My Math Forum Find the order of the cyclic subgroup of D2n generated by r

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 February 3rd, 2011, 05:54 PM #1 Newbie   Joined: Oct 2010 Posts: 25 Thanks: 0 Find the order of the cyclic subgroup of D2n generated by r Find the order of the cyclic subgroup of D2n generated by r. The order of an element r is the smallest positive integer n such that r^n = 1. Here is the representation of Dihedral group D2n = The elements that are in D2n = {1, r, r^2, ... , r^n-1, s, sr, sr^2, ... , s(r^n-1)} Dihedral group is non-abelian (cannot commute), but cyclic group is abelian (can commute) Okay it basically ask us to use the generator r from dihedral group to form a subgroup of D2n that is cyclic group. So obviously we can't choose the term that has s(r^i) for i=1, ... , n-1 since it's not power of r. that leaves us the set of choices {1, r, r^2, ... , r^n-1}, identity 1 has to be there and it commutes with all the elements in the group. Since the order of D2n=2n, now the subgroup has half of its entries, and the property of D2n such that r^n=1, therefore the order of r is n. Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

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### normal subgroup of d2n

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