TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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If $char k ∣ ∣ G ∣$ , the augmentation ideal of $k [G]$ has no $k [G]$ -module complement in the regular representation

Statement

Let $G$ be a finite group and let $k$ be a field with $char k ∣ ∣ G ∣$ . If $I_{G} = \ker ε$ is the augmentation ideal of $k [G]$ , then there is no left $k [G]$ -submodule $J$ with

$k [G] = I_{G} \oplus J .$

Facts & Assumptions

Given: A finite group $G$ , a field $k$ with $char k ∣ ∣ G ∣$ , the augmentation map $ε : k [G] \to k$ , and the augmentation ideal $I_{G} = \ker ε$ .

[L1]

The augmentation map satisfies $ε ([g]) = 1_{k}$ for every $g \in G$ , is a ring homomorphism, and has kernel $I_{G}$ (The augmentation map $ε : R [G] \to R$ and the augmentation ideal $I_{G} = \ker ε$ ).

[L2]

In the regular representation, $g \cdot x = [g] x$ for $g \in G$ and $x \in k [G]$ (The trivial representation, the regular representation, and permutation representations from finite $G$ -sets).

[A1]

Because $char k ∣ ∣ G ∣$ , the scalar $∣ G ∣ \cdot 1_{k}$ is $0$ in $k$ .

Proof

technique · contradiction

1.1

Assume, for contradiction, that $k [G] = I_{G} \oplus J$ for some left $k [G]$ -submodule $J$ . Since $ε ([e]) = 1$ by [L1], one has $[e] \notin I_{G}$ , so $J \neq 0$ . The restriction $ε ∣_{J}$ is injective because $\ker (ε ∣_{J}) = J \cap I_{G} = 0$ . It is also nonzero, for otherwise $J \subseteq I_{G}$ . Choose $x \in J$ with $ε (x) \neq 0$ , and replace $x$ by $ε (x)^{- 1} x$ so that $ε (x) = 1$ .

L1givenassume-contrachoosealgebra

2.1

For each $g \in G$ , the element $g x = [g] x$ lies in $J$ because $J$ is a $k [G]$ -submodule by [L2]. Also $ε (g x) = ε ([g]) ε (x) = 1 \cdot 1 = 1$ by [L1], so $ε (g x - x) = 0$ and therefore $g x - x \in I_{G}$ . Since both $g x$ and $x$ lie in $J$ , one also has $g x - x \in J$ . Thus $g x - x \in I_{G} \cap J = 0$ , so $g x = x$ for every $g \in G$ .

step 1.1L1L2givenalgebra

3.1

Write $x = \sum_{h \in G} a_{h} [h] .$ For any $g \in G$ , step 2.1 gives $\sum_{h \in G} a_{h} [h] = x = g x = \sum_{h \in G} a_{h} [g h] = \sum_{h \in G} a_{g^{- 1} h} [h] .$ Comparing coefficients in the basis of $k [G]$ shows $a_{h} = a_{g^{- 1} h}$ for all $g, h$ , so all coefficients are equal to one scalar $a \in k$ . Hence $1 = ε (x) = \sum_{h \in G} a = ∣ G ∣ a = 0$ by [L1] and [A1], a contradiction. Therefore no such complement $J$ exists.

step 2.1L1A1givendischarge-contradiction∎

Depends on

Used by

If char k ∣ |G|, then k[G] is not semisimple Corollary

Dependency tree · two levels

9 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, Example 1.1.7 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Proposition 3.2 (standard reference, not scraped)

If char⁡k∣∣G∣chark∣∣G∣, the augmentation ideal of k[G]k[G] has no k[G]k[G]-module complement in the regular representation