6. Abstract Algebra

Fundamental theorem of homomorphism

The Fundamental Theorem of Homomorphism for groups bridges group structures and quotient groups by establishing a canonical isomorphism between the image of a homomorphism and the domain modulo the kernel. This topic covers this theorem and its ring-theoretic counterpart.

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

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Definition and statement

Fundamental Theorem of Group Homomorphism (First Isomorphism Theorem)

Let $\phi: G \to H$ be a group homomorphism. Then the kernel $\ker(\phi)$ is a normal subgroup of $G$, the image $\text{Im}(\phi)$ is a subgroup of $H$, and the quotient group $G/\ker(\phi)$ is isomorphic to $\text{Im}(\phi)$. Explicitly, the induced mapping $\bar{\phi}: G/\ker(\phi) \to \text{Im}(\phi)$ defined by $\bar{\phi}(x\ker(\phi)) = \phi(x)$ is a well-defined group isomorphism.

📐 Geometric Interpretation

Geometrically, a homomorphism collapses certain sub-structures (the kernel) to the identity element. The quotient group $G/\ker(\phi)$ treats each coset as a single point, effectively flattening the domain by factoring out all null-directions, leaving an exact replica (isomorphic copy) of the image space $\text{Im}(\phi)$.

⚙️ Algebraic & Structural Meaning

It provides a universal algebraic mechanism to decompose complex group structures into simpler quotient parts and homomorphic images, serving as the foundational engine for classification theorems in abstract algebra.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Mapping Property ($G$ and $H$ are groups)	`Algebraic structures ($G, H$)`	Strictly Mandatory / Necessary	If operations are non-associative or lack inverses, the notion of kernel and coset quotient breaks down completely.
Homomorphism Condition (\phi(xy) = \phi(x)\phi(y))	`Binary operation preservation`	Strictly Mandatory	Without structure preservation, $\ker(\phi)$ fails to be a normal subgroup, rendering the quotient group $G/\ker(\phi)$ ill-defined.
Normality of Kernel (\ker(\phi) \trianglelefteq G)	`Subgroup property`	Essential for conclusion	If the kernel is not normal, the set of left cosets $G/\ker(\phi)$ does not form a group under coset multiplication.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Order Calculation via Homomorphism

$$|G| = |\ker(\phi)| \cdot |\text{Im}(\phi)|$$

Rule: If a homomorphism $\phi: G \to H$ is given, the cardinality of the domain group equals the product of the order of the kernel and the order of the image.

Application: Frequently asked in UP PGT to quickly determine kernel size or image size without explicit coset listing.

Cyclic Group Image Rule

$$G \text{ is cyclic} \implies \text{Im}(\phi) \text{ is cyclic}$$

Rule: Homomorphic images preserve cyclicity. If $G = \langle a \rangle$, then $\text{Im}(\phi) = \langle \phi(a) \rangle$.

Application: Directly tested in DSSSB and KVS when identifying possible homomorphic images of $(\mathbb{Z}, +)$.

Abelian Quotient Deduction

$$G \text{ is abelian} \implies G/N \text{ is abelian}$$

Rule: Every quotient group of an abelian group is abelian. Since $\text{Im}(\phi) \cong G/\ker(\phi)$, the image is also abelian.

Application: Used for rapid elimination in multiple-choice questions concerning non-abelian target groups.

Index of Kernel for Finite Groups

$$[G : \ker(\phi)] = |\text{Im}(\phi)|$$

Rule: The index of the kernel in $G$ equals the exact order of the image subgroup in $H$.

Application: Standard shortcut for Lagrange-based order constraints in UP TGT/PGT.

Modular Arithmetic Homomorphisms (\mathbb{Z}_n)

$$\phi: \mathbb{Z}_n \to \mathbb{Z}_m \implies \text{order of } \phi(1) \text{ divides } \gcd(n, m)$$

Rule: The number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is equal to $\gcd(n, m)$.

Rapid Elimination via Cardinality

$$|\text{Im}(\phi)| \text{ must divide } |H| \text{ and } |G|_d$$

Rule: Eliminate options where the claimed image order does not divide the order of the codomain group $H$.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming Every Subgroup is a Kernel

Common Pitfall: Candidates assume that for any subgroup $H \leq G$, there exists a homomorphism $\phi$ such that $\ker(\phi) = H$.

Counterexample: Subgroup $2\mathbb{Z}$ in $(\mathbb{Z}, +)$ is normal, but not every arbitrary subgroup in a non-abelian group is a kernel unless it is normal.

