Artin's induction theorem
This article states an induction theorem: a result relating the linear characters and linear representations of a group with the characters/representations induced from the linear characters/representations of subgroups
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This fact is related to: linear representation theory
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Let be a finite group and a family of subgroups of . Then the following are equivalent:
- The union of conjugates of elements of cover the whole of
- Every character of over is a rational linear combination of characters induced from characters of members of
Further, these equivalent conditions hold if is the collection of all cyclic subgroups of .
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With a little linear algebra, we can show that if a character of is a complex linear combination of characters induced from members of , all the coefficients are in fact rational. Thus, the problem reduces to showing that the class functions induced from members of span the space of all class functions on .
This proof follows by using Frobenius reciprocity, and the fact that the only class function on which restricts to the zero function on every member of , is the zero function on the whole of .
Proof details: (1) implies (2)
Given: A finite group with a family of subgroups such that the union of conjugates of members of is .
To prove: Every character of is a rational linear combination of characters induced from characters of members of .
|Step no.||Assertion/construction||Given data used||Facts used||Previous steps used||Explanation|
|1||The (rational) vector space span of the irreducible characters contain all the characters.||[SHOW MORE]|
|2||A subset of the -span of characters with the property that its -span contains all class functions also has the property that its -span contains all characters.||Something about linear algebra; fields are linearly closed||(1)|
|3||To show what we need to prove, it suffices to show that every class function is a -linear combination of characters induced from members of .||(1), (2)||[SHOW MORE]|
|4||To show what we need to prove, it suffices to show that every class function is a -linear combination of class functions induced from members of .||(3)||[SHOW MORE]|
|5||Let be the -span of all the class functions of . is also the space of class functions of . Let be the span of all class functions induced from characters of members of . In other words: .|
|6||Let be the orthogonal complement to in with respect to the inner product of class functions:||(5)|
|7||Suppose . Then, for any and any class function of , we have: .||Fact (1) (Frobenius reciprocity)|
|8||(5), (6)||[SHOW MORE]|
|9||(7), (8)||[SHOW MORE]|
|10||is orthogonal to every class function of .||(9)||[SHOW MORE]|
|11||for every||The inner product is an inner product.||(10)||[SHOW MORE]|
|12||for every element of in the union of conjugates of members of .||(6), (7): is a class function; (11)||[SHOW MORE]|
|13||is identically the zero function||is the union of all conjugates of members of||(12)||[SHOW MORE]|
|14||(13) and (7)||[SHOW MORE]|
|15||, i.e., every class function on is in the -span of those induced from class functions on members of .||(14), (5), and (6)||[SHOW MORE]|
|16||We are done.||(4) and (15)||Follows directly.|
Proof details: (2) implies (1)
The proof here is essentially the same; it uses Frobenius reciprocity to reason in the opposite direction.
Given: A finite group with a family of subgroups such that every character of is a rational linear combination of characters induced from .
To prove: is the union of conjugates of members of .
Proof: Since every character of is a rational linear combination of the characters induced from , it is in particular true that the -span of class functions induced from class functions of , is the whole space of class functions on .
Taking the usual inner product of class functions:
Now, suppose is a class function of that takes the value on the union of conjugates of and is outside. Then we have that for every and every class function of :
By Frobenius reciprocity, we get:
In other words, is orthogonal to all the class functions induced from members of . By assumption, is thus orthogonal to every class functino of , forcing . By te way we defined , we obtain that the union of conjugates of must be the whole group .
Proof details for the additional observation
The additional observation follows from fact (2): every group is a union of cyclic subgroups.