6. Abstract Algebra

Properties of homomorphism

Group homomorphisms preserve algebraic structures, mapping identities, inverses, and subgroups neatly from domain to codomain.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Fundamental Properties of Group Homomorphism

Let $\phi: G \to G'$ be a group homomorphism. Then: (1) $\phi(e) = e'$, where $e$ and $e'$ are identity elements of $G$ and $G'$ respectively. (2) $\phi(x^{-1}) = [\phi(x)]^{-1}$ for all $x \in G$. (3) If $H \le G$, then $\phi(H) \le G'$. (4) If $K' \le G'$, then $\phi^{-1}(K') \le G$.

📐 Geometric Interpretation

A homomorphism acts as an algebraic projection or structural folding, collapsing parts of the domain (the kernel) into the identity while preserving the relational network of the remaining elements.

⚙️ Algebraic & Structural Meaning

Establishes the foundational bridge between algebraic systems, allowing complex group structures to be studied through simpler structural images via the First Isomorphism Theorem.

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	`Entire group domain $G$ x $G$`	Strictly Mandatory / Necessary	If $\phi(xy) \neq \phi(x)\phi(y)$ for even a single pair, the mapping is not a homomorphism.
Identity Mapping	`Identity element $e \in G$`	Strictly Mandatory	Failure to map identity to identity implies the map fails the basic group-preserving criteria.
Inverse Compatibility	`Arbitrary element $x \in G$`	Essential structural consequence	If inverse property fails under mapping, subgroup preservation and kernel properties collapse.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Order Division Rule

$$ o(\phi(x)) \text{ divides } o(x) $$

Rule: The order of the image element divides the order of the pre-image element in finite groups.

Application: Frequently asked in UP PGT to quickly eliminate incorrect homomorphism options.

Kernel Triviality & Injectivity

$$ \ker(\phi) = \{e\} \iff \phi \text{ is injective} $$

Rule: To prove injectivity of a homomorphism, check only if the kernel contains solely the identity element.

Cyclic Group Homomorphism Count

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

Rule: Instantly compute the number of possible homomorphisms between finite cyclic groups using the greatest common divisor.

Subgroup Order Divisibility

$$ |\phi(H)| \text{ divides } |H| \text{ and } |G'| $$

Rule: The cardinality of the homomorphic image divides both the size of the subgroup and the codomain.

Application: Used for rapid elimination in advanced abstract algebra MCQs.

Abelian Domain Transmission

$$ G \text{ is abelian} \implies \phi(G) \text{ is abelian} $$

Rule: Homomorphic images of abelian groups are always abelian, though the converse is false.

Rapid Elimination via Exponents

$$ \phi(x^n) = [\phi(x)]^n $$

Rule: Check exponent equations to instantly test whether a given formula defines a valid homomorphism.

Application: Saves 2 minutes during heavy algebraic matching questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming Ker($\phi$) is always in the Codomain

Common Pitfall: Candidates confuse the kernel with the image, looking for kernel elements inside the codomain $G'$.

Counterexample: $\ker(\phi) \subset G \text{ while } \text{Im}(\phi) \subset G'$

Why it fails: Kernel consists of elements from the domain $G$ that map to the identity $e'$ of $G'$.

🛡️ Exam Defense: Always remember: Kernel lives in the domain ($G$), Image lives in the codomain ($G'$).

⚠️ COMMON ERROR #02

Converse of Lagrange-type Homomorphism Properties

Common Pitfall: Assuming that if $o(y)$ divides $o(x)$, a homomorphism mapping $x$ to $y$ must always exist.

Counterexample: Mapping from $\mathbb{Z}_4$ to $\mathbb{Z}_2$

Why it fails: Divisibility of orders is a necessary condition, but structural constraints across generators can prevent existence.

🛡️ Exam Defense: Verify generator mapping explicitly before confirming homomorphism existence.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Group Homomorphism	$\phi(xy) = \phi(x)\phi(y)$	$\phi(e) = e'$	Structural folding / projection	Check operation preservation
Kernel of Homomorphism	$\phi: G \to G'$	$\ker(\phi) = \{x \in G : \phi(x) = e'\}$	Collapse points to identity	Look for elements mapping to $0$ or $1$
First Isomorphism Theorem	$\phi$ is a homomorphism	$G/\ker(\phi) \cong \text{Im}(\phi)$	Quotient space isomorphism	Relating quotient groups to images
Automorphism	Isomorphism from $G$ to $G$	$\text{Aut}(G)$ forms a group	Symmetric self-mapping	Check bijective endomorphism

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the number of homomorphisms from the cyclic group $\mathbb{Z}_{12}$ to $\mathbb{Z}_{30}$.

(A) 3(B) 6(C) 12(D) 30

📚 Method A: Formal Textbook Derivation

Step 1: Identify group orders $n = 12$ and $m = 30$. Step 2: Apply the formula for the number of homomorphisms between cyclic groups: $\gcd(n, m)$. Step 3: Calculate $\gcd(12, 30) = 6$. Step 4: Select option (B).

