# Brauer's Theorems

Brauer proved two seemingly different theorems, both with important applications. In 1955 Brauer and Tate gave a single short proof that yields both theorems simultaneously! We cannot improve on their exposition in Annals of Mathematics, which is available on-line at the following link:

http://www.jstor.org/stable/2007097

Although we will not repeat their proof, we will discuss these two important theorems.

Let $p$ be a prime. A finite group $E$ is called $p$-elementary if $E \cong A \times B$ where $A$ is cyclic and $B$ is a $p$-group. If we make use of the internal direct product, we could write $E = A \times B$ instead of $\cong$. Also $E$ is called elementary if it is $p$ elementary for some $p$. The next exercise shows that in the definition of elementary we may assume that $\gcd (|A|, |B|) = 1$.

Exercise 4.1.1: Show that if $E = A \times B$ (internal direct product) where $A$ is cyclic and $B$ is a $p$-group, then there exist other subgroups $A'$ and $B'$ such that $E = A' \times B'$ where $A'$ is cyclic and $B'$ is a $p$-group and \ $\gcd (|A' |, |B' |) = 1$. Hint: take $B' = B_1 B$ where $B_1$ is a $p$-Sylow subgroup of $A$.

Exercise 4.1.2: Show that if $G = A \times B$ with $\gcd (|A|, |B|) = 1$, then any subgroup of $G$ is of the form $A_1 \times B_1$ where $A_1$ is a subgroup of $A$ and $B_1$ is a subgroup of $B$. Deduce that a subgroup of an elementary group is elementary.

One of Brauer's theorems has the following statement:

Theorem 4.1.1: (Brauer.) Let $G$ be a finite group and $\chi$ a class function on $G$. Then $\chi$ is a generalized character if and only if the restriction of $\chi$ to every elementary subgroup is a generalized character.

This result has had many applications. A typical one is this result of Green (1956) which he used to give a complete representation theory of $\operatorname{GL} (n, \mathbb{F}_q)$. To introduce Green's result, we recall some facts about finite fields.

Galois proved that for every prime power $q$ there is a field $\mathbb{F}_q$, unique up to isomorphism, with $q$ elements. We can regard $\mathbb{F}_q$ as a subfield of $\mathbb{F}_{q^r}$ for any $r$. If $f (X) \in \mathbb{F}_q [X]$ is an irreducible polynomial of degree $r$, then all the roots of $f$ are in $\mathbb{F}_{q^r}$, and in fact $\mathbb{F}_{q^r}$ can be constructed as $\mathbb{F}_q [X]$ modulo the ideal generated by $f$. If $r$ and $s$ are given, then $\mathbb{F}_{q^r}$ and $\mathbb{F}_{q^s}$ are both subfields of $\mathbb{F}_{q^{r s}}$. It follows that given any finite set $S$ of polynomials with coefficients in we can find an $r$ such that all the polynomials in $S$ have all their roots in $\mathbb{F}_{q^r}$.

Theorem 4.1.2: (Green.) Let $\mathbb{F}_q$ be a finite field, and $r$ be sufficiently large that all the roots of irreducible polynomials of degree $n$ lie in $\mathbb{F}^{q^r}$. Let $\theta : (\mathbb{F}^{q^r})^{\times} \longrightarrow \mathbb{C}^{\times}$ be a linear character. Let $G$ be a finite group, and let $\rho : G \longrightarrow \operatorname{GL} (n, \mathbb{F}_q)$ be a homomorphism. Let $\chi : G \longrightarrow \mathbb{C}$ be the function defined by $\chi (g) = \theta (\alpha_1) + \ldots + \theta (\alpha_n)$, where $\alpha_i$ are the eigenvalues of $\rho (g)$. Then $\chi$ is a generalized character.

I won't give the proof, but it just consists of checking that the statement is true in the case that $G$ is elementary, and is hence a typical application of Brauer's theorem. It may not be true that $\chi$ is a character. For example, consider the case where $n = 2$, $G = \operatorname{GL} (2, \mathbb{F}_q)$ and $\rho : G \longrightarrow \operatorname{GL} (n, \mathbb{F}_q)$ is the identity map. The group $\operatorname{GL} (2, \mathbb{F}_2)$ has irreducible representations of degrees $1, q, q - 1$ and $q + 2$. Since $\chi (1) = 2$, it is not a character, but only a generalized character; for most choices of $\theta$ this $\chi$ has the form $\chi_1 - \chi_2$, where $\chi_1$ and $\chi_2$ are irreducible characters of degrees $q + 1$ and $q - 1$.

Theorem 4.1.3: (Brauer.) Let $G$ be a finite group, and let $\chi$ be a generalized character. Then there exist elementary subgroups $E_1, E_2, \cdots$ and irreducible characters $\psi_1, \psi_2, \cdots$ of the $E_i$, and integers $a_i \in \mathbb{Z}$ such that $\chi = \sum a_i \psi_i^G$.

This result had an immediate major application, which in fact was its motivation, which is the meromorphic continuation of Artin L-functions. It is possible (and necessary for this application) to replace the word "irreducible'' in the above proof by "linear,'' and we now discuss this point.

Proposition 4.1.1: Let $G$ be a finite group and $\chi$ an irreducible character of $G$. Then the degree of $\chi$ divides $|G|$.

Proof. (Click to Expand/Collapse)

Unfortunately this requires a small amount of algebraic number theory, and I'm not going to prove it. Proofs may be found an any book on group representation theory or character theory.

