This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In the previous post, we examined the definition of a monoid; in particular, the natural numbers as defined in set theory form a commutative monoid under addition. In this post, we introduce a method for constructing an abelian group from a commutative semigroup.
First, consider the category \(\Ab\) of abelian groups. Since any abelian group can be viewed as a commutative monoid by forgetting the information about inverses, there is a forgetful functor \(U: \Ab \rightarrow \cMon\). It is known that this functor has a left adjoint \(K:\cMon \rightarrow \Ab\), and writing out this adjunction yields the formula
\[\Hom_\Ab(K(M), G)\cong\Hom_\cMon(M, U(G))\]That is, given a commutative monoid \(M\) and a monoid homomorphism \(M\rightarrow U(G)\), there must exist a unique group homomorphism \(K(M)\rightarrow G\).
Universal mapping problem
We now show the existence of the left adjoint \(K\) described above. More generally, following our main reference [Bou], we examine the process of constructing an abelian group from a commutative semigroup. Meanwhile, using the unit of the adjunction, we can spell out the properties that \(K\) must satisfy.
The abelian group \(K(S)\) and the semigroup homomorphism \(\eta_S:S\rightarrow K(S)\) constitute a pair satisfying the following property.
(Universal mapping problem) Given any abelian group \(G\) and any semigroup homomorphism \(f:S\rightarrow G\), there exists a unique group homomorphism \(\bar{f}:K(S)\rightarrow G\) such that \(f=\bar{f}\circ\eta_S\).
Intuitively, \(K(S)\) may be thought of as the smallest abelian group containing \((S,+)\).
Any \(K(S)\) satisfying the above property is unique up to isomorphism.
Proposition 1 If an abelian group \(H\) and a semigroup homomorphism \(\eta_S'\) satisfy the above universal mapping problem, then \(K(S)\cong H\).
Proof
First, consider the following diagram.
Then by the universal property, there exists a map \(\bar{\eta}_S': K(S)\rightarrow H\) such that \(\eta_S'= \bar{\eta}_S'\circ\eta_S\). On the other hand, from the following diagram
using the universal property for \(H\), there exists a map \(\bar{\eta}_S:H\rightarrow K(S)\) such that \(\eta_S=\bar{\eta}_S\circ\eta_S'\). Then
\[\bar{\eta}_S'\circ\bar{\eta}_S\circ\eta_S'=\bar{\eta}_S'\circ \eta_S=\eta=\id_{H}\circ \eta_S'\]and by the universal property again, the map \(f\) satisfying \(f\circ \eta_S'=\eta_S'\) is unique, so \(f=\id_H=\bar{\eta}_S'\circ \bar{\eta}_S\). Or, in the language of diagrams, the unique map \(H\rightarrow H\) making the following diagram commute must be \(\id_H=\bar{\eta}_S'\circ \bar{\eta}_S\).
Similarly, one can show that \(\id_{K(S)}=\bar{\eta}_S\circ \bar{\eta}_S'\), and therefore \(K(S)\cong H\).
On the other hand, if \(S\) is already an abelian group, then \(K(S)\) should be \(S\) itself without adjoining any new elements.
Proposition 2 If \(S\) is an abelian group, then the abelian group \(K(S)\) satisfying the above universal mapping problem is isomorphic to \(S\).
Proof
Since \(S\) and \(\id_S\) trivially satisfy the universal property, by Proposition 1, any abelian group satisfying the universal property must be isomorphic to \(S\).
The above propositions show that a \(K(S)\) satisfying the universal mapping problem is the abelian group we seek, but they do not establish that such a \(K(S)\) actually exists.
Definition of \(K(S)\)
The reason why \(S\) cannot be an abelian group is that inverses may not exist for all elements. Intuitively, this can be resolved by adding negatives.
Given a commutative semigroup \((S,+)\), consider the product semigroup \(S\times S\). (§Algebraic Structures, ⁋Example 5) Interpreting the second component of \(S\times S\) as a negative, the equation
\[(a_1, b_1)+(a_2, b_2)=(a_1+a_2, b_1+b_2)\]can be viewed as representing
\[(a_1+a_2)-(b_1+b_2)=(a_1-b_1)+(a_2-b_2)\]Of course, in general the formal difference \(a-b\) is far from well-defined, so we define an equivalence relation \(R\) on \(S\times S\) as follows.
\[(a_1, b_1)\equiv (a_2, b_2)\pmod{R}\iff a_1+b_2+c=a_2+b_1+c\text{ for some $c\in S$}\]First, we show that this relation is an equivalence relation.