Why it fails: Only *normal* subgroups can act as kernels of group homomorphisms.

🛡️ Exam Defense: Never link non-normal subgroups to kernels of group homomorphisms.

⚠️ COMMON ERROR #02

Confusing Isomorphism with Equality

Common Pitfall: Conclude that $G/\ker(\phi) = \text{Im}(\phi)$ in a literal set-theoretic sense.

Counterexample: Cosets $x\ker(\phi)$ are sets, whereas elements of $\text{Im}(\phi)$ are group elements in $H$.

Why it fails: They are structurally isomorphic ($\cong$), not identically equal sets.

🛡️ Exam Defense: Look for the symbol $\cong$ rather than $=$ when dealing with quotient structures and images.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
First Isomorphism Theorem	$\phi: G \to H$ is a homomorphism	$G/\ker(\phi) \cong \text{Im}(\phi)$	Collapsing null space to yield an exact isomorphic copy of the image	Keywords: quotient group, kernel, homomorphism image
Second Isomorphism Theorem	$H \leq G$ and $N \trianglelefteq G$	$HN/N \cong H/(H \cap N)$	Intersection scaling inside quotient spaces	Keywords: subgroup, normal subgroup, product of subgroups
Third Isomorphism Theorem	$K \leq H \trianglelefteq G$ with $K \trianglelefteq G$	$(G/K)/(H/K) \cong G/H$	Nested quotient reduction	Keywords: quotient of quotient groups
Ring Homomorphism Theorem	$R, S$ rings, $\phi: R \to S$ ring homomorphism	$R/\ker(\phi) \cong \text{Im}(\phi)$	Ideal factoring in ring structures	Keywords: ring, ideal, kernel of ring map

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $\phi: (\mathbb{Z}, +) \to (\mathbb{Z}_n, +_n)$ be defined by $\phi(x) = x \pmod n$. What is the kernel of $\phi$?

(A) $\mathbb{Z}$(B) $n\mathbb{Z}$(C) \{0\}(D) $\mathbb{Z}_n$

📚 Method A: Formal Textbook Derivation

Step 1: Set $\phi(x) = 0$ in $\mathbb{Z}_n$. Step 2: This implies $x \equiv 0 \pmod n$. Step 3: Therefore, $x$ is a multiple of $n$. Step 4: $\ker(\phi) = \{nx : x \in \mathbb{Z}\} = n\mathbb{Z}$.

⚡ Method B: 25-Second Exam Shortcut

By the First Isomorphism Theorem, $\mathbb{Z}/\ker(\phi) \cong \mathbb{Z}_n$. Since order of $\mathbb{Z}_n$ is $n$, the index of $\ker(\phi)$ in $\mathbb{Z}$ must be $n$, giving $n\mathbb{Z}$.

💡 Tip: Standard canonical question in UP PGT. Always check the modulus.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $\phi: (\mathbb{R}^*, \cdot) \to (\{1, -1\}, \cdot)$ be defined by $\phi(x) = \frac{x}{|x|}$. What is the quotient group $\mathbb{R}^*/\ker(\phi)$?

(A) $(\mathbb{R}^+, \cdot)$(B) $(\{1, -1\}, \cdot)$(C) $(\mathbb{Z}_2, +)$(D) Both (B) and (C) abstractly

📚 Method A: Formal Textbook Derivation

Step 1: Find kernel: $\phi(x) = 1 \implies x > 0$, so $\ker(\phi) = \mathbb{R}^+$. Step 2: Find image: $\text{Im}(\phi) = \{1, -1\}$. Step 3: By First Isomorphism Theorem, $\mathbb{R}^+/\mathbb{R}^+ \cong \{1, -1\} \cong \mathbb{Z}_2$.

⚡ Method B: 25-Second Exam Shortcut

Recognize that the image has order 2, and any group modulo its kernel is isomorphic to the image.

💡 Tip: Note that $(\{1, -1\}, \cdot)$ is isomorphic to $(\mathbb{Z}_2, +)$.

Practice questions

Choose an answer to check it and see the solution:

If $\phi: G \to H$ is a group homomorphism, then $\ker(\phi)$ is always:

A A cyclic subgroup of $G$

B A normal subgroup of $G$

C An abelian subgroup of $G$

D Equal to $H$

Derivation & Analysis:

The kernel of any group homomorphism is always a normal subgroup of the domain group $G$ by the fundamental properties of homomorphisms.

