Monoids and groups: identities and inverses
Objectives and prerequisites
Classify a stated set with a stated operation, check closure and associativity, find an identity and inverses, and compute an element's order in a small finite group. Prerequisites: sets, functions, basic integer arithmetic and remainders. This foundation does not complete group theory: subgroup tests, Lagrange's theorem, homomorphisms and advanced counting with groups need separate lessons.
The object being studied is the pair (S,★), not the set S alone. The same set can be a group under one operation and not a group under another. We use nonempty carrier sets and two-sided identities and inverses.
1. Begin with a binary operation
A binary operation on S assigns every ordered pair (a,b)∈S×S exactly one result a★b in S. The requirement that the result stays in S is closure. Ordinary addition is closed on integers; ordinary division is not a binary operation on all integers, since division by zero is undefined and 1/2 is not an integer.
Associativity says (a★b)★c=a★(b★c) for every a,b,c∈S. It permits regrouping without changing the order of the elements. Commutativity says a★b=b★a and permits exchanging their order. These are different properties: an associative operation need not be commutative.
Subtraction on integers is closed but not associative. With a=3,b=2,c=1, (3−2)−1=0 whereas 3−(2−1)=2. One counterexample disproves a universal property. Closure alone therefore does not make a group or even a semigroup.
2. The hierarchy and what each step adds
A semigroup is a nonempty set with an associative binary operation. A monoid is a semigroup with an identity element e satisfying e★a=a★e=a for every a∈S. A group is a monoid in which every element a has an inverse b∈S satisfying a★b=b★a=e. An abelian group is a group whose operation is also commutative.
Read the definitions cumulatively: a group still needs closure and associativity, not just an identity and an inverse-looking expression. Every group is a monoid; not every monoid is a group. Commutativity is extra for an abelian group, not part of the definition of every group.
If e and f were both two-sided identities, then e★f=f because e is an identity, and e★f=e because f is an identity. Hence e=f. A two-sided identity is unique. In a group, an element's inverse is also unique: if b and c are inverses of a, then b=b★e=b★(a★c)=(b★a)★c=e★c=c, using associativity.
3. Worked examples: change the set or the operation
Let N₀={0,1,2,…}. Under addition, N₀ is closed and associative and has identity 0. It is a commutative monoid, but not a group: 1 would need additive inverse −1, which is outside N₀. The element 0 having an inverse does not rescue the other elements; every element needs one.
Positive integers {1,2,3,…} under addition are a semigroup but not a monoid: additive identity 0 is missing. Do not import 0 from a larger set just because it is familiar.
Integers Z under addition form an abelian group. Identity is 0 and the inverse of a is −a. Integers under multiplication form a commutative monoid with identity 1, but not a group: 2 has no integer multiplicative inverse. An expression 1/2 is a rational number, not an element of the stated integer carrier.
The term inverse here refers to an element with respect to ★. It is not the inverse function f⁻¹ from the earlier functions lesson, and its notation depends on the operation. In an additive group the inverse is commonly written −a; in a multiplicative group it is a⁻¹.
4. Modular addition gives a finite group
Z₆={0,1,2,3,4,5} with addition modulo 6 means add ordinary integers and keep the remainder in this set. Thus 4+5≡3 modulo 6. Closure holds because the result is a residue in Z₆; ordinary addition and reduction preserve associativity. Identity is 0 and addition is commutative.
Find an additive inverse by solving a+b≡0 modulo 6:
| a | Additive inverse in Z₆ |
|---|---|
| 0 | 0 |
| 1 | 5 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 1 |
Every element has an inverse, so (Z₆,+ modulo 6) is an abelian group. Notice that an element may be its own inverse, as 3 is here: 3+3≡0. An inverse need not be a different element.
In a group, the order of an element is the smallest positive number of repeated applications needed to reach the identity, if such a number exists. For additive notation, repeatedly add the element. The multiples of 2 modulo 6 are 2,4,0, so its order is 3. The identity 0 has order 1, not 0, because the definition asks for a positive number of repetitions.
The multiples of 1 run through all six residues, so 1 generates this group. An element whose powers or additive multiples generate the whole group is a generator; a group with a generator is cyclic. Element 2 does not generate Z₆, since it reaches only {0,2,4}. Do not confuse the order of an element with its numeric label or with the size of the whole group.
5. Modular multiplication changes the answer
Now use the same carrier Z₆ but multiplication modulo 6. It is closed and associative, with identity 1. However, 0 has no inverse because 0·b≡0 for every b. Element 2 also has no inverse: its products modulo 6 are only 0,2,4, never 1. Thus this structure is a monoid but not a group.
Restrict instead to U={1,5}, keeping multiplication modulo 6:
| ★ | 1 | 5 |
|---|---|---|
| 1 | 1 | 5 |
| 5 | 5 | 1 |
The table shows closure, identity 1 and an inverse for every element: 1 and 5 are each their own inverse. Associativity follows from multiplication modulo 6, so this restricted structure is an abelian group. Changing the carrier changed the classification; it was not enough merely to rename the operation.
