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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30
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Induction is left adjoint to restriction for finite-group modules over a commutative ring

Statement

Let R be a commutative ring, let G be a finite group, let H≤G, let W be an R-linear H-module, and let V be an R-linear G-module. Then there is a natural isomorphism

Hom⁡G(Ind⁡HGW,V)≅Hom⁡H(W,Res⁡HGV).

Facts & Assumptions

Given: A commutative ring R, a finite group G, a subgroup H≤G, an R-linear H-module W, and an R-linear G-module V.

[F1]

The induced module consists of the functions f:G→W satisfying f(gh)=h−1⋅f(g), with G acting by (x⋅f)(g)=f(x−1g) (The induced R-linear G-module Ind⁡HGW as H-covariant functions on G).

[F2]

For modules, Hom⁡ is an abelian group under pointwise addition, with maps induced by composition (The abelian group Hom⁡R(M,N) and maps induced by pre- and postcomposition).

[F3]

G-equivariant maps are exactly the R[G]-module maps, and likewise for H (For a commutative ring R, R-linear G-actions are exactly the compatible left R[G]-module structures).

Proof

technique · constructive
1.1F1givenconstruct

For w∈W, define ηw:G→W by ηw(h):=h−1⋅w for h∈H and ηw(g):=0 for g∉H. If x∉H, then xh∉H for every h∈H, so ηw(xh)=0=h−1⋅0; if x∈H, then ηw(xh)=h−1x−1⋅w=h−1⋅ηw(x). Thus ηw∈Ind⁡HGW.

1.2F1F2givenchooseconstruct

Choose a left transversal T for G/H with e∈T. If ψ:W→Res⁡HGV is H-equivariant, define β(ψ)(f):=∑t∈Tt⋅ψ(f(t)). The sum is finite because T is finite.

2.1F1F2step 1.1construct

If Φ:Ind⁡HGW→V is G-equivariant, define α(Φ)(w):=Φ(ηw). For h∈H, one has ηh⋅w=h⋅ηw by the action formula in [F1], so α(Φ)(h⋅w)=Φ(h⋅ηw)=h⋅Φ(ηw)=h⋅α(Φ)(w). Hence α(Φ)∈Hom⁡H(W,Res⁡HGV).

2.2F1F2step 1.2algebra

The map β(ψ) is G-equivariant. Indeed, for x∈G and each t∈T, write x−1t=t′h with t′∈T and h∈H. Then t⋅ψ((x⋅f)(t))=t⋅ψ(f(x−1t))=t⋅ψ(f(t′h))=t⋅ψ(h−1⋅f(t′))=th−1⋅ψ(f(t′))=xt′⋅ψ(f(t′)), using the covariance from [F1] and the H-equivariance of ψ. Reindexing the finite sum by t′ shows β(ψ)(x⋅f)=x⋅β(ψ)(f).

3.1step 1.1step 2.1step 1.2algebra

For ψ∈Hom⁡H(W,Res⁡HGV), β(ψ)(ηw)=ψ(w) because ηw(t)=0 for t≠e and ηw(e)=w. Hence α(β(ψ))=ψ.

3.2F1step 2.1step 1.2algebra

For Φ∈Hom⁡G(Ind⁡HGW,V) and f∈Ind⁡HGW, one has β(α(Φ))(f)=∑t∈Tt⋅α(Φ)(f(t))=∑t∈Tt⋅Φ(ηf(t))=Φ(∑t∈Tt⋅ηf(t)). The function inside Φ equals f, because at a point th it takes the value h−1⋅f(t)=f(th) by covariance. So β(α(Φ))=Φ.

4.1F3step 3.1step 3.2discharge-construct∎

Steps 3.1 and 3.2 show that α and β are inverse group isomorphisms, giving the stated adjunction. Via [F3], this is equally the usual R[G]-module adjunction.

Depends on

Used by

Dependency tree · two levels

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Sources