Set Theory & Elementary Group Theory

Subgroup generated by subsets

The subgroup generated by a subset $S$ of a group $G$, denoted by $\langle S \rangle$, is the smallest subgroup of $G$ containing $S$, constructed as the intersection of all subgroups containing $S$. This topic covers this concept.

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

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Definition and statement

Characterization of the Generated Subgroup

Let $G$ be a group and let $S$ be a subset of $G$. The subgroup generated by $S$, denoted by $\langle S \rangle$, is defined as $\langle S \rangle = \bigcap \{ H \mid H \le G \text{ and } S \subseteq H \}$. Alternatively, if $S \neq \emptyset$, $\langle S \rangle$ consists of all finite products of elements of $S$ and their inverses: $\langle S \rangle = \left\{ x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k} \mid x_i \in S \cup S^{-1}, n_i \in \mathbb{Z}, k \in \mathbb{N} \right\}$, with $\langle \emptyset \rangle = \{ e \}$.

📐 Geometric Interpretation

Visualize $S$ as a set of generators (like vectors spanning a vector space or fundamental transformations). The generated subgroup $\langle S \rangle$ represents the entire closure of the space under group operations, acting like a discrete lattice generated by fundamental periods.

⚙️ Algebraic & Structural Meaning

It provides the bridge between arbitrary subsets and algebraic structures, allowing us to define cyclic groups, finitely generated groups, and normal closures, forming the foundation of combinatorial group theory.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Group Structure on $G$	`N/A`	Strictly Mandatory	If associative binary operation, identity, and inverses do not exist globally, the intersection of superset subgroups collapses and the algebraic word construction fails.
Non-emptiness Convention for Empty Set	`N/A`	Strictly Mandatory	If $S = \emptyset$, the intersection over an empty family of subgroups is undefined unless conventionally defined as the trivial subgroup $\{ e \}$.
Closure under Group Operation	`N/A`	Essential for subgroup test	Without including finite products of inverses, the resulting set fails closure and does not form a valid algebraic subgroup.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Single Element Generation (Cyclic Subgroup)

$$ \langle a \rangle = \{ a^n \mid n \in \mathbb{Z} \} $$

Rule: If $S = \{ a \}$ consists of a single element, the generated subgroup is simply the cyclic group generated by $a$. Its order equals the order of $a$ in $G$.

Application: Directly tested in UP PGT to find the order of elements and generated subgroups in $(\mathbb{Z}_n, +)$ or $(U(n), \cdot)$.

Generated Subgroup of Additive Integers

$$ \langle a, b \rangle = \gcd(a, b)\mathbb{Z} $$

Rule: In $(\mathbb{Z}, +)$, the subgroup generated by any two integers $a$ and $b$ is the principal ideal generated by their greatest common divisor.

Application: Standard recurring question type in TGT/PGT asking for $\langle 12, 18 \rangle$ in $\mathbb{Z}$.

Abelianization & Commutative Subsets

$$ \text{If } xy = yx \forall x, y \in S, \text{ then } \langle S \rangle = \left\{ \prod x_i^{n_i} \right\} $$

Rule: If all elements of $S$ commute with each other, arbitrary word products simplify to products of powers of individual elements without needing alternating inverse products.

Application: Saves massive calculation time when dealing with abelian matrix groups or modular arithmetic groups.

Subgroup Generated by Union

$$ \langle A \cup B \rangle = \langle \langle A \rangle \cup \langle B \rangle \rangle $$

Rule: The subgroup generated by the union of two subsets contains all products of elements from both subsets. It equals $A$ combined with $B$ only if one subgroup contains the other.

Application: Frequent conceptual trap in DSSSB and KVS asking whether $\langle A \cup B \rangle = \langle A \rangle \cup \langle B \rangle$.

Finite Generation in Permutation Groups $S_n$

$$ \langle (1\ 2), (1\ 2\ \cdots\ n) \rangle = S_n $$

Rule: Any symmetric group $S_n$ is generated by a single transposition and an $n$-cycle.

Elimination via Lagrange's Theorem

$$ |\langle S \rangle| \text{ divides } |G| $$

Rule: The order of any generated subgroup $\langle S \rangle$ must divide the order of the finite group $G$. Instantly eliminate any option whose order does not divide $|G|$.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Union of Subgroups Fallacy

Common Pitfall: Assuming that the union of two subgroups is always a subgroup, or that $\langle A \cup B \rangle = \langle A \rangle \cup \langle B \rangle$.

