Set Theory & Elementary Group Theory

Coset decomposition

Coset decomposition partitions a group into disjoint, equal-sized subsets using a subgroup, forming the cornerstone for proving Lagrange's Theorem and constructing quotient groups.

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

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Definition and statement

Partition Theorem via Left/Right Cosets

Let $H$ be a subgroup of a group $G$. For any $a, b \in G$, the left coset $aH = \{ah \mid h \in H\}$ either is identical to $bH$ or is disjoint from $bH$. Furthermore, $G$ can be expressed as a disjoint union of distinct left cosets of $H$ in $G$, i.e., $G = \bigcup_{i \in I} a_i H$, where $|a_i H| = |H|$ for all $i$.

📐 Geometric Interpretation

Visually, think of $G$ as a large geometric space and $H$ as a linear subspace passing through the origin. Left cosets $aH$ act as parallel affine translates that tile the entire space without overlapping, each having identical cardinality and structure.

⚙️ Algebraic & Structural Meaning

Coset decomposition establishes an equivalence relation on $G$ modulo $H$, proving that subgroups partition groups into equal partitions, which directly yields Lagrange's Theorem: $|G| = [G:H] \cdot |H|$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Subgroup property of $H$	`N/A`	Strictly Mandatory / Necessary	If $H$ is merely a subset or subgroupoid without identity or closure, $aH = bH$ equivalence relation properties collapse, and subsets fail to partition $G$ uniformly.
Group operation associativity and inverses	`N/A`	Strictly Mandatory	Without inverses, mapping $h \mapsto ah$ is not a bijection, meaning cosets $aH$ and $bH$ can have unequal cardinalities, destroying the index counting argument.
Choice of representative elements	`N/A`	Essential for partition completeness	Failing to select representatives from outside previously generated cosets leads to redundant overlapping sets instead of a strict disjoint partition.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Index Calculation via Lagrange's Theorem

$$ [G:H] = \frac{|G|}{|H|} $$

Rule: For finite groups, the number of distinct left (or right) cosets of $H$ in $G$ is precisely the ratio of their orders.

Application: Directly tested in UP PGT to find the number of distinct left cosets of a cyclic subgroup in dihedral or symmetric groups.

Coset Equality Condition

$$ aH = bH \iff b^{-1}a \in H \iff a \in bH $$

Rule: Two left cosets are identical if and only if the representative of one belongs to the other.

Application: Used in TGT/PGT to instantly verify whether two algebraic expressions generate the same coset.

Order of Coset equals Order of Subgroup

$$ |aH| = |H| \quad \forall a \in G $$

Rule: Every left or right coset has the exact same cardinality as the subgroup $H$, regardless of the choice of $a$.

Application: Crucial for counting elements satisfying specific equations or verifying finite group structures.

Self-Inverse / Index 2 Subgroups

$$ [G:H] = 2 \implies aH = Ha \text{ and } a^2 \in H \; \forall a \in G $$

Rule: Every subgroup of index 2 is automatically normal, meaning left and right cosets coincide entirely.

Application: Instant elimination rule in questions involving alternating groups $A_n$ inside $S_n$ or index 2 subgroups.

Lagrange Divisibility Rule for Elements

$$ o(a) \mid |G| \quad \forall a \in G $$

Rule: The order of any element in a finite group must divide the order of the group, derived directly from cyclic subgroup coset decomposition.

Application: Extensively used in UP PGT to rule out impossible element orders.

Rapid Elimination via Coprime Orders

$$ \gcd(|H|, [G:H]) = 1 \implies \text{Unique structural properties} $$

Rule: Use prime factorization of group orders to eliminate options that violate coset product cardinalities.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Left Coset with Right Coset in Non-Abelian Groups

Common Pitfall: Assuming $aH = Ha$ for all groups, which is only true if $H$ is a normal subgroup.

Counterexample: Let $G = S_3, H = \{(1), (1 2)\}, a = (1 3)$

Why it fails: Left coset $aH = \{(1 3), (1 2 3)\}$ while right coset $Ha = \{(1 3), (1 3 2)\}$. They are distinct because $H$ is not normal in $S_3$.

