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Group Homomorphisms and the Isomorphism Theorems: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For $n\ge2$ , reduction $\mathbb Z\to\mathbb Z/n$ has kernel $n\mathbb Z$ and realises $\mathbb Z/n$ by the first isomorphism theorem

Example

For $n\ge2$ , reduction $\mathbb Z\to\mathbb Z/n$ has kernel $n\mathbb Z$ and realises $\mathbb Z/n$ by the first isomorphism theorem.

Facts & Assumptions

Given: An integer $n\ge2$ and $\rho_n:\mathbb Z\to\mathbb Z/n$ , $\rho_n(a)=[a]_n$ .

[L1]

The first isomorphism theorem identifies a group modulo a homomorphism kernel with its image (First isomorphism theorem for groups: $G/\ker f\cong\operatorname{im}f$ ).

[L2]

$(\mathbb Z,+)/n\mathbb Z$ is the congruence-class group $(\mathbb Z/n,+)$ (For every $n\in\mathbb N$ , the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$ ).

[L3]

A group homomorphism preserves the operation (Monoid homomorphism and group homomorphism).

[L4]

The kernel is the inverse image of the identity (The kernel and image of a group homomorphism).

[L5]

The integers form a commutative ring, hence an additive group (The integers form a commutative ring).

Verification

technique · direct

1.1

Since $[a+b]_n=[a]_n+[b]_n$ , $\rho_n$ is a homomorphism of additive groups.

L1L2L3L4L5givenalgebra

2.1

Its kernel is $\{a:[a]_n=[0]_n\}=n\mathbb Z$ , and every residue class is $\rho_n(a)$ .

step 1.1L1L2L3L4L5givenalgebra

3.1

Therefore the kernel and image calculation yields $\mathbb Z/n\mathbb Z\cong\mathbb Z/n$ .

step 2.1∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

The doubling endomorphism of $(\mathbb Z,+)$ has trivial kernel but is not surjective

Statement refuted

An additive group homomorphism with trivial kernel must be surjective.

Facts & Assumptions

Given: The map $d:(\mathbb Z,+)\to(\mathbb Z,+)$ defined by $d(m)=2m$ .

[L1]

A group homomorphism is injective exactly when its kernel is trivial (A group homomorphism is injective if and only if its kernel is trivial).

[L2]

A group homomorphism preserves the group operation (Monoid homomorphism and group homomorphism).

[L3]

Surjectivity means that every codomain element is an image value (Injection, surjection, bijection).

[L4]

$\mathbb Z$ is an additive group with cancellation (The integers form a commutative ring).

Counterexample

technique · direct

1.1

The equality $d(a+b)=2a+2b=d(a)+d(b)$ makes $d$ a group homomorphism.

L1L2L3L4givenalgebra

2.1

If $d(m)=0$ , then $2m=0$ and integer cancellation gives $m=0$ , so $\ker d=\{0\}$ .

step 1.1L1L2L3L4givenalgebra

3.1

But $1$ is not even and hence is not in the image of $d$ , refuting the stated implication.

step 2.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The trivial homomorphism $G\to H$ has kernel $G$ and image $\{e_H\}$

Example

The trivial homomorphism $G\to H$ has kernel $G$ and image $\{e_H\}$ .

Facts & Assumptions

Given: Groups $G,H$ and $t:G\to H$ defined by $t(g)=e_H$ .

[L1]

A group homomorphism preserves products (Monoid homomorphism and group homomorphism).

[L2]

Kernels and images are defined by inverse images and value sets (The kernel and image of a group homomorphism).

[L3]

Every group has an identity element (Group and abelian group).

Verification

technique · direct

1.1

For $g,h\in G$ , $t(gh)=e_H=e_He_H=t(g)t(h)$ , so $t$ is a homomorphism.

L1L2L3givenalgebra

2.1

Every $g\in G$ satisfies $t(g)=e_H$ , while the only value of $t$ is $e_H$ .

step 1.1L1L2L3givenalgebra

3.1

Thus $\ker t=G$ and $\operatorname{im}t=\{e_H\}$ .

step 2.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Every inner automorphism of an abelian group is the identity

Example

Every inner automorphism of an abelian group is the identity.

