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✓ 5 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

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 $n \geq 2$ , reduction $Z \to Z / n$ has kernel $n Z$ and realises $Z / n$ by the first isomorphism theorem

Example

For $n \geq 2$ , reduction $Z \to Z / n$ has kernel $n Z$ and realises $Z / n$ by the first isomorphism theorem.

Facts & Assumptions

Given: An integer $n \geq 2$ and $ρ_{n} : Z \to Z / n$ , $ρ_{n} (a) = [a]_{n}$ .

[L1]

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

[L2]

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

[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 $[a + b]_{n} = [a]_{n} + [b]_{n}$ , $ρ_{n}$ is a homomorphism of additive groups.

L1L2L3L4L5givenalgebra

2.1

Its kernel is ${a : [a]_{n} = [0]_{n}} = n Z$ , and every residue class is $ρ_{n} (a)$ .

step 1.1L1L2L3L4L5givenalgebra

3.1

Therefore the kernel and image calculation yields $Z / n Z ≅ Z / n$ .

step 2.1∎

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

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

Statement refuted

An additive group homomorphism with trivial kernel must be surjective.

Facts & Assumptions

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

[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]

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

Counterexample

technique · direct

1.1

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

L1L2L3L4givenalgebra

2.1

If $d (m) = 0$ , then $2 m = 0$ 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 $g, h \in G$ , $t (g h) = e_{H} = e_{H} e_{H} = t (g) t (h)$ , 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 $im t = {e_{H}}$ .

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 $c_{g} (x) = g x g^{- 1}$ (Inner automorphisms and $Inn (G)$ ).

[L2]

In an abelian group $g x = x g$ 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 $g x g^{- 1} = x g g^{- 1} = x$ .

L1L2L3givenalgebra

2.1

Hence $c_{g} = {id}_{G}$ 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 $Sym ({1, 2, 3})$ exchanges the transpositions $(1 3)$ and $(2 3)$

Example

Conjugation by $(1 2)$ in $Sym ({1, 2, 3})$ 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 $x \mapsto g x g^{- 1}$ is an automorphism).

[L2]

Inner automorphisms are conjugations (Inner automorphisms and $Inn (G)$ ).

[L3]

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

[L4]

These bijections form a group ( $Sym (X)$ 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∎

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≥2, reduction Z→Z/nZ→Z/n has kernel nZnZ and realises Z/nZ/n by the first isomorphism theorem

Example

Facts & Assumptions

Verification

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

Statement refuted

Facts & Assumptions

Counterexample

The trivial homomorphism G→HG→H has kernel GG and image {eH}{eH​}

Example

Facts & Assumptions

Verification

Every inner automorphism of an abelian group is the identity

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

Sources

For $n \geq 2$ , reduction $Z \to Z / n$ has kernel $n Z$ and realises $Z / n$ by the first isomorphism theorem

The doubling endomorphism of $(Z, +)$ 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 $Sym ({1, 2, 3})$ exchanges the transpositions $(1 3)$ and $(2 3)$