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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Statement

Let $G$ be a finite group and $X$ a family of subgroups of $G$. Then the following are equivalent:

1. The union of conjugates of elements of $X$ cover the whole of $G$
2. Every character of $G$ over $\mathbb{C}$ is a rational linear combination of characters induced from characters of members of $X$

Further, these equivalent conditions hold if $X$ is the collection of all cyclic subgroups of $G$.

Proof

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Proof idea

With a little linear algebra, we can show that if a character of $G$ is a complex linear combination of characters induced from members of $X$, all the coefficients are in fact rational. Thus, the problem reduces to showing that the class functions induced from members of $X$ span the space of all class functions on $G$.

This proof follows by using Frobenius reciprocity, and the fact that the only class function on $G$ which restricts to the zero function on every member of $X$, is the zero function on the whole of $G$.

Proof details: (1) implies (2)

Given: A finite group $G$ with a family of subgroups $X$ such that the union of conjugates of members of $X$ is $G$.

To prove: Every character of $G$ is a rational linear combination of characters induced from characters of members of $X$.

Proof:

Step no. Assertion/construction Given data used Facts used Previous steps used Explanation
1 The $\mathbb{Q}$ (rational) vector space span of the irreducible characters contain all the characters. [SHOW MORE]
2 A subset of the $\mathbb{Q}$-span of characters with the property that its $\mathbb{C}$-span contains all class functions also has the property that its $\mathbb{Q}$-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 $\mathbb{C}$-linear combination of characters induced from members of $X$. (1), (2) [SHOW MORE]
4 To show what we need to prove, it suffices to show that every class function is a $\mathbb{C}$-linear combination of class functions induced from members of $X$. (3) [SHOW MORE]
5 Let $V$ be the $\mathbb{C}$-span of all the class functions of $G$. $V$ is also the space of class functions of $G$. Let $W$ be the span of all class functions induced from characters of members of $X$. In other words: $\! W = \langle \operatorname{Ind}_H^G \psi \mid H \in X \rangle$.
6 Let $U = W^\perp$ be the orthogonal complement to $W$ in $V$ with respect to the inner product of class functions: $\! \langle \alpha,\beta \rangle = \frac{1}{|G|} \sum_{g \in G} \alpha(g) \overline{\beta(g)}$ (5)
7 Suppose $f \in U$. Then, for any $H \in X$ and any class function $\psi$ of $H$, we have: $\! \langle \operatorname{Ind}_H^G (\psi), f \rangle = \langle \psi, \operatorname{Res}_H^G(f) \rangle$. Fact (1) (Frobenius reciprocity)
8 $\langle \operatorname{Ind}_H^G (\psi), f \rangle = 0$ (5), (6) [SHOW MORE]
9 $\langle \psi, \operatorname{Res}_H^G(f) \rangle = 0$ (7), (8) [SHOW MORE]
10 $\operatorname{Res}_H^G(f)$ is orthogonal to every class function of $H$. (9) [SHOW MORE]
11 $\operatorname{Res}_H^G(f) = 0$ for every $H \in X$ The inner product is an inner product. (10) [SHOW MORE]
12 $f(g) = 0$ for every element of $g$ in the union of conjugates of members of $X$. (6), (7): $f$ is a class function; (11) [SHOW MORE]
13 $f$ is identically the zero function $G$ is the union of all conjugates of members of $X$ (12) [SHOW MORE]
14 $W^\perp = 0$ (13) and (7) [SHOW MORE]
15 $W = V$, i.e., every class function on $G$ is in the $\mathbb{C}$-span of those induced from class functions on members of $X$. (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 $G$ with a family of subgroups $X$ such that every character of $G$ is a rational linear combination of characters induced from $X$.

To prove: $G$ is the union of conjugates of members of $X$.

Proof: Since every character of $G$ is a rational linear combination of the characters induced from $X$, it is in particular true that the $\mathbb{C}$-span of class functions induced from class functions of $X$, is the whole space of class functions on $G$.

Taking the usual inner product of class functions: $\langle \alpha,\beta \rangle = \frac{1}{|G|} \sum_{g \in G} \alpha(g) \overline{\beta(g)}$.

Now, suppose $f$ is a class function of $G$ that takes the value $0$ on the union of conjugates of $H$ and is $1$ outside. Then we have that for every $H \in X$ and every class function $\psi$ of $H$: $\langle \psi, \operatorname{Res}_H^G f \rangle = 0$.

By Frobenius reciprocity, we get: $\langle \operatorname{Ind}_H^G (\psi), f \rangle = 0$.

In other words, $f$ is orthogonal to all the class functions induced from members of $H$. By assumption, $f$ is thus orthogonal to every class functino of $G$, forcing $f = 0$. By te way we defined $f$, we obtain that the union of conjugates of $H \in X$ must be the whole group $G$.

Proof details for the additional observation

The additional observation follows from fact (2): every group is a union of cyclic subgroups.