The Central Idea
Semigroups, monoids, and groups are distinguished by which structural properties their operations satisfy.
The progression is:
\[ \boxed{ \text{semigroup} \longrightarrow \text{monoid} \longrightarrow \text{group} }. \]Each step adds one new requirement.
- A semigroup has an associative binary operation.
- A monoid is a semigroup with an identity element.
- A group is a monoid in which every element has an inverse.
Start with a Binary Operation
Let \(S\) be a set with a binary operation
\[ \star:S\times S\to S. \]Because \(\star\) is a binary operation on \(S\), closure is already built into the notation:
\[ a,b\in S \quad\Longrightarrow\quad a\star b\in S. \]From there, we ask whether the operation is associative, whether an identity exists, and whether inverses exist.
What Is a Semigroup?
A semigroup is a set equipped with an associative binary operation.
Thus \((S,\star)\) is a semigroup if:
- \(\star\) is a binary operation on \(S\);
- \(\star\) is associative.
Associativity means
\[ \boxed{ (a\star b)\star c = a\star(b\star c) } \]for all
\[ a,b,c\in S. \]A Semigroup Need Not Have an Identity
The definition of semigroup does not require an identity element.
Therefore a semigroup may have one, but it does not have to.
This is the first important distinction in the hierarchy:
\[ \text{associativity alone} \]is enough for a semigroup.
Example: Positive Integers Under Addition
Let
\[ \mathbb{Z}_{>0}=\{1,2,3,\ldots\}. \]Addition is closed on this set:
\[ a,b\in\mathbb{Z}_{>0} \Longrightarrow a+b\in\mathbb{Z}_{>0}. \]Addition is also associative:
\[ (a+b)+c=a+(b+c). \]Therefore,
\[ \boxed{(\mathbb{Z}_{>0},+)} \]is a semigroup.
Why Positive Integers Under Addition Are Not a Monoid
The additive identity would need to be
\[ 0. \]But
\[ 0\notin\mathbb{Z}_{>0}. \]Therefore the positive integers under addition have no identity element within the set.
So
\[ \boxed{(\mathbb{Z}_{>0},+)} \]is a semigroup but not a monoid.
What Is a Monoid?
A monoid is a semigroup that has an identity element.
Therefore \((S,\star)\) is a monoid if:
- \(\star\) is a binary operation on \(S\);
- \(\star\) is associative;
- there exists an identity element \(e\in S\).
The identity satisfies
\[ \boxed{ e\star a=a\star e=a } \]for every \(a\in S\).
Every Monoid Is a Semigroup
A monoid satisfies every requirement of a semigroup and adds an identity.
Therefore,
\[ \boxed{\text{every monoid is a semigroup}}. \]The converse is not true.
A semigroup without an identity is not a monoid.
Example: Nonnegative Integers Under Addition
Consider
\[ \mathbb{Z}_{\ge0}=\{0,1,2,3,\ldots\}. \]Addition is closed and associative.
The identity element is
\[ 0, \]because
\[ 0+a=a+0=a. \]Therefore,
\[ \boxed{(\mathbb{Z}_{\ge0},+)} \]is a monoid.
Why Nonnegative Integers Under Addition Are Not a Group
A group requires every element to have an inverse.
Under addition, the inverse of \(a\) is
\[ -a. \]But if \(a>0\), then
\[ -a\notin\mathbb{Z}_{\ge0}. \]For example, \(5\) would need the inverse \(-5\), which is outside the set.
Therefore
\[ \boxed{(\mathbb{Z}_{\ge0},+)} \]is a monoid but not a group.
What Is a Group?
A group is a monoid in which every element has an inverse.
Thus \((G,\star)\) is a group if:
- \(\star\) is a binary operation on \(G\);
- \(\star\) is associative;
- there is an identity element \(e\in G\);
- every element \(a\in G\) has an inverse \(a^{-1}\in G\).
The inverse condition is
\[ \boxed{ a\star a^{-1} = a^{-1}\star a = e }. \]Every Group Is a Monoid
A group satisfies every monoid requirement and adds inverses.
Therefore,
\[ \boxed{\text{every group is a monoid}}. \]Since every monoid is a semigroup,
\[ \boxed{\text{every group is also a semigroup}}. \]The implications run downward through the hierarchy.
The Hierarchy
The structure can be summarized as
\[ \boxed{ \text{group} \subset \text{monoid} \subset \text{semigroup} }. \]Here the subset notation means that every group is a monoid and every monoid is a semigroup.