🛡️ Exam Defense: Never assume $aH = Ha$ unless the group is abelian or $H$ is explicitly stated to be normal.

⚠️ COMMON ERROR #02

Assuming Every Subgroup has a Complementary Subgroup

Common Pitfall: Believing that if $[G:H] = k$, there exists another subgroup $K$ of order $k$ such that $G = HK$.

Counterexample: Subgroup $H = \{(1), (1 2)\}$ in $S_3$ has index 3, but there is no subgroup of order 3 normal complement in $S_3$ for that specific $H$.

Why it fails: Index measures the number of cosets, not necessarily the existence of a transversal subgroup.

🛡️ Exam Defense: Distinguish carefully between coset spaces $G/H$ (which are sets of cosets) and quotient groups (which require normality).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Left Coset ($aH$)	$H \le G, a \in G$	$aH = \{ah \mid h \in H\}$	Parallel affine translate of $H$	Appears when operating elements from the left
Right Coset ($Ha$)	$H \le G, a \in G$	$Ha = \{ha \mid h \in H\}$	Parallel translate from the right	Checked for normality ($aH = Ha$)
Lagrange's Theorem	$G$ is finite, $H \le G$	$\|G\| = [G:H] \cdot \|H\|$	Total space tiles into equal disjoint blocks	Asked to find order of subgroup or index
Index of Subgroup ($[G:H]$)	$H \le G$	$[G:H] = \|G\| / \|H\|$	Number of distinct coset partitions	Questions asking 'find the number of distinct left cosets'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the number of distinct left cosets of the subgroup $H = 4\mathbb{Z}$ in the additive group $G = \mathbb{Z}$.

(A) 2(B) 4(C) Infinite(D) 8

📚 Method A: Formal Textbook Derivation

Step 1: Identify group $G = \mathbb{Z}$ and subgroup $H = 4\mathbb{Z} = \{..., -8, -4, 0, 4, 8, ...\}$. Step 2: Form distinct left cosets by adding representatives $0, 1, 2, 3$: $0+4\mathbb{Z}, 1+4\mathbb{Z}, 2+4\mathbb{Z}, 3+4\mathbb{Z}$. Step 3: Total number of distinct cosets is 4.

⚡ Method B: 25-Second Exam Shortcut

Recognize that $\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}_4$, which has cardinality 4. Instant tick.

💡 Tip: For infinite groups like $(\mathbb{Z}, +)$, finite index occurs when the subgroup is formed by multiples $n\mathbb{Z}$, yielding index $n$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $H$ be a subgroup of a finite group $G$ such that $|G| = 48$ and $|H| = 8$. What is the index $[G:H]$?

(A) 4(B) 6(C) 8(D) 384

📚 Method A: Formal Textbook Derivation

Step 1: State Lagrange's Theorem formula $|G| = [G:H] \cdot |H|$. Step 2: Substitute values: $48 = [G:H] \cdot 8$. Step 3: Solve for $[G:H] = 48 / 8 = 6$.

⚡ Method B: 25-Second Exam Shortcut

Direct division: $48 / 8 = 6$. Takes 3 seconds in UP PGT.

💡 Tip: Always verify whether the question asks for group order, subgroup order, or index to avoid reciprocal errors.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $H$ is a subgroup of $G$ and $a, b \in G$, then $aH = bH$ if and only if:

A $ab^{-1} \in H$

B $b^{-1}a \in H$

C $ab \in H$

D $a^{-1}b \in H$

Derivation & Analysis:

By fundamental coset properties, $aH = bH \iff a \in bH \iff b^{-1}a \in H$.

MCQ 2: The number of distinct left cosets of subgroup $H$ in $G$ is called:

A Order of $G$

B Order of $H$

C Index of $H$ in $G$

D Degree of $G$

Derivation & Analysis:

By definition, the index $[G:H]$ is the number of distinct left (or right) cosets of $H$ in $G$.

MCQ 3: If $G = S_3$ and $H = \{(1), (1 2)\}$, what is the index $[G:H]$?

