6. Abstract Algebra

Group and examples

A group is a non-empty set equipped with a single binary operation that is associative, possesses an identity element, and contains an inverse for every element. This topic covers standard groups like $(\mathbb{Z}, +)$, $(\mathbb{R}^*, \cdot)$, modulo arithmetic groups, and matrix groups.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Definition of a Group

A non-empty set $G$ together with a binary operation $*$, denoted by $(G, *)$, is called a group if the following axioms hold: (i) Closure: $\forall a, b \in G, a * b \in G$. (ii) Associativity: $\forall a, b, c \in G, (a * b) * c = a * (b * c)$. (iii) Identity Element: $\exists e \in G$ such that $\forall a \in G, a * e = e * a = a$. (iv) Inverse Element: $\forall a \in G, \exists a^{-1} \in G$ such that $a * a^{-1} = a^{-1} * a = e$.

📐 Geometric Interpretation

Geometrically, groups represent fundamental symmetries of geometric objects, where the binary operation corresponds to the composition of rigid transformations (such as rotations and reflections) that leave the object invariant.

⚙️ Algebraic & Structural Meaning

It serves as the foundational algebraic structure in abstract algebra, unifying diverse mathematical systems under a single set of axioms and enabling the systematic study of symmetry, solvability of equations, and structural classification.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-emptiness of the underlying set $G$	`Set condition`	Strictly Mandatory / Necessary	An empty set cannot possess an identity element or binary operations satisfying the group axioms.
Closure under the binary operation $*$	`Algebraic closure`	Strictly Mandatory	If $a * b \notin G$ for some $a, b \in G$, the operation is not binary on $G$, rendering the structure invalid.
Existence of unique identity and inverse	`Uniqueness axioms`	Essential for group structure	Without unique inverses, cancellation laws fail, and the algebraic system collapses into a mere monoid or semigroup.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Finite Semigroup Cancellation

$$ \text{If } S \text{ is a finite semigroup satisfying left and right cancellation laws, then } S \text{ is a group.} $$

Rule: In finite sets, checking closure, associativity, and cancellation is sufficient to guarantee a group, bypassing explicit inverse checks.

Application: Use this to quickly identify finite groups among multiplicative or matrix subsets.

Order of Elements in Modular Arithmetic

$$ o(a) \text{ divides } |G| \text{ (Lagrange's Theorem corollary).} $$

Rule: The order of any element in a finite group $G$ must divide the order of the group $|G|$.

Application: Directly applied in TGT/PGT questions asking for the order of elements in $(\mathbb{Z}_n, +)$ or $(\mathbb{U}(n), \cdot)$.

Matrix Group Determinants

$$ \det(AB) = \det(A)\det(B) \text{ and } \det(A^{-1}) = \frac{1}{\det(A)} $$

Rule: To check if a set of matrices forms a group under multiplication, verify that the determinant is non-zero and closed under multiplication.

Application: Standard pattern in KVS PGT and DSSSB for $GL(n, \mathbb{R})$ and $SL(n, \mathbb{R})$.

Klein Four-Group & Cyclic Group Identification

$$ \text{If every non-identity element has order 2, the group is isomorphic to } V_4. $$

Rule: If $a^2 = e$ for all elements in a group, the group is abelian and isomorphic to the Klein four-group (if of order 4).

Direct Product Orders

$$ o((a,b)) = \text{lcm}(o(a), o(b)) \text{ in } G_1 \times G_2 $$

Rule: Use the least common multiple of component element orders to find the order of elements in direct product groups.

Rapid Elimination via Identity Check

$$ e * a = a * e = a $$

Rule: Always test the candidate identity element first. If a proposed set lacks the standard or modified identity, eliminate the option instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing $(\mathbb{Z}, \cdot)$ with $(\mathbb{Z}, +)$

Common Pitfall: Candidates assume integers form a group under multiplication, forgetting that non-zero integers lack multiplicative inverses within $\mathbb{Z}$.

