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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Contents
Statement
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
.
Related facts
Facts used
Proof
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Proof idea
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
.
Proof:
Step no. | Assertion/construction | Given data used | Facts used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | The ![]() |
[SHOW MORE] | |||
2 | A subset of the ![]() ![]() ![]() |
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 ![]() ![]() |
(1), (2) | [SHOW MORE] | ||
4 | To show what we need to prove, it suffices to show that every class function is a ![]() ![]() |
(3) | [SHOW MORE] | ||
5 | Let ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
||||
6 | Let ![]() ![]() ![]() ![]() |
(5) | |||
7 | Suppose ![]() ![]() ![]() ![]() ![]() |
Fact (1) (Frobenius reciprocity) | |||
8 | ![]() |
(5), (6) | [SHOW MORE] | ||
9 | ![]() |
(7), (8) | [SHOW MORE] | ||
10 | ![]() ![]() |
(9) | [SHOW MORE] | ||
11 | ![]() ![]() |
The inner product is an inner product. | (10) | [SHOW MORE] | |
12 | ![]() ![]() ![]() |
(6), (7): ![]() |
[SHOW MORE] | ||
13 | ![]() |
![]() ![]() |
(12) | [SHOW MORE] | |
14 | ![]() |
(13) and (7) | [SHOW MORE] | ||
15 | ![]() ![]() ![]() ![]() |
(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.