A 1

B 2

C 3

D 6

Derivation & Analysis:

$|S_3| = 6$ and $|H| = 2$. By Lagrange's theorem, $[S_3:H] = 6 / 2 = 3$.

MCQ 4: Which of the following statements is always true for any subgroup $H$ of a group $G$?

A $aH = Ha$ for all $a \in G$

B $|aH| = |H|$ for all $a \in G$

C $aH \cap bH = \emptyset$ for all $a, b \in G$

D $aH$ is always a subgroup of $G$

Derivation & Analysis:

The mapping $h \mapsto ah$ is a bijection from $H$ to $aH$, ensuring every coset has the exact same cardinality as $H$.

MCQ 5: If $G$ is a group of prime order $p$, then the number of proper non-trivial subgroups of $G$ is:

A 0

B 1

C p

D p-1

Derivation & Analysis:

By Lagrange's theorem, any subgroup order must divide $p$. Since $p$ is prime, only divisors are 1 and $p$, meaning no proper non-trivial subgroups exist.

MCQ 6: If $H$ and $K$ are subgroups of $G$ with $|H| = 12$ and $|K| = 18$, and $H \le K$, what can be said about $|G|$ if $G$ is finite?

A 36 divides $|G|$

B 12 divides $|G|$ only

C 18 divides $|G|$, hence 36 must divide $|G|$ if $H$ and $K$ are properly nested

D 6 divides $|G|$

Derivation & Analysis:

Since $H \le K \le G$, $|H|$ divides $|K|$ and $|K|$ divides $|G|$. Thus 18 divides $|G|$ (and since 12 divides 18 is false unless stated, wait: 12 does not divide 18, so $H$ cannot be a subgroup of $K$ unless $|H|$ divides $|K|$. If $H \le K$, then 12 must divide 18, which is false! Thus the premise $H \le K$ with these orders is impossible in a single chain unless stated otherwise. Wait, let's look at LCM: if $H, K \le G$, then $|HK| = \frac{|H||K|}{|H \cap K|}$, which divides $|G|$. Since $|K|=18$ divides $|G|$, 18 divides $|G|$.

MCQ 7: If $[G:H] = 1$, then:

A $G = \{e_G\}$

B $H = \{e_G\}$

C $H = G$

D $H$ is a maximal subgroup

Derivation & Analysis:

If the index is 1, there is only one distinct coset, which must be $G$ itself, meaning $H = G$.

MCQ 8: Assertion (A): Every subgroup of index 2 is normal. Reason (R): Left and right cosets partition the group into 2 parts.

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, (R) is false

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

Derivation & Analysis:

If $[G:H] = 2$, the two left cosets are $H$ and $G \setminus H$. Similarly, the two right cosets are $H$ and $G \setminus H$. Thus $aH = Ha$ for all $a$, proving normality.

MCQ 9: In the symmetric group $S_3$, the number of right cosets of $A_3$ is:

A 1

B 2

C 3

D 6

Derivation & Analysis:

$|S_3| = 6$ and $|A_3| = 3$. The index $[S_3:A_3] = 6 / 3 = 2$. Thus there are 2 right cosets.

Q10

MCQ 10: Which of the following cannot be the order of a subgroup of a group of order 30?

A 1

B 5

C 8

D 15

Derivation & Analysis:

By Lagrange's theorem, subgroup orders must divide $|G| = 30$. Since 8 does not divide 30, 8 cannot be the order of any subgroup.

Common questions

Left and right cosets coincide if and only if the subgroup is normal ($gHg^{-1} = H$ for all $g$). In non-abelian groups, arbitrary subgroups lack this symmetry, resulting in distinct left and right partition structures.

Candidates often confuse the index $[G:H]$ with the order of the subgroup or incorrectly assume that every subgroup of index $k$ has elements of order $k$.

Memorize standard group orders (like $S_n, A_n, \mathbb{Z}_n, D_n$) and apply Lagrange's divisibility and index formulas directly without constructing full Cayley tables or explicit coset lists.