When is D_{n} abelian? What's wrong with the proof?

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In summary, the conversation discussed the commutativity of the dihedral group Dn and how it is only commutative when n is less than or equal to 2. This is because two non-equal cycles of length 2 only commute if they are disjoint, and this is not the case when the length is larger than 2. It was also suggested that using reflections or two-cycles as an argument would make the proof simpler.
  • #1
AdamsJoK
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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.
 
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  • #2
AdamsJoK said:
... two non equal cycles of length 2 commute if and only if they are disjoint ...
So far so good.
... and also that this isn't the case when the length is larger than 2 ...
What do you mean by this? Why shouldn't, e.g. ##(1,2,3)(4,5,6,7)(8,9,10,11,12)## don't be commutative?

The dihedral groups contain rotations and reflections. Wouldn't it be a lot easier to use the reflections as an argument, or if you will two-cycles?
 
  • #3
##e,\ (ab),\ (cd)## and ##(ab)(cd)## gives an Abelian group that contains two-cycles from more than two elements and that are disjoint.
 

1. What does it mean for Dn to be abelian?

Dn refers to the dihedral group of order n, which is a group of symmetries of a regular n-gon. A group is considered abelian if its elements commute, meaning that the order in which the operations are performed does not change the result. In other words, for any two elements a and b in an abelian group, a*b = b*a.

2. How do we determine if Dn is abelian?

To determine if Dn is abelian, we can use the fact that the group is generated by two elements, r and s, where r represents a rotation and s represents a reflection. If rs = sr for all elements in the group, then Dn is abelian. This can be shown by considering the possible combinations of r and s and their resulting products.

3. What is the proof for Dn being abelian?

The proof for Dn being abelian involves showing that rs = sr for all elements in the group. This can be done by considering the possible combinations of r and s and their resulting products, and then showing that they are all equal to each other.

4. What is the most common mistake made in the proof for Dn being abelian?

The most common mistake made in the proof for Dn being abelian is assuming that rs = sr for all elements without considering the order of the elements. This is a common misconception as the order of operations can change the result in non-abelian groups.

5. How can we fix the mistake in the proof for Dn being abelian?

To fix the mistake in the proof for Dn being abelian, we need to carefully consider the order of the elements and their resulting products. This includes considering cases where the order of operations may change the result, and ensuring that rs = sr holds for all elements in the group.

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