6. Abstract Algebra

Group homomorphism

A group homomorphism is a structure-preserving map between two groups that bridges algebraic operations, forming the foundation of quotient groups and isomorphism theorems.

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

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Definition and statement

Fundamental Homomorphism Theorem for Groups (First Isomorphism Theorem)

Let $(G, \cdot)$ and $(G', *)$ be groups, and let $\phi: G \to G'$ be a group homomorphism. Then the kernel of $\phi$, denoted $\text{Ker}(\phi) = \{g \in G \mid \phi(g) = e_{G'}\}$, is a normal subgroup of $G$, and the image $\text{Im}(\phi)$ is a subgroup of $G'$. Furthermore, the quotient group $G / \text{Ker}(\phi)$ is isomorphic to $\text{Im}(\phi)$, i.e., $G / \text{Ker}(\phi) \cong \text{Im}(\phi)$.

📐 Geometric Interpretation

A homomorphism acts like a structural projection or folding of a large group onto a smaller target group, where entire cosets of the kernel are collapsed into single identity elements of the image.

⚙️ Algebraic & Structural Meaning

It establishes the canonical bridge between structural reduction, normal subgroups, quotient groups, and bijective isomorphisms, serving as the most powerful computational tool 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
Binary Operation Preservation (Homomorphism Property)	`Entire Domain $G \times G$`	Strictly Mandatory / Necessary	If $\phi(ab) \neq \phi(a)\phi(b)$ for some $a, b \in G$, the map fails to be a homomorphism, and concepts like kernel and induced isomorphisms are undefined.
Identity Preservation	`Identity Element $e_G \in G$`	Strictly Mandatory (Derived property)	If $\phi(e_G) \neq e_{G'}$, the mapping violates fundamental group structure axioms.
Inverse Preservation	`Arbitrary element $g \in G$`	Strictly Mandatory (Derived property)	If $\phi(g^{-1}) \neq [\phi(g)]^{-1}$, the algebraic symmetry between domain and codomain breaks.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Number of Homomorphisms from Cyclic Groups

$$ \text{Number of homomorphisms } \mathbb{Z}_m \to \mathbb{Z}_n \text{ is } \gcd(m, n) $$

Rule: The number of group homomorphisms from a finite cyclic group $\mathbb{Z}_m$ to another finite cyclic group $\mathbb{Z}_n$ equals the greatest common divisor $\gcd(m, n)$.

Order of Image Divides Orders

$$ |\text{Im}(\phi)| \mid |G| \quad \text{and} \quad |\text{Im}(\phi)| \mid |G'| $$

Rule: The order of the image of any homomorphism from a finite group $G$ to $G'$ must divide both the order of $G$ and the order of $G'$.

Application: Used to instantly eliminate impossible sizes of images and kernels in multiple-choice questions.

Trivial Homomorphism Check

$$ \phi(g) = e_{G'} \quad \forall g \in G $$

Rule: The zero or trivial map is always a valid group homomorphism for any pair of groups.

Homomorphisms from $(\mathbb{Q}, +)$ to $(\mathbb{Z}, +)

$$ \text{Hom}(\mathbb{Q}, \mathbb{Z}) = \{0\} $$

Rule: Any homomorphism from a divisible group like $(\mathbb{Q}, +)$ to a torsion-free finitely generated group like $(\mathbb{Z}, +)$ must be identically zero.

Application: Advanced PGT question pattern testing mapping between infinite groups.

Kernel Size and Injectivity

$$ \text{Ker}(\phi) = \{e_G\} \iff \phi \text{ is injective (one-to-one)} $$

Rule: A group homomorphism is injective if and only if its kernel is trivial.

Rapid Elimination via Exponents

$$ o(\phi(g)) \mid o(g) $$

Rule: The order of the element $\phi(g)$ in $G'$ must divide the order of $g$ in $G$.

Application: Instantly eliminates options where element orders are mismatched under the mapping.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming $\phi(0) = 0$ is sufficient

Common Pitfall: Candidates check only that the identity maps to identity and forget to verify the multiplicative/additive preservation law for all pairs.

