Prove that if a belongs to cyclic group G , then a(inverse) generates G... that is <a_inverse> = G
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a ∈ G. Let there be an identity element e in G.
Then a * e = a or a * a⁻¹ = a⁻¹ * a = e,
where a⁻¹ ∈ A and is called the inverse of a.
For every element a⁻¹ ∈ G, (a⁻¹)⁻¹ = a ∈ G since (a⁻¹) * (a⁻¹)⁻¹ = a⁻¹ * a = e
Thus if < a > = G , then < a⁻¹ > = G
Then a * e = a or a * a⁻¹ = a⁻¹ * a = e,
where a⁻¹ ∈ A and is called the inverse of a.
For every element a⁻¹ ∈ G, (a⁻¹)⁻¹ = a ∈ G since (a⁻¹) * (a⁻¹)⁻¹ = a⁻¹ * a = e
Thus if < a > = G , then < a⁻¹ > = G
kvnmurty:
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