# Proof generation of [Dn,Dn]

• May 18th 2012, 04:49 AM
ILikeSerena
Proof generation of [Dn,Dn]
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 $f: G \to A$ 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.
Here Gab is the group that is made abelian from G.

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.
• May 18th 2012, 12:00 PM
TDA120
Re: Proof generation of [Dn,Dn]
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.

Quote:

Originally Posted by ILikeSerena
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 $f: G \to A$ 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.
Here Gab is the group that is made abelian from G.

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.

• May 18th 2012, 01:05 PM
ILikeSerena
Re: Proof generation of [Dn,Dn]
Aha! Thanks for explaining! ;)
• May 18th 2012, 09:26 PM
Deveno
Re: Proof generation of [Dn,Dn]
although TDA's post gets the "idea" right, it is not quite correct.

for example, we must also consider commutators of the following form:

kσ)(ρlσ)(ρkσ)-1lσ)-1

= (ρkσ)(ρlσ)(ρkσ)(ρlσ)

= (ρkσ)(σρ-l)(ρkσ)(σρ-l)

= (ρk-l)2 = (ρ2)k-l.

it should be clear that k-l takes on every possible value from 0 to n-1, for the possible values of k and l.

this shows that [Dn,Dn] contains every power of ρ2 (which, to be fair, we already knew).

however, we do NOT yet know that EVERY commutator is a power of ρ2.

for we also need to consider commutators of the form:

kσ)(ρl)(ρkσ)-1l)-1

= (ρkσ)(ρl)(ρkσ)(ρ-l)

= (ρk-lσ)(σρ-k-1) = ρ-2l = (ρ2)-l.

(commutators consisting solely of powers of ρ are clearly trivial, as <ρ> is abelian).

thus not only does [Dn,Dn] contain <ρ2>, but is also contained within <ρ2>.

if n is even, then ρ2 has order n/2, and we get <ρ2> is cyclic of order n/2.

if n is odd, then p2 has order n (since gcd(2,n) = 1), so that <ρ2> = <ρ>, is cyclic of order n.

it is the order of Dn/[Dn,Dn] that equals 2 or 4, not the commutator group.

TDA's post does not address "which" group we get in the case where n is even, which i do now:

consider the coset ρ<ρ2> in Dn/[Dn,Dn], this has order 2 since ρ2 is in <ρ2>.

next, consider the coset σ<ρ2>, which also has order 2, since σ2 = e.

since Dn/[Dn,Dn] has 2 elements of order 2, it is not cyclic.
• May 19th 2012, 08:53 AM
ILikeSerena
Re: Proof generation of [Dn,Dn]
Good points!

So we need to check all different combinations of commutators (rotation with rotation, rotation with reflection, reflection with rotation, and reflection with reflection).
A rotation with a rotation is commutative.
A reflection with a rotation turns out to be the inverse of the same rotation with the reflection.
And you just showed what a reflection with a reflection does.

And since the order of a non-identity element in Dn/[Dn,Dn] is 2, it follows that it is isomorphic with V4 and not with C4.
• May 19th 2012, 12:13 PM
TDA120
Re: Proof generation of [Dn,Dn]
Thanks for all your extra explanation & corrections!!!

But what if n=1?
How can Dn/[Dn,Dn] ever be isomorphic with {+-1} if the group Dn only consists of id?
• May 19th 2012, 12:36 PM
ILikeSerena
Re: Proof generation of [Dn,Dn]
From the course material I have, I've got in §3:

"In §1 we studied the symmetry group D4 of the square.
More generally we have for any n ≥ 2 the symmetry group Dn of the regular n-polygon."

So in the definition I've got n=1 is not an option!

As it is D1 can be defined in different ways.
One of them is to put identity and a reflection in it, which would be isomorphic with {+-1}.
This choice is consistent with Dn for n ≥ 2.

Taken literally, D1 should have infinitely many rotations and infinitely many reflections.
In this case the group [D1,D1] would still consist of all rotations, and D1/[D1,D1] would still be isomorphic with {+-1}.
• May 19th 2012, 12:53 PM
Deveno
Re: Proof generation of [Dn,Dn]
most descriptions of Dn require than n > 1 (a 1-gon makes no sense). the dihedral group of order 4 (symmetries of the degenerate "2-gon", a line segment) is V, and since this group is abelian, in this (somewhat special case) we have V/[V,V] = V/{e} = V. this conforms to our pattern:

4/(2/2) = 4/1 = 4.

there is one interpretation of D1 that still "conforms to the pattern":

D1 = <ρ,σ>, where ρ1 = e, σ2 = e, and σρ = ρ-1σ.

in this view, it is clear that D1 = <σ> = {e,σ}, which conforms to the observed pattern for "odd n":

2/1 = 2.

(EDIT: geometrically, D1 is just the orthogonal group of R2, if we pick the "symmetric point" to be the origin. if we do so, the "rotation group" is then the special linear group (= special orthogonal, since our vector space is real), and arguing by determinants shows us the quotient is cyclic of order 2).
• May 19th 2012, 12:55 PM
TDA120
Re: Proof generation of [Dn,Dn]
Aha! So that’ll be smart; correcting the teachers’ material ;)

But in the second option, how can you state that [D1,D1] still consists of all rotations as it is generated by ρ^2 and contains infinitely many rotations? These odd powers of ρ will never be „reached”?
• May 19th 2012, 01:28 PM
TDA120
Re: Proof generation of [Dn,Dn]
Thanks Deveno! In another question I had to calculate the order of Hom (Sn, C*). As C* is abelian, I knew it would be 2 (using I like Serena’s corollary)
But now I start wondering about S1. Should I interpret that similar to D1 then?
• May 19th 2012, 01:50 PM
ILikeSerena
Re: Proof generation of [Dn,Dn]
Quote:

Originally Posted by TDA120
Aha! So that’ll be smart; correcting the teachers’ material ;)

But in the second option, how can you state that [D1,D1] still consists of all rotations as it is generated by ρ^2 and contains infinitely many rotations? These odd powers of ρ will never be „reached”?

Consider a ρ that may never be reached.
Now take the rotation around half its angle.
What is the commutator of this rotation with a reflection?

Quote:

Originally Posted by TDA120
Thanks Deveno! In another question I had to calculate the order of Hom (Sn, C*). As C* is abelian, I knew it would be 2 (using I like Serena’s corollary)
But now I start wondering about S1. Should I interpret that similar to D1 then?

S1 is well defined: the permutations of 1 element.
This is only the identity.

So the corollary says S1->{+-1}->A
Since S1 only has one element, that element has to be mapped to 1.
So how many homomorphisms does that make?
• May 19th 2012, 01:56 PM
Deveno
Re: Proof generation of [Dn,Dn]
S1 is an entirely different matter.

By definition, S1 is (isomorphic to) the set of all bijective functions on a 1-element set {a}. there is clearly only one such function, given by:

f(a) = a (there is another function that one could define on {a}, the "empty function", but this is not bijective, since a in {a} has no pre-image).

put another way, if you only have one object to permute, there's only one possible permutation (you can't play 3-card monte with just one card, and make any money).

so S1 is a trivial group. there is, therefore, only ONE possible homomorphism, e must map to 1.
• Mar 29th 2013, 11:42 AM
johng
Re: Proof generation of [Dn,Dn]
Here's what I think is an easy solution to the problem:

Attachment 27732