# Isom. of group Dn abelian

1. May 18, 2012

### TDA120

1. The problem statement, all variables and given/known data

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. Relevant equations

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

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

of $\pi: G \to G_{ab}$ with fab.

Corollary: Every homomorphism f: Sn->A to an abelian group A is the composition of $S_n \to \{\pm 1\} \overset{h}{\to} A$ 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.

Last edited: May 18, 2012
2. May 18, 2012

### I like Serena

Hey TDA120!

Shall we try to find the commutators of Dn?
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 [ρklσ]...
And any other possible commutators (which are those?).

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

3. May 18, 2012

### TDA120

Thanks! I think this meant I could make some new steps.!?
ρkρlσρ-klσ)-1=
ρkρlσρ-kσρ-l=
ρk+lσσρk-l=
ρ2k

So, every commutator will be either id or ρ2k
As 2k is a multiple of ρ2, [Dn,Dn] is generated by ρ2.
So [Dn,Dn] = $\left\langle$ ρ2k $\right\rangle$ = {id, ρ2, ρ4, …, ρn-2} if n is even and

[Dn,Dn] = $\left\langle$ ρ2k $\right\rangle$ = Dn+ if n is odd.

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

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

Last edited: May 18, 2012
4. May 18, 2012

### I like Serena

Yep!

Btw, are you allowed to use that ρkσ=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?

Last edited: May 18, 2012
5. May 19, 2012

### TDA120

Hi I like Serena!,

Thanks again!

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

Is this enough proof for ρkσ = σρ-k
ρkσ *ρkσ = id as ρkσ is a reflection
Then ρkσ *ρk = σ
And ρkσ = σρ-k

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

Does this hold?

6. May 19, 2012

### 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: $< \rho^{2k} >$.

Last edited: May 19, 2012
7. May 19, 2012

### TDA120

$\langle$Right. I wasn’t thinking the right way; thanks!$\rangle$

Last edited: May 19, 2012
8. May 19, 2012