# The Grothendienck Group

## Free group

Let \(A\) be an abelian group. Let
\((e_i)_{i \in I}\) be a family of
elements of \(A\). We say that this
family is a **basis** for \(A\) if the family is not empty, and if
every element of \(A\) has a unique
expression as a **linear expression** \[
x = \sum_{i \in I} x_i e_i
\] where \(x_i \in \mathbb{Z}\)
and almost all \(x_i\) are equal to
\(0\). This means that the sum is
actually finite. An abelian group is said to be **free** if
it has a basis. Alternatively, we may write \(A\) as a direct sum by \[
A \cong \bigoplus_{i \in I}\mathbb{Z}e_i.
\]

## Free abelian group generated by a set

Let \(S\) be a set. Say we want to
get a group out of this for some reason, so how? It is not a good idea
to endow \(S\) with a binary operation
beforehead since overall \(S\) is
merely a set. We shall **generate** a group out of \(S\) in the most **freely**
way.

Let \(\mathbb{Z}\langle S \rangle\)
be the set of all **maps** \(\varphi:S \to \mathbb{Z}\) such that, for
only a **finite** number of \(x
\in S\), we have \(\varphi(x) \neq
0\). For simplicity, we denote \(k
\cdot x\) to be some \(\varphi_0 \in
\mathbb{Z}\langle S \rangle\) such that \(\varphi_0(x)=k\) but \(\varphi_0(y) = 0\) if \(y \neq x\). For any \(\varphi\), we claim that \(\varphi\) has a unique expression \[
\varphi=k_1 \cdot x_1 + k_2 \cdot x_2 + \cdots + k_n \cdot x_n.
\] One can consider these integers \(k_i\) as the order of \(x_i\), or simply the time that \(x_i\) appears (may be negative). For \(\varphi\in\mathbb{Z}\langle S \rangle\),
let \(I=\{x_1,x_2,\cdots,x_n\}\) be the
set of elements of \(S\) such that
\(\varphi(x_i) \neq 0\). If we denote
\(k_i=\varphi(x_i)\), we can show that
\(\psi=k_1 \cdot x_1 + k_2 \cdot x_2 + \cdots
+ k_n \cdot x_n\) is equal to \(\varphi\). For \(x \in I\), we have \(\psi(x)=k\) for some \(k=k_i\neq 0\) by definition of the '\(\cdot\)'; if \(y
\notin I\) however, we then have \(\psi(y)=0\). This coincides with \(\varphi\). \(\blacksquare\)

By definition the zero map \(\mathcal{O}=0 \cdot x \in \mathbb{Z}\langle S \rangle\) and therefore we may write any \(\varphi\) by \[ \varphi=\sum_{x \in S}k_x\cdot x \] where \(k_x \in \mathbb{Z}\) and can be zero. Suppose now we have two expressions, for example \[ \varphi=\sum_{x \in S}k_x \cdot x=\sum_{x \in S}k_x'\cdot x \] Then \[ \varphi-\varphi=\mathcal{O}=\sum_{x \in S}(k_x-k'_x)\cdot x \] Suppose \(k_y - k_y' \neq 0\) for some \(y \in S\), then \[ \mathcal{O}(y)=k_y-k_y'\neq 0 \] which is a contradiction. Therefore the expression is unique. \(\blacksquare\)

This \(\mathbb{Z}\langle S \rangle\)
is what we are looking for. It is an additive group (which can be proved
immediately) and, what is more important, every element can be expressed
as a 'sum' associated with finite number of elements of \(S\). We shall write \(F_{ab}(S)=\mathbb{Z}\langle S \rangle\),
and call it the **free abelian group generated by \(S\)**. For elements in \(S\), we say they are **free
generators** of \(F_{ab}(S)\).
If \(S\) is a finite set, we say \(F_{ab}(S)\) is **finitely
generated**.

An abelian group is

freeif and only if it is isomorphic to a free abelian group \(F_{ab}(S)\) for some set \(S\).

**Proof.** First we shall show that \(F_{ab}(S)\) is free. For \(x \in M\), we denote \(\varphi = 1 \cdot x\) by \([x]\). Then for any \(k \in \mathbb{Z}\), we have \(k[x]=k \cdot x\) and \(k[x]+k'[y] = k\cdot x + k' \cdot
y\). By definition of \(F_{ab}(S)\), any element \(\varphi \in F_{ab}(S)\) has a unique
expression \[
\varphi = k_1 \cdot x_1 + \cdots + k_n \cdot x_n
=k_1[x_1]+\cdots+k_n[x_n]
\] Therefore \(F_{ab}(S)\) is
free since we have found the basis \(([x])_{x
\in S}\).

