Answer to Question #123771 in Abstract Algebra for Nii Laryea

Question #123771
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
1
Expert's answer
2020-06-26T16:40:24-0400

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



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