Prove any subgroup (H) of a cyclic group (G) is itself cyclic
Approach:
Let
Assume H is non-trivial
There is a such that k is the smallest +ve power among all 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
Let
where
obviously (closure property of H)
Thus r can't be a +ve number (as k is the smallest). Thus r=0.
Hence
And, (By definition)
Is this proof 'rigorous' enough?