Tensors of Free Groups and Abelian groups

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SUMMARY

The discussion centers on the relationship between the tensor product of free groups and Abelian groups, specifically demonstrating that for any set S, the equation ##\mathbb{Z}\{S\} \otimes G \sim |S|G = \bigoplus_{s \in S} G## holds true. The author establishes a bilinear map T from ##\mathbb{Z}\{S\} \times G## to ##\bigoplus_{s \in S} G##, confirming its bijectiveness through the universal property of tensor products. This approach effectively illustrates the structure of the tensor product in relation to direct sums of Abelian groups.

PREREQUISITES
  • Understanding of free groups, specifically ##\mathbb{Z}\{S\}##
  • Familiarity with tensor products of Abelian groups
  • Knowledge of bilinear maps and their properties
  • Basic concepts of direct sums in group theory
NEXT STEPS
  • Study the properties of tensor products in Abelian group theory
  • Explore the universal property of tensor products in depth
  • Investigate bilinear maps and their applications in algebra
  • Learn about the structure and properties of free groups
USEFUL FOR

Mathematicians, algebraists, and students studying group theory, particularly those interested in the properties of free groups and tensor products in the context of Abelian groups.

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##.
 
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Define a map

T:\mathbb{Z}\{S\}\times G\rightarrow \bigoplus_{s\in S} G

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

T:\mathbb{Z}\{S\}\otimes G\rightarrow \bigoplus_{s\in S} G

by the universal property of tensor products. Check that it is a bijection.
 
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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