According to the First Isomorphism Theorem, if $\phi: G \to H$ is a homomorphism, then $G/\ker(\phi)$ is isomorphic to:

A $G$

B $H$

C $\text{Im}(\phi)$

D $\ker(\phi)$

Derivation & Analysis:

The fundamental theorem of homomorphism states that $G/\ker(\phi) \cong \text{Im}(\phi)$.

Let $G$ be a group of order 60 and $\phi: G \to H$ be a homomorphism with $|\text{Im}(\phi)| = 12$. What is the order of $\ker(\phi)$?

A 12

B 60

C 5

D 720

Derivation & Analysis:

Using $|G| = |\ker(\phi)| \cdot |\text{Im}(\phi)|$, we get $60 = |\ker(\phi)| \times 12$, which yields $|\ker(\phi)| = 5$.

How many homomorphisms exist from $(\mathbb{Z}_4, +)$ to $(\mathbb{Z}_2, +)$?

A 1

B 2

C 4

D 0

Derivation & Analysis:

The number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is $\gcd(n, m)$. Here $\gcd(4, 2) = 2$.

If $\phi: G \to H$ is an injective homomorphism, then:

A $\ker(\phi) = G$

B $\ker(\phi) = \{e_G\}$

C $\text{Im}(\phi) = \{e_H\}$

D $G \cong H$

Derivation & Analysis:

A homomorphism is injective if and only if its kernel is trivial, i.e., $\ker(\phi) = \{e_G\}$.

The quotient group $\mathbb{Z}/5\mathbb{Z}$ is isomorphic to:

A $\mathbb{Z}$

B $5\mathbb{Z}$

C $\mathbb{Z}_5$

D The trivial group

Derivation & Analysis:

By definition, the quotient group of integers modulo $n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$. Thus $\mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}_5$.

Let $\phi: G \to H$ be a surjective group homomorphism. If $G$ is abelian, then $H$ must be:

A Non-abelian

B Abelian

C Cyclic

D Trivial

Derivation & Analysis:

Since $H = \text{Im}(\phi) \cong G/\ker(\phi)$, and quotient groups of abelian groups are abelian, $H$ is abelian.

Which of the following mapping conditions is mandatory for $\phi: G \to H$ to be a group homomorphism?

A $\phi(x+y) = \phi(x) + \phi(y)$

B $\phi(xy) = \phi(x)\phi(y)$

C $\phi(xy) = \phi(y)\phi(x)$

D $\phi(x^{-1}) = \phi(x)$

Derivation & Analysis:

The defining property of a group homomorphism is preservation of the group operation: $\phi(xy) = \phi(x)\phi(y)$. Note that group operations are often written multiplicatively.

If $G$ is a cyclic group of order 12, what is the maximum possible order of a homomorphic image of $G$?

A 6

B 12

C 1

D Infinite

Derivation & Analysis:

The image $\text{Im}(\phi)$ is a subgroup of the codomain $H$, but since $G$ is cyclic, $\text{Im}(\phi)$ is also cyclic and must divide $|G| = 12$. The maximum possible order is 12.

Q10

Let $\phi: \mathbb{R} \to \mathbb{C}^*$ be defined by $\phi(x) = e^{ix}$. What is the kernel of $\phi$?

A \{0\}

B $2\pi\mathbb{Z}$

C $\mathbb{Z}$

D $\pi\mathbb{Z}$

Derivation & Analysis:

$\phi(x) = 1 \implies e^{ix} = 1 \implies x = 2k\pi$ for $k \in \mathbb{Z}$. Thus $\ker(\phi) = 2\pi\mathbb{Z}$.

Common questions

For the quotient set $G/H$ to form a valid group under coset multiplication $(aH)(bH) = (ab)H$, the subgroup $H$ must be normal ($gHg^{-1} = H$ for all $g \in G$). Since the kernel of a homomorphism satisfies this condition naturally via $\phi(xgx^{-1}) = \phi(x)\phi(g)\phi(x^{-1}) = \phi(g)$, only normal subgroups can serve as kernels.

Candidates often confuse the order formula $|G| = |\ker(\phi)| \cdot |\text{Im}(\phi)|$ with Lagrange's theorem involving arbitrary subgroups. Remember that the image order is tied directly to the quotient group cardinality, not an arbitrary subgroup of the codomain.

Use the order relation $|G| = |\ker(\phi)| \cdot |\text{Im}(\phi)|$ combined with the fact that $\text{Im}(\phi)$ must divide the order of the codomain. Checking divisibility eliminates wrong options instantly without calculating cosets.

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