Artin's induction theorem

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 $$G$$.

Related facts

 * Frobenius reciprocity
 * Brauer's induction theorem

Facts used

 * 1) uses::Frobenius reciprocity
 * 2) uses::Every group is a union of cyclic subgroups

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:

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.