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PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04
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Restriction and induction along a subgroup preserve projective modules

Statement

Let HG be finite groups and let k be a field. Then restriction ResHG and induction IndHG both send projective modules to projective modules.

Facts & Assumptions

Given: A subgroup HG and a field k.

[F1]

Projective modules are characterized by lifting and by being direct summands of free modules (Projective modules and the lifting property, Equivalent characterizations of projective modules).

[L2]

A left transversal identifies IndHGW with a finite direct sum of copies of W (A left transversal identifies IndHGW with a direct sum of [G:H] copies of W).

Proof

technique · direct
1.1

Let P be a projective kG-module. By [F1], it is a direct summand of a free module (kG)n. Restricting to H preserves direct sums and summands. By [L2] with W=kH, the restricted regular module ResHG(kG) is a finite direct sum of copies of kH, hence is free as a kH-module. Therefore ResHGP is a direct summand of a free kH-module and is projective by [F1].

F1L2givenalgebra
2.1

Let Q be a projective kH-module. To prove that IndHGQ is projective, use the lifting characterization in [F1]. Given a surjection u:XY of kG-modules and a map ϕ:IndHGQY, adjunction [L1] turns ϕ into a map ψ:QResHGY. The restriction of u is still surjective, so projectivity of Q lifts ψ to ψ~:QResHGX. Applying [L1] again yields a lift ϕ~:IndHGQX of ϕ. Thus IndHGQ is projective.

F1L1step 1.1algebra
3.1

Steps 1.1 and 2.1 prove that both restriction and induction preserve projectives.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources