Every Subgroup of a Cyclic Group is Cyclic | Abstract Algebra

We prove that all subgroups of cyclic groups are themselves cyclic. We will need Euclid's division algorithm/Euclid's division lemma for this proof. We take an arbitrary subgroup H from our Cyclic group G, then we take an arbitrary element a^t from H. Certainly, all powers of a^t are in H, since H is closed. Then, it only remains to prove that all elements of H are in fact powers of a^t. #AbstractAlgebra

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awesome, well explained.
What if you started with G = <e>, which implies that G = {e} since e^n = e for any integer n. Then it is trivial to show that any subgroup of G is cyclic, since the only subgroup of G is {e}, and {e} is generated by e. So we have the subgroups of G in the special case that G = {e} are cyclic.
So we can assume G = <a> where a ≠ e, and start the proof from there, and then it makes sense a^m is the smallest positive power of a in H.

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So in Euclid's Division Lemma, t can be negative, it's just that m has to be positive?

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