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1st Year Math Notes: Ex 2.8 Group Theory MCQs

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1st Year Math Notes: Ex 2.8 Group Theory MCQs

Official Academic Study Notes • Published: September 06, 2026 • Free Printable Resource

What are the Core Formulas and Definitions for this Exercise?

Group theory forms the bedrock of Abstract Algebra in 1st year mathematics and higher-level competitive exams (PPSC/FPSC Lecturer Mathematics). Exercise 2.8 focuses on fundamental algebraic structures defined by binary operations on sets.

A non-empty set $G$ together with a binary operation $*$ denoted by $(G, *)$ is a Group if it satisfies the following four fundamental axioms:

Axiom Name	Mathematical Condition	Explanation
1. Closure Property	$\forall a, b \in G, \quad a * b \in G$	The result of operating on any two elements remains inside the set $G$.
2. Associative Law	$\forall a, b, c \in G, \quad (a * b) * c = a * (b * c)$	The grouping of elements does not affect the final result.
3. Identity Element	$\exists e \in G \text{ such that } a * e = e * a = a, \quad \forall a \in G$	There exists a unique element $e$ that leaves every element unchanged.
4. Inverse Element	$\forall a \in G, \, \exists a^{-1} \in G \text{ such that } a * a^{-1} = a^{-1} * a = e$	Every element has a unique inverse that yields the identity element.
5. Commutative Law (Abelian Group)	$\forall a, b \in G, \quad a * b = b * a$	If a group satisfies this additional 5th property, it is called an Abelian Group.

Key algebraic identities required for solving Exercise 2.8:

Reversal Law for Inverses: $(a * b)^{-1} = b^{-1} * a^{-1}$
Uniqueness of Identity: The identity element $e$ in a group $G$ is strictly unique.
Cancellation Laws: If $a * b = a * c$, then $b = c$ (Left Cancellation); if $b * a = c * a$, then $b = c$ (Right Cancellation).

How to Solve All Exercise Questions Step-by-Step?

Question 1: Show that the set $G = \{1, -1, i, -i\}$ forms an abelian group under usual multiplication.

Solution:

To prove $G = \{1, -1, i, -i\}$ is an abelian group under multiplication $(\cdot)$, we construct a Cayley Multiplication Table:

$\cdot$	$1$	$-1$	$i$	$-i$
$1$	$1$	$-1$	$i$	$-i$
$-1$	$-1$	$1$	$-i$	$i$
$i$	$i$	$-i$	$-1$	$1$
$-i$	$-i$	$i$	$1$	$-1$

Step-by-Step Verification:

Closure Property: Every entry in the Cayley table belongs to $G = \{1, -1, i, -i\}$. Thus, $G$ is closed under multiplication.
Associative Law: Complex number multiplication is universally associative: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ for all $a, b, c \in G$.
Identity Element: The element $1 \in G$ acts as the multiplicative identity since $a \cdot 1 = 1 \cdot a = a$ for all $a \in G$.
Inverse Element:
- $1^{-1} = 1$
- $(-1)^{-1} = -1$
- $i^{-1} = -i$ (since $i \cdot (-i) = -i^2 = -(-1) = 1$)
- $(-i)^{-1} = i$
Since every element has an inverse in $G$, the inverse axiom holds.
Commutative Property: The table is symmetric across its main diagonal. Thus, $a \cdot b = b \cdot a$ holds for all elements.

Conclusion: $G = \{1, -1, i, -i\}$ is a finite abelian group of order $4$.

Question 2: Prove that in any group $(G, *)$, the inverse of a product is given by $(a * b)^{-1} = b^{-1} * a^{-1}$ for all $a, b \in G$.

