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Symmetries of 6-gon

  1. Feb 20, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    We were given the following argument showing that ##D_6 \cong D_3 \times \mathbb{Z}_2##.
    What we did was take two subgroups H and K of G such that their intersection was trivial and that ##hk = kh## for all h and k. H = {symmetries of 6-gon that preserved a triangle inside} and K = <g3>. By some theorem, we proved this shows that HK is isomorphic to H x K which in turn is isomorphic to ##D_3 \times \mathbb{Z}_2##

    The question is: Does a similar argument apply to ##D_n## where n is even ##\geq 6##?

    3. The attempt at a solution

    I presume the question is asking if we can write ##D_n \cong D_x \times \mathbb{Z}_y## for some x and y?

    What I have done so far is the following:
    Let G = Dn. Choose H and K such that |H| . |K| = 2n. For the above argument to work, necessarily H and K are subgroups of G. So by Lagrange, the orders of H and K have to divide the order of Dn. This means: $$|D_n| = a|H|, |D_n| = b|K|\,\Rightarrow\,a|H| = b|K|.$$Since H and K are subgroups, they are subsets so H,K contained in G. But since we want the intersection to be trivial (i.e ##H \cap K = \left\{e\right\}##, we have that ##|H| \neq |K|##. I am not sure if this helps me at all and I am unsure of how to proceed.

    Many thanks.
     
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  3. Feb 21, 2013 #2

    CAF123

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    Any ideas? Or is my question difficult to understand?
     
  4. Feb 21, 2013 #3

    micromass

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    I understand that [itex]D_6[/itex] is the dihedral group of order 6?
    But what do you mean with [itex]D_3[/itex]?
     
  5. Feb 21, 2013 #4

    micromass

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    Oh, you're using the notation that the order of [itex]D_n[/itex] is 2n. I get it.
     
  6. Feb 21, 2013 #5

    micromass

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    Maybe we should focus first on whether

    [tex]D_{2n}\sim D_n\times \mathbb{Z}_2[/tex]

    Can you determine centers of both groups? Thus, can you tell me what [itex]Z(D_{2n})[/itex] and what [itex]Z(D_n\times \mathbb{Z}_2)[/itex]is?
     
  7. Feb 21, 2013 #6

    CAF123

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    H turned out to have the same number of elements as ##D_3##. So we then said there existed an isomorphism ##H \cong D_3## and this led to the conclusion.
     
  8. Feb 21, 2013 #7

    CAF123

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    Hi micromass,

    We covered centres right at the end of our course and I am not familiar with that Z notation. I am doing some revision and this question came quite early on, just after subgroups, cyclic groups, homormorphisms. Is there a way to do it using this?
     
  9. Feb 21, 2013 #8

    micromass

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    OK, so let's say that [itex]D_{2n}\sim D_n\times \mathbb{Z}_2[/itex].

    So then we can write [itex]D_{2n}=HK[/itex] with [itex]H\cap K=\{0\}[/itex] and where H and K are both normal. Furthermore, [itex]H\sim D_n[/itex]and [itex]K\sim \mathbb{Z}_2[/itex].

    So if it were true that we could write this, then in particular, we could find a normal subgroup of [itex]D_{2n}[/itex] that has order 2. Can you find me all normal subgroups of order 2 in [itex]D_{2n}[/itex]?? Maybe you can start by listing all elements of order 2, and then see if they generate something normal.
     
  10. Feb 21, 2013 #9

    CAF123

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    Why do they have to be normal?

    Thanks for your reply, but we didn't really do much with normal subgroups, so I am struggling even to understand why what you said above is true. We had a sort of introductory lecture on things like centres/normal subgroups/class eqn/conjugacy etc..but didn't go into too much detail. Perhaps the element ##h## would have order 2, but I cannot be sure whether this would hold for all even n
     
  11. Feb 21, 2013 #10

    micromass

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    Because both [itex]D_n[/itex] (or more precisely: [itex]D_n\times\{0\}[/itex]) and [itex]\mathbb{Z}_2[/itex] (or more precisely: [itex]\{e\}\times \mathbb{Z}_2[/itex]) are normal in [itex]D_n\times \mathbb{Z}_2[/itex]. The subgroups of [itex]D_{2n}[/itex] corresponding to [itex]D_n[/itex] (respectively [itex]\mathbb{Z}_2[/itex]) are H (respectively K). So they need to be normal.

    But let's say we have the following isomorphism: [itex]T:D_n\times \mathbb{Z}_2\rightarrow D_{2n}[/itex], then (e,1) has order 2 in [itex]D_n\times \mathbb{Z}_2[/itex] and thus so has T(e,1). And T(e,1) exactly generates H.

    Since you seem to be struggling with basic concepts, I think you have misinterpreted the question. You seem to have interpreted the question as: is [itex]D_{2n}\cong D_x\times \mathbb{Z}_y[/itex]. I don't think this is the way you should interpret it.

    I think the question asks you to generalize the exact proof of [itex]D_6\sim D_3\times \mathbb{Z}_2[/itex] to the context [itex]D_{2n}\sim D_n\times \mathbb{Z}_2[/itex]. And they ask whether that proof is still valid. I don't think they want you to actually show the result in general. They just want you to see if the generalization of the proof is valid.
    It might be worth contacting your professor and asking him for more clarificiations on the question.
     
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