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Thread: 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 $\displaystyle \{\ldots -2, -1, 0, 1, 2, 3 \ldots\}$ under addition, or you can think of them as $\displaystyle \{\ldots a^{-2}, a^{-1}, 1, a, a^2 \ldots\}$ under multiplication. They are still the same group!

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

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