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Why not embed F/R into F?

  1. Aug 28, 2008 #1

    tgt

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    Why doesn't such an embedding exist? When R is a normal subgroup of F.
     
  2. jcsd
  3. Aug 28, 2008 #2

    morphism

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    Why would such an embedding exist? Conceptually it doesn't make sense.

    By way of example, can you embed Z/2Z into Z?
     
  4. Aug 28, 2008 #3

    tgt

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    Sure, I see the exampe but can you explain the conceptual part? How is it conceptual to you?
     
  5. Aug 28, 2008 #4

    morphism

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    Well, if you think about it for a bit, what does it mean to be able to embed F/R into F? Intuitively, modding out by a subgroup means you collapse that subgroup to zero. So for F/R to be isomorphic to some subgroup of F, it would be necessary for R to be 'complemented' inside of F by another subgroup (think of complementation here as you would in the setting of vector spaces). I personally don't see any reason why one would believe every normal subgroup of F to have this property. (But I admit to being prejudiced: functional analysis has made me very suspicious of the process of complementation!)

    To make this a bit more precise, we can use semidirect products: If F/R embeds into F, then F is essentially [itex]R \rtimes F/R[/itex]. It's easy to see that a necessary and sufficient condition for this to happen is that the canonical exact sequence

    [tex]1 \longrightarrow R \longrightarrow F \longrightarrow F/R \longrightarrow 1[/tex]

    splits.

    By the way, is there any specific reason you're using F and R to denote groups?!
     
  6. Aug 28, 2008 #5

    tgt

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    Some good points raised. Although what does split mean?

    Those notations naturally fit into the notation of a presentation.
     
    Last edited: Aug 28, 2008
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