1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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.!?
    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] = [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

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Isom. of group Dn abelian
  1. Abelian group (Replies: 9)

  2. Abelian group (Replies: 1)

  3. Abelian Groups (Replies: 5)

  4. Abelian Groups (Replies: 9)

  5. Abelian group (Replies: 2)

Loading...