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6. Abstract Algebra

Symmetric group Sn

The symmetric group $S_n$ comprises all bijective mappings from a finite set of $n$ elements to itself under composition, forming a non-abelian group of order $n!$ for $n \ge 3$. This topic covers its cycle structures, conjugacy classes, alternating subgroups $A_n$, and simplicity theorems.

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Definition and statement

Cayley's Theorem & Structure of Symmetric Groups
Every group $G$ is isomorphic to a subgroup of some symmetric group $S_G$. Furthermore, for any finite set $X$ of cardinality $n$, the symmetric group $S_n$ has order $|S_n| = n!$ and is generated by transpositions, and for $n \ge 5$, the alternating group $A_n$ is simple.

📐 Geometric Interpretation

Visualized as the group of rigid and non-rigid symmetries or permutations of $n$ distinct labeled vertices of an $n$-simplex in $\mathbb{R}^{n-1}$, where group multiplication corresponds to successive geometric re-indexing.

⚙️ Algebraic & Structural Meaning

Serves as the universal embedding framework for finite groups via Cayley's representation, and provides core examples of non-abelian structures, commutator subgroups ($A_n$), and inner automorphism groups.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Finiteness of underlying set $X$ Finite set of cardinality $n \in \mathbb{N}$ Strictly Mandatory If the set is infinite (e.g., $\mathbb{R}$), the symmetric group lacks a well-defined factorial order and ordinary cycle decomposition breaks down.
Degree constraint for Simplicity ($n \ge 5$) Integer $n \ge 5$ Strictly Mandatory for non-solvability/simplicity For $n < 5$, $A_n$ is either trivial ($n=1,2$), abelian ($n=3$), or solvable ($n=4$), failing the simplicity condition.
Disjoint Cycle Decomposition Condition Products of disjoint cycles Essential for determining orders and parities Failure to ensure disjointness invalidates the rule that the order of a permutation is the least common multiple of its cycle lengths.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Order of a Permutation

$$ \text{O}(\sigma) = \text{lcm}(l_1, l_2, \dots, l_k) $$

Rule: Express any permutation $\sigma \in S_n$ as a product of disjoint cycles of lengths $l_1, l_2, \dots, l_k$. The order of $\sigma$ is instantly the least common multiple of these lengths.

Application: Frequently asked in UP PGT and KVS to find the order of elements like $(1 2 3)(4 5)$.
02

Parity and Sign of a Permutation

$$ \text{sgn}(\sigma) = (-1)^{n - k} $$

Rule: Where $n$ is the total number of symbols and $k$ is the number of disjoint cycles (including 1-cycles). Alternatively, $\text{sgn} = (-1)^{\sum (l_i - 1)}$.

Application: Directly applied to classify permutations into $A_n$ (even permutations) or $S_n \setminus A_n$ (odd permutations).
03

Number of Elements of a Given Cycle Type

$$ N_{(l_1, l_2, \dots)} = \frac{n!}{1^{k_1} 2^{k_2} \dots k_1! k_2! \dots} $$

Rule: To find the number of permutations with $k_1$ cycles of length 1, $k_2$ cycles of length 2, etc., apply the Cauchy formula instantly without listing.

Application: Standard question type in DSSSB PGT and UP PGT asking for the number of elements of order 2 in $S_4$ or $S_5$.
04

Maximum Order in $S_n$

$$ \max \{\text{lcm}(l_1 + \dots + l_k = n) \} $$

Rule: Partition $n$ into integers whose sum is $\le n$ to maximize the LCM.

05

Center of $S_n$

$$ Z(S_n) = \{e\} \quad \text{for } n \ge 3 $$

Rule: For $n=1,2$, $S_n$ is abelian so $Z(S_n) = S_n$. For $n \ge 3$, the center is trivial.

06

Number of Subgroups of Order 2

$$ \text{Count of transpositions} = \binom{n}{2} $$

Rule: Every element of order 2 in $S_n$ generated by a transposition gives a subgroup of order 2. Total subgroups of order 2 equals number of 2-cycles plus products of disjoint 2-cycles.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting Non-Disjoint Cycles Must Be Multiplied First

Common Pitfall: Candidates take the LCM of lengths of cycles when they are NOT disjoint (e.g., calculating order of $(1 2)(1 2 3)$ by taking $\text{lcm}(2,3)=6$ directly).

Counterexample: Let $\sigma = (1 2)(1 2 3) = (2 3)$

Why it fails: Multiplying $(1 2)$ and $(1 2 3)$ yields $(2 3)$, which is a single 2-cycle of actual order 2, not 6! Disjointness is mandatory.

🛡️ Exam Defense: Always reduce a product of overlapping cycles to a single disjoint cycle decomposition before calculating order or parity.
⚠️ COMMON ERROR #02

Miscalculating Alternating Group Order

Common Pitfall: Assuming $|A_n| = n!$ or $n! / n$.

Counterexample: $|A_n| = \frac{n!}{2}$ for $n \ge 2$

Why it fails: Half the permutations in $S_n$ are even and half are odd, meaning the index of $A_n$ in $S_n$ is 2.

🛡️ Exam Defense: Remember $|A_n| = \frac{n!}{2}$ for all $n \ge 2$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Symmetric Group $S_n$Finite set of $n$ elements$|S_n| = n!$Full permutation symmetry of $n$ objectsKeywords: 'symmetric group', 'permutations'
Alternating Group $A_n$Even permutations in $S_n$$|A_n| = \frac{n!}{2}$, simple for $n \ge 5$Orientation-preserving symmetriesKeywords: 'alternating group', 'even permutations'
Order of PermutationDisjoint cycle form$\text{lcm}(l_1, l_2, \dots, l_k)$Periodic recurrence under compositionKeywords: 'order of element', 'cycle notation'
Center $Z(S_n)$$n \ge 3$$Z(S_n) = \{e\}$Elements commuting with all othersKeywords: 'center', 'abelian', 'normal subgroup'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the order of the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 5 & 1 & 3 \end{pmatrix}$ in $S_5$.

