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Math Help - Isomorphic to the product of the subgroups H and K.

  1. #1
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    Isomorphic to the product of the subgroups H and K.

    Here's the problem:



    I cannot figure this out. So if H is not normal in G, does that mean that HxK is not isomorphic to G? Or could I define some other map that would be isomorphic?
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  2. #2
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    Opalg's Avatar
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    Is H commutative?
    Is K commutative?
    Is HxK commutative?
    Is G commutative?
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  3. #3
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    Yes.
    Yes.
    Yes.
    No.

    I think I'm on the brink of seeing the solution. Are you saying that if HxK is commutative then G must also be commutative in order for HxK to be isomorphic to G? And if so, why is this?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by CropDuster View Post
    Yes.
    Yes.
    Yes.
    No.

    I think I'm on the brink of seeing the solution. Are you saying that if HxK is commutative then G must also be commutative in order for HxK to be isomorphic to G? And if so, why is this?
    Two groups are isomorphic if they are the same `up to labelling'. They are essentially the same, just every element has a different name. For example, you can think of the integers as \{\ldots -2, -1, 0, 1, 2, 3 \ldots\} under addition, or you can think of them as \{\ldots a^{-2}, a^{-1}, 1, a, a^2 \ldots\} under multiplication. They are still the same group!

    Basically, if H \times K and G are isomorphic then let \phi:H\times K \mapsto G be the isomorphism. As aba^{-1}b^{-1}=1 for all a, b \in H\times K we must have that (aba^{-1}b^{-1})\phi=1 in G. Aka, (a\phi)(b\phi)(a^{-1}\phi)(b^{-1}\phi)=1. As \phi is onto, you are done.

    (Note that aba^{-1}b^{-1}=1 \Leftrightarrow ab=ba)
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