6. Abstract Algebra

Normal subgroups

Normal subgroups are foundational substructures in abstract algebra that allow the construction of well-defined quotient groups through the preservation of conjugation. This topic covers normal subgroups, their equivalent characterizations, and tests.

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

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Definition and statement

Definition and Fundamental Characterizations of Normal Subgroups

A subgroup $N$ of a group $G$ is called a normal subgroup (denoted by $N \trianglelefteq G$) if it is invariant under inner automorphisms of $G$, i.e., \forall g \in G, \forall n \in N, gng^{-1} \in N. Equivalently, left cosets and right cosets coincide: \forall g \in G, gN = Ng.

📐 Geometric Interpretation

Symmetry preservation under conjugation; visualizing group actions where internal substructures are left invariant by external transformations.

⚙️ Algebraic & Structural Meaning

Normal subgroups are precisely the kernels of group homomorphisms and form the exact algebraic condition required to define a valid group structure on the set of cosets (quotient group $G/N$).

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 ($N \le G$)	`Algebraic substructure`	Strictly Mandatory	If $N$ is not a subgroup, coset operations and quotient structures fail entirely.
Closure under Conjugation ($gng^{-1} \in N$)	`Universal quantification over $G$`	Strictly Mandatory	Failing this condition means left and right cosets do not partition the group compatibly, rendering $G/N$ non-multiplicative.
Index 2 Condition ($[G:N] = 2$)	`Cardinality condition`	Sufficient (not strictly necessary)	If index is not 2, normality cannot be automatically concluded solely from index without checking conjugation or kernel properties.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Index 2 Subgroup Rule

$$\text{If } H \le G \text{ and } [G:H] = 2, \text{ then } H \trianglelefteq G.$$

Rule: Any subgroup of index 2 in a group $G$ is automatically a normal subgroup.

Application: Frequently asked in UP PGT and KVS to instantly identify normal subgroups of symmetric groups $S_n$ (like $A_n$) or dihedral groups.

Center of a Group

$$Z(G) = \{z \in G \mid zg = gz, \forall g \in G\} \implies Z(G) \trianglelefteq G$$

Rule: The center of any group $G$ is always a normal subgroup of $G$. Furthermore, every subgroup of an abelian group is normal.

Application: Direct theoretical questions in DSSSB PGT and UP PGT regarding abelian groups and centers.

Kernel of Homomorphism

$$\text{If } \phi: G \to H \text{ is a group homomorphism, then } \ker(\phi) \trianglelefteq G.$$

Rule: Every normal subgroup arises as the kernel of some group homomorphism, and vice versa.

Sylow $p$-Subgroup Uniqueness

$$\text{If } P \text{ is the unique Sylow } p\text{-subgroup of } G, \text{ then } P \trianglelefteq G.$$

Rule: A unique Sylow $p$-subgroup for a given prime $p$ is always normal in $G$.

Direct Product Components

$$G = G_1 \times G_2 \implies G_1 \times \{e_2\} \trianglelefteq G$$

Rule: Direct product factors are normal subgroups of the direct product group.

Application: Appears in structural classification of finite abelian groups.

Simple Groups Elimination

$$|G| = p \text{ (prime) } \implies \text{ only normal subgroups are } \{e\} \text{ and } G$$

Rule: Groups of prime order have no proper non-trivial normal subgroups (they are simple).

Application: Rapid elimination trick for options involving cyclic groups of prime order.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Subgroup of Normal Subgroup is Normal

Common Pitfall: Candidates assume that if $H \trianglelefteq N$ and $N \trianglelefteq G$, then $H \trianglelefteq G (normality is transitive).

Counterexample: Alternating group $A_4$ has a normal subgroup $V_4$ (Klein 4-group), and $V_4$ has subgroups of order 2 which are normal in $V_4$ but not normal in $A_4$.

Why it fails: Normality is not a transitive relation in general.

🛡️ Exam Defense: Never assume transitivity for normal subgroups unless specified (e.g., characteristic subgroups).

⚠️ COMMON ERROR #02

Every Subgroup of Abelian Group vs Non-Abelian Group

Common Pitfall: Assuming non-abelian groups have no normal subgroups other than trivial ones, or forgetting that all subgroups of abelian groups are normal.

Counterexample: Dihedral group $D_4$ has non-normal subgroups of order 2.

Why it fails: Non-abelian groups possess a mix of normal and non-normal subgroups.

🛡️ Exam Defense: Always verify conjugation $gng^{-1} \in N$ or index for non-abelian groups.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Normal Subgroup	$gN = Ng \text{ for all } g \in G$	$gng^{-1} \in N	Symmetric invariant substructure	Quotient group formation
Center $Z(G)$	Commutes with all elements	$zg = gz \forall g	Central fixed core	Always normal in any group
Kernel $\ker(\phi)$	Homomorphism $\phi: G \to H$	$f(x) = e_H$	Collapse to identity element	First Isomorphism Theorem
Index 2 Subgroup	$[G:H] = 2$	$G = H \cup gH$	Half-split partition	Instant normality deduction

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $H$ is a subgroup of $G$ of index 2, then which of the following is always true?

