nLab Artin induction theorem

Idea

Artin’s induction theorem says that working over the ground field of complex numbers, the character of any finite-dimensional representation of a finite group $G$ is a rational linear combination of characters of representations induced from cyclic subgroups of $G$.

Historically this is a precursor to the Brauer induction theorem, but neither result follows immediately from the other. Brauer showed that every character is an integral linear combination of characters of representations of a larger class of subgroups of $G$, the so-called elementary subgroups.

One thing elementary groups and cyclic groups have in common is that all their irreducible representations are one-dimensional, so in each case the result shows that every character of $G$ is a linear combination (either rational or integral) of characters of representations induced from one-dimensional representations of subgroups.

Proof

In Linear Representations of Finite Groups, Serre states Artin’s induction theorem in the following more general way:

Theorem

(Artin induction theorem (Serre’s version))

Let $G$ be a finite group and $X$ any family of subgroups.

Then the following are equivalent:

1. $G$ is the union of conjugates of the subgroups in $X$

2. for every character $\chi$ of $G$ there exist characters $\chi_H$ of $H$ for each $H \in X$ and $n \in \N$ such that $n \chi = \sum_{H\in X} ind_H^G(\chi_H)$.

This in turn implies Artin’s original version, by choosing $X$ to be the set of all cyclic subgroups of $G$.

We begin by recalling some preliminary concepts and notation.

The representation ring $R(G)$ of $G$ is the set of formal differences of isomorphism classes of (finite-dimensional, complex) representations of $G$, made into a ring using the tensor product of representations. Using the fact that any representation of $G$ has a character which is a class function, we can treat $R(G)$ as a sub-ring of the $\mathbb{C}$-algebra of class functions on $G$. Indeed that algebra is isomorphic to $\mathbb{C}\otimes R(G)$.

The functor of restricting representations of $G$ to a subgroup $H \subseteq G$ and the adjoint functor of inducing representations from $H$ to $G$ give abelian group homomorphisms:

$$res_H^G: R(G) \to R(H)$$

$$ind_H^G: R(H) \to R(G)$$

With these notations, Serre’s theorem can be equivalently rewritten as follows:

Theorem

If $X$ is a family of subgroups of $G$, the following properties are equivalent:

1. $G$ is the union of the conjugates of the subgroups in $X$
2. The cokernel of $ind \colon \bigoplus_{H\in X}R(H) \to R(G)$ is finite.

To prove this we use the following lemma:

Lemma

Let $H$ be a subgroup of the finite group $G$ and let $f\in R(H)$. If $g \in G$ is not conjugate to any element of $H$, then the character $ind_H^G(f)$ vanishes on $g$.

Proof

It is enough to prove this lemma for the character $\phi$ of a representation $\theta \colon H \to \mathrm{GL}(W)$, since any $f \in R(H)$ is a difference of two such characters. So, let $\rho \colon G\to \mathrm{GL}(V)$ be the representation of $G$ induced from the representation $\theta$ of $H$, and let $(r_i)$ be a set of representatives of the cosets of $H$ in $G$, which are the points of $G/H$. By definition of induced representation, $V$ is the direct sum of its subspaces $\rho(r_i) W$, and the linear transformation $\rho(g)$ permute these subspaces, since

$$\rho(g) \circ \rho(r_i)W=\rho(g r_i) W=\rho(r_{i'})W$$

where $g r_i=r_{i'}h$ for some $h\in H$. To show that $ind(\phi)(g)=\text{tr}_V(\rho(g))$ vanishes, we now choose a basis for $V$ that is a union of the bases of the subspaces $\rho(r_i)W$. In this basis for $V$, the diagonal matrix entry of $\rho(g)$ vanishes for each basis vector in $\rho(r_i)W$ if $r_i\neq r_{i'}$. But $r_i=r_{i'}$ would imply $r_i^{-1} g r_i \in H$, which is ruled out by our assumption that $x$ is not conjugate to any $g \in H$. Thus, all the diagonal matrix entries of $\rho(g)$ vanish, and $\text{tr}_V(\rho_x)=0$ as desired.

Proof of Theorem

First we prove 2 $\implies$ 1. The lemma implies that all elements in the image of $ind \colon \bigoplus_{H\in X}R(H) \to R(G)$ vanish on every $g \in G$ in the set

$$S := G - \bigcup_{g\in G, H \in X} g^{-1}H g .$$

The same therefore holds for all elements in the image of the $\mathbb{C}$-linear map

$$\mathbb{C}\otimes ind \colon \mathbb{C}\otimes\bigoplus_{H\in X} R(H) \to \mathbb{C}\otimes R(G)$$

On the other hand, this map is surjective, because otherwise $ind$ would have an infinite cokernel, contradicting assumption 2. Thus, every element of $\mathbb{C}\otimes R(G)$ vanishes on $S$, insuring $S= \emptyset$, so that every element of $G$ is conjugate to an element of some subgroup $H \in X$, as was to be shown.

Next we prove 1 $\implies$ 2. First, note that it is enough to show 1 implies the linear map

$$\mathbb{C}\otimes ind\colon \mathbb{C}\otimes\bigoplus_{H\in X}R(H) \to \mathbb{C}\otimes R(G)$$

is surjective. Indeed, this surjectivity implies that $\mathbb{C}\otimes R(G)$ has a basis $(e_i)$ composed of elements of the image $A$ of $ind$. Since this basis must have the same cardinality $n$ as $(\chi_i)$, the quotient $R(G)/A$ is isomorphic to some quotient $\mathbb{Z}^n/\prod_i^n H_i\cong \prod_i^n\mathbb{Z}/H_i$ where the $H_i$ are non-trivial ideals of $\mathbb{Z}$, and this quotient is clearly finite, giving 2.

By Frobenius reciprocity, proving the surjectivity of $\mathbb{C}\otimes ind$ is equivalent to proving the injectivity of

$$\mathbb{C}\otimes res \colon \mathbb{C}\otimes R(G) \to \mathbb{C}\otimes\bigoplus_{H\in X}R(H).$$

But this is clearly true, because it says that if a character vanishes on every conjugacy class of $G$ it vanishes, which holds because characters are constant on each conjugacy class.