The commutator subgroup of Dn: Is it generated by ρ2?

[TDA120]

1. Homework Statement

My challenge is as follows:
Let Dn be the dihedral group (symmetries of the regular n-polygon) of order 2n and let ρ be a rotation of Dn with order n.

(a) Proof that the commutator subgroup [Dn,Dn] is generated by ρ2.

(b) Deduce that the abelian made Dn,ab is isomorphic with {±1} in case n is odd, and with V4 (the Klein four-group) in case n is even.


2. Homework Equations

The Fundamental theorem on homomorphisms
Fundamental theorem on homomorphisms - Wikipedia, the free encyclopedia


Proposition: Let [itex]f:[/itex] G [itex]\rightarrow[/itex] [itex]A[/itex] be a homomorphism to an abelian group A.
Then there exists a homomorphism [itex]f_{ab}: G_{ab}=G/[G,G] \to A[/itex] so that f can be created as a composition
[itex]G \overset{\pi}{\to} G_{ab} \overset{f_{ab}}{\to}A[/itex]

of [itex]\pi: G \to G_{ab}[/itex] with fab.


Corollary: Every homomorphism f: Sn->A to an abelian group A is the composition of [itex]S_n \to \{\pm 1\} \overset{h}{\to} A[/itex] of the sign function with a homomorphism h: {±1} -> A


3. The Attempt at a Solution

I have worked out [Dn, Dn] for n=3,4,5 and 6 and have noticed the above described pattern. I just cannot proof it.

 

[I like Serena]

Hey TDA120!

Shall we try to find the commutators of D_n?
Let's start with [ρ,σ], where ρ is a rotation over an angle α=2π/n and σ is a reflection in the x-axis.
Do you know how to find that?
Perhaps with the use of complex numbered functions?
Did you perhaps already do an exercise where you combined a rotation with a reflection, and/or a reflection with another reflection?

Once you have that, we can try to generalize to [ρ^k,ρ^lσ]...
And any other possible commutators (which are those?).

Btw, did you know that ρ^kσ=σρ^-k?
That might make it easier to calculate the commutator.

 

[TDA120]

Thanks! I think this meant I could make some new steps.!?
ρ^kρ^lσρ^-k(ρ^lσ)^-1=
ρ^kρ^lσρ^-kσρ^-l=
ρ^k+lσσρ^k-l=
ρ^2k

So, every commutator will be either id or ρ^2k
As 2k is a multiple of ρ², [D_n,D_n] is generated by ρ².
So [D_n,D_n] = [itex]\left\langle[/itex] ρ^2k [itex]\right\rangle[/itex] = {id, ρ², ρ⁴, …, ρ^n-2} if n is even and

[D_n,D_n] = [itex]\left\langle[/itex] ρ^2k [itex]\right\rangle[/itex] = D_n⁺ if n is odd.

Then the order of [D_n,D_n] is 2n/n = 2 if n is odd.
And the order of [D_n,D_n] is n/2 if n is even.
D_n/[D_n,D_n] = 2n/(n/2) = 4

Each group of order 4 is isomorphic with the Cyclic group C₄ or V₄.

 

[I like Serena]

Yep!

Btw, are you allowed to use that ρ^kσ=σρ^-k?
Or do you need to prove that?

And did you consider other commutators?
Such as a reflection with a reflection?

Are you clear on which of C4 or V4 you're isomorphic with?

 

[TDA120]

Hi I like Serena!,

Thanks again!

Yes I three-double checked with all sorts of commutators.

Is this enough proof for ρ^kσ = σρ^-k
ρ^kσ *ρ^kσ = id as ρ^kσ is a reflection
Then ρ^kσ *ρ^k = σ
And ρ^kσ = σρ^-k

If n is even, D_n /[D_n,D_n] consists of four elements, id and three others: ρ, σ and ρσ as D_n has been divided by all even powers of ρ. This group has two generators, just as V₄.

Does this hold?

 

[I like Serena]

Sounds good! ;)

I only think it's not trivial that the group requires 2 generators and not just 1.
How do you know that it requires at least 2 generators?

Oh and those generator angles should be written as just \langle and \rangle, so you get: ##\langle \rho^{2k} \rangle##.
You can also use < and >, although I usually feel they look less nice, getting: ##< \rho^{2k} >##.

 

[TDA120]

[itex]\langle[/itex]Right. I wasn’t thinking the right way; thanks![itex]\rangle[/itex]

 

[I like Serena]