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# A discrete mathemathics problem

inuyasha
How to prove subgroups of cyclic groups are cyclic? I searched it on the Internet and found out a common way is like this:
According to Lagrange Theorem, the vth subgroup of a pth group has v = p\t. So the subgroup is (a ^ t).
But what if the n of a nth cyclic group is infinite?
xtraco
 Quote: The following is a proof that all subgroups of a cyclic group are cyclic. Proof. Let G be a cyclic group and HG . If G is trivial, then H=G , and H is cyclic. If H is the trivial subgroup, then H=eG=eG , and H is cyclic. Thus, for the remainder of the proof, it will be assumed that both G and H are nontrivial. Let g be a generator of G . Let n be the smallest positive integer such that gnH . Claim: H=gn Let agn . Then there exists z with a=(gn)z . Since gnH , we have that (gn)zH . Thus, aH . Hence, gnH . Let hH . Then hG . Let x with h=gx . By the division algorithm, there exist qr with 0rn such that x=qn+r . Thus, h=gx=gqn+r=gqngr=(gn)qgr . Therefore, gr=h(gn)−q . Recall that hgnH . Hence, grH . By choice of n , r cannot be positive. Thus, r=0 . Therefore, h=(gn)qg0=(gn)qeG=(gn)qgn . Hence, Hgn . This proves the claim. It follows that every subgroup of G is cyclic

Very nice but you didn't thank PlanetMath from where you copied it. That is a nono. Please use quote tags (I have put them in on this posting), say where the material came from, thus:
Source : http://planetmath.org/ProofThatAllSubgroupsOfACyclicGroupAreCyclic.html

And add something of your own. Otherwise don't post cut-paste. It is technically theft and non-technically just yucky.
Bikerman