It does not mean that the underlying sets themselves are necessarily subsets of one another.
Example: Integers Under Addition Form a Group
Consider
\[ (\mathbb{Z},+). \]Closure:
\[ a,b\in\mathbb{Z} \Longrightarrow a+b\in\mathbb{Z}. \]Associativity:
\[ (a+b)+c=a+(b+c). \]Identity:
\[ 0+a=a+0=a. \]Inverse:
\[ a+(-a)=(-a)+a=0. \]Therefore,
\[ \boxed{(\mathbb{Z},+)\text{ is a group}}. \]Example: Nonzero Rational Numbers Under Multiplication
Let
\[ \mathbb{Q}^{\times} = \mathbb{Q}\setminus\{0\}. \]Multiplication is closed on \(\mathbb{Q}^{\times}\), because the product of two nonzero rational numbers is nonzero and rational.
Multiplication is associative.
The identity is
\[ 1. \]Every \(a\in\mathbb{Q}^{\times}\) has inverse
\[ \frac1a\in\mathbb{Q}^{\times}. \]Therefore,
\[ \boxed{(\mathbb{Q}^{\times},\cdot)\text{ is a group}}. \]Why Zero Must Be Removed
If we tried to use all rational numbers under multiplication,
\[ (\mathbb{Q},\cdot), \]the identity \(1\) would still exist.
But \(0\) has no multiplicative inverse.
No rational number \(b\) satisfies
\[ 0\cdot b=1. \]Therefore
\[ (\mathbb{Q},\cdot) \]is a monoid but not a group.
Groups Need Not Be Commutative
The definition of a group does not require
\[ a\star b=b\star a. \]A group operation may be commutative, but it does not have to be.
This is one of the most important distinctions in group theory.
Abelian Groups
A group whose operation is commutative is called an abelian group.
Thus an abelian group satisfies
\[ \boxed{ a\star b=b\star a } \]for all
\[ a,b\in G. \]The name honors mathematician Niels Henrik Abel.
Example: Integers Under Addition Are Abelian
We already know
\[ (\mathbb{Z},+) \]is a group.
Addition is also commutative:
\[ a+b=b+a. \]Therefore,
\[ \boxed{(\mathbb{Z},+)\text{ is an abelian group}}. \]Example: Invertible Matrices Form a Group
Let
\[ GL_n(\mathbb{R}) \]denote the set of invertible \(n\times n\) real matrices.
Under matrix multiplication:
- the product of invertible matrices is invertible;
- matrix multiplication is associative;
- the identity matrix \(I\) is the identity element;
- every matrix \(A\in GL_n(\mathbb{R})\) has inverse \(A^{-1}\).
Therefore,
\[ \boxed{(GL_n(\mathbb{R}),\cdot)\text{ is a group}}. \]Matrix Groups Are Usually Nonabelian
For matrices,
\[ AB \]and
\[ BA \]are generally different.
Therefore, for \(n\ge2\),
\[ GL_n(\mathbb{R}) \]is generally a nonabelian group.
This shows that associativity, identity, and inverses do not force commutativity.
Functions Under Composition
Let \(X\) be a set, and consider all functions
\[ f:X\to X. \]Under composition, these functions form a monoid:
- composition is closed;
- composition is associative;
- the identity function \(\operatorname{id}_X\) is the identity.
But not every function has an inverse.
Therefore all self-maps of \(X\) form a monoid, not generally a group.
Bijective Functions Under Composition Form a Group
If we restrict to bijections
\[ f:X\to X, \]then every function has an inverse function.
Therefore the bijections from \(X\) to itself form a group under composition.
This group is called the symmetric group on \(X\) when \(X\) is finite.
Addition Modulo \(n\)
Let
\[ \mathbb{Z}_n=\{0,1,\ldots,n-1\} \]with addition modulo \(n\).
This operation is closed and associative.
The identity is
\[ 0. \]Every element \(a\) has an additive inverse modulo \(n\).
Therefore,
\[ \boxed{(\mathbb{Z}_n,+)\text{ is a group}}. \]Addition Modulo \(n\) Is Abelian
Modular addition satisfies
\[ a+b\equiv b+a\pmod n. \]Therefore
\[ (\mathbb{Z}_n,+) \]is not merely a group but an abelian group.
Example: Addition Modulo 4
In
\[ \mathbb{Z}_4=\{0,1,2,3\}, \]the identity is \(0\).
The inverses are
\[ 0^{-1}=0, \] \[ 1^{-1}=3, \] \[ 2^{-1}=2, \] \[ 3^{-1}=1, \]where the inverse notation is relative to addition modulo \(4\).