Facts & Assumptions

Given: An abelian group $G$ and $g\in G$ .

[L1]

An inner automorphism has the form $c_g(x)=gxg^{-1}$ (Inner automorphisms and $\operatorname{Inn}(G)$ ).

[L2]

In an abelian group $gx=xg$ for all $g,x\in G$ (Group and abelian group).

[L3]

The center consists of elements commuting with every element (The center $Z(G)$ of a group).

Verification

technique · direct

1.1

Since $G$ is abelian, $g\in Z(G)$ and $gxg^{-1}=xgg^{-1}=x$ .

L1L2L3givenalgebra

2.1

Hence $c_g=\operatorname{id}_G$ for the chosen $g$ .

step 1.1L1L2L3given

3.1

As $g$ was arbitrary, every inner automorphism is the identity.

step 2.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Conjugation by $(1\,2)$ in $\operatorname{Sym}(\{1,2,3\})$ exchanges the transpositions $(1\,3)$ and $(2\,3)$

Example

Conjugation by $(1\,2)$ in $\operatorname{Sym}(\{1,2,3\})$ exchanges the transpositions $(1\,3)$ and $(2\,3)$ .

Facts & Assumptions

Given: The symmetric group on $\{1,2,3\}$ , with composition acting right-to-left.

[L1]

Conjugation is an automorphism (Conjugation $x\mapsto gxg^{-1}$ is an automorphism).

[L2]

Inner automorphisms are conjugations (Inner automorphisms and $\operatorname{Inn}(G)$ ).

[L3]

Elements of $\operatorname{Sym}(\{1,2,3\})$ are bijections composed right-to-left (The symmetric group $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition).

[L4]

These bijections form a group ( $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Verification

technique · direct

1.1

Direct evaluation on $1,2,3$ gives $(1\,2)(1\,3)(1\,2)=(2\,3)$ .

L1L2L3L4givenalgebra

2.1

Since $(1\,2)^{-1}=(1\,2)$ , the same computation with the roles reversed gives $(1\,2)(2\,3)(1\,2)=(1\,3)$ .

step 1.1L1L2L3L4givenalgebra

3.1

Thus this inner automorphism exchanges the two stated transpositions.

step 2.1∎

Sources

Standard references

Recommended treatments; not extraction sources.

Group Homomorphisms and the Isomorphism Theorems: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

For n≥2n\ge2n≥2, reduction Z→Z/n\mathbb Z\to\mathbb Z/nZ→Z/n has kernel nZn\mathbb ZnZ and realises Z/n\mathbb Z/nZ/n by the first isomorphism theorem

Example

Facts & Assumptions

Verification

The doubling endomorphism of (Z,+)(\mathbb Z,+)(Z,+) has trivial kernel but is not surjective

Statement refuted

Facts & Assumptions

Counterexample

The trivial homomorphism G→HG\to HG→H has kernel GGG and image {eH}\{e_H\}{eH​}

Example

Facts & Assumptions

Verification

Every inner automorphism of an abelian group is the identity

Example

Facts & Assumptions

Verification

Conjugation by (1 2)(1\,2)(12) in Sym⁡({1,2,3})\operatorname{Sym}(\{1,2,3\})Sym({1,2,3}) exchanges the transpositions (1 3)(1\,3)(13) and (2 3)(2\,3)(23)

Example

Facts & Assumptions

Verification

Sources

Standard references

For $n\ge2$ , reduction $\mathbb Z\to\mathbb Z/n$ has kernel $n\mathbb Z$ and realises $\mathbb Z/n$ by the first isomorphism theorem

The doubling endomorphism of $(\mathbb Z,+)$ has trivial kernel but is not surjective

The trivial homomorphism $G\to H$ has kernel $G$ and image $\{e_H\}$

Conjugation by $(1\,2)$ in $\operatorname{Sym}(\{1,2,3\})$ exchanges the transpositions $(1\,3)$ and $(2\,3)$