Pinter’s Book of Abstract Algebra Chapter 11, Exercise D5

Let G be a group and let a, b \in G. Prove the following:
Let \operatorname{ord}(a)=n, and suppose a has a kth root, say a=b^k. Then \langle a\rangle=\langle b\rangle iff k and n are relatively prime.

Proof. Consider the cyclic subgroup of \mathbb{Z}_{10}. For a = 2 and b = 1, we have a = b^2, n = 5, and \gcd(2, 5) = 1. However, \langle 1 \rangle = \mathbb{Z}_{10} but \langle 2 \rangle consists of all even numbers of \mathbb{Z}_{10}. Hence \langle a \rangle \neq \langle b \rangle. A contradiction. \qed

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