Lemma 3 The relation \(R\) defined above is an equivalence relation on the product semigroup \(S\times S\) compatible with its operation.
Proof
First, we show that \(R\) is an equivalence relation. For any \((a,b)\in S\times S\),
\[a+b+c=a+b+c\]holds for any \(c\in S\); hence \((a,b)\equiv(a,b)\). Suppose \((a_1,b_1)\equiv (a_2,b_2)\). That is, for some \(c\in S\),
\[a_1+b_2+c=a_2+b_1+c\]holds. But this is precisely the condition for \((a_2,b_2)\equiv (a_1,b_1)\), so \(R\) is symmetric. Finally, suppose \((a_1,b_1)\equiv(a_2,b_2)\) and \((a_2,b_2)\equiv (a_3,b_3)\). Then for some \(c\) and \(c'\),
\[a_1+b_2+c=a_2+b_1+c,\qquad a_2+b_3+c'=a_3+b_2+c'\]hold. Adding the two equations,
\[a_1+b_3+(a_2+b_2+c+c')=a_3+b_1+(a_2+b_2+c+c')\]so \((a_1,b_1)\equiv(a_3,b_3)\). Thus \(R\) is an equivalence relation.
Now we must show that \(R\) is compatible with the operation on \(S\times S\). To this end, suppose \((a_1, b_1)\equiv(a_1',b_1')\) and \((a_2, b_2)\equiv (a_2',b_2')\). We must show that \((a_1+a_2, b_1+b_2)\equiv(a_1'+a_2', b_1'+b_2')\). By the given conditions, there exist \(c_1\) and \(c_2\) such that
\[a_1+b_1'+c_1=a_1'+b_1+c_1,\qquad a_2+b_2'+c_2=a_2'+b_2+c_2\]hold. Adding these two equations yields
\[(a_1+a_2)+(b_1'+b_2')+(c_1+c_2)=(a_1'+a_2')+(b_1+b_2)+(c_1+c_2)\]so by definition \((a_1+a_2, b_1+b_2)\equiv(a_1'+a_2', b_1'+b_2')\pmod{R}\). Thus \(R\) is compatible with the operation on \(S\times S\).
Therefore, \((S\times S)/R\) is a commutative semigroup. We denote this by \(K(S)\).
Lemma 4 \(K(S)\) is an abelian group.
Proof
It suffices to show that \(K(S)\) has an identity element and inverses. Since we regard \((a,b)\) as \(a-b\), the identity element should be of the form \((a,a)\), and the inverse of \((a,b)\) should be \(-(a-b)=b-a\), i.e. \((b,a)\). Let us prove this.
First, for any \(c\in S\), we show that \([(c,c)]\) is the identity element. For any \([(a,b)]\in K(S)\),
\[[(a,b)]+[(c,c)]=[(a+c, b+c)]\]holds. Since
\[(a+c)+b+d=(b+c)+a+d\]holds for any \(d\in S\), we have \((a+c, b+c)\equiv (a,b)\), and therefore \([(a+c, b+c)]=[(a,b)]\). By commutativity, \([(c,c)]+[(a,b)]=[(a,b)]\) also holds, so \([(c,c)]\) is the identity element of \(K(S)\).
On the other hand, for any \([(a,b)]\in K(S)\),
\[[(a,b)]+[(b,a)]=[(a+b,a+b)]\]so by the preceding argument, \([(a,b)]+[(b,a)]\) is the identity element of \(K(S)\), and the same holds for \([(b,a)]+[(a,b)]\). Thus every element of \(K(S)\) has an inverse, so \(K(S)\) carries a group structure.
Thus \(K(S)\) becomes the abelian group we were looking for. That is, \(K(S)\) satisfies the above universal mapping problem.
Proposition 5 For a commutative semigroup \((S, +)\), the abelian group \(K(S)\) constructed as above, together with the natural semigroup homomorphism \(\eta_S:S\rightarrow K(S)\), satisfies the universal property.
