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Semigroups and Monoids
Definition 1 A magma \((A, \star)\) satisfying the associative law is called a semigroup.
The homomorphism, substructure, and quotient structures defined on a magma all carry over to semigroups without any modification. In particular, if \(A\) is a semigroup, then any submagma \(S\) of \(A\) is also a semigroup.
Definition 2 For an arbitrary magma \((A,\star)\), if some element \(e\in A\) satisfies
\[x\star e=e\star x=x\]for all \(x\in A\), then \(e\) is called an identity element.
An arbitrary magma \(A\) has at most one identity element. Indeed, if both \(e\) and \(e'\) are identity elements of \(A\), then
\[e=e\star e'=e'\star e=e'\]holds.
On the other hand, an arbitrary element of a set \(A\) is the same as a function from a singleton \(\ast\) to \(A\). ([Category Theory] §Representable Functors, ⁋Example 2) From this perspective, the fact that \(e:\ast\rightarrow A\) is an identity element means that the following diagram
commutes.
Definition 3 If a semigroup \((A,\star)\) has an identity element, it is called a monoid.
Since a monoid is well-defined only when we have the set \(A\), the operation \(\star\) defined on it, and the identity element \(e\) for \(\star\), we denote a monoid by a tuple such as \((A,\star, e)\). From the above discussion, we see that a monoid is a monoid object in \(\Set\). ([Category Theory] §Monoid Objects, ⁋Definition 1)
Some care is needed when defining monoid homomorphism and submonoid. For example, for two monoids \((A,\star,e)\) and \((A',\star',e')\), a magma homomorphism \(f:A\rightarrow A'\) need not preserve the identity element, so a monoid homomorphism is defined to also preserve the data of the identity element \(e\).
Definition 4 For two monoids \((A, \star, e)\) and \((A',\star', e')\), a magma homomorphism satisfying \(f(e)=e'\) is called a monoid homomorphism.
Monoids and monoid homomorphisms defined in this way form a category.
Proposition 5 There exists a category \(\Mon\) whose objects are monoids and whose morphisms are monoid homomorphisms.
Proof
Suppose arbitrary monoid homomorphisms \(f:M_1\rightarrow M_2\) and \(g:M_2\rightarrow M_3\) are given. Then by §Algebraic Structures, ⁋Proposition 7, \(g\circ f\) is a magma homomorphism. Moreover, from the following equation
\[(g\circ f)(e_1)=g(f(e_1))=g(e_2)=e_3\]we also see that \(g\circ f\) is a monoid homomorphism.
Since monoid homomorphisms are functions, their composition satisfies the associative law. Also, for any monoid \(M\), the identity function \(\id_M\) is always a monoid homomorphism.
Also, a submagma of a monoid need not contain the identity element, so a new definition is needed as follows.
Definition 6 A submonoid of a monoid \((A,\star, e)\) is a submagma of \(A\) that contains the identity element \(e\).
However, if a family \((S_i)\) of submonoids of a monoid \((A,\star,e)\) is given, then the intersection \(S=\bigcap S_i\) is again a submonoid. This is because \(e\in S_i\) for all \(i\), and therefore \(e\in S\).
In the case of quotient structures, if a monoid \((A, \star,e)\) and an equivalence relation \(R\) compatible with \(\star\) are given, then \(A/R\) naturally inherits a monoid structure. Considering the equivalence class \([e]\) of \(e\) in the set \(A/R\), for any \([x]\in A/R\) we have
\[[x]\mathbin{\tiny\char"2606}[e]=[x\star e]=[x]=[e\star x]=[e]\mathbin{\tiny\char"2606}[x]\]so the identity property holds.
Assuming the existence of an identity element on a magma is one of the stronger conditions. For example, the following theorem shows that for any set \(X\), there is only one way to endow it with two compatible magma structures and identity elements, and the result is a commutative monoid.
Theorem 7 (Eckmann-Hilton) Suppose two operations \(\star_1\) and \(\star_2\) are defined on a set \(X\) such that \((X,\star_1,e_1)\) and \((X,\star_2,e_2)\) are both magmas with identity elements. If
\[(a\star_1 b)\star_2(c\star_1 d)=(a\star_2 c)\star_1(b\star_2 d)\]holds for all \(a,b,c,d\in X\), then \(\star=\star_1=\star_2\), \(e=e_1=e_2\), and \((X,\star,e)\) is a commutative monoid.
