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Over a splitting field, every $G$ -endomorphism of an irreducible representation is scalar

Statement

Let $G$ be a finite group, let $k$ be a splitting field for $G$ , and let $V$ be an irreducible representation of $G$ over $k$ . Then every endomorphism in ${End}_{G} (V)$ has the form $λ {id}_{V}$ with $λ \in k$ .

Facts & Assumptions

Given: A finite group $G$ , a splitting field $k$ for $G$ , and an irreducible representation $V$ of $G$ over $k$ .

[L1]

For an irreducible representation, ${End}_{G} (V)$ is a division ring (Schur's lemma for irreducible representations: a nonzero intertwiner is an isomorphism, and ${End}_{G} (V)$ is a division ring).

[L2]

By definition, a splitting field for $G$ is a field over which every irreducible representation has endomorphism ring exactly $k$ (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).

Proof

technique · direct

1.1

By [L2], the endomorphism ring ${End}_{G} (V)$ is exactly the scalar copy of $k$ inside ${End}_{k} (V)$ .

L2given

2.1

Therefore every $G$ -endomorphism of $V$ is $λ {id}_{V}$ for a unique $λ \in k$ . Step 1.1 is compatible with the division-ring conclusion of [L1], which is why the scalar copy is a field.

step 1.1L1∎

Depends on

Used by

Every irreducible representation of a finite abelian group over a splitting field is one-dimensional Theorem

Dependency tree · two levels

7 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, Proposition 9.2.5 (standard reference, not scraped)

Over a splitting field, every GG-endomorphism of an irreducible representation is scalar