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 $X$

Related facts

• Frobenius reciprocity
• Brauer's induction theorem

Facts used

1. Frobenius reciprocity

Proof

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: First, note that all characters of $G$ are integral linear combinations of irreducible characters of $G$. Thus, the $\mathbb{Q}$-vector space span of the irreducible characters contain all the characters. If there is a subset of this space whose $\mathbb{C}$-span contains this vector space, so does its $\mathbb{Q}$-span. Thus, to show that a collection of characters admits every character as a rational linear combination, it suffices to show that every character is a $\mathbb{C}$-linear combination of characters from that collection.

Further, since the irreducible characters of a finite group span the space of class functions to $\mathbb{C}$, this reduces to proving that any class function of $G$ is a $\mathbb{C}$-linear combination of class functions induced from members of $X$.

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:

$W=⟨{\mathrm{I}}_{\mathrm{n}}^{\mathrm{d}}\psi \mid H\in X⟩$.

Let $U=W^{\perp }$ be the orthogonal complement to $W$ with respect to the inner product of class functions:

$⟨\alpha ,\beta ⟩=\frac{1}{\left|G\right|}{\sum }_{g\in G}\alpha (g)\overline{beta(g)}$.

Suppose $f\in U$. Then, for any $H\in X$ and any class function $\psi$ of $H$, we have, by Frobenius reciprocity:

$⟨{\mathrm{I}}_{\mathrm{n}}^{\mathrm{d}}(\psi ),f⟩=⟨\psi ,{\mathrm{R}}_{\mathrm{e}}^{\mathrm{s}}(f)⟩$.

Since the left side is zero by assumption, so is the right side. Thus, ${\mathrm{R}}_{\mathrm{e}}^{\mathrm{s}}(f)$ is orthogonal to every class function of $H$, and thus, $\mathrm{R}\mathrm{e}\mathrm{s}(f)=0$. This applies to every $H\in G$, so $f$ is the zero function on each $H\in X$.

Now, since $f$ is a class function, $f$ is also the zero function on every conjugate subgroup to a member of $X$. Thus, $f(g)=0$ whenever $g$ lies in some conjugate of some $H\in X$. By our assumption that $G$ is the union of conjugates of members of $X$, we obtain that $f$ is the zero function. Thus, $U=0$, and we get $W=V$.

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:

$⟨\alpha ,\beta ⟩=\frac{1}{\left|G\right|}{\sum }_{g\in G}\alpha (g)\overline{beta(g)}$.

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 $H$:

$⟨\psi ,{\mathrm{R}}_{\mathrm{e}}^{\mathrm{s}}f⟩=0$.

By Frobenius reciprocity, we get:

$⟨{\mathrm{I}}_{\mathrm{n}}^{\mathrm{d}}(\psi ),f⟩=0$.

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 $G$.