Conversely, if \(A\) is free, then it is immediate that its basis \((e_i)_{i \in I}\) generates \(A\). Our statement is therefore proved. \(\blacksquare\)

### The connection between an arbitrary abelian group an a free abelian group

(Proposition 1)If \(A\) is an abelian group, then there is a free group \(F\) which has a subgroup \(H\) such that \(A \cong F/H\).

**Proof.** Let \(S\) be
any set containing \(A\). Then we get a
surjective map \(\gamma: S \to A\) and
a free group \(F_{ab}(S)\). We also get
a unique homomorphism \(\gamma_\ast:F_{ab}(S)
\to A\) by \[
\begin{aligned}
\gamma_\ast:F_{ab}(S) &\to A \\
\varphi=\sum_{x \in S}k_x\cdot x &\mapsto \sum_{x \in S}k_x\gamma(x)
\end{aligned}
\] which is also surjective. By the first isomorphism theorem, if
we set \(H=\ker(\gamma_\ast)\) and
\(F_{ab}(S)=F\), then \[
F/H \cong A.
\] \(\blacksquare\)

(Proposition 2)If \(A\) is finitely generated, then \(F\) can also be chosen to be finitely generated.

**Proof.** Let \(S\) be
the generator of \(A\), and \(S'\) is a set containing \(S\). Note if \(S\) is finite, which means \(A\) is finitely generated, then \(S'\) can also be finite by inserting
one or any finite number more of elements. We have a map from \(S\) and \(S'\) into \(F_{ab}(S)\) and \(F_{ab}(S')\) respectively by \(f_S(x)=1 \cdot x\) and \(f_{S'}(x')=1 \cdot x'\). Define
\(g=f_{S'} \circ \lambda:S' \to
F_{ab}(S)\) we get another homomorphism by \[
\begin{aligned}
g_\ast:F_{ab}(S') &\to F_{ab}(S) \\
\varphi'=\sum_{x \in S'}k_{x} \cdot x &\mapsto \sum_{x \in
S'}k_{x}\cdot g(x)
\end{aligned}
\] This defines a unique homomorphism such that \(g_\ast \circ f_{S'} = g\). As one can
also verify, this map is also surjective. Therefore by the first
isomorphism theorem we have \[
A \cong F_{ab}(S) \cong F_{ab}(S')/\ker(g_\ast)
\] \(\blacksquare\)

It's worth mentioning separately that we have implicitly proved two statements with commutative diagrams:

(Proposition 3 | Universal property)If \(g:S \to B\) is a mapping of \(S\) into some abelian group \(B\), then we can define a unique group-homomorphism making the following diagram commutative:

(Proposition 4)If \(\lambda:S \to S\) is a mapping of sets, there is a unique homomorphism \(\overline{\lambda}\) making the following diagram commutative:

(In the proof of Proposition 2 we exchanged \(S\) an \(S'\).)

## The Grothendieck group

(The Grothendieck group)Let \(M\) be a commutative monoid written additively. We shall prove that there exists a commutative group \(K(M)\) with a monoid homomorphism \[ \gamma:M \to K(M) \]satisfying the following universal property: If \(f:M \to A\) is a homomorphism from \(M\) into a abelian group \(A\), then there exists a unique homomorphism \(f_\gamma:K(M) \to A\) such that \(f=f_\gamma\circ\gamma\). This can be represented by a commutative diagram:

**Proof.** There is a commutative diagram describes what
we are doing.

Let \(F_{ab}(M)\) be the free abelian group generated by \(M\). For \(x \in M\), we denote \(1 \cdot x \in F_{ab}(M)\) by \([x]\). Let \(B\) be the group generated by all elements of the type \[ [x+y]-[x]-[y] \] where \(x,y \in M\). This can be considered as a subgroup of \(F_{ab}(M)\). We let \(K(M)=F_{ab}(M)/B\). Let \(i=x \to [x]\) and \(\pi\) be the canonical map \[ \pi:F_{ab}(M) \to F_{ab}(M)/B. \] We are done by defining \(\gamma: \pi \circ i\). Then we shall verify that \(\gamma\) is our desired homomorphism satisfying the universal property. For \(x,y \in M\), we have \(\gamma(x+y)=\pi([x+y])\) and \(\gamma(x)+\gamma(y) = \pi([x])+\pi([y])=\pi([x]+[y])\). However we have \[ [x+y]-[x]-[y] \in B, \] which implies that \[ \gamma(x)+\gamma(y)=\pi([x]+[y])=\pi([x+y]) = \gamma(x+y). \] Hence \(\gamma\) is a monoid-homomorphism. Finally the universal property. By proposition 3, we have a unique homomorphism \(f_\ast\) such that \(f_\ast \circ i = f\). Note if \(y \in B\), then \(f_\ast(y) =0\). Therefore \(B \subset \ker{f_\ast}\) Therefore we are done if we define \(f_\gamma(x+B)=f_\ast (x)\). \(\blacksquare\)

### Comments of the proof

Why such a \(B\)? Note in general
\([x+y]\) is not necessarily equal to
\([x]+[y]\) in \(F_{ab}(M)\), but we don't want it to be so.
So instead we create a new **equivalence relation**, by
factoring a subgroup generated by \([x+y]-[x]-[y]\). Therefore in \(K(M)\) we see \([x+y]+B = [x]+[y]+B\), which finally makes
\(\gamma\) a homomorphism. We use the
same strategy to generate the **tensor product** of two
modules later. But at that time we have more than one relation to take
care of.

