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 $G$.

Related facts

• Frobenius reciprocity
• Brauer's induction theorem

Facts used

1. Frobenius reciprocity
2. Every group is a union of cyclic subgroups

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] All characters of $G$ are nonnegative integral linear combinations of irreducible characters of $G$. Integers are all in $\mathbb{Q}$
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] By step (1), all characters induced from characters of $X$ are characters, hence they are in the $\mathbb{Q}$-span of the space of all characters. By step (2), if we show that every class function is in the $\mathbb{C}$-span of these, we would show that every character is in the $\mathbb{Q}$-span of these. This would show what we need to prove.
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] This is the same as Step (3), except that we've replaced characters induced by class functions induced. This replacement is valid because class functions induced are in turn linear combinations of characters induced.
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] By definition, $f \in U = W^\perp$, so its inner product with anything in $W$ is zero.
9	$\langle \psi, \operatorname{Res}_H^G(f) \rangle = 0$			(7), (8)	[SHOW MORE] By Step (8), the left side of the expression in (7) is 0. Hence, so is the right side.
10	$\operatorname{Res}_H^G(f)$ is orthogonal to every class function of $H$.			(9)	[SHOW MORE] This is just Step (9), and the observation that $\psi$ was arbitrary.
11	$\operatorname{Res}_H^G(f) = 0$ for every $H \in X$		The inner product is an inner product.	(10)	[SHOW MORE] By Step (10), and the fact that we are dealing with an inner product, $\operatorname{Res}_H^G(f)$ is the zero function. Further, since $H$ was an arbitrary member of $X$, this result holds for all $H \in X$.
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] By Step (11), $f(g) = 0$ for every $g$ contained in any member of $X$. Since $f$ is a class function, $f$ also takes the value $0$ on all elements of conjugates of members of $X$.
13	$f$ is identically the zero function	$G$ is the union of all conjugates of members of $X$		(12)	[SHOW MORE] Follows directly from (12) and the assumption that $G$ is the union of all conjugates of members of $X$.
14	$W^\perp = 0$			(13) and (7)	[SHOW MORE] $f$ was an arbitrary element of $W^\perp$, so its being zero means $W^\perp$ is the zero space.
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] Since $W^\perp = 0$, and we are dealing with an inner product, we must have that $W$ equals the whole space $V$.
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.