Answer to Question #123771 in Abstract Algebra for Nii Laryea

Question

Let G be a group. Suppose that there are two elements a, b ϵ G with b ≠ ℮ satisfying
aba⁻¹ = b²,
where ℮ is the identity of the group. Prove that
a⁴ba⁻⁴ = bⁱ⁶

Answer 1

Q#123771

G is given to be group and a,b element of G with b is NOT an identity

Now given to us aba^-1=b² .....(1)

Now squaring both side of equation (1),we get

(aba^-1)²=b⁴

So, (aba^-1)(aba^-1)=b⁴ (because aa^-1=e)

$\implies$ (ab²a^-1)=b⁴

Put the value of b² into above equation ,we get

a(aba^-1)a^-1=b⁴ $\implies$ a²ba^-2=b⁴ .....(2)

Squaring both side of equation (2), we get

(a²ba^-2)(a²ba^-2)=b⁸$\implies$ (a²b²a^-2)=b⁸

Put the value of b² into above equation,we get

a²(aba^-1)a^-2=b⁸$\implies$a³ba^-3=b⁸ ...(3)

Squaring both side of equation (3), we get

(a³ba^-3)(a³ba^-3)=b¹⁶$\implies$ a³b²a^-3=b⁸

put the value of b² into above equation,we get

a³(aba^-1)a^-3=b¹⁶$\implies$ a⁴ba^-4 =b¹⁶

Hence proved