PropositionStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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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.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

${Hom}_{G} (V, W)$ is a $k$ -vector space and ${End}_{G} (V)$ is a $k$ -algebra

Statement

Let $V$ and $W$ be representations of a group $G$ over a field $k$ .

The intertwiner space ${Hom}_{G} (V, W)$ is a $k$ -vector space.
The endomorphism space ${End}_{G} (V)$ is a $k$ -algebra.

Facts & Assumptions

Given: Representations $ρ : G \to GL (V)$ and $σ : G \to GL (W)$ over a field $k$ .

[L1]

An intertwiner $f : V \to W$ satisfies $f ρ (g) = σ (g) f$ for every $g \in G$ , and ${End}_{G} (V) = {Hom}_{G} (V, V)$ (Intertwiners, the spaces ${Hom}_{G} (V, W)$ and ${End}_{G} (V)$ , equivalent representations, and faithful representations).

[L2]

The space of all linear maps $L (V, W)$ is a $k$ -vector space ( $L (V, W)$ is a vector space over the common scalar field).

[L3]

Pointwise addition and composition make ${End}_{k} (V)$ a unital ring (Module endomorphisms form a ring under pointwise addition and composition, The endomorphism ring ${End}_{R} (M)$ under addition and composition).

[L4]

A $k$ -algebra is a unital ring whose multiplication is $k$ -bilinear and whose scalar copy of $k$ is central (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

Proof

technique · direct

1.1

The zero map satisfies $0 ρ (g) = 0 = σ (g) 0$ for every $g \in G$ . If $f, h \in {Hom}_{G} (V, W)$ and $λ \in k$ , then $(f + h) ρ (g) = f ρ (g) + h ρ (g) = σ (g) f + σ (g) h = σ (g) (f + h)$ and $(λ f) ρ (g) = λ (f ρ (g)) = λ (σ (g) f) = σ (g) (λ f)$ . So ${Hom}_{G} (V, W)$ contains $0$ and is closed under the pointwise operations of $L (V, W)$ .

L1L2givenalgebra

1.2

When $V = W$ , the identity map satisfies ${id}_{V} ρ (g) = ρ (g) = ρ (g) {id}_{V}$ , and if $f, h \in {End}_{G} (V)$ then $(f h) ρ (g) = f (h ρ (g)) = f (ρ (g) h) = ρ (g) f h$ . Thus ${End}_{G} (V)$ is closed under composition and contains the identity.

L1L3givenalgebra

2.1

Step 1.1 exhibits ${Hom}_{G} (V, W)$ as a linear subspace of the vector space $L (V, W)$ of [L2]. Therefore ${Hom}_{G} (V, W)$ is itself a $k$ -vector space.

step 1.1L2

3.1

By step 2.1, ${End}_{G} (V)$ is a $k$ -vector space. By step 1.2 and [L3], it is also a unital subring of ${End}_{k} (V)$ . For $λ \in k$ and $f, h \in {End}_{G} (V)$ , the usual identities $(λ f) h = λ (f h) = f (λ h)$ show that multiplication is $k$ -bilinear and the scalar copy of $k$ is central. Therefore [L4] makes ${End}_{G} (V)$ a $k$ -algebra.

step 2.1step 1.2L3L4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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 2 Section 2.1 (standard reference, not scraped)

Hom⁡G(V,W)HomG​(V,W) is a kk-vector space and End⁡G(V)EndG​(V) is a kk-algebra