Algebraic Structures and Groups PYQs for GATE CS

Solve 6+ Algebraic Structures and Groups previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Algebraic Structures and Groups

Orientation

Algebraic Structures and Groups

A structured route from binary operations to algebraic laws, then to monoids, groups, cyclicity, and subgroups.

Binary Operations and Algebraic Laws

Custom operators, closure, commutativity, associativity, distributivity.

Foundation

Monoids of Functions under Pointwise Operations

Pointwise operations, identity elements on function spaces.

Moderate

Groups from Set Operations and Direct Products

Symmetric difference, power sets, Cartesian products of groups.

High Yield

Group Properties, Cyclicity and Subgroups

Generators, Lagrange's theorem, subgroup tests.

High Yield

What is a Binary Operation?

Concept

What is a Binary Operation?

The core idea

A binary operation is a rule that combines two elements from a set and returns exactly one element.

The important part is not the symbol. The important part is whether the output stays inside the same set.

Formal definition

* : S \times S \to S

It accepts an ordered pair $(a, b)$ with $a, b \in S$ .
It gives exactly one result.
We write the result as $a * b = c$ .

Golden rule: closure The result

c

must belong to

S

. If even one valid input pair produces an output outside

S

, the rule is not a binary operation on

S

Example: on positive integers,

3 - 5 = - 2

. Since

- 2

is not positive, subtraction is not closed on the positive integers.

Algebraic Structures and Groups: Solved Questions with Step-by-Step Explanations (6 Problems)

Question 1 · Engineering Mathematics · 2025_Set1 MSQ

A = {0, 1, 2, 3, \dots}

is the set of non-negative integers. Let

F

be the set of functions from

A

to itself. For any two functions,

f_{1}, f_{2} \in F

, we define

(f_{1} ⨀ f_{2}) (n) = f_{1} (n) + f_{2} (n)

for every number

n

A

. Which of the following is/are CORRECT about the mathematical structure

(F, ⨀)

A.
$(F, ⨀)$ is an Abelian group.
B.
$(F, ⨀)$ is an Abelian monoid.
C.
$(F, ⨀)$ is a non-Abelian group.
D.
$(F, ⨀)$ is a non-Abelian monoid.

Correct Answer:

["B"]

Step-by-Step Solution

Insight: Pointwise operations inherit algebraic properties from the base set. The base set here is non-negative integers under addition, which forms a monoid but not a group.

Exam route: Check the axioms in order. Closure: sum of non-negative integers is non-negative (Yes). Associativity: inherited from integer addition (Yes). Identity: the zero function $z (n) = 0$ maps $A \to A$ (Yes). Inverses: for $f (n) = 1$ , the inverse would need $g (n) = - 1 \in / A$ (No). Commutativity: integer addition is commutative (Yes). Thus, it is an Abelian monoid.

Learning route:

The set $F$ contains functions $f : A \to A$ where $A = {0, 1, 2, \dots}$ .
The operation is $(f_{1} ⨀ f_{2}) (n) = f_{1} (n) + f_{2} (n)$ .
Since $f_{1} (n) \geq 0$ and $f_{2} (n) \geq 0$ , their sum is $\geq 0$ , so the result is in $F$ . Closure holds.
Associativity and commutativity follow directly from the properties of standard addition.
The identity is the constant function $e (n) = 0$ . Since $0 \in A$ , $e \in F$ .
For inverses, consider $f (n) = 1$ . We need $g (n)$ such that $1 + g (n) = 0 ⟹ g (n) = - 1$ . But $- 1 \in / A$ , so $g \in / F$ . Inverses fail.

Conclusion: It is an Abelian monoid.

Question 2 · Engineering Mathematics · 2024_Set2 NAT

Let $Z_{n}$ be the group of integers ${0, 1, 2, \dots, n - 1}$ with addition modulo $n$ as the group operation. The number of elements in the group $Z_{2} \times Z_{3} \times Z_{4}$ that are their own inverses is __________

Correct Answer:

4.00

Step-by-Step Solution

Insight: In a direct product of groups, an element is its own inverse if and only if each of its components is its own inverse in the respective factor group.

Exam route: For $Z_{n}$ under addition, an element $x$ is its own inverse if $x + x \equiv 0 (mod n) ⟹ 2 x \equiv 0 (mod n)$ . Count the solutions for $n = 2, 3, 4$ independently and multiply them.

Learning route:

The group is $Z_{2} \times Z_{3} \times Z_{4}$ under component-wise addition modulo $n$ . The identity is $(0, 0, 0)$ .
An element $(x, y, z)$ is its own inverse if $(x, y, z) + (x, y, z) = (0, 0, 0)$ , which means $2 x \equiv 0 (mod 2)$ , $2 y \equiv 0 (mod 3)$ , and $2 z \equiv 0 (mod 4)$ .
In $Z_{2}$ : $2 x \equiv 0 (mod 2)$ is satisfied by $x \in {0, 1}$ . Count = 2.
In $Z_{3}$ : $2 y \equiv 0 (mod 3)$ is satisfied only by $y = 0$ (since $g cd (2, 3) = 1$ ). Count = 1.
In $Z_{4}$ : $2 z \equiv 0 (mod 4)$ is satisfied by $z \in {0, 2}$ . Count = 2.
Total number of self-inverse elements = $2 \times 1 \times 2 = 4$ .