Counterexample: Map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi(x) = x + 1$. Here $\phi(0) = 1 \neq 0$, and $\phi(x+y) = x+y+1 \neq \phi(x)+\phi(y)$.

Why it fails: Fails the core homomorphism condition $\phi(a+b) = \phi(a) + \phi(b)$.

🛡️ Exam Defense: Always test $\phi(a * b) = \phi(a) \circ \phi(b)$ explicitly before checking identities.

⚠️ COMMON ERROR #02

Confusing Normal Subgroups with Kernels

Common Pitfall: Assuming every subgroup of $G$ can be the kernel of some homomorphism.

Counterexample: Any non-normal subgroup $H$ of a non-Abelian group $G$ (e.g., subgroup generated by a reflection in $D_4$).

Why it fails: The kernel of any group homomorphism is always a *normal* subgroup. Non-normal subgroups can never be kernels.

🛡️ Exam Defense: Remember: Kernel is always normal. If a subset is not a normal subgroup, it cannot be a kernel.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Monomorphism	Ker(phi) = {e}	phi is injective	Embedding without dimensional collapse	Tested via kernel triviality
Epimorphism	Im(phi) = G'	phi is surjective	Full covering projection	Tested via target coverage
Isomorphism	Bijective homomorphism	phi^-1 exists	Perfect structural structural duplication	Tested via structural equivalence
Automorphism	Isomorphism G -> G	phi in Aut(G)	Symmetric self-mapping	Tested via conjugation/permutations

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the total number of homomorphisms from the cyclic group $\mathbb{Z}_{10}$ to $\mathbb{Z}_{15}$.

(A) 1(B) 5(C) 10(D) 15

📚 Method A: Formal Textbook Derivation

Step 1: Identify domain as $\mathbb{Z}_{10}$ and codomain as $\mathbb{Z}_{15}$. Step 2: Recall that the number of homomorphisms from $\mathbb{Z}_m$ to $\mathbb{Z}_n$ is given by $\gcd(m, n)$. Step 3: Calculate $\gcd(10, 15) = 5$. Step 4: Conclude there are exactly 5 homomorphisms.

⚡ Method B: 25-Second Exam Shortcut

Directly apply formula $\gcd(10, 15) = 5$ in 2 seconds.

💡 Tip: Whenever cyclic groups are involved in domain and codomain, instantly check for $\gcd$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $\phi: (\mathbb{Z}, +) \to (\mathbb{Z}, +)$ be a group homomorphism such that $\phi(1) = 3$. What is $\phi(n)$ for any $n \in \mathbb{Z}$?

(A) $n$(B) $3n$(C) $n+3$(D) $3^n$

📚 Method A: Formal Textbook Derivation

Step 1: For any integer $n > 0$, $\phi(n) = \phi(1 + 1 + \dots + 1 \text{ ($n$ times)})$. Step 2: Use homomorphism property to write $\phi(1) + \phi(1) + \dots + \phi(1)$. Step 3: This equals $n \cdot \phi(1) = n(3) = 3n$. Step 4: Extends similarly for negative integers and zero.

⚡ Method B: 25-Second Exam Shortcut

For infinite cyclic group $\mathbb{Z}$, homomorphism is completely determined by the image of 1: $\phi(n) = n \cdot \phi(1)$.

💡 Tip: Check generator mapping; in cyclic groups, the entire homomorphism is fixed by the image of the generator.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the kernel of the homomorphism $\phi: (\mathbb{R}, +) \to (\mathbb{C}^{\times}, \cdot)$ defined by $\phi(x) = e^{ix}$?

A \{0\}

B $\{2n\pi \mid n \in \mathbb{Z}\}$

C $\{n\pi \mid n \in \mathbb{Z}\}$

D $\mathbb{Z}$

Derivation & Analysis:

Kernel consists of $x$ such that $e^{ix} = 1$, which occurs when $x = 2n\pi$ for $n \in \mathbb{Z}$.

MCQ 2: The number of non-trivial homomorphisms from $\mathbb{Z}_{12}$ to $\mathbb{Z}_{8}$ is:

A 2

B 3

C 4

D 7

Derivation & Analysis:

Total homomorphisms = $\gcd(12, 8) = 4$. Since 1 is the trivial homomorphism, non-trivial homomorphisms equal $4 - 1 = 3$ (Wait, $\gcd(12,8)=4$, total is 4. Non-trivial is 3).

