Question
Prove that every non-trivial subgroup of a cyclic group has finite index. Hence prove that is not cyclic.
Answer :
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Let (G = \langle g \rangle) be a cyclic group, and let (H) be a non-trivial subgroup of (G). Since (G) is cyclic, every element of (G) can be written as (g^n) for some integer (n). Because (H) is non-trivial, there exists a smallest positive integer (m) such that (g^m \in H). Then (H = \langle g^m \rangle), which is also cyclic. To find the index ([G : H]), consider the cosets of (H) in (G):
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