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Home / Modern Abstract Algebra / Every subgroup of a cyclic group is cyclic

Every subgroup of a cyclic group is cyclic

Proof: Let G = (a) be a cyclic group generated by a. Its trivial subgroup {e} containing the identity in of G, is cyclic. Let H be any nontrivial subgroup G. Every element of H is some integral power of a, i.e., ak where k is an integer. Let m be the smallest positive integer such that am ϵ H.

Since H is a subgroup, it contains a-m, the inverse of am. Let al be any element of H.

Then by divisor algorithm, ℓ = mk + r, where 0≤r<m and k are integers. We have

r = ℓ – mk, and so ar = al-mk. But al – mk =al a-mk = al (a-m)k ∊H.

[Since al, a-mk ∊H ⟹ al ,(a-m)k ∊H ⟹ al (a-m)k ∊H by closure property in H]

Thus al-mk = ar ∊H where 0≤r<m.

Since m is the smallest positive integer such that am ∊H. r must be equal to zero.

i.e., l-mk = 0 ⟹ l= mk.

Thus al = amk = (am)k.

We find every element in H as some integral power of am. Hence H is a cyclic subgroup generated by am. (Proved)

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