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**aman_cc** Prove any subgroup (H) of a cyclic group (G) is itself cyclic

Approach:

Let $\displaystyle G = (a)$

Assume H is non-trivial

There is a $\displaystyle a^k$ such that k is the smallest +ve power among all $\displaystyle a^i $ that are in H. Well order principle of natural numbers ensures there will always be a unique such k as long as H is non-trivial

Claim $\displaystyle H = (a^k)$

Let $\displaystyle a^i \in H$

$\displaystyle i = mk+r$ where $\displaystyle r<k$

obviously $\displaystyle a^r$$\displaystyle \in H$ (closure property of H)

Thus r can't be a +ve number (as k is the smallest). Thus r=0.

Hence $\displaystyle a^i = (a^k)^m$

And, $\displaystyle H = (a^k)$ (By definition)

Is this proof 'rigorous' enough?