Proof generation of [Dn,Dn]

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

My challenge is as follows:

Let D_{n} be the dihedral group (symmetries of the regular n-polygon) of order 2n and let ρ be a rotation of D_{n} with order n.

(a) Proof that the commutator subgroup [D_{n},D_{n}] is generated by ρ^{2}.

(b) Deduce that the abelian made D_{n,ab} is isomorphic with {±1} in case n is odd, and with V_{4} (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 $\displaystyle f: G \to A$ be a homomorphism to an abelian group A.

Then there exists a homomorphism $\displaystyle f_{ab}: G_{ab}=G/[G,G] \to A$ so that f can be created as a composition$\displaystyle G \overset{\pi}{\to} G_{ab} \overset{f_{ab}}{\to}A$

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

Here G_{ab} is the group that is made abelian from G.

**Corollary**: Every homomorphism f: S_{n}->A to an abelian group A is the composition of $\displaystyle S_n \to \{\pm 1\} \overset{h}{\to} A$ of the sign function with a homomorphism h: {±1} -> A

**3. The attempt at a solution**

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

Re: Proof generation of [Dn,Dn]

Thanks! I think this meant I could make some new steps.!?

ρ^{k}ρ^{l}σρ^{-k}(ρ^{l}σ)^{-1}=

ρ^{k}ρ^{l}σρ^{-k}σρ^{-l}=

ρ^{k+l}σσρ^{k-l}=

ρ^{2k}

So, every commutator will be either id or ρ^{2k}

As 2k is a multiple of ρ^{2}, [D_{n},D_{n}] is generated by ρ^{2}.

So [D_{n},D_{n}] = $\displaystyle \left\langle$ ρ^{2k} $\displaystyle \right\rangle$ = {id, ρ^{2}, ρ^{4}, …, ρ^{n-2}} if n is even and

[D_{n},D_{n}] = $\displaystyle \left\langle$ ρ^{2k} $\displaystyle \right\rangle$ = 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 C_{4} or V_{4}.

Quote:

Originally Posted by

**ILikeSerena** **1. The problem statement, all variables and given/known data**
My challenge is as follows:

Let D

_{n} be the dihedral group (symmetries of the regular n-polygon) of order 2n and let ρ be a rotation of D

_{n} with order n.

(a) Proof that the commutator subgroup [D

_{n},D

_{n}] is generated by ρ

^{2}.

(b) Deduce that the abelian made D

_{n,ab} is isomorphic with {±1} in case n is odd, and with V

_{4} (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 $\displaystyle f: G \to A$ be a homomorphism to an abelian group A.

Then there exists a homomorphism $\displaystyle f_{ab}: G_{ab}=G/[G,G] \to A$ so that f can be created as a composition

$\displaystyle G \overset{\pi}{\to} G_{ab} \overset{f_{ab}}{\to}A$

of $\displaystyle \pi: G \to G_{ab}$ with f

_{ab}.

Here G

_{ab} is the group that is made abelian from G.

**Corollary**: Every homomorphism f: S

_{n}->A to an abelian group A is the composition of $\displaystyle S_n \to \{\pm 1\} \overset{h}{\to} A$ of the sign function with a homomorphism h: {±1} -> A

**3. The attempt at a solution**
I have worked out [D

_{n}, D

_{n}] for n=3,4,5 and 6 and have noticed the above described pattern. I just cannot proof it.

Re: Proof generation of [Dn,Dn]

Aha! Thanks for explaining! ;)

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:

(ρ^{k}σ)(ρ^{l}σ)(ρ^{k}σ)^{-1}(ρ^{l}σ)^{-1}

= (ρ^{k}σ)(ρ^{l}σ)(ρ^{k}σ)(ρ^{l}σ)

= (ρ^{k}σ)(σρ^{-l})(ρ^{k}σ)(σρ^{-l})

= (ρ^{k-l})^{2} = (ρ^{2})^{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 [D_{n},D_{n}] contains every power of ρ^{2} (which, to be fair, we already knew).

however, we do NOT yet know that EVERY commutator is a power of ρ^{2}.

for we also need to consider commutators of the form:

(ρ^{k}σ)(ρ^{l})(ρ^{k}σ)^{-1}(ρ^{l})^{-1}

= (ρ^{k}σ)(ρ^{l})(ρ^{k}σ)(ρ^{-l})

= (ρ^{k-l}σ)(σρ^{-k-1}) = ρ^{-2l} = (ρ^{2})^{-l}.

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

thus not only does [D_{n},D_{n}] contain <ρ^{2}>, but is also contained within <ρ^{2}>.

if n is even, then ρ^{2} has order n/2, and we get <ρ^{2}> is cyclic of order n/2.

if n is odd, then p^{2} has order n (since gcd(2,n) = 1), so that <ρ^{2}> = <ρ>, is cyclic of order n.

it is the order of D_{n}/[D_{n},D_{n}] 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<ρ^{2}> in D_{n}/[D_{n},D_{n}], this has order 2 since ρ^{2} is in <ρ^{2}>.

next, consider the coset σ<ρ^{2}>, which also has order 2, since σ^{2} = e.

since D_{n}/[D_{n},D_{n}] has 2 elements of order 2, it is not cyclic.

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.

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?

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}.

Re: Proof generation of [Dn,Dn]

most descriptions of D_{n} 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 D_{1} that still "conforms to the pattern":

D_{1} = <ρ,σ>, where ρ^{1} = e, σ^{2} = e, and σρ = ρ^{-1}σ.

in this view, it is clear that D_{1} = <σ> = {e,σ}, which conforms to the observed pattern for "odd n":

2/1 = 2.

(EDIT: geometrically, D_{1} is just the orthogonal group of R^{2}, 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).

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”?

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?

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?

Re: Proof generation of [Dn,Dn]

S_{1} is an entirely different matter.

By definition, S_{1} 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 S_{1} is a trivial group. there is, therefore, only ONE possible homomorphism, e must map to 1.

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Re: Proof generation of [Dn,Dn]

Here's what I think is an easy solution to the problem:

Attachment 27732