Multiplication Modulo \(n\) Is Different
Under multiplication modulo \(n\), not every nonzero residue necessarily has an inverse.
For example, modulo \(4\),
\[ 2\cdot0\equiv0, \] \[ 2\cdot1\equiv2, \] \[ 2\cdot2\equiv0, \] \[ 2\cdot3\equiv2 \pmod4. \]The result is never \(1\).
Thus \(2\) has no multiplicative inverse modulo \(4\).
Units Modulo \(n\)
The elements of \(\mathbb{Z}_n\) that do have multiplicative inverses are called units modulo \(n\).
These units form a group under multiplication modulo \(n\).
This group is commonly written
\[ \mathbb{Z}_n^{\times} \]or
\[ U(n). \]This provides an early example of forming a group by selecting only the invertible elements of a larger monoid.
Group Axioms in Compact Form
A group \((G,\star)\) satisfies:
Closure
\[ a,b\in G \Longrightarrow a\star b\in G. \]Associativity
\[ (a\star b)\star c = a\star(b\star c). \]Identity
\[ e\star a=a\star e=a. \]Inverses
\[ a\star a^{-1} = a^{-1}\star a = e. \]If commutativity is added,
\[ a\star b=b\star a, \]the group is abelian.
Group Laws Produce Consequences
The group axioms are enough to derive many useful facts.
For example:
- the identity is unique;
- each inverse is unique;
- cancellation laws hold;
- equations of the form \(a\star x=b\) have unique solutions;
- the inverse of the identity is the identity;
- the inverse of an inverse returns the original element.
These are theorems, not additional axioms.
Cancellation in a Group
Suppose
\[ a\star b=a\star c. \]Multiply on the left by \(a^{-1}\):
\[ a^{-1}\star(a\star b) = a^{-1}\star(a\star c). \]By associativity,
\[ (a^{-1}\star a)\star b = (a^{-1}\star a)\star c. \]Since
\[ a^{-1}\star a=e, \]we get
\[ e\star b=e\star c. \]Therefore,
\[ \boxed{b=c}. \]Solving Group Equations
Suppose
\[ a\star x=b. \]Apply \(a^{-1}\) on the left:
\[ a^{-1}\star(a\star x) = a^{-1}\star b. \]By associativity,
\[ (a^{-1}\star a)\star x = a^{-1}\star b. \]Therefore,
\[ e\star x = a^{-1}\star b, \]so
\[ \boxed{x=a^{-1}\star b}. \]Order Matters in Nonabelian Groups
If the group is not commutative, then the equation
\[ x\star a=b \]is solved differently.
We multiply on the right by \(a^{-1}\):
\[ (x\star a)\star a^{-1} = b\star a^{-1}. \]Therefore,
\[ \boxed{x=b\star a^{-1}}. \]In a nonabelian group,
\[ a^{-1}\star b \]and
\[ b\star a^{-1} \]need not be equal.
The Inverse of a Product
In a group,
\[ \boxed{ (a\star b)^{-1} = b^{-1}\star a^{-1} }. \]The order reverses.
To verify,
\[ (a\star b)\star(b^{-1}\star a^{-1}) \]becomes, by associativity,
\[ a\star(b\star b^{-1})\star a^{-1}. \]Since
\[ b\star b^{-1}=e, \]this reduces to
\[ a\star e\star a^{-1}=e. \]Why the Order Reverses
Suppose two actions are performed in the order
\[ a \quad\text{then}\quad b. \]To undo the combined action, we first undo \(b\), then undo \(a\).
This explains the reversed order
\[ b^{-1}\star a^{-1}. \]The same principle appears in inverse functions, matrix inverses, and transformations.
Finite Groups and Operation Tables
A finite group can be represented by an operation table.
Because every element has an inverse and cancellation holds, each row and each column contains every group element exactly once.
Thus a finite group table has the structure of a Latin square.
This property is stronger than the requirements for an arbitrary binary operation table.
Why Group Rows Cannot Repeat an Entry
Suppose a row labeled \(a\) contained the same result twice:
\[ a\star b=a\star c. \]Left cancellation would imply
\[ b=c. \]Therefore two different columns cannot produce the same output in one row.
Since the row has the same number of entries as the group has elements, every element must appear exactly once.
Finite Group Tables and Inverses
In a finite group table, every row and every column contains the identity exactly once.
The position of the identity in row \(a\) identifies the inverse of \(a\).
This makes inverse relationships easy to locate visually.