For a new arbitrary operation table, do not assume associativity because the table is symmetric. Symmetry across the diagonal checks commutativity. Associativity is a statement about triples; for a carrier of n elements, a direct complete check has n³ triples. A valid algebraic argument can replace that enumeration when the operation is already known.
6. A monoid that is not commutative
Let S be all finite strings over {a,b}, including the empty string ε. Use concatenation as ★: put one string after another. Concatenating two allowed strings stays in S. It is associative: (uv)w and u(vw) form the same ordered sequence of symbols. The empty string is the identity because εu=uε=u.
This monoid is not commutative: a★b=ab whereas b★a=ba. It is not a group: no finite string can concatenate with a to produce ε, since the result would have length at least 1. The empty string alone has an inverse. This example separates associativity, commutativity and invertibility without relying only on number operations.
A classification checklist
- Write the carrier and operation, including any modulo rule.
- Check the operation is defined on every pair and its result remains inside the carrier.
- Prove associativity or find one failing triple.
- Find a two-sided identity that actually belongs to the carrier.
- For a group claim, find a two-sided inverse inside the carrier for every element.
- Check commutativity separately; do not assume it from the word group.
When rejecting a property, give a specific witness: a result outside the carrier, a failing triple, a missing identity, or an element without an inverse. When confirming a property, checking one favorable example is insufficient. A complete small table or a general argument is needed.
Original practice: six single-correct MCQs
Original questions, not previous-year questions. Choose exactly one option. All stated domains and operations matter.
C01. How is (N₀,+), where N₀={0,1,2,…}, classified?
A. A commutative monoid, but not a group B. An abelian group C. A semigroup with no identity D. Not closed under addition
Answer: A
A: Addition is closed and associative with identity 0, but 1 has no additive inverse in N₀.
B: The required inverse −1 of 1 is outside the carrier, so it is not a group.
C: 0 belongs to N₀ and is a two-sided identity.
D: Adding two nonnegative integers gives another nonnegative integer, so closure holds.
C02. What is the additive inverse of 2 in Z₆ under addition modulo 6?
A. 2 B. 3 C. 4 D. 1/2
Answer: C
A: 2+2≡4, not the additive identity 0.
B: 2+3≡5, not 0.
C: 2+4≡0 modulo 6, and the same holds in the reverse order.
D: This suggests a multiplicative inverse over rationals; it is not a residue in Z₆ and not an additive inverse.
C03. What is the order of element 2 in the additive group Z₆?
A. 2 B. 3 C. 6 D. 0
Answer: B
A: The numeric label 2 is not its order; two additions give 4, not 0.
B: The successive positive multiples are 2,4,0 modulo 6; the first return to identity occurs after three.
C: Six is the group size, but the element returns to identity after only three repetitions.
D: Order is defined using a positive repetition count, so 0 cannot be the answer.
C04. Why is Z₆ with multiplication modulo 6 not a group?
A. The operation is not closed. B. The identity 1 is absent. C. Multiplication modulo 6 is not associative. D. The element 0 has no multiplicative inverse.
Answer: D
A: Reduction modulo 6 keeps every product inside Z₆, so it is closed.
B: 1 is in Z₆ and acts as the identity.
C: The modular operation inherits associativity from integer multiplication.
D: For every b, 0·b≡0, never identity 1; lacking even one inverse defeats the group condition.
C05. All finite strings over {a,b}, including ε, use concatenation. Which description is correct?
A. An abelian group with identity ε B. A semigroup with no identity C. A noncommutative monoid with identity ε, not a group D. Not a semigroup because ab≠ba
Answer: C
A: Nonempty strings lack inverses, and concatenation is not commutative.
B: The empty string is a two-sided identity, so the structure is a monoid.
C: Concatenation is associative, ε is identity, ab≠ba shows noncommutativity, and a has no inverse.
D: ab≠ba disproves commutativity, not associativity; semigroups need not be commutative.
C06. On integers define a★b=a−b. Which statement correctly explains why this is not a semigroup?
A. Closure fails because 3−2 is not an integer. B. Associativity fails: (3−2)−1=0 but 3−(2−1)=2. C. It is a group because every integer has a negative. D. Every semigroup must be commutative.
Answer: B
A: 3−2=1 is an integer; subtraction is closed on integers.
B: One unequal result for the two groupings disproves associativity, a required semigroup property.
C: Having a negative concerns addition and does not repair subtraction’s nonassociativity.
D: Semigroups require associativity, not commutativity.
Sources and scope
This original foundation lesson aligns with Discrete Mathematics in Section 1 of the official GATE 2027 CS syllabus. It is not full-topic or full-syllabus coverage. The syllabus confirms relevance; the NPTEL course is a primary institutional further-study reference. Examples and practice are original, not copied examination questions.
Sources checked 2026-10-14.
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