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Math Help - Isomorphism

  1. #1
    Senior Member slevvio's Avatar
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    Isomorphism

    Prove that any two cyclic groups of the same order are isomorphic.

    Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that  \phi : G \rightarrow G' defined by  \phi(a^m) = (a')^m for  m \in \mathbb{Z} is an isomorphism. If G and G' are both of finite order n, use  \phi again but note that it is necessary to prove that  \phi is consistently defined because it is possible to have  a^i = a^j when i does not equal j.

    It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Prove that any two cyclic groups of the same order are isomorphic.

    Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that  \phi : G \rightarrow G' defined by  \phi(a^m) = (a')^m for  m \in \mathbb{Z} is an isomorphism. If G and G' are both of finite order n, use  \phi again but note that it is necessary to prove that  \phi is consistently defined because it is possible to have  a^i = a^j when i does not equal j.

    It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone
    a'\,^m=1\Longrightarrow n\mid m \Longrightarrow m=kn\Longrightarrow a^m=(a^n)^k=1^k=1\Longrightarrow Ker\,\phi = 1

    Tonio
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  3. #3
    Senior Member slevvio's Avatar
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    thanks
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  4. #4
    MHF Contributor Drexel28's Avatar
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    I am not sure what consistently defined means? Most likely well-defined?
    But maybe the following is along the lines of what you want?

    Problem: Let G,G' be cyclic groups of order n. Prove that G\cong G'

    Proof: Let \mathcal{G}=\langle g\rangle be any cyclic group of order n. Define a mapping \Phi:\mathbb{Z}_n\mapsto \mathcal{G} by \Phi(k)=g^k. Clearly this is well defined. To see that it is an injection note that g^{\kappa}=g^{\kappa-\kappa_1}=e, which means that n|\kappa-\kappa_1 but since -(n-1)\le \kappa-\kappa_1\le n-1 this will only be true if \kappa-\kappa_1=0\implies \kappa=\kappa_1. To see that this mapping is surjective let g'\in\mathcal{G}, then g'=g^{m} for some 0\le m \le n-1. Therefore g'=\Phi(m). Lastly, we must show that \Phi is a homomorphism, but this is trivial since \Phi(\tau+\tau_1)=g^{\tau+\tau_1}=g^{\tau}g^{\tau_  1}=\Phi(\tau)\Phi(\tau_1). Therefore any cyclic grou of order n is isomorphic to \mathbb{Z}_n, but since \cong is an equivalence relation (you can check this yourself) we see that G\cong \mathbb{Z}_n\text{ and }G'\cong\mathbb{Z}_n\implies G\cong G' by transivity.
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  5. #5
    Senior Member slevvio's Avatar
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    yeah thanks guys, it's very strange as it says well defined in all my notes and examples then suddenly it starts talking about consistently definedeness
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  6. #6
    Senior Member slevvio's Avatar
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    Two Isomorphism Questions

    Hello I am struggling with these 2 questions.

    1) Prove that  S_3 and  D_6 (symmetry group of the equilateral triangle) are isomorphic.

    It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in  S_3 would give  \phi (fg) = \phi (f) \phi (g)

    I have this theorem in my notes: Every group G is isomorphic to a subgroup of  S_x for some choice of set X. [X = { bijections f: f: X -> X }]
    In particular, if  |G| < \infty then G and H are isomorphic for some H \le S_n for some  n \in \mathbb{N} .

    Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.

    2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:

    (a) G = HK, (b)  H \cap K = {1} , (c)  H \lhd G , (d)  K \lhd G .

    Show also that   G \cong H x K .

    HINT: Show first that any element of G can be uniquely expressed in the form hk, where  h \in H, k \in K . Then, by considering elements of the form  h^{-1}k^{-1}hk where  h \in H, k \in K show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.

    OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that  h^{-1}k^{-1}hk was in H and in K and therefore in  H \cap K = {1}

    And of course if  h^{-1}k^{-1}hk = 1  then  hk = kh

    Any help and discussion would be appreciated.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Hello I am struggling with these 2 questions.

    1) Prove that  S_3 and  D_6 (symmetry group of the equilateral triangle) are isomorphic.

    It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in  S_3 would give  \phi (fg) = \phi (f) \phi (g)

    I have this theorem in my notes: Every group G is isomorphic to a subgroup of  S_x for some choice of set X. [X = { bijections f: f: X -> X }]
    In particular, if  |G| < \infty then G and H are isomorphic for some H \le S_n for some  n \in \mathbb{N} .

    Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.

    2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:

    (a) G = HK, (b)  H \cap K = {1} , (c)  H \lhd G , (d)  K \lhd G .

    Show also that  G \cong H x K .

    HINT: Show first that any element of G can be uniquely expressed in the form hk, where  h \in H, k \in K . Then, by considering elements of the form  h^{-1}k^{-1}hk where  h \in H, k \in K show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.

    OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that  h^{-1}k^{-1}hk was in H and in K and therefore in  H \cap K = {1}

    And of course if  h^{-1}k^{-1}hk = 1 then  hk = kh

    Any help and discussion would be appreciated.
    1. D_{2n}\subseteq S_n. Also, \left|D_{2n}\right|=2n, \left|S_{n}\right|=n!. So \left|S_{3}\right|=3!=6 and \left|D_{6}\right|=6 and since D_{6}\subseteq S_{3} the conclusion follows.

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  8. #8
    Senior Member slevvio's Avatar
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    why is  D_{2n} \subset S_n ? And why does their order being the same make them isomorphic?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    why is  D_{2n} \subset S_n ? And why does their order being the same make them isomorphic?
    First of all, I used bad notation. I should have put D_{2n}\le S_n. Secondly, D_{2n} is a permutation group, in other words it is isomorphic to a subgroup of the symmetric group on n letters. Think of it as S_{n} is "how many ways can I permute the numbers \left\{1,\cdots,n\right\}" and D_{2n} as "how many ways can I permute \left\{1,\cdots,n\right\} so that each number still shares it's original neighbors?". So we can easily define an isomorphism between the subsetof S_n that's "neighbor preserving" and D_{2n}. But a subgroup with exactly the same elements as the group cannot be proper, which forces it to be improper.
    Last edited by Chris L T521; November 19th 2009 at 08:02 AM.
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