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Freely Indecomposable

  1. Sep 22, 2006 #1
    Can anyone explain why a freely indecomposable group is important or handy?

    I can't seem to understand the definition. It says that a group G is free indecomposable if

    [tex]G = A \star B \Rightarrow G = A \vee G = B[/tex]

    So is it really saying that a free product A * B is freely indecomposable if

    [tex]A \star B \Rightarrow A \star B = A \vee A \star B = B[/tex]?
     
  2. jcsd
  3. Sep 22, 2006 #2

    matt grime

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    I've never seen it before, but I think you're wrong to use equals signs, and they should be isomorphisms. Anyway, if G is free indecomposable it cannot be written as the direct product of strictly smaller free groups. And it is always important to know what the indecomposable objects in your theory are since they are by definition the building blocks of all objects in your theory.
     
  4. Sep 26, 2006 #3
    This sounds like prime numbers! "If an integer is prime it cannot be written as a product of strictly smaller integers." Is there meant to be an analogy here? Besides, prime numbers are thought of as building blocks for the integers.
     
  5. Sep 26, 2006 #4
    What is a finite group? Is it one which is simply generated by finitely many generators? I would say that a finite group is indecomposable. I have no clue as to why at this stage though. This is just a gut feeling.

    What is a simple group? Is it simply a non-trivial group with no non-trivial normal subgroups? I would guess that a simple group is not indecomposable.

    EDIT: Actually I take that back. I think that if [itex]G \equiv A \star B[/itex] is simple then it is freely indecomposable. Because if G is simple then it has no non-trivial normal subgroups.
     
    Last edited: Sep 26, 2006
  6. Sep 26, 2006 #5

    matt grime

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    A finite group is one with finitely many elements (*not* generators).

    Anyway, it is clearly impossible to write a finite group as the free product of two non-trivial groups, so a finite group is, I suppose, free indecomposable.

    The integers are also freely indecomosable with that definition.

    Since there are finite simple groups your two gut feelings are contrdictory.

    Z^2 is also free indecomposable, I believe. F_2 is not, nor is any free group with more than 2 generators.


    It is probably easier to describe what a 'free decomposition' of a group is.

    G is freely decomposable if we can pick a partition of a set of generators of G into I and J, with G= <I>*<J> (<I> is the group generated by I).

    I.e. we can split generators for G into two subsets which have no relations between them.
     
    Last edited: Sep 26, 2006
  7. Sep 26, 2006 #6
    Why is this clear? Why will I be unable to find two non-trivial groups whose free product will be finite? Is it because of how the free product is constructed?

    Did you post this after my edit? My gut now thinks that both simple and finite groups are freely indecomposable. And thanks to you I almost understand why the finite groups are freely indecomposable.
     
  8. Sep 26, 2006 #7

    matt grime

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    Just look at the definition of a free product. If G=A*B and A and B are non-trival, then there are nonidentity a and b in A and B. Now what can you say about ab,abab,ababab,ababababab, etc?
     
  9. Sep 26, 2006 #8

    matt grime

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    I don't follow your reasoning about simple groups. If G=A*B, then that does not say that A is normal in G. Indeed, it is clearly not normal *by the definition of free product*. I am almost forced to the conclusion that you do not know what a free product really is.
     
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