(A) $H$ is not a subgroup(B) $H$ is a normal subgroup of $G$(C) $H$ is equal to $G$(D) $H$ is abelian

📚 Method A: Formal Textbook Derivation

Step 1: Note index $[G:H] = 2$. Step 2: Left and right cosets partition $G$ into two sets: $H$ and $G \setminus H$. Step 3: Since $G = H \cup gH = H \cup Hg$ for any $g \notin H$, left and right cosets coincide. Step 4: By definition, $H \trianglelefteq G$.

⚡ Method B: 25-Second Exam Shortcut

Apply Shortcut 1 directly: Index 2 implies normality instantly.

💡 Tip: Direct theorem statement frequently asked in UP TGT/PGT.

CANONICAL PROBLEM 02 Dual-Method Breakdown

What is the number of proper non-trivial normal subgroups of the symmetric group $S_3$?

(A) 0(B) 1(C) 2(D) 3

📚 Method A: Formal Textbook Derivation

Step 1: List all subgroups of $S_3$ (orders 1, 2, 3, 6). Step 2: Subgroups of order 2 have index $6/2 = 3$ (not index 2, hence not normal). Step 3: Subgroup of order 3 is $A_3$, which has index $6/3 = 2$, hence normal. Step 4: Only $A_3$ is a proper non-trivial normal subgroup.

⚡ Method B: 25-Second Exam Shortcut

Use index 2 rule: $A_3$ has index 2 in $S_3$, so it is normal. Subgroups of order 2 are not normal because they are not unique in their conjugacy class.

💡 Tip: Remember $A_n$ is always normal in $S_n$ with index 2.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: A subgroup $N$ of $G$ is normal if and only if for every $g \in G$:

A $gN \neq Ng$

B $gN = Ng$

C $gN \subset Ng$

D $Ng \subset gN$

Derivation & Analysis:

By definition, a subgroup $N$ is normal in $G$ if and only if its left cosets and right cosets coincide, i.e., $gN = Ng$ for all $g \in G$.

MCQ 2: If $G$ is an abelian group, then every subgroup of $G$ is:

A Not a subgroup

B Normal

C Infinite

D Cyclic

Derivation & Analysis:

In an abelian group, $gh = hg$ for all $g, h \in G$. Thus $gNg^{-1} = gg^{-1}N = N$, so every subgroup is normal.

MCQ 3: The index of the alternating group $A_n$ in the symmetric group $S_n$ ($n \ge 2$) is:

A $n$

B $n!$

C 2

D $n/2$

Derivation & Analysis:

$A_n$ consists of all even permutations, which has order $n!/2$. Thus the index $[S_n : A_n] = \frac{n!}{n!/2} = 2$.

MCQ 4: Which of the following is always a normal subgroup of any group $G$?

A Any subgroup of prime order

B The center $Z(G)$

C Any cyclic subgroup

D Any subgroup of order 2

Derivation & Analysis:

The center $Z(G) = \{z \in G \mid zg = gz, \forall g \in G\}$ commutes with every element of $G$ and is always a normal subgroup.

MCQ 5: If $H$ is a subgroup of index 2 in $G$, then $H$ is:

A Always normal

B Never normal

C Abelian only

D Trivial

Derivation & Analysis:

Any subgroup of index 2 in any group $G$ is automatically a normal subgroup of $G$.

MCQ 6: Kernel of any group homomorphism $\phi: G \to H$ is always a:

A Non-normal subgroup

B Normal subgroup of $G$

C Normal subgroup of $H$

D Cyclic group

Derivation & Analysis:

The kernel of a homomorphism $\phi: G \to H$ satisfies the subgroup property and normality condition, hence $\ker(\phi) \trianglelefteq G$.

MCQ 7: If $N$ and $M$ are normal subgroups of $G$, then their intersection $N \cap M$ is:

A Not a subgroup

B Always a normal subgroup of $G$

C Always trivial

D Always of index 2

Derivation & Analysis:

The intersection of any family of normal subgroups is a normal subgroup.

MCQ 8: Assertion (A): Every subgroup of a cyclic group is normal. Reason (R): Every cyclic group is abelian.

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 abelian, and every subgroup of an abelian group is normal. Thus both are true and (R) explains (A).

MCQ 9: In the symmetric group $S_3$, the subgroup $H = \{(1), (1 2)\}$ is:

A A normal subgroup

B Not a normal subgroup

C equal to $A_3$

D The center of $S_3$

Derivation & Analysis:

H has order 2 and index 3 in $S_3$. Conjugating $(1 2)$ by $(1 3)$ yields $(2 3) \notin H$, hence $H$ is not normal.

Q10

MCQ 10: If $G$ is a simple group of order 60, its proper non-trivial normal subgroups number is:

A 0

B 1

C 2

D 5

Derivation & Analysis:

By definition, a simple group has no proper non-trivial normal subgroups, so the count is 0.

Common questions

Without normality ($gN = Ng$), coset multiplication $(aN)(bN) = abN$ is not well-defined, as the product would depend on the choice of coset representatives.

Assuming normality is transitive (i.e., $H \trianglelefteq N$ and $N \trianglelefteq G \implies H \trianglelefteq G$), which is false in general.

Memorize high-yield shortcuts such as the index 2 rule, center normality, and homomorphism kernel properties to bypass lengthy conjugation tests.