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). Let (h \in H) be a non-identity element. Then (h = g^m) for some nonzero integer (m). The subgroup (H) is itself cyclic, generated by (g^m), so (H =
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