Closure, Associativity, Semigroup and Monoid MCQs: 12 Solved Questions

Students often remember the words closure, semigroup and monoid, but apply the properties in the wrong order when an unfamiliar operation appears. The hard cases are operations with no familiar name: x⊕y = x² + y², max(a, b), a + b + 3, and gcd on the divisors of 12. Choose an option before reading each explanation, then use the solution as a property-by-property audit instead of another fact to memorise.

Closure, associativity and identity: use the structure ladder

The ladder runs in one direction:

1. A binary operation is closed if combining any two elements of the set returns an element of the same set.

2. Closure plus associativity gives a semigroup.

3. A semigroup with a two-sided identity is a monoid.

4. A monoid in which every element has an inverse is a group.

Commutativity is not required for a semigroup, monoid or group. Check in this order: set and operation, closure, associativity, two-sided identity, inverses, then commutativity if asked. For example, 4 + 5 = 9 stays in {4,5,6,...}, but the required additive identity 0 is outside that set. The full algebraic-structure ladder develops these distinctions further.

Questions 1 to 3: recognise the defining properties

Question 1

Which of the following is a necessary condition for a set to be a group?

• A. Closure

• B. Associativity

• C. Existence of Identity

• D. All of the above

Answer: D. All of the above. A group needs closure, associativity and an identity, along with an inverse for every element. The option does not say that the three listed conditions are sufficient by themselves. It says, correctly, that each one is necessary.

Question 2

A semigroup must satisfy:

• A. Closure and Associativity

• B. Closure and Existence of Identity

• C. Closure and Existence of Inverse

• D. Associativity and Existence of Identity

Answer: A. Closure and Associativity. For every a,b in the set, a*b must also lie in the set. For every a,b,c, the operation must satisfy (a*b)*c = a*(b*c). Identity and inverses belong to later rungs of the structure ladder.

Question 3

A monoid is a semigroup with:

• A. Commutative Property

• B. Associative Property

• C. Identity Element

• D. Inverse Element

Answer: C. Identity Element. Associativity is already part of being a semigroup. A monoid adds an element e such that e*a = a and a*e = a for every a. Commutativity is optional. Requiring an inverse for every element would produce a group.

Questions 4 to 6: missing associativity, identity and idempotence

Question 4

A set with an operation that satisfies closure, but not associativity, is called:

• A. Group

• B. Semigroup

• C. Monoid

• D. None of these

Answer: D. None of these. Failure of associativity rules out every named structure in the options. Closure alone is not enough to make a semigroup, so it cannot make a monoid or group either.

Question 5

What is the identity element in the group of integers under addition?

• A. 0

• B. 1

• C. -1

• D. Any integer

Answer: A. 0. Solve a+e=a. Subtracting a gives e=0. It works from both sides: 37+0=37 and 0+37=37. Under multiplication the identity is 1, which shows why the operation, not merely the set, determines the identity.

Question 6

In a group (G, *), for an element 'a' in G, if 'a * a = a', what is 'a'?

• A. Identity element

• B. Inverse element

• C. Finite element

• D. Infinite element

Answer: A. Identity element. Start with a*a=a and left-multiply both sides by a⁻¹. By associativity,

a⁻¹*(a*a) = (a⁻¹*a)*a = e*a = a,

while the right side becomes a⁻¹*a=e. Therefore a=e. A semigroup can contain other idempotent elements, but the identity is the only idempotent element in a group.

Questions 7 to 9: test the operation, not its symbol

Question 7

A binary operation ⊕ on the set of integers is defined as

x⊕y = x² + y²

Which one of the following statements is TRUE about ⊕?

• A. Commutative but not associative

• B. Both commutative and associative

• C. Associative but not commutative

• D. Neither commutative nor associative

Answer: A. Commutative but not associative. It is commutative because x²+y²=y²+x². To test associativity, take x=1, y=1 and z=0:

• (1⊕1)⊕0 = 2⊕0 = 2²+0² = 4

• 1⊕(1⊕0) = 1⊕1 = 1²+1² = 2

Since 4 != 2, associativity fails. One counterexample is enough to disprove a property claimed for every triple.

Question 8

The algebraic structure ({4, 5, 6, ...}, +) is

• A. Not a semigroup

• B. A semigroup, but not a commutative semigroup

• C. A commutative semigroup, but not monoid

• D. A monoid

Answer: C. A commutative semigroup, but not monoid. Addition is closed here because, for example, 4+5=9 remains in the set. Also, (4+5)+6=15=4+(5+6), and 4+5=5+4. The structure is therefore a commutative semigroup. Its additive identity would have to be 0, but 0 does not belong to {4,5,6,...}, so it is not a monoid.

