- #1
AdamsJoK
- 4
- 0
I agree that this could have been done more simply(i'm not looking for an alternative proof), but I don't understand how it is wrong, any insight?
Since Dn is an dihedral group, we know its elements are symmetries, Dn = (R1,R2,R3...Ri) and since R is a symmetry, we know it's a permutation, so, each Ri can be written as a product of disjoint cycles (w1,w2...wk), now since each element in w represents the vertices's of Dn, which has n vertices's, it follows that w has a total length of n, therefore by problem 4,which states, two non equal cycles of length 2 commute if and only if they are disjoint and also that this isn't the case when the length is larger than 2, we know that Dn is commutative when n <= 2 and therefore abelian when n <= 2.
I'm not sure if this site has latex, thanks.
Since Dn is an dihedral group, we know its elements are symmetries, Dn = (R1,R2,R3...Ri) and since R is a symmetry, we know it's a permutation, so, each Ri can be written as a product of disjoint cycles (w1,w2...wk), now since each element in w represents the vertices's of Dn, which has n vertices's, it follows that w has a total length of n, therefore by problem 4,which states, two non equal cycles of length 2 commute if and only if they are disjoint and also that this isn't the case when the length is larger than 2, we know that Dn is commutative when n <= 2 and therefore abelian when n <= 2.
I'm not sure if this site has latex, thanks.