Proof
First, we show that \(e_1=e_2\). This follows from the following equation:
\[e_1=e_1\star_1 e_1=(e_1\star_2e_2)\star_1(e_2\star_2e_1)=(e_1\star_1 e_2)\star_2(e_2\star_1 e_1)=e_2\star_2 e_2=e_2\]Now for arbitrary \(a,b\) we have
\[a\star_1 b=(a\star_2 e_2)\star_1(e_2\star_2b)=(a\star_1 e_2)\star_2(e_2\star_1b)=a\star_2b\]so \(\star=\star_1=\star_2\), and
\[a\star b=(e\star a)\star(b\star e)=(e\star b)\star(a\star e)=b\star a\]and
\[a\star(b\star c)=(a\star 1)\star(b\star c)=(a\star b)\star(1\star c)=(a\star b)\star c\]so \((X,\star,e)\) is a commutative monoid.
Groups
We finally define a group. Intuitively, it can be thought of as a monoid in which every element has an inverse.
Definition 8 For a monoid \((A,\star,e)\), an element \(x\) is left cancellable if for any \(a,b\in A\), \(x\star a=x\star b\) implies \(a=b\). Similarly, we can define an element that is right cancellable. Also, a left inverse of \(y\) is an element \(x\) such that \(x\star y=e\). Similarly, we can define that \(x\) is a right inverse of \(y\).
If \(x\) is both a left inverse and a right inverse of \(y\), then \(x\) is called an inverse of \(y\), and in this case \(y\) is called invertible.
A general monoid may have a left inverse but not a right inverse, or conversely have a right inverse but not a left inverse. It is common to write the inverse of \(x\) as \(x^{-1}\), but if the operation is denoted by \(+\), we write \(-x\) instead. For such notation to make sense, the inverse must be uniquely determined.
Proposition 9 For a monoid \((A, \star, e)\), if \(x\in A\) is an invertible element of \(A\), then the inverse of \(x\) is unique.
Proof
If both \(x'\) and \(x''\) were inverses of \(x\), then
\[x'=x'\star e=x'\star( x\star x'')=(x'\star x)\star x''=e\star x''=x''\]so \(x'=x''\).
Using this, we obtain the following corollary.
Corollary 10 For invertible elements \(a,b\) of a monoid \((A,\star,e)\), the following hold:
- $(a^{-1})^{-1}=a$
- $(a\star b)^{-1}=b^{-1}\star a^{-1}$.
Proof
By the previous proposition, the inverse is unique, so it suffices to verify directly that the right-hand sides of the given equations satisfy the condition for being an inverse.
First, we check whether \(a\) is the inverse of \(a^{-1}\). The inverse of \(a^{-1}\) is an element \(x\) satisfying the two equations
\[a^{-1}\star x=x\star a^{-1}=e\]However, since
\[a^{-1}\star a=a\star a^{-1}=e\]holds by the definition of \(a^{-1}\), we see that \(x=a\) satisfies the above equations. Since the inverse of \(a^{-1}\) is now unique, the inverse of \(a^{-1}\), namely \((a^{-1})^{-1}\), must be \(a\).
Similarly, the second claim also follows immediately from the following two equations:
\[\begin{aligned}(a\star b)\star(b^{-1}\star a^{-1})&=a\star(b\star b^{-1})\star a^{-1}=a\star e\star a^{-1}=a\star a^{-1}=e,\\(b^{-1}\star a^{-1})\star(a\star b)&=b^{-1}\star(a^{-1}\star a)\star b=b^{-1}\star e\star b=b^{-1}\star b=e.\end{aligned}\]A group is now defined as follows.
Definition 11 A monoid in which every element is invertible is called a group. If \(\star\) satisfies the commutative law, it is called an abelian group (or commutative group).
