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xsw001

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Registered: 2010-10-23

Posts: 7

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 = <r, s|r^n=s^2=1, rs=s^-1>
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?

Last edited by xsw001 (2011-02-03 16:27:31)

Alg Num Theory

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Registered: 2017-11-24

Posts: 693

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Re: Find the order of the cyclic subgroup of D2n generated by r

Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

Yes.

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Me, or the ugly man, whatever (3,3,6)