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Tensors of Free Groups and Abelian groups

  1. May 27, 2014 #1

    WWGD

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    Hi, let S be any set and let ##Z\{S\}## be the free group on ##S##, i.e., ##Z\{S\}## is
    the collection of all functions of finite support on ##S##. I am trying to show
    that for an Abelian group ##G## , we have that :

    ## \mathbb Z\{S\}\otimes G \sim |S|G = \bigoplus_{ s \in S} G ##, i.e., the direct sum of ##|S|## copies of ##G## , where ## \otimes ## is the tensor product of Abelian groups.


    I know that for ## \mathbb Z ## the integers , then ##\mathbb Z \otimes G \sim G ##( tensor over the integers ) by, e.g., the map ##(z,g)\rightarrow zg =g+g+...+g## ##z## times. I know two that any
    basis for ## \mathbb Z\{S\} ## has cardinality that of ##S##.

    But I don't see why ## \mathbb Z\{S\}\otimes G = \bigoplus_{ s \in S} G## , the direct sum of ##|S|## copies of ##G##.
     
    Last edited: May 27, 2014
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  3. May 27, 2014 #2

    micromass

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    Define a map

    [tex]T:\mathbb{Z}\{S\}\times G\rightarrow \bigoplus_{s\in S} G[/tex]

    such that for ##s_k\in S## and ##c_k\in \mathbb{Z}##, we have ##T(c_1 s_1 + ... + c_n s_n, g) = (x_s)_{s\in S}## with ##x_{s_k} = c_kg##and ##x_s = 0_G## otherwise.

    For example, we define ##T:\mathbb{Z}\{a,b,c\}\times G\rightarrow G^3## by ##T(na + mb + kc, g) = (ng, mg, kg)##.

    Check that this map is bilinear as a map between ##\mathbb{Z}##-modules. Thus it gives rise to a map

    [tex]T:\mathbb{Z}\{S\}\otimes G\rightarrow \bigoplus_{s\in S} G[/tex]

    by the universal property of tensor products. Check that it is a bijection.
     
  4. May 27, 2014 #3

    WWGD

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    I see, every bilinear map factors thru the quotient of the free module on S by the module of the bilinearity relations , and this quotient is the tensor product, so we get a linear map from the tensor product to ##\bigoplus _{ s\in S} G . I can see it is a bijection. This is a useful approach I have seen used in some results. My algebra obviously needs work. Thanks.
     
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