In: Math
let G = D2n =
{e,r,r^2,...,rn-1,s,sr,sr2,..,srn-1}
a diedergroup of order 2n, where n >=3
(a) prove that [G,G] = <r2>
(b) prove that G/[G,G] consists of two elements if n is uneven and
4 elements if n is even
The proof for (a) is provided as follows: First we show that every element of the cyclic group <r2> is a commutator.....(1) and later we show that every commutator of G is in the group <r2>...(2), so that (1) and (2) completes the proof.
So let us start with the proof of the fact (1)..The commutator [r,s] is defined as rsr-1s-1 which equals rrss-1. [REASON- In any group (ab)-1=b-1a-1]. That implies [r,s]=r2. So r2 is indeed a commutator. More generally the above logic can be extended as [rm,s]=rmsr-ms-1=rmrmss-1=r2m. Thus elements of the type r2m are commutators which establishes fact 1.
For the other way argument (to establish fact (2) we shall consider elements of the type [a,b] where 'a' and 'b' are either rotations or reflections. [REASON-Dihedral groups are made of rotations and reflections]
If both a and b are rotations then it is a trivial fact that the commutator[a,b] is in <r2>.
If a is a rotation and b is a reflection then a=rm, b=rls for some integral powers m and l. Using the fact that b-1=b we compute the power aba-1b-1=aba-1b=rmrlsr-mrls=r2m ....[By proper rearrangement]
Similarly if a is a reflection, b is a rotation
(aba-1b-1)-1= bab-1a-1. and by the previously made observation this element is also of the type r2m. and so is (aba-1b-1).
Finally if we set both a and b as reflections we get again (aba-1b-1) simplified of the type r2k.
So all the above cases show that every commutator is an element of <r2>.
PROBLEM (b): We are trying to understand the quotient G/[G,G]. If n is uneven, i.e 'n' is odd then by part (a) we see that [G,G] consists of every power 'r' and hence the quotient consists exactly of 2 elements ...It is in fact always abelian. If n is even then the quotient G/[G,G] will be the klein 4 group and hence consists of 4 4 elements.