Results 1 to 13 of 13
Like Tree3Thanks
  • 1 Post By TDA120
  • 1 Post By Deveno
  • 1 Post By ILikeSerena

Math Help - Proof generation of [Dn,Dn]

  1. #1
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    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.
    Last edited by ILikeSerena; May 18th 2012 at 04:17 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2012
    From
    Netherlands
    Posts
    8
    Thanks
    1

    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 View Post
    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.
    Last edited by TDA120; May 18th 2012 at 11:02 AM.
    Thanks from ILikeSerena
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Proof generation of [Dn,Dn]

    Aha! Thanks for explaining!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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.
    Thanks from ILikeSerena
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2012
    From
    Netherlands
    Posts
    8
    Thanks
    1

    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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    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}.
    Last edited by ILikeSerena; May 19th 2012 at 11:50 AM.
    Thanks from Deveno
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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).
    Last edited by Deveno; May 19th 2012 at 12:07 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Apr 2012
    From
    Netherlands
    Posts
    8
    Thanks
    1

    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”?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2012
    From
    Netherlands
    Posts
    8
    Thanks
    1

    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?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Proof generation of [Dn,Dn]

    Quote Originally Posted by TDA120 View Post
    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 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    659
    Thanks
    271

    Re: Proof generation of [Dn,Dn]

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

    Proof generation of [Dn,Dn]-mhfdihedralgp.png
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Generation of Sinusoids
    Posted in the Math Software Forum
    Replies: 1
    Last Post: December 18th 2011, 07:45 PM
  2. Polynomial Generation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 5th 2010, 04:34 AM
  3. sequence generation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 6th 2009, 06:04 AM
  4. Automatic Proof Generation
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: April 14th 2008, 03:57 PM

Search Tags


/mathhelpforum @mathhelpforum