Maximum degree of irreducible representation of subgroup is less than or equal to maximum degree of irreducible representation of whole group
Suppose is a finite group, is a subgroup, and is a field whose characteristic does not divide the order of (we do not require to be a splitting field, though the splitting field case is of particular interest).
Then, the Maximum degree of irreducible representation (?) of over is at least as much as the maximum degree of irreducible representation of over .
Proof in characteristic zero
Given: is a finite group, is a subgroup, and is a field of characteristic zero. is the maximum of the degrees of irreducible representations of over .
To prove: has an irreducible representation over of degree at least .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Let be the character of an irreducible representation of of degree .||is the maximum of degrees of irreducible representations of .|
|2||Let be the character of an irreducible representation of arising as a subrepresentation of||is a subgroup of the finite group||Step (1)|
|3||The inner product is a positive integer, because is the character of a subrepresentation of the representation affording . (Note: Over a splitting field, the inner product is equal to the multiplicity, but in general, it could be bigger if is not absolutely irreducible).||Fact (1)||To make sense of positive integer, we need to have characteristic zero.||Step (2)|
|5||is a positive integer, indicating that contains at least one copy of . (Note again: Over a splitting field, the inner product yields the multiplicity, but in general, it could be bigger if is not absolutely irreducible).||Steps (3), (4)|
|6||The degree of is at least as much as that of||Step (5)||[SHOW MORE]|
|7||The proof is done: is the desired character and the representation realizing it is the desired representation.||Steps (2), (6)||Step-combination-direct:|