 5)   =(1 2 7 9 4)(3 5 8)  so here option (b) is true

alenahelenash · Accepted Answer

Step 1 We have α=(1 7 3)(5 4 9 2) β=(2 3)(7 4)(5 1 8) Step 2 α2=(1 7 3)(5 4 9 2)(1 7 3)(5 4 9 2) =(1 3 7)(2 4)(5 9) β=(2 3)(7 4)(5 1 8) Inwerse of β can be calculated by rewersing elements, β−1=(8 1 5)(4 7)(3 2) So, α2β−1=(1 3 7)(2 4)(5 9)(8 1 5)(4|7)(3 2) =(1 9 5 8 3 4)(2 7) Noneofthegivenoptionsiscorrect.

Let \alpha=(1\ 7\ 3)(5\ 4\ 2\ 9) and \beta=(2\ 3)(7\

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