TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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The regular representation is faithful

Statement

Let $G$ be a finite group and let $k$ be a field. Then the regular representation of $G$ over $k$ is faithful.

Facts & Assumptions

Given: A finite group $G$ and its regular representation on $k [G]$ .

[L1]

In the regular representation, $g \cdot [h] = [g h]$ for all $g, h \in G$ (The trivial representation, the regular representation, and permutation representations from finite $G$ -sets).

[L2]

A representation is faithful when the only group element acting as the identity linear map is the identity element of the group (Intertwiners, the spaces ${Hom}_{G} (V, W)$ and ${End}_{G} (V)$ , equivalent representations, and faithful representations).

Proof

technique · direct

1.1L1L2given

Suppose $g \in G$ acts as the identity in the regular representation. Applying that operator to the basis vector $[e]$ gives $[g] = g \cdot [e] = [e]$ by [L1].

2.1step 1.1L2∎

The basis vectors of $k [G]$ are indexed by the elements of $G$ , so $[g] = [e]$ forces $g = e$ . By [L2], this is exactly faithfulness.

Depends on

Used by

The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters Theorem

Dependency tree · two levels

9 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, Example 4.3.4 (standard reference, not scraped)