CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-01

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The product set $H K$ of two subgroups need not be a subgroup

Statement refuted

For any two subgroups $H, K \leq G$ , the product set $H K = {h k : h \in H, k \in K}$ is a subgroup of $G$ .

Facts & Assumptions

Given: The group $S_{3} = Sym ({1, 2, 3})$ , and the subgroups $H = {e, (12)}$ and $K = {e, (23)}$ .

[F1]

$S_{3}$ is a group under rightmost-first composition, and $H, K$ are subgroups because each transposition squares to $e$ (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, Subgroup).

[L1]

If $H$ is a subgroup of a finite group $G$ , then under the canonical embedding $ι : N \to Z$ one has $ι (∣ H ∣) ∣ ι (∣ G ∣)$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Counterexample

technique · direct

1.1

The six elements of $S_{3}$ are $e, (12), (13), (23), (123), (132)$ , so $∣ S_{3} ∣ = 6$ .

1.2

Direct multiplication gives $H K = {e, (12), (23), (12) (23)} = {e, (12), (23), (123)}$ , a set of four distinct elements.

2.1

If $H K$ were a subgroup, [L1] would force $ι (4) ∣ ι (6)$ in $Z$ , that is $4 ∣ 6$ , which is false. Hence $H K$ is not a subgroup.

step 1.1step 1.2L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Thomas W. Judson, Abstract Algebra: Theory and Applications, Cosets and Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.1: Cosets (standard reference, not scraped)

The product set HKHK of two subgroups need not be a subgroup