MCQ 3: If $\phi: G \to G'$ is a group homomorphism, then for any $g \in G$, the order of $\phi(g)$:

A Divides the order of $g$

B Equals the order of $g$

C Is a multiple of the order of $g$

D Is always 1

Derivation & Analysis:

By group theory properties, if $g^k = e$, then $\phi(g)^k = \phi(g^k) = \phi(e) = e'$, hence $o(\phi(g)) \mid o(g)$.

MCQ 4: Which of the following mappings from $(\mathbb{R}, +)$ to $(\mathbb{R}^+, \cdot)$ is a homomorphism?

A $\phi(x) = x^2$

B $\phi(x) = e^x$

C $\phi(x) = \ln x$

D $\phi(x) = 2x$

Derivation & Analysis:

$\phi(x+y) = e^{x+y} = e^x e^y = \phi(x)\phi(y)$, satisfying the homomorphism axiom.

MCQ 5: Let $\phi: G \to G'$ be a group homomorphism. The kernel $\text{Ker}(\phi)$ is always:

A A cyclic subgroup

B An abelian subgroup

C A normal subgroup of $G$

D A normal subgroup of $G'$

Derivation & Analysis:

The kernel of any group homomorphism is a normal subgroup of the domain group $G$.

MCQ 6: If $\phi: G \to G'$ is an epimorphism and $G$ is finite, then:

A $|G| = |G'|$

B $|G| \ge |G'|$

C $|G| \le |G'|$

D No relation between orders

Derivation & Analysis:

Since $\phi$ is surjective (epimorphism), $G'$ is a homomorphic image of $G$, implying $|G'| \le |G|$, hence $|G| \ge |G'|$.

MCQ 7: The kernel of the determinant homomorphism $\det: GL_n(\mathbb{R}) \to (\mathbb{R}^{\times}, \cdot)$ is:

A $SL_n(\mathbb{R})$

B Identity matrix $I_n$

C Center of $GL_n(\mathbb{R})$

D Zero matrix

Derivation & Analysis:

By definition, $\text{Ker}(\det) = \{A \in GL_n(\mathbb{R}) \mid \det(A) = 1\}$, which is the special linear group $SL_n(\mathbb{R})$.

MCQ 8: Assertion (A): Every group homomorphism maps identity to identity. Reason (R): $\phi(e_G) = e_{G'}$ is derived from cancellation laws.

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:

$\phi(e) = \phi(e e) = \phi(e)\phi(e)$, multiplying by inverse gives $\phi(e) = e'$.

MCQ 9: If $G$ is a simple group and $\phi: G \to G'$ is a non-trivial homomorphism, then $\phi$ is:

A The zero map

B An isomorphism onto its image

C Not well-defined

D Surjective

Derivation & Analysis:

Since $\text{Ker}(\phi)$ is a normal subgroup of $G$ and $G$ is simple, $\text{Ker}(\phi)$ must be either $\{e\}$ or $G$. Since $\phi$ is non-trivial, $\text{Ker}(\phi) = \{e\}$, making $\phi$ injective (an isomorphism onto $\text{Im}(\phi)$).

Q10

MCQ 10: What is the number of homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}_{n}$?

A 0

B 1

C $n$

D Infinite

Derivation & Analysis:

Since $\mathbb{Z}$ is generated by 1, $\phi(1)$ can be chosen as any of the $n$ elements in $\mathbb{Z}_n$. Thus there are $n$ distinct homomorphisms.

Common questions

No. Only normal subgroups can serve as kernels of group homomorphisms. If a subgroup is not normal, it cannot be the kernel of any homomorphism.

Candidates often confuse the condition for ring homomorphisms (preserving addition and multiplication) with group homomorphisms (preserving a single binary operation).

Memorize standard shortcuts like $\gcd(m, n)$ for cyclic groups $\mathbb{Z}_m \to \mathbb{Z}_n$ and recognize that the image size divides both group orders.

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Group isomorphism