Common trap: Assuming only the identity element is self-inverse. This is true for odd-order cyclic groups, but for even-order cyclic groups, the element $n /2$ is also its own inverse.

Question 3 · Engineering Mathematics · 2024_Set1 MSQ

Consider the operators $⋄$ and $□$ defined by $a ⋄ b = a + 2 b$ , $a □ b = ab$ , for positive integers. Which of the following statements is/are TRUE?

A.
Operator $⋄$ obeys the associative law
B.
Operator $□$ obeys the associative law
C.
Operator $⋄$ over the operator $□$ obeys the distributive law
D.
Operator $□$ over the operator $⋄$ obeys the distributive law

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Substitute the definitions directly into the associativity and distributivity equations. Do not assume properties based on the symbols.

Exam route:

Option A: $(a ⋄ b) ⋄ c = (a + 2 b) + 2 c = a + 2 b + 2 c$ . $a ⋄ (b ⋄ c) = a + 2 (b + 2 c) = a + 2 b + 4 c$ . Not equal. False.

Option B: $□$ is standard multiplication. $(ab) c = a (b c)$ . True.

Option C: $a ⋄ (b □ c) = a + 2 b c$ . $(a ⋄ b) □ (a ⋄ c) = (a + 2 b) (a + 2 c) = a^{2} + 2 a c + 2 ab + 4 b c$ . Not equal. False.

Option D: $a □ (b ⋄ c) = a (b + 2 c) = ab + 2 a c$ . $(a □ b) ⋄ (a □ c) = (ab) + 2 (a c) = ab + 2 a c$ . Equal. True.

Learning route:

For associativity, evaluate Left = $(a ⋄ b) ⋄ c$ and Right = $a ⋄ (b ⋄ c)$ .
For distributivity, carefully identify which operation is distributing and which is being distributed over.
$⋄$ over $□$ means $a ⋄ (b □ c) = (a ⋄ b) □ (a ⋄ c)$ .
$□$ over $⋄$ means $a □ (b ⋄ c) = (a □ b) ⋄ (a □ c)$ .
Plug in the definitions $a ⋄ b = a + 2 b$ and $a □ b = ab$ and expand.

Question 4 · Engineering Mathematics · 2023 MSQ

Let

X

be a set and

2^{X}

denote the powerset of

X

.
Define a binary operation

Δ

2^{X}

as follows:

A Δ B = (A - B) \cup (B - A) .

Let

H = (2^{X}, Δ)

. Which of the following statements about

H

is/are correct?

A.
$H$ is a group.
B.
Every element in $H$ has an inverse, but $H$ is NOT a group.
C.
For every $A \in 2^{X}$ , the inverse of $A$ is the complement of $A$ .
D.
For every $A \in 2^{X}$ , the inverse of $A$ is $A$ .

Correct Answer:

["A","D"]

Step-by-Step Solution

Insight: This is a "verify group axioms + identify the inverse" question. The operation $Δ$ is symmetric difference, a classic group structure on power sets where every element turns out to be its own inverse.

Exam route:

Verify the four group axioms for $H = (2^{X}, Δ)$ :

Closure: For any $A, B \subseteq X$ , $A Δ B \subseteq X$ , so $A Δ B \in 2^{X}$ . ✓
Associativity: Symmetric difference is associative: $(A Δ B) Δ C = A Δ (B Δ C)$ . ✓
Identity: We need $E$ such that $A Δ E = A$ for all $A$ . Since $A Δ\emptyset = (A - \emptyset) \cup (\emptyset - A) = A$ , the identity is $\emptyset$ . ✓
Inverse: We need $B$ such that $A Δ B = \emptyset$ . $(A - B) \cup (B - A) = \emptyset ⟹ A - B = \emptyset$ and $B - A = \emptyset ⟹ A \subseteq B$ and $B \subseteq A ⟹ B = A$ . So every element is its own inverse. ✓

All axioms hold, so $H$ is a group. Option A is correct; Option B is wrong.

Check the proposed inverse formulas:

Complement $A^{c} = X - A$ : $A Δ A^{c} = (A - A^{c}) \cup (A^{c} - A) = A \cup A^{c} = X$ . Since $X \neq = \emptyset$ (for non-trivial $X$ ), the complement is NOT the inverse. Option C is wrong.
Self $A$ : $A Δ A = (A - A) \cup (A - A) = \emptyset$ . This matches the identity, so $A^{- 1} = A$ . Option D is correct.

Answer: A, D

Learning route:

The operation $A Δ B = (A - B) \cup (B - A)$ picks elements that are in exactly one of $A$ or $B$ . To verify a group, we must check closure, associativity, identity, and inverses.