Proof
First, let us consider what the natural semigroup homomorphism from \(S\) to \(K(S)\) should be. Since we regard an element \((a,b)\) in \(K(S)\) as \(a-b\), we see that \(a\) corresponds to \((a+b)-b\) in \(K(S)\), i.e. \([(a+b, b)]\). Therefore, we define \(\eta_S\) by \(a\mapsto[(a+a, a)]\). Of course, choosing any \(b\) and defining \(a\mapsto[(a+b,b)]\) yields the same element.
To prove the universal property, let an arbitrary abelian group \(G\) and a semigroup homomorphism \(f:S\rightarrow G\) be given.
First, if \(\bar{f}:K(S)\rightarrow S\) satisfying the given property exists, then \(\bar{f}\) must be unique. For any \([(a,b)]\),
\[\begin{aligned}\bar{f}\left([(a,b)]\right)&=\bar{f}\left([(a+(a+b), b+(a+b))]\right)=\bar{f}\left([(a+a,a)]+[(b, b+b)]\right)\\ &\bar{f}\left([(a+a, a)]\right)+\bar{f}\left([(b,b+b)]\right)=\bar{f}\left(\eta_S(a)\right)-\bar{f}\left(\eta_S(b)\right)\\ &=f(a)-f(b)\end{aligned}\]so its values on individual elements are uniquely determined.
Guided by the uniqueness argument, we define \(\bar{f}([(a,b)])=f(a)-f(b)\). First, this definition is well defined. That is, if \((a_1,b_1)\equiv(a_2,b_2)\), then \(f(a_2)-f(b_2)=f(a_1)-f(b_1)\). Since \((a_1,b_1)\equiv(a_2,b_2)\), there exists \(c\in S\) such that \(a_1+b_2+c=a_2+b_1+c\); therefore
\[f(a_1)+f(b_2)+f(c)=f(a_1+b_2+c)=f(a_2+b_1+c)=f(a_2)+f(b_1)+f(c)\]Subtracting \(f(c)\) from both sides and rearranging, we obtain
\[f(a_1)-f(b_1)=f(a_2)-f(b_2)\]Moreover, \(\bar{f}\) is a group homomorphism. For any \([(a_1, b_1)]\) and \([(a_2,b_2)]\),
\[\begin{aligned}\bar{f}\left([(a_1,b_1)]+[(a_2, b_2)]\right)&=\bar{f}\left([(a_1+a_2, b_1+b_2)]\right)=f(a_1+a_2)-f(b_1+b_2)\\&=f(a_1)+f(a_2)-f(b_1)-f(b_2)=(f(a_1)-f(b_1))+(f(a_2)-f(b_2))\\&=\bar{f}\left([(a_1, b_1)]\right)+\bar{f}\left([(a_2,b_2)]\right)\end{aligned}\]holds.
Finally, that \(\bar{f}\) satisfies the required condition \(f=\bar{f}\circ\eta_S\) follows by a direct computation.
Thus, we have obtained the desired abelian group \(K(S)\). In particular, this yields a rigorous definition of the integers.
Definition 6 For the monoid \((\mathbb{N},+)\), the abelian group obtained through the above process is denoted by \((\mathbb{Z},+)\).
Monoid of fractions
In the above discussion, we obtained \(K(S)\) by adjoining inverses for all elements of \(S\). On the other hand, looking at Definition 6, what we are actually doing is adjoining inverses only for the elements of the subset \(\mathbb{N}\setminus\{0\}\) of \(\mathbb{N}\). This can also be achieved with slight modifications to the above discussion; we omit the proofs and only sketch the process.
Consider a commutative monoid \(E\), a subset \(S\) of \(E\), and the submonoid \(S'\) of \(E\) generated by \(S\). We also assume that the operation of \(E\) is written as multiplication. Define the following relation on \(E\times S'\):
\[(a,p)\equiv (b,q)\pmod{R}\iff aqs=bps\text{ for some $s\in S'$}\]Then this relation is an equivalence relation on \(E\times S'\) compatible with its operation, so \((E\times S')/R\) is a monoid.
Definition 7 The monoid \((E\times S')/R\) obtained as above is called the monoid of fractions of \(E\) with denominator \(S\), denoted by \(E_S\). Elements \((a,p)\) of this monoid are denoted by \(a/p\).
In this case, since \(E\) is a monoid, unlike in the above discussion, it has an identity element \(1\). Then the homomorphism \(\eta_S\) in Proposition 5 can be described explicitly as \(a\mapsto a/1\).
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
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