(A) 3(B) 4(C) 6(D) 12
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Convert to cycle notation: $\sigma = (1 2 4)(3 5)$. Step 2: Cycles are disjoint with lengths 3 and 2. Step 3: Compute order $\text{lcm}(3,2) = 6$.
⚡ Method B: 25-Second Exam Shortcut
Write cycles directly from row mapping, spot lengths 3 and 2, and take $\text{lcm}(3,2)=6$ in 10 seconds.
💡 Tip: Always double-check cycle disjointness before taking LCM.
CANONICAL PROBLEM 02 Dual-Method Breakdown

What is the number of elements of order 2 in the symmetric group $S_3$?

(A) 1(B) 2(C) 3(D) 4
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Elements of order 2 must be self-inverse and not the identity. In $S_3$ (order 6), elements can have order 1 (identity), 2 (transpositions), or 3 (3-cycles). The transpositions are $(1 2), (2 3), (1 3)$, totaling 3 elements.
⚡ Method B: 25-Second Exam Shortcut
Elements of order 2 in $S_n$ are products of disjoint transpositions. For $S_3$, only single transpositions exist, numbering $\binom{3}{2} = 3$.
💡 Tip: Memorize small group element distributions for $S_3$ and $S_4$.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: What is the order of the symmetric group $S_4$?
A 12
B 24
C 6
D 4
Derivation & Analysis:

The order of $S_n$ is $n!$. For $n=4$, $4! = 24$.

Q2
MCQ 2: The alternating group $A_n$ is a simple group for:
A All $n \ge 3$
B All $n \ge 4$
C $n \ge 5$ only
D Prime $n$ only
Derivation & Analysis:

By Jordan's Theorem, $A_n$ is simple if and only if $n \ge 5$.

Q3
MCQ 3: What is the parity of the permutation $(1 2 3 4)$ in $S_4$?
A Even
B Odd
C Both even and odd
D Neither
Derivation & Analysis:

A single cycle of length $k$ can be written as $k-1$ transpositions. Here length is 4, giving 3 transpositions (odd), making it an odd permutation.

Q4
MCQ 4: What is the center $Z(S_4)$ of the symmetric group $S_4$?
A $S_4$
B $A_4$
C $\{(1)\}$
D Klein 4-group
Derivation & Analysis:

For any $n \ge 3$, the center of $S_n$ is trivial, i.e., $Z(S_n) = \{(1\}$.

Q5
MCQ 5: What is the maximum order of an element in $S_5$?
A 4
B 5
C 6
D 10
Derivation & Analysis:

Partitions of 5 that give maximal LCM are $3+2$, giving $\text{lcm}(3,2) = 6$.

Q6
MCQ 6: The index of $A_n$ in $S_n$ for $n \ge 2$ is:
A 1
B 2
C $n$
D $n!/2$
Derivation & Analysis:

The sign homomorphism $\text{sgn}: S_n \to \{1, -1\}$ has kernel $A_n$, so $|S_n : A_n| = 2$.

Q7
MCQ 7: How many elements of cycle type $(2, 2)$ exist in $S_4$?
A 3
B 6
C 8
D 4
Derivation & Analysis:

Using Cauchy's formula for cycle type $2^1 2^1$: $\frac{4!}{2^2 \cdot 2!} = \frac{24}{4 \times 2} = 3$.

Q8
MCQ 8: Every permutation can be uniquely expressed as a product of:
A Disjoint cycles
B Transpositions only
C 3-cycles
D Even permutations
Derivation & Analysis:

Fundamental theorem of symmetric groups states every permutation has a unique disjoint cycle decomposition (up to order).

Q9
MCQ 9: If $\sigma = (1 2 3)(2 3 4)$, what is $\sigma^{-1}$?
A $(1 3 4)$
B $(1 4 3)$
C $(1 2)(3 4)$
D $(2 4 3)$
Derivation & Analysis:

First compute $\sigma = (1 2)(3 4)$. Its inverse is itself since $(1 2)(3 4)$ squared is identity, wait: $(1 2 3)(2 3 4) = (1 3)(2 4)$, so inverse is $(1 3)(2 4) = (1 4 3)$? Let's check: $(1 2 3)(2 3 4)(1 4 3) = 1$. Option B is correct.

Q10
MCQ 10: Which of the following groups is isomorphic to a subgroup of $S_3$?
A $\mathbb{Z}_7$
B $\mathbb{Z}_6$
C Klein 4-group $V_4$
D $\mathbb{Z}_3$
Derivation & Analysis:

By Lagrange's theorem, any subgroup of $S_3$ (order 6) must divide 6. $\mathbb{Z}_3$ has order 3, which divides 6 and is cyclic, hence realizable.

Common questions

$A_n$ is the kernel of the group homomorphism $\text{sgn}: S_n \to \{1, -1\}$. Since the kernel of any homomorphism is a normal subgroup, $A_n \lhd S_n$ with index 2.

Failing to convert overlapping cycles into disjoint cycles before applying the LCM formula. Always multiply non-disjoint cycles first.

Memorize standard partitions and maximum orders for $S_3, S_4, S_5, S_6$, and practice rapid cycle-notation reading without writing out full Cayley tables.

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