Counterexample: For $2 \in \mathbb{Z}$, there is no $b \in \mathbb{Z}$ such that $2 \cdot b = 1$.

Why it fails: The inverse axiom fails for all integers except $1$ and $-1$.

🛡️ Exam Defense: Always specify the operation explicitly; integers form a monoid, not a group, under multiplication.

⚠️ COMMON ERROR #02

Assuming Matrix Multiplication is Always Commutative

Common Pitfall: Assuming general linear groups $GL(n, \mathbb{R})$ for $n \ge 2$ are abelian.

Counterexample: For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, $AB \neq BA$.

Why it fails: Matrix multiplication is non-commutative for $n \ge 2$, so $GL(n, \mathbb{R})$ is non-abelian.

🛡️ Exam Defense: Never apply commutativity rules to matrix groups unless restricted to diagonal or $n=1$ cases.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Group $(G, \cdot)$	Closure, Associativity, Identity, Inverse	$a * a^{-1} = e$	Symmetry transformations	Involves invertible structures
Abelian Group	Group + Commutativity ($ab = ba$)	$a * b = b * a$	Symmetric commutative actions	Keywords: commutative, symmetric
Cyclic Group	Generated by a single element $a$	$G = \langle a \rangle = \{a^n \mid n \in \mathbb{Z}\}	Discrete rotational symmetry	Keywords: generator, modulo $n$
Subgroup Test	Non-empty subset $H \subseteq G$	$ab^{-1} \in H \text{ for all } a, b \in H$	Nested geometric subspaces	Keywords: subset, closed under operation

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Which of the following sets forms a group under multiplication modulo $5$?

(A) $\{0, 1, 2, 3, 4\}$(B) $\{1, 2, 3, 4\}$(C) $\{2, 3, 4\}$(D) $\{1, 3, 4\}$

📚 Method A: Formal Textbook Derivation

Step 1: Check multiplicative identity $1$. Step 2: Verify non-zero elements have multiplicative inverses modulo 5 ($1^{-1}=1, 2^{-1}=3, 3^{-1}=2, 4^{-1}=4$). Step 3: Confirm closure. Thus, $\mathbb{Z}_5^*$ forms a group.

⚡ Method B: 25-Second Exam Shortcut

Exclude $0$ immediately for multiplicative groups, and check if remaining non-zero residues coprime to $5$ form $\mathbb{Z}_5^*$.

💡 Tip: For modulo $n$ multiplication, the group is always formed by integers coprime to $n$, of size $\phi(n)$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

What is the order of the element $3$ in the group $(\mathbb{Z}_7^*, \cdot_{7})$?

(A) $3$(B) $6$(C) $2$(D) $4$

📚 Method A: Formal Textbook Derivation

Step 1: Compute powers of $3$ modulo $7$. $3^1 \equiv 3$, $3^2 \equiv 2$, $3^3 \equiv 6$, $3^4 \equiv 4$, $3^5 \equiv 5$, $3^6 \equiv 1$. Step 2: The smallest positive integer $n$ such that $3^n \equiv 1 \pmod 7$ is $6$.

⚡ Method B: 25-Second Exam Shortcut

By Fermat's Little Theorem, $a^{p-1} \equiv 1 \pmod p$. The order must divide $\phi(7) = 6$. Test divisors $1, 2, 3, 6$.

💡 Tip: Always test divisors of the order of the group when seeking element orders.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following is not a group under the given operation?

A $(\mathbb{Z}, +)$

B $(\mathbb{Q}, +)$

C $(\mathbb{R}, \cdot)$

D $(\mathbb{C}^*, \cdot)$

Derivation & Analysis:

In $(\mathbb{R}, \cdot)$, the element $0$ does not possess a multiplicative inverse ($0 \cdot x = 1$ has no solution in $\mathbb{R}$), hence it is not a group.

