# 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 free if 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 $f_\gamma(x+B)=f_\ast (x)$. $\blacksquare$

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.