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Math Help - How to find a proper mapping in order to prove isomorphism of groups

  1. #1
    Junior Member rubix's Avatar
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    How to find a proper mapping in order to prove isomorphism of groups

    as the title says, given two groups, and if i'm asked to prove they're isomorphic...how do i find the mapping that satisfies? is there a general way of doing this? if not, what kind of thought process you do?

    ex: show isomorphism from nZ to mZ
    n and m are some integers
    nZ and mZ are collection of multiples of n and m respectively
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rubix View Post
    as the title says, given two groups, and if i'm asked to prove they're isomorphic...how do i find the mapping that satisfies? is there a general way of doing this? if not, what kind of thought process you do?

    ex: show isomorphism from nZ to mZ
    n and m are some integers
    nZ and mZ are collection of multiples of n and m respectively
    I'm afraid to say that there is no `general rule'. I mean, for your example I would prove that every subgroup of \mathbb{Z} is isomorphic to \mathbb{Z} (can you see why this is sufficient?). But if I was to prove that D_4, the dihedral group of order 4, was isomorphic to the Klein 4-group I would point out that it is not cyclic, and there are only two groups of order 4, one of which is cyclic, the other if the Klein 4-group.

    However, when trying to find an isomorphism there are a few things to note,

    If you map to the generators then your homomorphism will be surjective,

    If the kernel is trivial then your homomorphism will be injective,

    Try to find a `common' group which both groups are isomorphic to (as in the example),

    Often there exists some well-known function which actually turns out to be an isomorphism of groups (for example, (\mathbb{R}, +) \cong (\mathbb{R}^{+}, \times), and the isomorphism is given by e^x).

    A homomorphism is an isomorphism if and only if there exists an inverse function. Sometimes this can be easily found.
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  3. #3
    Junior Member rubix's Avatar
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    transitive property? if subset of Z is isomorphic to Z then nZ and mZ are isomorphic?

    but how would one actually go about proving isomorphism? in particular i'm thinking what the mapping would be?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rubix View Post
    transitive property? if subset of Z is isomorphic to Z then nZ and mZ are isomorphic?

    but how would one actually go about proving isomorphism? in particular i'm thinking what the mapping would be?
    Well, I'll leave you to figure the isomorphism out. But as a helping hand, and another tip, remember that a homomorphism is defined precisely by where the generators are mapped to.

    So, for example, you know that if H \leq \mathbb{Z} then H=<i> for some i \in \mathbb{Z}, by the Euclidean algorithm. So, \phi: \mathbb{Z} \rightarrow H defined by 1 \mapsto i defines an isomorphism (can you see why it defines an isomorphism?).
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