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Math Help - Does this uniquely determine G?

  1. #1
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    Does this uniquely determine G?

    (up to isomorphism)

     1\longrightarrow \mathbb{Z}_4 \longrightarrow^{\!\!\!\!\!\!\!\!\!\beta}\ \, G \longrightarrow^{\!\!\!\!\!\!\!\!\!\alpha}\ \,  \mathbb{Z}_2 \longrightarrow 1

    \mathbb{Z}_2\longrightarrow^{\!\!\!\!\!\!\!\gamma}  \ \, G

    \gamma \circ \alpha = 1_{\mathbb{Z}_2}

    (if my latex were better i'd try to put it all on one line, this is a right-split short exact sequence).

    if so, is there a "smaller" diagram which conveys the same information?
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    MHF Contributor Drexel28's Avatar
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    Re: Does this uniquely determine G?

    Quote Originally Posted by Deveno View Post
    (up to isomorphism)

     1\longrightarrow \mathbb{Z}_4 \longrightarrow^{\!\!\!\!\!\!\!\!\!\beta}\ \, G \longrightarrow^{\!\!\!\!\!\!\!\!\!\alpha}\ \,  \mathbb{Z}_2 \longrightarrow 1

    \mathbb{Z}_2\longrightarrow^{\!\!\!\!\!\!\!\gamma}  \ \, G

    \gamma \circ \alpha = 1_{\mathbb{Z}_2}

    (if my latex were better i'd try to put it all on one line, this is a right-split short exact sequence).

    if so, is there a "smaller" diagram which conveys the same information?
    Presumably you meant \alpha\circ\gamma=\text{id}_{\mathbb{Z}_2}. If so, we can definitely say that G\cong \mathbb{Z}_4\rtimes_\varphi\mathbb{Z}_2 for some homomorphism \varphi:\mathbb{Z}_2\to\text{Aut}(\mathbb{Z}_4). But, since \text{Aut}(\mathbb{Z}_4)=\mathbb{Z}_2 there is precisely one non-trivial homomorphism. And so, in particular, up to isomorphism the only semidirect product gives D_4. Thus, G\cong D_4.
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  3. #3
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    Re: Does this uniquely determine G?

     G \cong \mathbb{Z}_4 \times \mathbb{Z}_2 satisfies the conditions too. so G is not uniquely determined.
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    MHF Contributor Drexel28's Avatar
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    Re: Does this uniquely determine G?

    Quote Originally Posted by NonCommAlg View Post
     G \cong \mathbb{Z}_4 \times \mathbb{Z}_2 satisfies the conditions too. so G is not uniquely determined.
    Right, I forgot the trivial homomorphism \varphi:\mathbb{Z}_2\to\mathbb{Z}_2.
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  5. #5
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    Re: Does this uniquely determine G?

    yes i did mean the other composition ( can i claim it was because i was looking at herstein last night? no? ok, i admit it, i'm just stupid).

    so...how would i indicate there is no left-split with a picture, to rule out the trivial semi-direct (that is direct) product?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Does this uniquely determine G?

    Quote Originally Posted by Deveno View Post
    yes i did mean the other composition ( can i claim it was because i was looking at herstein last night? no? ok, i admit it, i'm just stupid).

    so...how would i indicate there is no left-split with a picture, to rule out the trivial semi-direct (that is direct) product?
    You mean you just want to indicate, diagramatically that there is no left split?
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  7. #7
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    Re: Does this uniquely determine G?

    correct. what got me thinking about this, is thinking about identifying D4 solely in terms of homomorphisms to and from it (although i admit including the extra information of the short exact sequence is cheating a little) and wondering how well this can be done, for an arbitrary group.
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