# Find the order of the cyclic subgroup of D2n generated by r

xsw001

## Homework Statement

Find the order of the cyclic subgroup of D2n generated by r.

## Homework Equations

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)

## The Attempt at a Solution

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?

## Answers and Replies

gamma5772
Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

Yes, the order is n.

Okay it basically ask us to use the generator r from dihedral group to form a subgroup of D2n that is cyclic group.

I'm not quite sure, but it's possible you are slightly mistaken. To be absolutely clear, you are asked to find not any cyclic group containing r, but to find the cyclic group generated by r. That is, $$<r> = \{ \ldots, r^{-2}, r^{-1}, 1=r^0, r^1, r^2, r^3, \ldots \}$$. In a finite group we may simplify slightly to $$<r> = \{1, r, r^2, r^3, \ldots \}$$ because inverses are automatically included in this set.

In this particular case, by the relation r^n=1, we find that <r> = {1, r, r^2, ... r^n-1}, the order of which is n.