Counterexample: In $(\mathbb{Z}, +)$, let $A = 2\mathbb{Z}$ and $B = 3\mathbb{Z}$. $2\mathbb{Z} \cup 3\mathbb{Z}$ is not a subgroup because $2 + 3 = 5$ is not in the union.

Why it fails: The union of two subgroups is a subgroup if and only if one is contained in the other.

🛡️ Exam Defense: Never confuse set union with generated subgroup. $\langle A \cup B \rangle$ is the smallest subgroup containing both, which is strictly larger than $A \cup B$.

⚠️ COMMON ERROR #02

Assuming Non-Commutative Word Closure

Common Pitfall: In non-abelian groups, assuming $\langle x, y \rangle = \{ x^n y^m \mid n, m \in \mathbb{Z} \}$.

Counterexample: In $D_4$ or $GL_2(\mathbb{R})$, words must include alternating products like $xyx^{-1}y^2$.

Why it fails: Non-commuting elements generate words involving arbitrary alternating products of generators and their inverses, not just simple exponent combinations.

🛡️ Exam Defense: Check if the group is abelian before combining exponents of different generators.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Cyclic Subgroup $\langle a \rangle$	$a \in G$, $G$ is any group	$\langle a \rangle = \{ a^n \mid n \in \mathbb{Z} \}$	Orbit of an element under repeated operation	Single element given
Subgroup Generated by Subset	$S \subseteq G$, $G$ is a group	$\langle S \rangle = \bigcap \{ H \le G \mid S \subseteq H \}$	Smallest closed substructure containing seeds	Subset $S$ or generators specified
Additive Integer Subgroup	$a, b \in \mathbb{Z}$	$\langle a, b \rangle = \gcd(a, b)\mathbb{Z}$	Lattice of common multi-step increments	Numbers given in $(\mathbb{Z}, +)$
Normal Closure $\langle\langle S \rangle\rangle$	$S \subseteq G$, Normal subgroup required	Smallest normal subgroup containing $S$	Conjugation-invariant structural closure	Keywords 'normal subgroup generated by'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the subgroup generated by the subset $S = \{ 4, 6 \}$ in the additive group of integers $(\mathbb{Z}, +)$.

(A) $2\mathbb{Z}$(B) $4\mathbb{Z}$(C) $12\mathbb{Z}$(D) $\mathbb{Z}$

📚 Method A: Formal Textbook Derivation

Step 1: Recognize that $(\mathbb{Z}, +)$ is an abelian group. Step 2: By property, the subgroup generated by two integers $a$ and $b$ is $\gcd(a, b)\mathbb{Z}$. Step 3: Compute $\gcd(4, 6) = 2$. Step 4: Therefore, $\langle 4, 6 \rangle = 2\mathbb{Z}$.

⚡ Method B: 25-Second Exam Shortcut

Instantly calculate $\gcd(4, 6) = 2$ and match with $2\mathbb{Z}$. Takes 5 seconds.

💡 Tip: Always remember that in $(\mathbb{Z}, +)$, $\langle a_1, a_2, \dots, a_k \rangle = \gcd(a_1, a_2, \dots, a_k)\mathbb{Z}$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

In the symmetric group $S_3$, what is the order of the subgroup generated by the transposition $S = \{(1\ 2)\}$?

(A) $1$(B) $2$(C) $3$(D) $6$

📚 Method A: Formal Textbook Derivation

Step 1: The subset is $S = \{(1\ 2)\}$. Step 2: The generated subgroup is $\langle (1\ 2) \rangle = \{ id, (1\ 2) \}$. Step 3: Counting the elements, the order is 2, which also matches the order of the element $(1\ 2)$ in $S_3$.

⚡ Method B: 25-Second Exam Shortcut

The order of a subgroup generated by a single element equals the order of that element. Transpositions in $S_n$ always have order 2.

💡 Tip: Do not confuse the order of the element with the order of the entire group $S_3$ (which is $3! = 6$).

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the subgroup generated by the empty set $\emptyset$ in any group $G$?

A $G$

B \{ e \}

C Undefined

D The center of $G$

Derivation & Analysis:

By definition, the subgroup generated by the empty set is the trivial subgroup containing only the identity element, $\langle \emptyset \rangle = \{ e \}$.

MCQ 2: In $(\mathbb{Z}, +)$, what is $\langle 15, 25, 35 \rangle$?

A $5\mathbb{Z}$

B $15\mathbb{Z}$

C $35\mathbb{Z}$

D $\mathbb{Z}$

Derivation & Analysis:

The subgroup generated by a set of integers in $(\mathbb{Z}, +)$ is generated by their greatest common divisor. $\gcd(15, 25, 35) = 5$, so the generated subgroup is $5\mathbb{Z}$.