Semigroups Can Be Commutative
Commutativity is optional at every level of this hierarchy.
A semigroup may be commutative or noncommutative.
If it is commutative, it is sometimes called a commutative semigroup.
The same idea applies to monoids.
Commutative Monoids
A monoid whose operation is commutative is called a commutative monoid.
For example,
\[ (\mathbb{Z}_{\ge0},+) \]is a commutative monoid.
It is not a group because positive elements do not have additive inverses within the set.
Structure Comparison
| Structure | Binary Operation | Associative | Identity | Every Element Invertible |
|---|---|---|---|---|
| Semigroup | Yes | Yes | Not required | Not required |
| Monoid | Yes | Yes | Yes | Not required |
| Group | Yes | Yes | Yes | Yes |
| Abelian Group | Yes | Yes | Yes | Yes, and operation is commutative |
Example Classification: \((\mathbb{Z}_{>0},+)\)
Addition is closed and associative.
But \(0\), the additive identity, is not in the set.
Therefore,
\[ \boxed{ (\mathbb{Z}_{>0},+) \text{ is a semigroup but not a monoid} }. \]Example Classification: \((\mathbb{Z}_{\ge0},+)\)
Addition is closed and associative.
The identity \(0\) belongs to the set.
But positive integers do not have additive inverses in the set.
Therefore,
\[ \boxed{ (\mathbb{Z}_{\ge0},+) \text{ is a monoid but not a group} }. \]Example Classification: \((\mathbb{Z},+)\)
Addition is closed and associative.
The identity is \(0\).
Every integer \(a\) has inverse \(-a\).
Addition is also commutative.
Therefore,
\[ \boxed{ (\mathbb{Z},+) \text{ is an abelian group} }. \]Example Classification: \((\mathbb{Z},\cdot)\)
Integer multiplication is closed and associative.
The identity is \(1\).
But most integers do not have multiplicative inverses in \(\mathbb{Z}\).
For example, \(2^{-1}=\frac12\notin\mathbb{Z}\).
Therefore,
\[ \boxed{ (\mathbb{Z},\cdot) \text{ is a commutative monoid but not a group} }. \]Example Classification: \((\mathbb{R}^{\times},\cdot)\)
Let
\[ \mathbb{R}^{\times}=\mathbb{R}\setminus\{0\}. \]Multiplication is closed and associative.
The identity is \(1\).
Every element \(a\ne0\) has inverse \(\frac1a\).
Multiplication is commutative.
Therefore,
\[ \boxed{ (\mathbb{R}^{\times},\cdot) \text{ is an abelian group} }. \]Common Error: Requiring Commutativity for Every Group
A group does not have to be commutative.
The group axioms require:
- closure;
- associativity;
- an identity;
- inverses.
Commutativity is an additional condition that makes the group abelian.
Common Error: Assuming a Monoid Has Inverses
A monoid requires an identity but does not require every element to be invertible.
For example,
\[ (\mathbb{Z}_{\ge0},+) \]is a monoid, but \(5\) has no additive inverse in the set.
Common Error: Assuming Every Semigroup Has an Identity
An identity is not part of the semigroup definition.
The positive integers under addition form a semigroup even though \(0\) is missing.
Adding an identity is exactly what promotes a semigroup to a monoid.
Common Error: Forgetting Closure
Before testing associativity, identity, or inverses, confirm that the operation is actually internal to the set.
For example, subtraction on positive integers fails closure:
\[ 3-5=-2\notin\mathbb{Z}_{>0}. \]Therefore it cannot define a semigroup on that set.
Common Error: Finding an Inverse Outside the Set
An inverse must belong to the structure.
In
\[ (\mathbb{Z},\cdot), \]the number \(2\) has reciprocal
\[ \frac12, \]but
\[ \frac12\notin\mathbb{Z}. \]Therefore \(2\) is not invertible in this monoid.
A Reliable Classification Procedure
- Identify the set and operation.
- Verify that the operation is closed on the set.
- Check associativity.
- If associativity fails, the structure is not a semigroup.
- If associativity holds, look for an identity.
- If no identity exists, the structure may be a semigroup but not a monoid.
- If an identity exists, test whether every element has an inverse.
- If every element is invertible, the structure is a group.
- Finally, test commutativity to determine whether the group is abelian.
Example: Full Classification Workflow
Consider
\[ S=\{0,1,2\} \]with addition modulo \(3\).
Closure
Every modular sum is one of
\[ 0,1,2. \]So the operation is closed.
Associativity
Modular addition inherits associativity from ordinary integer addition.
