Over the complex numbers
Over a splitting field of characteristic zero
The proof as presented here works only over the complex numbers, but it can be generalized to any splitting field for that has characteristic zero.
- Conjugacy class of prime power size implies not simple
- Order has only two prime factors implies solvable, also called Burnside's -theorem (proved via conjugacy class of prime power size implies not simple)
The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
|Fact no.||Statement||Steps in the proof where it is used||Qualitative description of how it is used||What does it rely on?||Difficulty level||Other applications|
|1||Size-degree-weighted characters are algebraic integers: For an irreducible linear representation over , multiplying any character value by the size of the conjugacy class and then dividing by the degree of the representation gives an algebraic integer.||Step (1) (in turn used in Step (4), leading to Step (5))||Helps in showing that is an algebraic integer.||Algebraic number theory + linear representation theory||click here|
|2||Characters are algebraic integers: The character of a linear representation is an algebraic integer.||Step (4), leading to Step (5)||Helps in showing that is an algebraic integer.||Basic linear representation theory||2||click here|
|3||Element of finite order is semisimple and eigenvalues are roots of unity||Step (6), which in turn is critical to later steps||Critical to understanding and , when combined with the triangle inequality and other facts.||Basic linear representation theory||click here|
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Given: A finite group , a nontrivial irreducible linear representation of over with character . An element with conjugacy class . The degree of and the size of are relatively prime.
To prove: Either or is a scalar.
Proof: Note that when the symbol appears as an input to a representation or a character, it refers to the identity element of . When it appears as the output of a character, or in another context, it refers to the real number .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The number is an algebraic integer.||Fact (1)||is finite, is an irreducible representation of over with character||Given+Fact direct|
|2||There exist integers and such that||and (the degree of ) are relatively prime.||By definition of relatively prime.|
|3||We get||Step (2)||Multiply both sides of Step (2) by .|
|4||The expression gives an algebraic integer.||Fact (2)||Step (1)||[SHOW MORE]|
|5||is an algebraic integer.||Steps (3), (4)||[SHOW MORE]|
|6||is the sum of many roots of unity (not necessarily all distinct), namely, the eigenvalues of the corresponding element .||Fact (3)||is finite.||[SHOW MORE]|
|7||Every algebraic conjugate of is also a sum of roots of unity.||Step (6)||[SHOW MORE]|
|8||Every algebraic conjugate of has modulus less than or equal to .||Step (7)||[SHOW MORE]|
|9||The modulus of the algebraic norm of in a Galois extension containing it is either 0 or 1.||Steps (5), (8)||[SHOW MORE]|
|10||If the modulus of the algebraic norm of is , then||[SHOW MORE]|
|11||If the modulus of the algebraic norm of is , then is a scalar matrix.||Steps (6), (8)||[SHOW MORE]|
|12||Either or is scalar.||Steps (9), (10), (11)||Step-combination direct.|