⚡ Method B: 25-Second Exam Shortcut

Instantly compute $\gcd(12, 30) = 6$ using prime factorization.

💡 Tip: Whenever cyclic groups are involved in homomorphisms, immediately think of $\gcd$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

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

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

📚 Method A: Formal Textbook Derivation

Step 1: For additive groups, $\phi(nx) = n\phi(x)$. Step 2: Substitute $x = 1$, giving $\phi(n\cdot 1) = n\phi(1)$. Step 3: Since $\phi(1) = 3$, we get $\phi(n) = 3n$. Step 4: Confirm with option (B).

⚡ Method B: 25-Second Exam Shortcut

Additive homomorphism behaves like scalar multiplication: $\phi(n) = n \cdot \phi(1)$.

💡 Tip: Watch out for multiplicative vs. additive notation traps.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $\phi: G \to G'$ is a group homomorphism, then $\phi(x^{-1})$ is equal to:

A $\phi(x)$

B $[\phi(x)]^{-1}$

C $e'$

D Undefined

Derivation & Analysis:

By fundamental properties of homomorphisms, inverses are preserved: $\phi(x^{-1}) = [\phi(x)]^{-1}$.

MCQ 2: The kernel of any group homomorphism $\phi: G \to G'$ is always a:

A Normal subgroup of $G'$

B Normal subgroup of $G$

C Cyclic group

D Field

Derivation & Analysis:

The kernel $\ker(\phi)$ is always a normal subgroup of the domain group $G$.

MCQ 3: The number of homomorphisms from $\mathbb{Z}_10$ to $\mathbb{Z}_15$ is:

A 1

B 5

C 10

D 15

Derivation & Analysis:

The number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is $\gcd(n, m) = \gcd(10, 15) = 5$.

MCQ 4: If $\phi: G \to G'$ is an injective homomorphism, then:

A $\ker(\phi) = G$

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

C $\text{Im}(\phi) = \{e'\}$

D $G \cong G'$

Derivation & Analysis:

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

MCQ 5: Let $\phi: (\mathbb{R}, +) \to (\mathbb{R}^+, \cdot)$ be defined by $\phi(x) = 2^x$. This mapping is:

A Not a homomorphism

B A homomorphism but not isomorphism

C An isomorphism

D None of these

Derivation & Analysis:

$\phi(x+y) = 2^{x+y} = 2^x \cdot 2^y = \phi(x)\phi(y)$. It is bijective, hence an isomorphism.

MCQ 6: If $G$ is a finite group and $\phi: G \to G'$ is a homomorphism, then the order of $\phi(x)$ divides:

A Only $o(x)$

B Only $|G'|$

C Both $o(x)$ and $|G'|$

D Neither

Derivation & Analysis:

The order of the image element divides both the order of the element $o(x)$ and the order of the codomain group $|G'|$.

MCQ 7: Which of the following maps from $(\mathbb{Z}, +)$ to $(\mathbb{Z}, +)$ is a homomorphism?

A $\phi(x) = x + 1$

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

C $\phi(x) = 5x$

D $\phi(x) = 0$

Derivation & Analysis:

Only $\phi(x) = 5x$ satisfies $\phi(x+y) = 5(x+y) = 5x + 5y = \phi(x) + \phi(y)$.

MCQ 8: According to the First Isomorphism Theorem, if $\phi: G \to G'$ is a homomorphism, then:

A $G \cong G'$

B $G/\ker(\phi) \cong \text{Im}(\phi)$

C $\ker(\phi) \cong G'$

D $G' / \text{Im}(\phi) \cong G$

Derivation & Analysis:

The First Isomorphism Theorem states that $G/\ker(\phi) \cong \text{Im}(\phi)$.

MCQ 9: If $\phi: G \to G'$ is a homomorphism and $H$ is a subgroup of $G$, then $\phi(H)$ is:

A Always a normal subgroup of $G'$

B A subgroup of $G'$

C Not necessarily a subgroup

D Always equal to $G'$

Derivation & Analysis:

The homomorphic image of a subgroup is always a subgroup of the codomain.

Q10

MCQ 10: If $\phi: \mathbb{Z}_8 \to \mathbb{Z}_8$ is a group homomorphism with $\phi(1) = 4$, then $\text{Im}(\phi)$ has order:

A 1

B 2

C 4

D 8

Derivation & Analysis:

Generated by 4 in $\mathbb{Z}_8$, the elements are $\{0, 4\}$, which has order 2? Wait: $\phi(1) = 4$, so $\text{Im}(\phi) = \langle 4 \rangle = \{0, 4\}$, order 2. Let's verify options: order 2 corresponds to subgroup size. Correct option index adjusted in logic or structure.

Common questions

No. By definition, $\phi(e) = e'$ is a mandatory property derived directly from the cancellation law in the codomain group.

Check the operation preservation property $\phi(xy) = \phi(x)\phi(y)$ and verify that $\phi(e) = e'$ instantly.

Remember the mnemonic: 'Kernel kills to identity' (lives in domain) while 'Image is the footprint' (lives in codomain).

← Previous Topic

Order of an element

Next Topic →

Subgroup generated by subsets