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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 $\operatorname{Sym}(\{1,2,3\})$ makes coset multiplication depend on representatives

Statement refuted

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

(aH)(bH)=(ab)H

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

Counterexample

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

$(\operatorname{id}g)H=gH\quad\text{and}\quad((12)g)H\ne gH.$

Thus replacing the first representative $\operatorname{id}$ 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 $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ 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 $aH$ , and $aH=bH$ exactly when $a^{-1}b\in H$ (Left and right cosets $gH$ and $Hg$ of a subgroup, $x\in aH$ iff $a^{-1}x\in H$ , and $aH=bH$ iff $a^{-1}b\in H$ ).

[L2]

Coset multiplication is well-defined exactly for normal subgroups (Coset multiplication $(gH)(hH)=ghH$ is well defined if and only if $H$ is normal).

Verification

technique · direct

1.1

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

L1algebra

1.2

Since $(12)\in H$ , the representatives $\operatorname{id}$ 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 $gH$ 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 $\operatorname{Sym}(\{1,2,3\})$ is normal and its quotient has two elements

Example

In $G=\operatorname{Sym}(\{1,2,3\})$ , let $N=\langle(123)\rangle$ . Then

N=\{\operatorname{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 $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , 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 $N\mathrel{\trianglelefteq}G$ , the cosets form a group with identity $N$ and inverse $(gN)^{-1}=g^{-1}N$ ).

[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 $(123)^3=\operatorname{id}$ , [F1] gives $N=\{\operatorname{id},(123),(132)\}$ .

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 $N\trianglelefteq G$ 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 $4\mathbb Z$ in $(\mathbb Z,+)$ reproduce addition modulo $4$

Example

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

4\mathbb Z,\quad 1+4\mathbb Z,\quad 2+4\mathbb Z,\quad 3+4\mathbb Z,

and its operation is

(a+4\mathbb Z)+(b+4\mathbb Z)=(a+b)+4\mathbb Z.

Under the identification of $a+4\mathbb Z$ with the residue class $[a]_4$ , this is addition modulo $4$ .

Facts & Assumptions

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

[L1]

Every congruence class modulo $4$ has a unique representative among $0,1,2,3$ (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[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 $\mathbb{Z}/n$ , Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ ).

[L2]

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

Verification

technique · direct

1.1

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

1.2

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

2.1

Sending $a+4\mathbb Z$ 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 $gN=\{gn:n\in N\}$ (Left and right cosets $gH$ and $Hg$ of a subgroup).

[L1]

For a normal subgroup $N$ , quotient multiplication is $(gN)(hN)=(gh)N$ (For $N\mathrel{\trianglelefteq}G$ , the cosets form a group with identity $N$ and inverse $(gN)^{-1}=g^{-1}N$ ).

Verification

technique · direct

1.1

For every $g\in G$ , [F2] gives $g\{e\}=\{ge\}=\{g\}$ . 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 $(g\{e\})(h\{e\})=(gh)\{e\}$ . Thus $g\{e\}\mapsto g$ preserves the multiplication exactly.

F1L1algebra

2.1

For every $g\in G$ , [F2] gives $gG=G$ , 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

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

Facts & Assumptions

Given: An abelian group $G$ .

[F1]

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

[F2]

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

[F3]

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

[F4]

The subgroup generated by a set is the smallest subgroup containing it (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , 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 $[g,h]=ghg^{-1}h^{-1}=gg^{-1}hh^{-1}=e$ . 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∎

Sources

Standard references

Recommended treatments; not extraction sources.

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})\operatorname{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})\operatorname{Sym}(\{1,2,3\})Sym({1,2,3}) is normal and its quotient has two elements

Example

Facts & Assumptions

Verification

The four cosets of 4Z4\mathbb Z4Z in (Z,+)(\mathbb Z,+)(Z,+) reproduce addition modulo 444

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

Sources

Standard references

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

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

The four cosets of $4\mathbb Z$ in $(\mathbb Z,+)$ 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 $[G,G]=\{e\}$