Maximum degree of irreducible representation of subgroup is less than or equal to maximum degree of irreducible representation of whole group

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Statement

Suppose G is a finite group, H is a subgroup, and K is a field whose characteristic does not divide the order of G (we do not require K to be a splitting field, though the splitting field case is of particular interest).

Then, the Maximum degree of irreducible representation (?) of G over K is at least as much as the maximum degree of irreducible representation of H over K.

Facts used

  1. Orthogonal projection formula
  2. Frobenius reciprocity: See also inner product of functions

Proof

Proof in characteristic zero

Given: G is a finite group, H is a subgroup, and K is a field of characteristic zero. m is the maximum of the degrees of irreducible representations of H over K.

To prove: G has an irreducible representation over K of degree at least m.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let \alpha be the character of an irreducible representation of H of degree m. m is the maximum of degrees of irreducible representations of H.
2 Let \beta be the character of an irreducible representation of G arising as a subrepresentation of \operatorname{Ind}_H^G\alpha. H is a subgroup of the finite group G Step (1)
3 The inner product \langle \beta, \operatorname{Ind}_H^G \alpha \rangle_G is a positive integer, because \beta is the character of a subrepresentation of the representation affording \operatorname{Ind}_H^G\alpha. (Note: Over a splitting field, the inner product is equal to the multiplicity, but in general, it could be bigger if \beta is not absolutely irreducible). Fact (1) To make sense of positive integer, we need K to have characteristic zero. Step (2)
4 \langle \operatorname{Res}^G_H \beta, \alpha \rangle_H = \langle \beta, \operatorname{Ind}_H^G \alpha \rangle_G. Fact (2) Fact-direct
5 \langle \operatorname{Res}^G_H \beta, \alpha \rangle_H is a positive integer, indicating that \operatorname{Res}^G_H \beta contains at least one copy of \alpha. (Note again: Over a splitting field, the inner product yields the multiplicity, but in general, it could be bigger if \alpha is not absolutely irreducible). Steps (3), (4)
6 The degree of \beta is at least as much as that of \alpha. Step (5) [SHOW MORE]
7 The proof is done: \beta is the desired character and the representation realizing it is the desired representation. Steps (2), (6) Step-combination-direct