**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., a^{k} where k is an integer. Let m be the smallest positive integer such that a^{m }ϵ H.

### Since H is a subgroup, it contains a^{-m}, the inverse of a^{m}. Let a_{l} 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 a^{r} = a^{l-mk}. But a^{l – mk} =a^{l} a^{-mk} = a^{l} (a^{-m})^{k} ∊H.

[Since a^{l}, a^{-mk}∊H ⟹ a^{l},(a^{-m})^{k}∊H ⟹ a^{l}(a^{-m})^{k}∊H by closure property in H]

Thus a^{l-mk} = a^{r} ∊H where 0≤r<m.

Since m is the smallest positive integer such that a^{m} ∊H. r must be equal to zero.

i.e., l-mk = 0 ⟹ l= mk.

Thus a^{l} = a^{mk} = (a^{m})^{k}.

*We find every element in H as some integral power of a ^{m}. Hence H is a cyclic subgroup generated by a^{m}. (Proved)*