CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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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∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources

UCL lecture notes, Normal subgroups and quotients (standard reference, not scraped)

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