Tensors of Free Groups and Abelian groups

[WWGD]

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 $G$.

 

[micromass]

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.

 

[WWGD]

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.