Closure is immediate since subsets of $X$ are closed under set difference and union.
Associativity is a standard set-theoretic identity (can be verified via characteristic functions: $Δ$ corresponds to XOR on ${0, 1}^{∣ X ∣}$ ).
The identity must satisfy $A Δ E = A$ . The only set that adds nothing and removes nothing is $\emptyset$ .
The inverse must satisfy $A Δ B = \emptyset$ . The symmetric difference is empty only when $A$ and $B$ have exactly the same elements, i.e., $B = A$ .
The complement $A^{c}$ contains all elements NOT in $A$ , so $A Δ A^{c}$ contains ALL elements of $X$ , not the empty set. This is a common confusion between "what's missing" (complement) and "what cancels out" (inverse under $Δ$ ).

Wrong-option walkthrough:

Option B fails because the student correctly finds inverses but incorrectly assumes some other axiom (usually associativity or closure) fails. In fact, all axioms hold.
Option C fails because the student confuses the set-theoretic complement with the group-theoretic inverse. The complement gives $X$ , not $\emptyset$ .

Question 5 · Engineering Mathematics · 2022 MSQ

Which of the following statements is/are TRUE for a group $G$ ?

A.
If for all $x, y \in G$ , $(x y)^{2} = x^{2} y^{2}$ , then $G$ is commutative.
B.
If for all $x \in G$ , $x^{2} = 1$ , then $G$ is commutative. Here, $1$ is the identity element of $G$ .
C.
If the order of $G$ is $2$ , then $G$ is commutative.
D.
If $G$ is commutative, then a subgroup of $G$ need not be commutative.

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Insight: Test each algebraic identity by expanding and cancelling. For small orders, use group classification.

Exam route:

Option A: $(x y)^{2} = x^{2} y^{2} ⟹ x y x y = xx y y$ . Left-multiply by $x^{- 1}$ and right-multiply by $y^{- 1}$ to get $y x = x y$ . True.

Option B: $x^{2} = 1 ⟹ x = x^{- 1}$ . Then $(x y)^{- 1} = y^{- 1} x^{- 1} = y x$ . But since $x y \in G$ , $(x y)^{2} = 1 ⟹ (x y)^{- 1} = x y$ . Thus $x y = y x$ . True.

Option C: $∣ G ∣ = 2 ⟹ G = {e, a}$ . The only products are $ee, e a, a e, aa$ , all of which commute. True.

Option D: Subgroups of Abelian groups are always Abelian. The statement "need not be" is False.

Learning route:

For A, expand $(x y) (x y) = xx y y$ . Cancel $x$ on the left: $y x y = x y y$ . Cancel $y$ on the right: $y x = x y$ .
For B, the condition means every element is its own inverse. The inverse of $x y$ is $y^{- 1} x^{- 1} = y x$ . But $x y$ is also its own inverse, so $x y = y x$ .
For C, a group of order 2 is isomorphic to $Z_{2}$ , which is Abelian.
For D, if $ab = ba$ for all $a, b \in G$ , then for any $a, b \in H \subseteq G$ , $ab = ba$ still holds.

Question 6 · Engineering Mathematics · 2021_Set1 MCQ

Let $G$ be a group of order 6, and $H$ be a subgroup of $G$ such that $1 < ∣ H ∣ < 6$ . Which one of the following options is correct?

A.
Both $G$ and $H$ are always cyclic.
B.
$G$ may not be cyclic, but $H$ is always cyclic.
C.
$G$ is always cyclic, but $H$ may not be cyclic.
D.
Both $G$ and $H$ may not be cyclic.

Correct Answer:

Step-by-Step Solution

Insight: Lagrange's theorem restricts subgroup orders to divisors of the group order. Groups of prime order are always cyclic.

Exam route:

$∣ G ∣ = 6$ . By Lagrange, $∣ H ∣$ divides 6, so $∣ H ∣ \in {1, 2, 3, 6}$ .
The constraint $1 < ∣ H ∣ < 6$ leaves $∣ H ∣ \in {2, 3}$ .
Both 2 and 3 are prime. Any group of prime order $p$ is cyclic (isomorphic to $Z_{p}$ ). So $H$ is ALWAYS cyclic.
Is $G$ always cyclic? Groups of order 6 are $Z_{6}$ (cyclic) and $S_{3}$ (non-cyclic). Since $S_{3}$ exists, $G$ MAY NOT be cyclic.
Match with options: $G$ may not be cyclic, but $H$ is always cyclic. This is option B.

Learning route:

List divisors of 6: 1, 2, 3, 6.
Apply the strict inequality $1 < ∣ H ∣ < 6$ to filter the list to 2 and 3.
Recall the theorem: Every group of prime order is cyclic. Since 2 and 3 are prime, $H$ must be cyclic.
Recall the classification of groups of order 6: $Z_{6}$ and $S_{3}$ . $S_{3}$ is the symmetric group on 3 elements, which is non-Abelian and thus not cyclic.
Conclude that $G$ is not necessarily cyclic, but $H$ is.

Try a question

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Algebraic Structures and Groups PYQs for GATE CS

Chapter Roadmap: Algebraic Structures and Groups

Algebraic Structures and Groups

What is a Binary Operation?

What is a Binary Operation?

The core idea

Formal definition

Algebraic Structures and Groups: Solved Questions with Step-by-Step Explanations (6 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit