ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-28

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.
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Any nontrivial finite group algebra has zero divisors coming from a nonidentity cyclic subgroup

Example

Let $G$ be a nontrivial finite group, let $k$ be a field, and choose $g \in G ∖ {e}$ . Then $k [G]$ has nonzero zero divisors: $([e] - [g]) ([e] + [g] + \dots + [g^{n - 1}]) = 0,$ where $n \geq 2$ is the order of $g$ .

Facts & Assumptions

Given: A nontrivial finite group $G$ , a field $k$ , and $g \in G ∖ {e}$ .

[L1]

The group algebra has basis vectors $[x]$ indexed by the elements of $G$ , and they multiply by $[x] [y] = [x y]$ (The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements, The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ ).

[L2]

The powers $g^{m}$ are defined for all integers, with $g^{0} = e$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

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

Verification

technique · direct

1.1L1L2L3given

The subgroup $⟨ g ⟩$ of [L3] is finite because it is a subset of the finite group $G$ , so some least integer $n \geq 1$ satisfies $g^{n} = e$ . Because $g \neq e$ , this least $n$ is at least $2$ . The vectors $[e] - [g]$ and $[e] + [g] + \dots + [g^{n - 1}]$ are both nonzero, since they are sums of distinct basis vectors from [L1].

2.1step 1.1L1L2algebra∎

Using [L1] and [L2], $([e] - [g]) ([e] + [g] + \dots + [g^{n - 1}]) = ([e] + [g] + \dots + [g^{n - 1}]) - ([g] + [g^{2}] + \dots + [g^{n}]) = [e] - [g^{n}] = 0.$ So $k [G]$ has nonzero zero divisors.

Depends on

Used by

FALSE: if |G|>1, then k[G] is a field False statement

Dependency tree · two levels

21 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

Peter Webb, A Course in Finite Group Representation Theory, Chapter 1 Section 1.1 (standard reference, not scraped)