Artin's Theorem of Induced Characters

Introduction

When studying a linear space, when some subspaces are known, we are interested in the contribution of these subspaces, by studying their sum or (inner) direct sum if possible. This philosophy can be applied to many other fields.

In the context of representation theory, say, we are given a finite group $G$, with a subgroup $H$, we want to know how a character of $H$ is related to a character of $G$, through induction if anything. Next we state the content of this post more formally.

Let $G$ be a finite group with distinct irreducible characters $\chi_1,\dots,\chi_h$. A class function $f$ on $G$ is a character if and only if it is a linear combination of the $\chi_i$’s with non-negative integer coefficients. We denote the space of characters by $R^+(G)$. However, $R^+(G)$ lacks a satisfying algebraic structure, as one is not even allowed to freely do subtraction. For this reason, we extend the coefficients to all of integers, by defining

$$R(G)=\mathbb{Z}\chi_1 \oplus \cdots \oplus \mathbb{Z}\chi_h.$$

An element of $R(G)$ is called a virtual character because when one coefficient of some $\chi_i$ is negative, it cannot be a character in the usual sense. Note that $R(G)$ is a finitely generated free abelian group, hence we are free to do subtraction in the normal sense.

Besides, since the product of two characters are still a character, we see $R(G)$ is a ring (not necessarily commutative). To be precise, it is a subring of the ring $F_\mathbb{C}(G)$, the ring of class functions of $G$ over $\mathbb{C}$. Furthermore, we actually have $F_\mathbb{C}(G) \cong \mathbb{C} \otimes R(G)$.

Let $H$ be a subgroup of $G$. Then the operation of restriction and induction defines homomorphisms $\mathrm{Res}:R(G) \to R(H)$ and $\mathrm{Ind}:R(H) \to R(G)$. By extending Frobenius reciprocity linearly, still we find that $\mathrm{Res}$ and $\mathrm{Ind}$ are adjoints of each other. We also notice that the image of $\mathrm{Ind}:R(H) \to R(G)$ is a right ideal of $R(G)$. This is because, for any $\varphi \in R(H)$ and $\psi \in R(G)$, one has

$$\mathrm{Ind}(\varphi)\cdot \psi = \mathrm{Ind}(\psi \cdot \mathrm{Res}(\psi)) \in \operatorname{Im}(\mathrm{Ind}).$$

But being an ideal should not be the end of our story. We want to know what happens if we consider more than one subgroups. For example, since every group is the union of all of its cyclic groups, what if we consider all cyclic subgroups of $G$? We are also interested in how all these ideals work together. This is where Artin’s theorem comes in.

Artin’s Theorem - Statement and a Concrete Example

Artin’s Theorem. Let $X$ be a family of subgroups of a finite group $G$. Let $\mathrm{Ind}:\oplus_{H \in X}R(H) \to R(G)$ be the homomorphism defined by the family of $\mathrm{Ind}_H^G$, $H \in X$. Then the following statements are equivalent:

(i) $G$ is the union of the conjugates of all $H \in X$. Equivalently, for any $\sigma \in G$, there is some $H \in X$ such that $H$ contains a conjugate of $\sigma$.

(ii) The cokernel of $\mathrm{Ind}:\bigoplus_{H \in X}R(H) \to R(G)$ is finite.

Example. Put $G=D_4$, the dihedral group consists of rotations ($\sigma$) and flips ($\tau$) of the square. We write

$$G=\{1,\sigma,\sigma^2,\sigma^3,\tau,\tau\sigma,\tau\sigma^2,\tau\sigma^3\}.$$

In this example we take $X=\{\langle\sigma\rangle,\langle\tau\rangle,\langle\tau\sigma\rangle\}$. First of all we put down the character table of $G$:

$$\begin{array}{c|ccccc} G & 1 & \sigma^2 & \sigma & \tau & \sigma\tau \\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 \\ \chi_2 & 1 & 1 & -1 & 1 & -1 \\ \chi_3 & 1 & 1 & -1 & -1 & 1 \\ \chi_4 & 1 & 1 & 1 & -1 & -1 \\ \chi_5 & 2 & -2 & 0 & 0 & 0 \end{array}$$

The character table of elements of $X$ is not difficult to carry out as they are characters of cyclic groups.

$$\begin{array}{c|cccc} \langle\sigma\rangle & 1 & \sigma & \sigma^2 & \sigma^3 \\ \hline \chi_1^\sigma & 1 & 1 & 1 & 1 \\ \chi_2^\sigma & 1 & i & -1 & -i \\ \chi_3^\sigma & 1 & -1 & 1 & -1 \\ \chi_4^\sigma & 1 & -i & -1 & i \end{array}\quad\quad \begin{array}{c|cc} \langle\tau\rangle & 1 &\tau \\ \hline \chi_1^\tau & 1 & 1 \\ \chi_2^\tau & 1 & -1 \end{array}\quad\quad \begin{array}{c|cc} \langle\tau\sigma\rangle & 1 &\tau\sigma \\ \hline \chi_1^{\tau\sigma} & 1 & 1 \\ \chi_2^{\tau\sigma} & 1 & -1 \end{array}$$

