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

## 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 $K$.

## 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