Taking the inverse is a function from \(G\) to itself1, and therefore a group \(G\) is determined by the data \((G,\star,e, (-)^{-1})\). The inverse \((-)^{-1}\) can be expressed by the following diagram:
From this, we can verify that any group is a group object in \(\Set\). ([Category Theory] §Monoid Objects, ⁋Definition 1)
On the other hand, a monoid homomorphism \(f:G\rightarrow G'\) must preserve inverses:
\[f(x)\star'f(x^{-1})=f(x\star x^{-1})=f(e)=e',\qquad f(x^{-1})\star'f(x)=f(x^{-1}\star x)=f(e)=e'.\]Thus \(\Grp\) is a full subcategory of \(\Mon\). ([Category Theory] §Functors, ⁋Definition 10)
Moreover, for a magma homomorphism \(f:G\rightarrow G'\) between two groups, we have
\[e'\star' f(e)=f(e)=f(e\star e)=f(e)\star'f(e)\]and operating on both sides on the right by the inverse of \(f(e)\) gives \(e'=f(e)\). Thus by the preceding argument, \(\Grp\) is also a full subcategory of \(\Magma\).
The above argument used the following lemma.
Lemma 12 (Cancellation law) Any invertible element is cancellable.
Proof
Operate on the left or right of both sides by the inverse of \(a\).
On the other hand, for the same reason as in Proposition 5, groups and group homomorphisms also form a category.
Proposition 13 There exists a category \(\Grp\) whose objects are groups and whose morphisms are group homomorphisms. Also, there exists a full subcategory \(\Ab\) whose objects are abelian groups and whose morphisms are group homomorphisms.
We can verify that these categories have a zero object \(\{e\}\). Just as with submonoids, we can define a subgroup.
Definition 14 A subset \(S\) of a group \((G,\star, e, {}^{-1})\) is called a subgroup if \(S\) is a submonoid closed under taking inverses.
The following proposition tells us that we can determine whether a given subset is a subgroup using a single criterion, without having to check the existence of an identity element, closure under inverses, etc.
Proposition 15 A nonempty subset \(S\) of a group \((G, \star, e, {}^{-1})\) is a subgroup of \(G\) if and only if \(a\star b^{-1}\in S\) always holds for any \(a,b\in S\).
Proof
If \(S\) is a subgroup of \(G\), then since \(b\in S\) we have \(b^{-1}\in S\), and therefore \(a\star b^{-1}\in S\) holds trivially.
Thus it suffices to prove the converse. Since \(S\) is nonempty, there exists some \(a\in S\), and then \(a\star a^{-1}\in S\), so \(e\in S\). Now for any \(a\in S\), we have \(a^{-1}=e\star a^{-1}\in S\). Also, for any \(a,b\in S\), we have \(a\star b=a\star(b^{-1})^{-1}\in S\).
For a family \((S_i)\) of subgroups of a group \(G\), the intersection \(S=\bigcap S_i\) is a subgroup. This is because if we choose any \(a,b\in S\), then \(ab^{-1}\in S_i\) for all \(i\), and therefore \(ab^{-1}\in S\). In particular, for any subset \(S\) of \(G\), applying this discussion to the collection of subgroups of \(G\) containing \(S\) yields the smallest subgroup containing \(S\). We denote this by \(\langle S\rangle\). With a little effort, we can also prove that \(\langle S\rangle\) coincides with the set of all elements obtained by applying the group operation finitely many times to elements of \(S\).
On the other hand, for a group \((G, \star, e)\) and an equivalence relation \(R\) compatible with \(\star\), we verified that \(G/R\) has a monoid structure; moreover, \(G/R\) also has a group structure. To verify this, it suffices to show that any element \([x]\) of \(G/R\) is invertible. Since
\[[x]\mathbin{\tiny\char"2606}\bigl[x^{-1}\bigr]=\bigl[x\star (x^{-1})\bigr]=[e]=\bigl[x^{-1}\star x\bigr]=\bigl[x^{-1}\bigr]\mathbin{\tiny\char"2606}[x]\]holds, we see that every element of \(G/R\) is invertible.
From now on, when dealing with a general group we always denote the operation by multiplication, and thus write \(x\star y\) simply as \(xy\), the inverse of \(x\) as \(x^{-1}\), and the identity element as \(e\). However, if the group \(G\) is specifically abelian, we denote the operation by addition, the inverse of \(x\) by \(-x\), and the identity element by \(0\).
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
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Except when \(G\) is an abelian group, \((-)^{-1}\) is not a group homomorphism. ↩
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