LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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Group algebra bimodule is induced from the diagonal

Statement

With the double action above, $k G ≅ {Ind}_{Δ G}^{G \times G} k$ , where $k$ is the trivial diagonal module.

Facts & Assumptions

Given: A finite group $G$ over the field $k$ .

[F1]

The double action is $(x, y) v = x v y^{- 1}$ . (Block bimodule for the double group)

Proof

technique · direct

1.1

The map from left cosets $(x, y) Δ G \mapsto x y^{- 1}$ is well-defined, since $(x h) (y h)^{- 1} = x y^{- 1}$ . It is surjective using $(g, 1)$ . If $x y^{- 1} = x' (y')^{- 1}$ , then $x^{- 1} x' = y^{- 1} y' = h$ , so $(x', y') = (x, y) (h, h)$ and the cosets agree. Thus it is bijective.

2.1

The induced module $k [G \times G] \otimes_{k Δ G} k$ has the cosets as a basis: the tensor relation identifies precisely multiplication on the right by a diagonal element. Extending the bijection linearly gives an isomorphism; $(a, b)$ sends its image to $a x y^{- 1} b^{- 1}$ , the image of $(a x, b y) Δ G$ , proving equivariance.

F1step 1.1∎

Sources

Webb, A Course in Finite Group Representation Theory, §§5.2, 11.3, 11.6, 12.3–12.5; especially Lemma 12.4.4 and Theorem 12.4.5, pp.240–241. Local argument and conventions as displayed above.

Depends on

Block bimodule for the double group

Used by

Dependency tree · two levels

3 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

Webb, A Course in Finite Group Representation Theory, §§5.2, 11.3, 11.6, 12.3–12.5; especially Lemma 12.4.4 and Theorem 12.4.5, pp.240–241 (standard reference, not scraped)