MCQ 3: Let $G$ be a group and $S \subseteq G$. Which of the following is always true for $\langle S \rangle$?

A $\langle S \rangle$ is the union of all elements in $S$.

B $\langle S \rangle$ is the intersection of all subgroups of $G$ containing $S$.

C $\langle S \rangle$ is always a cyclic group.

D $\langle S \rangle$ is never equal to $G$.

Derivation & Analysis:

By definition, $\langle S \rangle$ is the intersection of all subgroups of $G$ that contain $S$, making it the smallest subgroup containing $S$.

MCQ 4: If $H$ and $K$ are subgroups of a group $G$, then $\langle H \cup K \rangle = G$ if:

A $H \subseteq K$

B $K \subseteq H$

C $H \cup K = G$

D $H$ and $K$ are disjoint

Derivation & Analysis:

If the union $H \cup K$ already equals $G$, then the subgroup generated by $G$ is $G$ itself. More generally, if $H \cup K$ is a subgroup, then $H \le K$ or $K \le H$, but $\langle H \cup K \rangle$ is the smallest subgroup containing the union.

MCQ 5: What is the order of the subgroup generated by $2$ in the multiplicative group $(\mathbb{Z}_{11} \setminus \{0\}, \cdot)$?

A 5

B 10

C 11

D 2

Derivation & Analysis:

In $\mathbb{Z}_{11}^*$, powers of 2 are: $2^1=2, 2^2=4, 2^3=8, 2^4=5, 2^5=10, 2^6=9, 2^7=7, 2^8=3, 2^9=6, 2^{10}=1$. Thus, the order of 2 is 10, generating the entire group of order 10.

MCQ 6: If $S = \{ a \}$ in a group $G$, then $\langle S \rangle$ is:

A A normal subgroup always

B A cyclic subgroup

C The trivial subgroup

D Equal to $G$

Derivation & Analysis:

The subgroup generated by a single element $a$ is $\langle a \rangle = \{ a^n \mid n \in \mathbb{Z} \}$, which is by definition a cyclic subgroup.

MCQ 7: Which of the following subsets of $\mathbb{R}$ under addition generates $\mathbb{Q}$?

A $\{ 1/n \mid n \in \mathbb{N} \}$

B $\{ \sqrt{2} \}$

C $\{ \pi \}$

D \{ 0 \}

Derivation & Analysis:

The set of all unit fractions $\{1/n \mid n \in \mathbb{N}\}$ generates the additive group of rational numbers $\mathbb{Q}$ because any rational can be formed by integer linear combinations of unit fractions.

MCQ 8: Assertion (A): The intersection of any collection of subgroups of a group $G$ is a subgroup of $G$. Reason (R): $\langle S \rangle$ is defined as the intersection of all subgroups containing $S$.

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 of (A).

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

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

Derivation & Analysis:

Both statements are fundamental theorems of group theory. The intersection property of subgroups justifies the well-definedness of $\langle S \rangle$.

MCQ 9: In the Klein 4-group $V_4 = \{e, a, b, c\}$, what is $\langle a, b \rangle$?

A \{e, a\}

B \{e, b\}

C \{e, a, b\}

D $V_4$

Derivation & Analysis:

In $V_4$, $ab = c$. Thus, the subset $\{a, b\}$ generates products including $ab = c$, yielding the entire group $V_4$.

Q10

MCQ 10: If $H$ is a subgroup of $G$ and $S \subseteq H$, what can we conclude about $\langle S \rangle$?

A $\langle S \rangle = H$

B $\langle S \rangle \subseteq H$

C $H \subseteq \langle S \rangle$

D $\langle S \rangle$ is disjoint from $H$

Derivation & Analysis:

Since $\langle S \rangle$ is the intersection of all subgroups containing $S$, and $H$ is one such subgroup, $\langle S \rangle$ must be a subset of $H$.

Common questions

Yes. For any non-empty subset $S$ of a group $G$, $\langle S \rangle$ consists precisely of all possible finite products of elements from $S$ and their inverses. For abelian groups, this simplifies to sums/products of integer powers of elements in $S$.

The union of two subgroups (or subsets) is generally not closed under the group operation. The generated subgroup must include all necessary products and inverses to satisfy the subgroup axioms, making it strictly larger than the union in most cases.

Identify whether the group is abelian or cyclic. Use gcd rules for additive integer groups, order formulas for cyclic groups, and closure properties under multiplication for permutation or matrix groups.

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