### Cancellative monoid

If for all \(x,y,z \in M\), \(x+y=x+z\) implies \(y=z\), then we say \(M\) is a cancellative monoid, or the cancellation law holds in \(M\). Note for the proof above we didn't use any property of cancellation. However we still have an interesting property for cancellation law.

(Theorem)The cancellation law holds in \(M\) if and only if \(\gamma\) is injective.

**Proof.** This proof involves another approach to the
Grothendieck group. We consider pairs \((x,y)
\in M \times M\) with \(x,y \in
M\). Define \[
(x,y) \sim (x',y') \iff \exists \ell \in M,
y+x'+\ell=x+y'+\ell.
\] Then we get a equivalence relation (try to prove it
yourself!). We define the addition component-wise, that is, \((x,y)+(x',y')=(x+x',y+y')\),
then the equivalence classes of pairs form a group \(A\), where the zero element is \([(0,0)]\). We have a monoid-homomorphism
\[
f:x \mapsto [(x,0)].
\] If cancellation law holds in \(M\), then \[
\begin{aligned}
f(x) = f(y) &\implies [(x,0)] = [(y,0)] \\
&\implies 0+y+\ell=x+0+\ell \\
&\implies x=y.
\end{aligned}
\] Hence \(f\) is injective. By
the universal property of the Grothendieck group, we get a unique
homomorphism \(f_\gamma\) such that
\(f_\gamma \circ \gamma = f\). If \(x \neq 0\) in \(M\), then \(f_\gamma \circ \gamma(x) \neq 0\) since
\(f\) is injective. This implies \(\gamma(x) \neq 0\). Hence \(\gamma\) is injective.

Conversely, if \(\gamma\) is injective, then \(i\) is injective (this can be verified by contradiction). Then we see \(f=f_\ast \circ i\) is injective. But \(f(x)=f(y)\) if and only if \(x+\ell = y+\ell\), hence \(x+ \ell = y+ \ell\) implies \(x=y\), the cancellation law holds on \(M\).

### Examples

Our first example is \(\mathbb{N}\). Elements of \(F_{ab}(\mathbb{N})\) are of the form \[ \varphi=k_1 \cdot n_1 + k_2 \cdot n_2+\cdots + k_m \cdot n_m. \] For elements in \(B\) they are generated by \[ \varphi=1\cdot (m+n)-1\cdot m - 1\cdot n \] which we wish to represent \(0\). Indeed, \(K(\mathbb{N}) \simeq \mathbb{Z}\) since if we have a homomorphism \[ \begin{aligned} f:K(\mathbb{N}) &\to \mathbb{Z} \\ \sum_{j=1}^{m}k_j \cdot n_j +B &\mapsto \sum_{j=1}^{m}k_j n_j. \end{aligned} \] For \(r \in \mathbb{Z}\), we see \(f(1 \cdot r+B)=r\). On the other hand, if \(\sum_{j=1}^{m}k_j \cdot n_j \not\in B\), then its image under \(f\) is not \(0\).

In the first example we 'granted' the natural numbers 'subtraction'. Next we grant the division on multiplicative monoid.

Consider \(M=\mathbb{Z} \setminus 0\). Now for \(F_{ab}(M)\) we write elements in the form \[ \varphi={}^{k_1}n_1{}^{k_2}n_2\cdots{}^{k_m}n_m \] which denotes that \(\varphi(n_j)=k_j\) and has no other differences. Then for elements in \(B\) they are generated by \[ \varphi = {}^1(n_1n_2){}^{-1}(n_1)^{-1}(n_2) \] which we wish to represent \(1\). Then we see \(K(M) \simeq \mathbb{Q} \setminus 0\) if we take the isomorphism \[ \begin{aligned} f:K(M) &\to \mathbb{Q} \setminus 0 \\ \left(\prod_{j=1}^{m}{}^{k_j}n_j\right)B &\mapsto \prod_{j=1}^{m}n_j^{k_j}. \end{aligned} \]

Of course this is not the end of the Grothendieck group. But for further example we may need a lot of topology background. For example, we have the topological \(K\)-theory group of a topological space to be the Grothendieck group of isomorphism classes of topological vector bundles. But I think it is not a good idea to post these examples at this timing.