Instead of writing something like $\mathrm{Ind}_{\langle\sigma\rangle}^{D_4}\chi_1^\sigma=\chi_1+\chi_4$ manually for all characters, we put all of them in an induction-restriction table:

$$\begin{array}{c|cccc|cc|cc} D_4 & \langle\sigma\rangle & & & &\langle\tau\rangle & & \langle\tau\sigma\rangle & \\ &\chi_1^\sigma & \chi_2^\sigma & \chi_3^\sigma &\chi_4^\sigma &\chi_1^\tau & \chi_2^\tau & \chi_1^{\sigma\tau} & \chi_2^{\sigma\tau} \\ \hline \chi_1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ \chi_2 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\ \chi_3 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ \chi_4 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ \chi_5 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \end{array}$$

which yields a matrix naturally:

$$T=\left(\begin{array}{cccc|cc|cc} 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \end{array} \right).$$

How to read the induction-restriction table? For example, the first column is $\langle \mathrm{Ind}_{\langle\sigma\rangle}^{D_4}\chi_1^\sigma,\chi_j\rangle$. Since $\mathrm{Ind}_{\langle\sigma\rangle}^{D_4}\chi_1^\sigma=\chi_1+\chi_4$, the column becomes $(1,0,0,1,0)$. On the other hand, the rows are indicated by the inner product with restriction. For example, since we have $\mathrm{Res}_{\langle\sigma\rangle}^{D_4}\chi_5=1$, thus $\langle\chi_4^\sigma,\mathrm{Res}_{\langle\sigma\rangle}^{D_4}\chi_5\rangle=1$ and therefore $T_{54}=1$. Induction and restriction coexist up to a transpose, which is another way to illustrate Frobenius reciprocity.

We obtain the induction map explicitly:

$$R(\langle\sigma\rangle) \oplus R(\langle\tau\rangle) \oplus R(\langle\tau\sigma\rangle) \xrightarrow{\mathrm{Ind}}R(D_4), \\ \mathbb{Z}^4 \oplus \mathbb{Z}^2 \oplus \mathbb{Z}^2 \xrightarrow{\text{multiplication by $T$}} \mathbb{Z}^5$$

where the basis of $R(D_4)$ is $\chi_1,\dots,\chi_5$ and the basis of $R(\langle\sigma\rangle) \oplus R(\langle\tau\rangle) \oplus R(\langle\tau\sigma\rangle)$ is given by the second row of the induction-restriction table. By doing Gaussian elimination of rows and columns of $T$ (over $\mathbb{Z}$), i.e. changing the basis for $\mathbb{Z}^5$ and $\mathbb{Z}^8$, the matrix $T$ is reduced to the form

$$U=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \end{pmatrix}$$

The image of $U$ is $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus 2\mathbb{Z}$, hence the cokernel of the induction map is

$$\mathbb{Z}^5/(\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus 2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z},$$

which is certainly finite. One can also verify that $X$ satisfies (i).

Proof of Artin’s Theorem

(i) => (ii)

Consider the exact sequence

$$\bigoplus_{H \in X}R(H) \xrightarrow{\mathrm{Ind}}R(G) \to \mathrm{coker}(\mathrm{Ind}) \to 0.$$

To show that $\mathrm{coker}(\mathrm{Ind})$ is finite (it is a finitely generated ring to begin with), it suffices to show that it suffices to see its result from tensoring with $\mathbb{Q}$, in other words, that

$$1_\mathbb{Q}\otimes\mathrm{Ind}:\bigoplus_{H \in X}\mathbb{Q} \otimes R(H) \to \mathbb{Q}\otimes R(G)$$

is a surjective map, i.e. it has trivial cokernel. This is equivalent to the surjectivity of the $\mathbb{C}$-linear map

$$1_\mathbb{C}\otimes\mathrm{Ind}:\bigoplus_{H\in X}\mathbb{C}\otimes R(H) \to \mathbb{C} \otimes R(G).$$

By Frobenius reciprocity, this is on the other hand equivalent to the injectivity of

$$1 \otimes \mathrm{Res}:\mathbb{C}\otimes R(G) \to \bigoplus_{H \in X}\mathbb{C} \otimes R(H).$$

Notice that $\mathbb{C} \otimes R(G)$ is the space of class functions of $G$. For a class function $f$ of $G$, if its restriction on each $H$ is $0$, according to (i), all values of $f$ have been determined, therefore $f$ is $0$ everywhere.

(ii) => (i)

Let $S$ be the union of the conjugates of the subgroups $H \in X$. Then we write elements in $\oplus_{H \in X}R(H)$ as $g=\sum_{H \in X}\mathrm{Ind}_H^G(f_H)$. It follows that $g$ always vanishes on $G \setminus S$. If (ii) holds, then

$$1_\mathbb{C}\otimes\mathrm{Ind}:\bigoplus_{H\in X}\mathbb{C}\otimes R(H) \to \mathbb{C} \otimes R(G)$$

is a surjective map. Therefore class functions of $G$, i.e. elements of $\mathbb{C} \otimes R(G)$ vanish on $G \setminus S$, which forces $G \setminus S$ to be empty, i.e. $G=S$.

References / Further Reading

• Jean-Pierre Serre, Linear Representations of Finite Groups.
• Kiyoshi Igusa, ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS