How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 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.

Normal Subgroups and Quotient Groups: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

A nonnormal two-element subgroup of $Sym ({1, 2, 3})$ makes coset multiplication depend on representatives

Statement refuted

For every subgroup $H$ of a group $G$ , the rule

$(a H) (b H) = (a b) H$

is independent of the representatives $a$ and $b$ .

Counterexample

Take $G = Sym ({1, 2, 3})$ and $H = {id, (12)}$ . Then $id H = (12) H = H$ , but with $g = (123)$ one has

$(id g) H = g H and ((12) g) H \neq g H .$

Thus replacing the first representative $id$ by the equivalent representative $(12)$ changes the proposed product.

Facts & Assumptions

Given: Permutations are composed from right to left, and $g = (123)$ .

[L1]

The symmetric group on ${1, 2, 3}$ is a group under composition (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[F1]

A left coset has the form $a H$ , and $a H = b H$ exactly when $a^{- 1} b \in H$ (Left and right cosets $g H$ and $H g$ of a subgroup, $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L2]

Coset multiplication is well-defined exactly for normal subgroups (Coset multiplication $(g H) (h H) = g h H$ is well defined if and only if $H$ is normal).

Verification

technique · direct

1.1

The set $H = {id, (12)}$ is a subgroup because $(12)^{2} = id$ , so both elements have their inverses in $H$ and $H$ is closed under composition.

L1algebra

1.2

Since $(12) \in H$ , the representatives $id$ and $(12)$ determine the same left coset $H$ .

1.3

These output cosets would be equal only if $g^{- 1} (12) g \in H$ . Direct composition gives $g^{- 1} (12) g = (13) \notin H$ , so they are unequal.

F1algebra

2.1

For this $g$ , the two proposed products obtained from the equal first input cosets are $g H$ and $((12) g) H$ .

step 1.2F1

3.1

Hence the displayed multiplication rule depends on the chosen representative for this subgroup, refuting the proposed statement and concretely realizing the nonnormal case in [L2].

step 2.1step 1.3L2∎

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

The three-cycle subgroup of $Sym ({1, 2, 3})$ is normal and its quotient has two elements

Example

In $G = Sym ({1, 2, 3})$ , let $N = ⟨ (123) ⟩$ . Then

$N = {id, (123), (132)}$

is normal, and $G / N$ is the two-element group ${N, (12) N}$ .

Facts & Assumptions

Given: Permutations are composed from right to left.

[L1]

[F1]

The subgroup generated by an element consists of its integral powers (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F2]

The index $[G : N]$ is the cardinality of the left-coset set (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L2]

A subgroup of index two is normal (Every subgroup of index two is normal).

[L3]

A normal subgroup gives a quotient group under coset multiplication (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[L4]

A quotient with finite index has order equal to that index (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

Verification

technique · direct

1.1

Since $(123)^{2} = (132)$ and $(123)^{3} = id$ , [F1] gives $N = {id, (123), (132)}$ .

L1F1algebra

2.1

The remaining three permutations are the transpositions, and $(12) N = {(12), (23), (13)}$ . Thus $N$ and $(12) N$ are the two left cosets of $N$ in $G$ .

step 1.1algebra

3.1

Consequently $[G : N] = 2$ by [F2], so $N ⊴ G$ by [L2].

step 2.1F2L2

4.1

The quotient group therefore exists by [L3], and its underlying coset set is precisely ${N, (12) N}$ . Its order is two by [L4].

step 2.1step 3.1L3L4∎

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

The four cosets of $4 Z$ in $(Z, +)$ reproduce addition modulo $4$

Example

The quotient group $Z / 4 Z$ has the four cosets

$4 Z, 1 + 4 Z, 2 + 4 Z, 3 + 4 Z,$

and its operation is

$(a + 4 Z) + (b + 4 Z) = (a + b) + 4 Z .$

Under the identification of $a + 4 Z$ with the residue class $[a]_{4}$ , this is addition modulo $4$ .

Facts & Assumptions

Given: The additive group $(Z, +)$ and its subgroup $4 Z$ .

[L1]

Every congruence class modulo $4$ has a unique representative among $0, 1, 2, 3$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[F1]

The integers modulo $4$ are congruence classes, and addition is defined by $[a]_{4} + [b]_{4} = [a + b]_{4}$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

[L2]

The quotient group $Z / 4 Z$ is literally the same set of classes with the same addition as the additive group of integers modulo $4$ (For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$ ).

Verification

technique · direct

1.1

By [L1], every coset $a + 4 Z$ equals exactly one of the four displayed cosets, and the four are distinct.

1.2

Quotient addition adds representatives, so the sum of $a + 4 Z$ and $b + 4 Z$ is $(a + b) + 4 Z$ .

2.1

Sending $a + 4 Z$ to $[a]_{4}$ therefore matches the four cosets with the four residue classes and carries the operation in step 1.2 to the modular addition in [F1].

step 1.1step 1.2F1L2∎

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

$G / {e}$ reproduces $G$ , while $G / G$ is the one-element quotient group

Example

For every group $G$ with identity $e$ , the quotient $G / {e}$ consists of the singleton cosets ${g}$ and has exactly the same multiplication as $G$ after identifying $g {e}$ with $g$ . At the other extreme, $G / G = {G}$ is the one-element quotient group.

Facts & Assumptions

Given: A group $G$ with identity $e$ .

[F1]

The sets ${e}$ and $G$ are subgroups of $G$ (Subgroup).

[F2]

A left coset is $g N = {g n : n \in N}$ (Left and right cosets $g H$ and $H g$ of a subgroup).

[L1]

For a normal subgroup $N$ , quotient multiplication is $(g N) (h N) = (g h) N$ (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

Verification

technique · direct

1.1

For every $g \in G$ , [F2] gives $g {e} = {g e} = {g}$ . Hence the cosets of ${e}$ are precisely the singleton subsets of $G$ .

1.2

The subgroup ${e}$ is normal because $g {e} g^{- 1} = {e}$ , and [L1] gives $(g {e}) (h {e}) = (g h) {e}$ . Thus $g {e} \mapsto g$ preserves the multiplication exactly.

F1L1algebra

2.1

For every $g \in G$ , [F2] gives $g G = G$ , so $G / G$ has the sole element $G$ . Since $G$ is normal in itself, [L1] makes this the one-element quotient group.

F1F2L1∎

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

For an abelian group $G$ , $Z (G) = G$ and $[G, G] = {e}$

Example

If $G$ is an abelian group with identity $e$ , then every element is central and every commutator is the identity. Consequently

$Z (G) = G and [G, G] = {e} .$

Facts & Assumptions

Given: An abelian group $G$ .

[F1]

The group axioms provide associativity, identity, and inverses (Group and abelian group).

[F2]

The center is $Z (G) = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

The commutator is $[g, h] = g h g^{- 1} h^{- 1}$ , and $[G, G]$ is the subgroup generated by all commutators (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F4]

The subgroup generated by a set is the smallest subgroup containing it (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Verification

technique · direct

1.1

Since $G$ is abelian, every $z \in G$ commutes with every $g \in G$ . Thus every element satisfies [F2], and $Z (G) = G$ .

F2given

1.2

For $g, h \in G$ , commutativity and the group laws give $[g, h] = g h g^{- 1} h^{- 1} = g g^{- 1} h h^{- 1} = e$ . Hence the set of all commutators is ${e}$ .

F1F3given

2.1

The set ${e}$ is itself a subgroup, so by the minimality in [F4] the subgroup it generates is ${e}$ . Therefore [F3] gives $[G, G] = {e}$ .

step 1.2F3F4∎

Normal Subgroups and Quotient Groups: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A nonnormal two-element subgroup of Sym⁡({1,2,3})Sym({1,2,3}) makes coset multiplication depend on representatives

Statement refuted

Counterexample

Facts & Assumptions

Verification

The three-cycle subgroup of Sym⁡({1,2,3})Sym({1,2,3}) is normal and its quotient has two elements

Example

Facts & Assumptions

Verification

The four cosets of 4Z4Z in (Z,+)(Z,+) reproduce addition modulo 44

Example

Facts & Assumptions

Verification

G/{e}G/{e} reproduces GG, while G/GG/G is the one-element quotient group

Example

Facts & Assumptions

Verification

For an abelian group GG, Z(G)=GZ(G)=G and [G,G]={e}[G,G]={e}

Example

Facts & Assumptions

Verification

Sources

A nonnormal two-element subgroup of $Sym ({1, 2, 3})$ makes coset multiplication depend on representatives

The three-cycle subgroup of $Sym ({1, 2, 3})$ is normal and its quotient has two elements

The four cosets of $4 Z$ in $(Z, +)$ reproduce addition modulo $4$

$G / {e}$ reproduces $G$ , while $G / G$ is the one-element quotient group

For an abelian group $G$ , $Z (G) = G$ and $[G, G] = {e}$