MCQ 2: The order of the additive group $\mathbb{Z}_n$ is:

A $n-1$

B $n$

C $\infty$

D $\phi(n)$

Derivation & Analysis:

The set $\mathbb{Z}_n = \{0, 1, 2, \dots, n-1\}$ contains exactly $n$ elements, so its order is $n$.

MCQ 3: If $G$ is a group such that $(ab)^2 = a^2 b^2$ for all $a, b \in G$, then $G$ is:

A Cyclic

B Infinite

C Abelian

D Trivial only

Derivation & Analysis:

Expanding $(ab)(ab) = aabb \implies abab = aabb$. By left and right cancellation of $a$ and $b$, $ba = ab$, proving $G$ is abelian.

MCQ 4: The number of generators of the cyclic group $(\mathbb{Z}_{10}, +)$ is:

A $1$

B $2$

C $4$

D $5$

Derivation & Analysis:

The number of generators of a finite cyclic group of order $n$ is $\phi(n)$. Here $\phi(10) = \phi(2) \cdot \phi(5) = 1 \cdot 4 = 4$.

MCQ 5: Which of the following matrix sets forms a group under matrix multiplication?

A All $2 \times 2$ real matrices

B All $2 \times 2$ matrices with $\det = 0$

C All $2 \times 2$ matrices with $\det = 1$

D All $2 \times 2$ symmetric matrices

Derivation & Analysis:

The set of $2 \times 2$ matrices with determinant $1$ forms the Special Linear Group $SL(2, \mathbb{R})$, which satisfies closure, associativity, identity, and inverse properties.

MCQ 6: In any group $G$, the solution of the equation $ax = b$ for $a, b \in G$ is:

A $x = ba^{-1}$

B $x = a^{-1}b$

C $x = ab^{-1}$

D $x = ba$

Derivation & Analysis:

Multiplying $ax = b$ on the left by $a^{-1}$ gives $a^{-1}(ax) = a^{-1}b \implies (a^{-1}a)x = a^{-1}b \implies x = a^{-1}b$.

MCQ 7: The inverse of the element $(2, 3)$ in the direct product group $\mathbb{Z}_5 \times \mathbb{Z}_4$ is:

A $(2, 3)$

B $(3, 1)$

C $(2, 1)$

D $(3, 2)$

Derivation & Analysis:

In $\mathbb{Z}_5$, $-2 \equiv 3 \pmod 5$. In $\mathbb{Z}_4$, $-3 \equiv 1 \pmod 4$. Thus, the inverse is $(3, 1)$.

MCQ 8: Assertion (A): Every cyclic group is abelian. Reason (R): If $a, b \in \langle x \rangle$, then $a = x^m, b = x^n$, so $ab = x^{m+n} = x^{n+m} = ba$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Every cyclic group is generated by a single element, and powers of the same element commute under addition of exponents, confirming both assertion and reason.

MCQ 9: The order of the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$ in $S_3$ is:

A $1$

B $2$

C $3$

D $6$

Derivation & Analysis:

Writing $\sigma$ in disjoint cycle notation gives $(1 2 3)$, which is a 3-cycle of length 3, hence its order is 3.

Q10

MCQ 10: If $H$ and $K$ are two subgroups of a group $G$, then $H \cap K$ is:

A Never a subgroup

B Always a subgroup of $G$

C A subgroup only if $|H| = |K|$

D A subgroup only if $H \cup K = G$

Derivation & Analysis:

The intersection of any arbitrary collection of subgroups of a group $G$ is always a subgroup of $G$.

Common questions

Yes. By Lagrange's theorem, any group whose order is a prime number $p$ has no proper non-trivial subgroups, and consequently, every non-identity element generates the entire group, making it cyclic and abelian.

Candidates frequently forget to verify whether the identity element and inverses belong to the specified subset when testing for subgroups or group structures.

Memorize standard properties like $\phi(n)$ for generator counts, LCM formulas for direct product element orders, and cycle decomposition lengths for permutation orders to solve problems.