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Homework Help: Isom. of group Dn abelian

  1. May 18, 2012 #1
    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 [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.
    Last edited: May 18, 2012
  2. jcsd
  3. May 18, 2012 #2

    I like Serena

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    Hey TDA120! :smile:

    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.
  4. May 18, 2012 #3
    Thanks! I think this meant I could make some new steps.!?

    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] = [itex]\left\langle[/itex] ρ2k [itex]\right\rangle[/itex] = {id, ρ2, ρ4, …, ρn-2} if n is even and

    [Dn,Dn] = [itex]\left\langle[/itex] ρ2k [itex]\right\rangle[/itex] = 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
  5. May 18, 2012 #4

    I like Serena

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    Yep! :approve:

    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
  6. May 19, 2012 #5
    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?
  7. May 19, 2012 #6

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    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
  8. May 19, 2012 #7
    [itex]\langle[/itex]Right. I wasn’t thinking the right way; thanks![itex]\rangle[/itex]
    Last edited: May 19, 2012
  9. May 19, 2012 #8

    I like Serena

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