Suppose the element a,b in a group satisfies aba-1=b2 for b<>e

Question

Comments

https://gateoverflow.in/250917/groups

Answer 1

(aba^-1)=b² 

(aba^-1)²=b⁴

(aba^-1)(aba^-1)=b⁴

ab(aa^-1)ba^-1=b⁴

ab²a^-1=b⁴

a(aba^-1)a^-1=b⁴

a²ba^-2=b⁴

 

(a²ba^-2)²=b⁸

(a²ba^-2)(a²ba^-2)=b⁸

a²b²a^-2=b⁸

a²(aba^-1)a^-2=b⁸

a³ba^-3=b⁸

 

(a³ba^-3)(a³ba^-3)=b¹⁶

a³b²a^-3=b¹⁶

a³(aba^-1)a^-3=b¹⁶

a⁴ba^-4=b¹⁶

 

(a⁴ba^-4)(a⁴ba^-4)=b³²

a⁴b²a^-4=b³²

a⁴(aba^-1)a^-4=b³²

a⁵ba^-5=b³²

Answer 2

b³² = (((((b²)²)²)²)² = ((((aba^-1)²)²)²)² = ............ //do recursively 

//(aba^-1)​² = aba^-1aba^-1 = ab²a^-1 = a(aba^-1)a^-1= a²ba^-2

= a⁵ba^-5

Comments

Hi , but , as a and b are in group

so , aa^-1 will give identify element , right ?

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yes..  & I*A = A*I =  A

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but , I can not understand afterwards.

like how (aba^-1)² is evaluated ? 

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(aba^-1)​² = aba^-1aba^-1 = ab²a^-1 = a(aba^-1)a^-1= a²ba^-2