Let G be a group and a, b *ϵ G* with O(a) = 5, a^{3}b = ba^{3} .Prove that G is an abelian group.

Proof:

Given a^{3}b = ba^{3}

Multiply left side by a^{2},

a^{2}.a^{3}.b = a^{2}.b.a^{3}

a^{5}.b = a^{2}.b.a^{3}

1.b = a^{2}.b.a^{3} [since O(a) = 5 i.e., a^{5} = 1]

Multiply right side by a^{2},

ba^{2} = a^{2}ba^{3}a^{2}

ba^{2} = a^{2}ba^{5 }[since O(a) = 5 i.e., a^{5} = 1]

ba^{2} = a^{2}b

Multiply right side by a

ba^{3} = a^{2}ba

a^{3}b = a^{2}ba [since a^{3}b = ba^{3}]

a^{2}.a.b = a^{2}b.a

ab = ba [by left cancellation law]

Therefore, G is abelian group.

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[…] Let G be a group and a, b ϵ G with O(a) = 5, a^3b = ba^3 .Prove that G is an abelian group. […]

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