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The dimension of an induced finite-dimensional representation is [G:H]dim⁡W

Statement

Let k be a field, let G be a finite group, let H≤G, and let W be a finite-dimensional representation of H over k. Then Ind⁡HGW is a finite-dimensional representation of G over k and

dim⁡kInd⁡HGW=[G:H]dim⁡kW.

Facts & Assumptions

Given: A field k, a finite group G, a subgroup H≤G, and a finite-dimensional representation W of H over k.

[F1]

A left transversal identifies Ind⁡HGW with a direct sum of one copy of W for each left coset of H in G (A left transversal identifies Ind⁡HGW with a direct sum of [G:H] copies of W).

[F2]

The dimension of a finite-dimensional vector space is the cardinality of any finite basis (Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis).

[F3]

A finite-dimensional representation is a finite-dimensional vector space with a linear group action (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree).

Proof

technique · direct
1.1

Choose a left transversal T={t1,…,tn} for G/H; since G is finite, n=[G:H]. By [F1], Ind⁡HGW≅⨁i=1nW as k-vector spaces.

F1givenchoose
2.1

Let B be a basis of W with ∣B∣=dim⁡kW by [F2]. The vectors supported in one summand and equal there to a basis element of B form a basis of ⨁i=1nW, so that direct sum has ndim⁡kW basis vectors. Hence dim⁡kInd⁡HGW=ndim⁡kW=[G:H]dim⁡kW.

F2step 1.1algebra
3.1

The induced module already carries a k-linear G-action by its definition, and step 2.1 shows that its underlying vector space is finite-dimensional. Therefore it is a finite-dimensional representation of G over k in the sense of [F3].

F3step 2.1
4.1

Steps 2.1 and 3.1 prove the stated dimension formula and finite-dimensionality claim.

step 2.1step 3.1∎

Depends on

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