CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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

1.1L1algebra

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.

1.2F1

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

1.3F1algebra

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.

2.1step 1.2F1

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

3.1step 2.1step 1.3L2∎

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].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two 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})Sym({1,2,3}) makes coset multiplication depend on representatives