Exercise 4.1.3: (Transitivity of Induction.) Let $G$ be a finite group and let $H \subset K \subset G$ be subgroups. Let $\psi$ be a character of $H$. First induce $\psi$ to obtain a character $\psi^K$ of $K$; then induce that to obtain a character $(\psi^K)^G$ of $G$. Prove that what you get is the same as if you induce $\psi^G$ all the way up to $G$ in one step. (Hint: use (\ref{inducedxform}).

Exercise 4.1.4: Suppose that $G$ is $p$-elementary and that $\chi$ is an irreducible character. Show that the degree of $\chi$ is a power of $p$. (Hint: let $E = A \times B$ where $A$ is cyclic and $B$ is a $p$-group. Remind us why we may write $\chi (a b) = \chi_1 (a) \chi_2 (b) w$here $\chi_1$ and $\chi_2$ are characters of $A$ and $B$, when $a \in A$ and $b \in B$. Remind us what we know about the degrees of $\chi_1$ and $\chi_2$.

Theorem 4.1.4: Let $G$ be elementary, and let $\chi$ be an irreducible character. Then there exists a subgroup $H$ of $G$ and a linear character $\psi$ of $H$ such that $\chi = \psi^G$.

The following argument can be found in many books. I think it's due to Brauer, though when $G$ is a $p$-group, I think the statement is older.

Proof. (Click to Expand/Collapse)

If $\chi$ has degree 1, then we may take $H = G$ and $\psi = \chi$. Thus the degree of $\chi$ is $p^r$ for some $r > 0$ (Exercise 4.1.4). Let $\Gamma$ be the set of linear characters $\lambda$ of $G$ such that $\chi = \chi \lambda$. Thus
 $\text{\lambda (g) = 1 if \chi (g) \neq 0} .$ (4.1.1)

It is clear that $\Gamma$ is a group. Now $\chi$ and $\chi \lambda$ are both irreducible characters, so we see that $\left\langle \chi, \chi \lambda \right\rangle = \left\{ \begin{array}{ll} 1 & \text{if \lambda \in \Gamma,}\\ 0 & \text{if \lambda \notin \Gamma.} \end{array} \right.$ Now we observe that $\left\langle \chi, \chi \lambda \right\rangle = \left\langle \chi \bar{\chi}, \lambda \right\rangle$ since both equal $\frac{1}{|G|} \sum_{g \in G} \chi (g) \overline{\chi (g)} \overline{\lambda (g)} .$ Thus
 $\left\langle \chi \bar{\chi}, \lambda \right\rangle = \left\{ \begin{array}{ll} 1 & \text{if \lambda \in \Gamma,}\\ 0 & \text{if \lambda \notin \Gamma.} \end{array} \right.$ (4.1.2)

Now $\chi \bar{\chi}$ is a character, so we can decompose it as a sum of irreducibles, and the multiplicity of the irreducible character $\phi_i$ of $G$ in $\chi \bar{\chi}$ is $\left\langle \chi \bar{\chi}, \phi_i \right\rangle$; in other words if we write $\chi \bar{\chi} = \sum a_i \phi_i$ where $\phi_i$ runs through the irreducible characters, then $a_i = \left\langle \chi \bar{\chi}, \phi_i \right\rangle$. In view of (4.1.2) we may therefore write $\chi \bar{\chi} = \left( \sum_{\lambda \in \Gamma} \lambda \right) + \sum_{\deg (\phi_i) > 1} a_i \phi_i .$ Now $\chi \bar{\chi} (1)$ is a power of $p$ greater than 1, and so are each of the $\phi_i$, and so $| \Gamma |$ is a multiple of $p$. Thus there exists an element $\lambda$ of the group $\Gamma$ whose order is a multiple of $p$. Let $K$ be the kernel of $\lambda$. Thus $[G : K] = p$.

We claim that the restriction of $\chi$ to $K$ is reducible. Indeed by (4.1.1) $\chi$ vanishes off $K$. Thus $\left\langle \chi, \chi \right\rangle_K = \frac{1}{|K|} \sum_{g \in K} | \chi (g) |^2 = \frac{1}{|K|} \sum_{g \in G} | \chi (g) |^2 = \frac{p}{|G|} \sum_{g \in G} | \chi (g) |^2 = p.$ Therefore $\chi |_K$ is reducible.

Let $\theta$ be an irreducible constituent of $\chi |_K$. We will prove that $\chi = \theta^G$. Since $\chi |_K$ is reducible, the degree of $\theta$ is $< p^r$; since $K$ is elementary (Exercise 4.1.2), the degree of $\theta$ is a power of $p$ (Exercise 4.1.4), say $\theta (1) = p^s$ where $s < r$. Now $\theta^G (1) = p^{s + 1}$, and since $\theta^G$ has an irreducible constituent $\chi$ of degree $r$ we get $p^{s + 1} \ge p^r$. Thus $s \ge r - 1$ and since $s < r$ we have $s = r - 1$. Thus $s = r - 1$ and the irreducible constituent $\chi$ of $\theta^G$ must be the only irreducible constituent, i.e. $\theta^G = \chi$.

Now by induction on $|G|$, the theorem is true for the smaller elementary group $K$, that is, $\theta = \psi^H$ for some linear character $\psi$ of an elementary subgroup $H$ of $K$. By transitivity of induction (Exercise 4.1.3), $\chi = \psi^G$.

Exercise 4.1.5: Explain why this result allows us to replace the word "irreducible'' by the word "linear'' in Brauer's theorem.