Question 9

Consider a set of integers Z, with respect to *, such that a * b = max(a, b). Which of the following is true?

• A. Algebraic structure

• B. semi-group

• C. Monoid

• D. group

Answer: B. semi-group. The maximum of two integers is an integer, so the operation is closed. It is associative because taking a maximum is unaffected by grouping. With actual values,

max(2,max(-5,7))=7=max(max(2,-5),7).

There is no identity in the integers for this operation. For any proposed identity e, choose a=e-1. Then max(a,e)=e, not a. Thus the structure is a semigroup, but not a monoid.

Questions 10 to 12: prove each claim before naming the structure

Question 10

Let (Z,*) be an algebraic structure where * is defined by a*b=a+b+3, ∀a,b∈Z. The algebraic structure is

• A. Not a semigroup

• B. A semigroup, but not monoid

• C. A monoid, but not a commutative monoid

• D. A commutative monoid.

Answer: D. A commutative monoid. The result is an integer, so closure holds. For associativity,

• (a*b)*c = (a+b+3)+c+3 = a+b+c+6

• a*(b*c) = a+(b+c+3)+3 = a+b+c+6

For the identity, solve a*e=a: a+e+3=a, so e=-3. The other side also works because e*a=-3+a+3=a. Finally, a*b=a+b+3=b+a+3=b*a, so the operation is commutative.

There is a stronger classification than the options ask for. Solving a*b=-3 gives b=-a-6, which is an integer for every integer a. Every element therefore has an inverse, so this is actually an abelian group. Option D is still true, but it is not the strongest possible label.

Question 11

Let ({a,b},*) be a semigroup, where a*a=b.

(A) a*b=b*a

(B) b*b=b

Choose the most appropriate answer from the options given below:

• A. (A) only true

• B. (B) only true

• C. Both (A) and (B) true

• D. Neither (A) nor (B) true

Answer: C. Both (A) and (B) true. Associativity gives

(a*a)*a = a*(a*a), so b*a=a*b. This proves (A).

Let p=a*b=b*a. Since the set is exactly {a,b}, either p=a or p=b. Now

(a*a)*b = a*(a*b), so b*b=a*p.

If p=a, then a*p=a*a=b. If p=b, then a*p=a*b=p=b. In both cases b*b=b, proving (B).

Question 12

(D₁₂, *) where a*b = g.c.d of (a, b) ∀a, b ∈ D₁₂ then (D₁₂, *) is

• A. a semigroup but not monoid

• B. a monoid but not a group

• C. a group

• D. not a semi group

Answer: B. a monoid but not a group. Here D₁₂={1,2,3,4,6,12}. The gcd of two divisors of 12 is also a divisor of 12, so closure holds, and gcd is associative. The identity is 12 because gcd(a,12)=a for every a in D₁₂.

It is not a group. For a=6, gcd(6,b) can never equal the identity 12, so 6 has no inverse.

Four traps: counterexamples, two-sided identity, stronger labels and commutativity

1. A counterexample and a proof do different jobs. The result 4 != 2 in Question 7 disproves associativity. Several examples that happen to work would not prove it. A proof of associativity must cover arbitrary a,b,c.

2. An identity must work from both sides and belong to the set. Check both a*e=a and e*a=a. The missing 0 in {4,5,6,...} and the absence of a smallest integer under max show why this membership check matters.

3. Do not stop at the first true label. Every group is also a monoid and semigroup. Question 10 is a commutative monoid, but a complete audit reveals that it is an abelian group.

4. Commutativity is an extra property. A semigroup or monoid need not be commutative. Keep it separate from closure, associativity and identity when working through mixed Discrete Mathematics MCQs.

Short version and the next practice step

For any new operation, ask five questions in order: Is it closed? Is it associative? Is there a two-sided identity? Does every element have an inverse? Is it commutative? Redo Questions 7, 9, 10 and 12 without looking. Together they cover a counterexample, a missing identity, a shifted operation and a gcd identity.

Use GATE Guidance by Sanchit Sir for a sequenced Discrete Mathematics path, or Zero to Hero for a broader CS fundamentals route. The CS Fundamentals category is the practical browsing route for related topics.