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Thread: 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 $\displaystyle \phi : G \rightarrow G' $ defined by $\displaystyle \phi(a^m) = (a')^m $ for $\displaystyle m \in \mathbb{Z} $ is an isomorphism. If G and G' are both of finite order n, use $\displaystyle \phi $ again but note that it is necessary to prove that $\displaystyle \phi $ is consistently defined because it is possible to have $\displaystyle 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 $\displaystyle \phi : G \rightarrow G' $ defined by $\displaystyle \phi(a^m) = (a')^m $ for $\displaystyle m \in \mathbb{Z} $ is an isomorphism. If G and G' are both of finite order n, use $\displaystyle \phi $ again but note that it is necessary to prove that $\displaystyle \phi $ is consistently defined because it is possible to have $\displaystyle 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
    $\displaystyle a'\,^m=1\Longrightarrow n\mid m \Longrightarrow m=kn\Longrightarrow a^m=(a^n)^k=1^k=1\Longrightarrow Ker\,\phi = 1$

    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 $\displaystyle G,G'$ be cyclic groups of order $\displaystyle n$. Prove that $\displaystyle G\cong G'$

    Proof: Let $\displaystyle \mathcal{G}=\langle g\rangle$ be any cyclic group of order $\displaystyle n$. Define a mapping $\displaystyle \Phi:\mathbb{Z}_n\mapsto \mathcal{G}$ by $\displaystyle \Phi(k)=g^k$. Clearly this is well defined. To see that it is an injection note that $\displaystyle g^{\kappa}=g^{\kappa-\kappa_1}=e$, which means that $\displaystyle n|\kappa-\kappa_1$ but since $\displaystyle -(n-1)\le \kappa-\kappa_1\le n-1$ this will only be true if $\displaystyle \kappa-\kappa_1=0\implies \kappa=\kappa_1$. To see that this mapping is surjective let $\displaystyle g'\in\mathcal{G}$, then $\displaystyle g'=g^{m}$ for some $\displaystyle 0\le m \le n-1$. Therefore $\displaystyle g'=\Phi(m)$. Lastly, we must show that $\displaystyle \Phi$ is a homomorphism, but this is trivial since $\displaystyle \Phi(\tau+\tau_1)=g^{\tau+\tau_1}=g^{\tau}g^{\tau_ 1}=\Phi(\tau)\Phi(\tau_1)$. Therefore any cyclic grou of order $\displaystyle n$ is isomorphic to $\displaystyle \mathbb{Z}_n$, but since $\displaystyle \cong$ is an equivalence relation (you can check this yourself) we see that $\displaystyle G\cong \mathbb{Z}_n\text{ and }G'\cong\mathbb{Z}_n\implies G\cong G'$ 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 $\displaystyle S_3 $ and $\displaystyle 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 $\displaystyle S_3 $ would give $\displaystyle \phi (fg) = \phi (f) \phi (g) $

    I have this theorem in my notes: Every group G is isomorphic to a subgroup of $\displaystyle S_x $ for some choice of set X. [X = { bijections f: f: X -> X }]
    In particular, if $\displaystyle |G| < \infty $ then G and H are isomorphic for some $\displaystyle H \le S_n $ for some $\displaystyle 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) $\displaystyle H \cap K = {1} $, (c) $\displaystyle H \lhd G $, (d) $\displaystyle K \lhd G $.

    Show also that $\displaystyle G \cong H x K $.

    HINT: Show first that any element of G can be uniquely expressed in the form hk, where $\displaystyle h \in H, k \in K $. Then, by considering elements of the form$\displaystyle h^{-1}k^{-1}hk $ where $\displaystyle 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 $\displaystyle h^{-1}k^{-1}hk $was in H and in K and therefore in $\displaystyle H \cap K = {1} $

    And of course if $\displaystyle h^{-1}k^{-1}hk = 1 $ then $\displaystyle 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 $\displaystyle S_3 $ and $\displaystyle 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 $\displaystyle S_3 $ would give $\displaystyle \phi (fg) = \phi (f) \phi (g) $

    I have this theorem in my notes: Every group G is isomorphic to a subgroup of $\displaystyle S_x $ for some choice of set X. [X = { bijections f: f: X -> X }]
    In particular, if $\displaystyle |G| < \infty $ then G and H are isomorphic for some $\displaystyle H \le S_n $ for some $\displaystyle 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) $\displaystyle H \cap K = {1} $, (c) $\displaystyle H \lhd G $, (d) $\displaystyle K \lhd G $.

    Show also that $\displaystyle G \cong H x K $.

    HINT: Show first that any element of G can be uniquely expressed in the form hk, where $\displaystyle h \in H, k \in K $. Then, by considering elements of the form$\displaystyle h^{-1}k^{-1}hk $ where $\displaystyle 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 $\displaystyle h^{-1}k^{-1}hk $was in H and in K and therefore in $\displaystyle H \cap K = {1} $

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

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

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  8. #8
    Senior Member slevvio's Avatar
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    why is $\displaystyle 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 $\displaystyle 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 $\displaystyle D_{2n}\le S_n$. Secondly, $\displaystyle D_{2n}$ is a permutation group, in other words it is isomorphic to a subgroup of the symmetric group on n letters. Think of it as $\displaystyle S_{n}$ is "how many ways can I permute the numbers $\displaystyle \left\{1,\cdots,n\right\}$" and $\displaystyle D_{2n}$ as "how many ways can I permute $\displaystyle \left\{1,\cdots,n\right\}$ so that each number still shares it's original neighbors?". So we can easily define an isomorphism between the subsetof $\displaystyle S_n$ that's "neighbor preserving" and $\displaystyle D_{2n}$. 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; Nov 19th 2009 at 07:02 AM.
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