Identity
The identity is
\[ 0. \]Inverses
The inverses are
\[ 0^{-1}=0, \] \[ 1^{-1}=2, \] \[ 2^{-1}=1. \]Commutativity
Modular addition is commutative.
Therefore,
\[ \boxed{ (\mathbb{Z}_3,+) \text{ is an abelian group} }. \]Practice 1: Semigroup or Not?
Determine whether
\[ (\mathbb{Z}_{>0},+) \]is a semigroup.
Practice 2: Monoid or Not?
Determine whether
\[ (\mathbb{Z}_{>0},+) \]is a monoid.
Practice 3: Classify Nonnegative Integer Addition
Classify
\[ (\mathbb{Z}_{\ge0},+) \]as precisely as possible among semigroup, monoid, and group.
Practice 4: Classify Integer Addition
Classify
\[ (\mathbb{Z},+). \]Practice 5: Classify Integer Multiplication
Classify
\[ (\mathbb{Z},\cdot) \]as precisely as possible among semigroup, monoid, and group.
Practice 6: Abelian or Nonabelian?
Is
\[ (\mathbb{Z},+) \]abelian?
Practice 7: Nonzero Reals Under Multiplication
Explain why
\[ (\mathbb{R}^{\times},\cdot) \]is a group.
Practice 8: Why All Real Numbers Fail
Explain why
\[ (\mathbb{R},\cdot) \]is not a group.
Practice 9: Solve a Group Equation
In a group \((G,\star)\), solve
\[ a\star x=b \]for \(x\).
Practice 10: Identify the Error
A student says:
“Every monoid is a group because every monoid has an identity.”
Explain the error.
Practice Answers
Practice 1
Yes. Positive integers are closed under addition, and addition is associative.
Therefore,
\[ \boxed{(\mathbb{Z}_{>0},+)\text{ is a semigroup}}. \]Practice 2
No. The additive identity is \(0\), but
\[ 0\notin\mathbb{Z}_{>0}. \]Therefore the structure is not a monoid.
Practice 3
Addition is associative and has identity \(0\), so
\[ (\mathbb{Z}_{\ge0},+) \]is a monoid.
It is not a group because positive elements have no additive inverses in the set.
Practice 4
The integers under addition form a group, and addition is commutative.
Therefore,
\[ \boxed{(\mathbb{Z},+)\text{ is an abelian group}}. \]Practice 5
Integer multiplication is associative and has identity \(1\), so
\[ (\mathbb{Z},\cdot) \]is a monoid.
It is not a group because most integers do not have multiplicative inverses in \(\mathbb{Z}\).
Practice 6
Yes. Integer addition satisfies
\[ a+b=b+a. \]Therefore \((\mathbb{Z},+)\) is abelian.
Practice 7
Nonzero real numbers are closed under multiplication, multiplication is associative, \(1\) is the identity, and every nonzero real number \(a\) has inverse
\[ \frac1a. \]Therefore,
\[ \boxed{(\mathbb{R}^{\times},\cdot)\text{ is a group}}. \]Practice 8
The element \(0\) has no multiplicative inverse.
Therefore not every element of \(\mathbb{R}\) is invertible under multiplication.
Hence \((\mathbb{R},\cdot)\) is not a group.
Practice 9
Multiply on the left by \(a^{-1}\):
\[ a^{-1}\star(a\star x) = a^{-1}\star b. \]Associativity and the identity law give
\[ \boxed{x=a^{-1}\star b}. \]Practice 10
An identity is not enough to make a monoid a group.
A group also requires every element to have an inverse.
For example,
\[ (\mathbb{Z}_{\ge0},+) \]is a monoid but not a group.
What to Remember
The hierarchy is
\[ \boxed{ \text{semigroup} \longrightarrow \text{monoid} \longrightarrow \text{group} }. \]- A semigroup has an associative binary operation.
- A monoid is a semigroup with an identity.
- A group is a monoid in which every element has an inverse.
- An abelian group is a group whose operation is commutative.
Therefore,
\[ \boxed{ \text{group} \subset \text{monoid} \subset \text{semigroup} }. \]Each step adds structure, and each added property produces stronger algebraic consequences.
Where We Go Next
Groups provide a reusable framework for studying invertible operations. They become especially powerful when we examine smaller groups contained within larger ones and groups arising from geometric symmetries.
The next lesson, Subgroups and Symmetry Groups, develops the subgroup idea, tests for subgroup structure, and connects group operations to rotations, reflections, permutations, and symmetry.