Artin's induction theorem: Difference between revisions

Revision as of 12:35, 9 October 2008

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:

The union of conjugates of elements of $X$ cover the whole of $G$
Every character of $G$ over $C$ is a rational linear combination of characters induced from characters of members of $X$

Related facts

Facts used

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 $Q$ -vector space span of the irreducible characters contain all the characters. If there is a subset of this space whose $C$ -span contains this vector space, so does its $Q$ -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 $C$ -linear combination of characters from that collection.

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

Let $V$ be the $C$ -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:

Failed to parse (unknown function "\oepratorname"): {\displaystyle W = \langle \oepratorname{Ind}_H^G \psi \mid H \in X \rangle} .

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

$⟨ α, β ⟩ = \frac{1}{| G |} \sum_{g \in G} α (g) \bar{b e t a (g)}$ .

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

$⟨ I_{n}^{d} (ψ), f ⟩ = ⟨ ψ, R_{e}^{s} (f) ⟩$ .

Since the left side is zero by assumption, so is the right side. Thus, $R_{e}^{s} (f)$ is orthogonal to every class function of $H$ , and thus, $R e s (f) = 0$ . 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 $C$ -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:

$⟨ α, β ⟩ = \frac{1}{| G |} \sum_{g \in G} α (g) \bar{b e t a (g)}$ .

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 $ψ$ of $H$ :

$⟨ ψ, R_{e}^{s} f ⟩ = 0$ .

By Frobenius reciprocity, we get:

$⟨ I_{n}^{d} (ψ), f ⟩ = 0$ .

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 $G$ .

@@ Line 1: / Line 1: @@
 {{induction theorem}}
-{{factrelatedto|linear representation theory}}
+{{fact related to|linear representation theory}}
 ==Statement==
@@ Line 7: / Line 7: @@
 Let <math>G</math> be a [[finite group]] and <math>X</math> a family of subgroups of <math>G</math>. Then the following are equivalent:
-* The union of conjugates of elements of <math>X</math> cover the whole of <math>G</math>
+# The union of conjugates of elements of <math>X</math> cover the whole of <math>G</math>
-* Every character of <math>G</math> is a rational linear combination of characters induced from characters of members of <math>X</math>
+# Every character of <math>G</math> over <math>\mathbb{C}</math> is a rational linear combination of characters induced from characters of members of <math>X</math>
+==Related facts==
+* [[Frobenius reciprocity]]
+* [[Brauer's induction theorem]]
+==Facts used==
+# [[uses::Frobenius reciprocity]]
 ==Proof==
-===Simplifying the result to complex linear combinations===
+===Proof idea===
 With a little linear algebra, we can show that if a character of <math>G</math> is a complex linear combination of characters induced from members of <math>X</math>, all the coefficients are in fact rational. Thus, the problem reduces to showing that the class functions induced from members of <math>X</math> span the space of all class functions on <math>G</math>.
@@ Line 18: / Line 27: @@
 This proof follows by using [[Frobenius reciprocity]], and the fact that the only class function on <math>G</math> which restricts to the zero function on every member of <math>X</math>, is the zero function on the whole of <math>G</math>.
-===Details===
+===Proof details: (1) implies (2)===
+'''Given''': A finite group <math>G</math> with a family of subgroups <math>X</math> such that the union of conjugates of members of <math>X</math> is <math>G</math>.
+'''To prove''': Every character of <math>G</math> is a rational linear combination of characters induced from characters of members of <math>X</math>.
+'''Proof''': First, note that all characters of <math>G</math> are integral linear combinations of irreducible characters of <math>G</math>. Thus, the <math>\mathbb{Q}</math>-vector space span of the irreducible characters contain all the characters. If there is a subset of this space whose <math>\mathbb{C}</math>-span contains this vector space, so does its <math>\mathbb{Q}</math>-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 <math>\mathbb{C}</math>-linear combination of characters from that collection.
+Further, since the irreducible characters of a finite group span the space of class functions to <math>\mathbb{C}</math>, this reduces to proving that any class function of <math>G</math> is a <math>\mathbb{C}</math>-linear combination of class functions induced from members of <math>X</math>.
+Let <math>V</math> be the <math>\mathbb{C}</math>-span of all the class functions of <math>G</math>. <math>V</math> is also the space of class functions of <math>G</math>. let <math>W</math> be the span of all class functions induced from characters of members of <math>X</math>. In other words:
+<math>W = \langle \oepratorname{Ind}_H^G \psi \mid H \in X \rangle</math>.
+Let <math>U = W^\perp</math> be the orthogonal complement to <math>W</math> with respect to the inner product of class functions:
+<math>\langle \alpha,\beta \rangle = \frac{1}{|G|} \sum_{g \in G} \alpha(g) \overline{beta(g)}</math>.
+Suppose <math>f \in U</math>. Then, for any <math>H \in X</math> and any class function <math>\psi</math> of <math>H</math>, we have, by Frobenius reciprocity:
+<math>\langle \operatorname{Ind}_H^G (\psi), f \rangle = \langle \psi, \operatorname{Res}_H^G(f) \rangle</math>.
+Since the left side is zero by assumption, so is the right side. Thus, <math>\operatorname{Res}_H^G(f)</math> is orthogonal to every class function of <math>H</math>, and thus, <math>\operatorname{Res}(f) = 0</math>. This applies to every <math>H \in G</math>, so <math>f</math> is the zero function on each <math>H \in X</math>.
+Now, since <math>f</math> is a class function, <math>f</math> is also the zero function on every conjugate subgroup to a member of <math>X</math>. Thus, <math>f(g) = 0</math> whenever <math>g</math> lies in some conjugate of some <math>H \in X</math>. By our assumption that <math>G</math> is the union of conjugates of members of <math>X</math>, we obtain that <math>f</math> is the zero function. Thus, <math>U = 0</math>, and we get <math>W = V</math>.
+===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 <math>G</math> with a family of subgroups <math>X</math> such that every character of <math>G</math> is a rational linear combination of characters induced from <math>X</math>.
+'''To prove''': <math>G</math> is the union of conjugates of members of <math>X</math>.
+'''Proof''': Since every character of <math>G</math> is a rational linear combination of the characters induced from <math>X</math>, it is in particular true that the <math>\mathbb{C}</math>-span of class functions induced from class functions of <math>X</math>, is the whole space of class functions on <math>G</math>.
+Taking the usual inner product of class functions:
+<math>\langle \alpha,\beta \rangle = \frac{1}{|G|} \sum_{g \in G} \alpha(g) \overline{beta(g)}</math>.
+Now, suppose <math>f</math> is a class function of <math>G</math> that takes the value <math>0</math> on the union of conjugates of <math>H</math> and is <math>1</math> outside. Then we have that for every <math>H \in X</math> and every class function <math>\psi</math> of <math>H</math>:
+<math>\langle \psi, \operatorname{Res}_H^G f \rangle = 0</math>.
+By Frobenius reciprocity, we get:
+<math>\langle \operatorname{Ind}_H^G (\psi), f \rangle = 0</math>.
-{{fillin}}
+In other words, <math>f</math> is orthogonal to all the class functions induced from members of <math>H</math>. By assumption, <math>f</math> is thus orthogonal to every class functino of <math>G</math>, forcing <math>f = 0</math>. By te way we defined <math>f</math>, we obtain that the union of conjugates of <math>H \in X</math> must be the whole group <math>G</math>.