Solution:

Let $a, b \in G$. By the definition of group inverses, an element $x$ is the inverse of $(a * b)$ if and only if:

$$(a * b) * x = e \quad \text{and} \quad x * (a * b) = e$$

Let us test $x = b^{-1} * a^{-1}$:

Check Right Identity:

$$(a * b) * (b^{-1} * a^{-1}) = a * (b * b^{-1}) * a^{-1} \quad \text{(by Associativity)}$$ $$= a * e * a^{-1} \quad \text{(since } b * b^{-1} = e\text{)}$$ $$= (a * e) * a^{-1} = a * a^{-1} = e$$

Check Left Identity:

$$(b^{-1} * a^{-1}) * (a * b) = b^{-1} * (a^{-1} * a) * b \quad \text{(by Associativity)}$$ $$= b^{-1} * e * b = b^{-1} * b = e$$

Since $(a * b) * (b^{-1} * a^{-1}) = e = (b^{-1} * a^{-1}) * (a * b)$, by uniqueness of group inverses, we conclude:

$$(a * b)^{-1} = b^{-1} * a^{-1}$$

Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)

Q1: What is the identity element of the group $(\mathbb{Z}, +)$ of integers under ordinary addition?

Explanation: Under addition, $a + 0 = 0 + a = a$ for any integer $a \in \mathbb{Z}$. Hence, $0$ is the additive identity.

Q2: For a group $G$, if $(a \cdot b)^2 = a^2 \cdot b^2$ holds for all $a, b \in G$, then $G$ must be:

Explanation: $(a b)(a b) = a^2 b^2 \implies a(ba)b = a(ab)b$. Applying cancellation laws on the left ($a$) and right ($b$) gives $ba = ab$, proving $G$ is abelian.

Q3: In the cube roots of unity group $G = \{1, \omega, \omega^2\}$ under multiplication, what is the inverse of $\omega$?

Explanation: Since $\omega \cdot \omega^2 = \omega^3 = 1$, the inverse of $\omega$ is $\omega^2$.

Q4: Which of the following non-empty sets forms a group under multiplication?

Explanation: $\mathbb{R} \setminus \{0\}$ excludes $0$ (which lacks a multiplicative inverse), ensuring every element $x$ has a valid multiplicative inverse $\frac{1}{x} \in \mathbb{R} \setminus \{0\}$.

Q5: If $x$ is an element of a group $G$ such that $x^2 = x$, then $x$ equals:

Explanation: Multiply both sides of $x \cdot x = x$ by $x^{-1}$: $x^{-1}(x \cdot x) = x^{-1} \cdot x \implies (x^{-1} \cdot x) x = e \implies e \cdot x = e \implies x = e$.

Q6: The order of the element $i$ in the multiplicative group $\{1, -1, i, -i\}$ is:

Explanation: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. The smallest positive integer $n$ such that $i^n = 1$ is $n = 4$.

Q7: A semi-group that possesses an identity element is called a:

Explanation: By algebraic definition: Closure = Groupoid; Closure + Associativity = Semi-group; Semi-group + Identity = Monoid; Monoid + Inverses = Group.

Q8: What is the inverse of $(a * b^{-1})^{-1}$ in any general group $G$?

Explanation: Applying reversal law: $(a * b^{-1})^{-1} = (b^{-1})^{-1} * a^{-1} = b * a^{-1}$.

Frequently Asked Questions: What are Common Student Errors in this Exercise?

Q1: Why is the set of natural numbers $\mathbb{N}$ under addition not considered a group?
Ans: Although $\mathbb{N} = \{1, 2, 3, \dots\}$ satisfies closure and associativity under addition, it fails the identity property (since $0 \notin \mathbb{N}$) and the inverse property (for $a \in \mathbb{N}$, $-a \notin \mathbb{N}$). Thus, $(\mathbb{N}, +)$ is only a semi-group, not a group.

Q2: Why is the order of elements in reversal laws written as $(a * b)^{-1} = b^{-1} * a^{-1}$ instead of $a^{-1} * b^{-1}$?
Ans: In non-abelian groups, operations are not commutative ($a * b \neq b * a$). Writing $(a * b) * (a^{-1} * b^{-1})$ does not allow inner terms to simplify to identity $e$. The reversal sequence ensures that $b * b^{-1} = e$ compresses first, leaving $a * e * a^{-1} = a * a^{-1} = e$.

Q3: How do board exams and competitive entry tests (PPSC/FPSC) test Cayley tables?
Ans: In 1st year board exams, students must construct full Cayley tables for finite sets like $\{1, -1, i, -i\}$ or $\{1, \omega, \omega^2\}$. In PPSC/FPSC exams, MCQs test properties directly from matrix symmetry—a table symmetric across its main diagonal signifies an abelian group, while any duplicate elements in a single row/column indicate a violation of group axioms.