A (Slightly) Different Proof of Burnside’s Theorem

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1. Introduction

Burnside’s theorem, first proved in the early 20th century by William Burnside, shows that a group of order ${p^aq^b}$, where ${p}$ and ${q}$ are primes and ${a, b \in \mathbf{N} \cup \{0\}}$, is solvable. The original proof of Burnside’s theorem utilized representation theory in an essential way. A proof using purely group theoretical tools was given by D. Goldschmidt around 1970. The aim of this note is to provide yet another proof of Burnside’s theorem, via representation theory. First we establish a fundamental definition: A group ${G}$, with identity ${1}$, is solvable, if there exists a chain of subgroups

$\displaystyle \begin{array}{rcl} G = G_0 \supseteq G_1 \supseteq \cdots \supseteq G_{k-1} \supseteq G_k = \{1\} \end{array}$

such that ${G_i}$ is normal in ${G_{i-1}}$ and ${G_i/G_{i-1}}$ is abelian for ${i = 1, \cdots, k}$. Such a chain is called an abelian tower. Thus, for instance, all abelian subgroups are solvable. We also get the easy result: If ${P}$ is a ${p}$-group, then ${P}$ is solvable and has a non-trivial center. Proof: Note that ${P}$ acts on itself by conjugation, so by the class equation:

$\displaystyle \begin{array}{rcl} |P| &=& |Z(P)| + \sum (P : \text{Stab}(x)). \end{array}$

where the sum is over distinct conjugacy classes, and ${Z(P)}$ is the center of ${P}$. Because each index ${(P:\text{Stab}(x))}$ is divisible ${p}$ and ${|Z(P)| \geq 1}$, it follows that ${|Z(P)| >1}$. Now, note that a group of order ${p}$ is abelian, so it is solvable. Thus, by induction, ${P/Z(P)}$ is solvable. We can lift any abelian tower in ${P/Z(P)}$ to get an abelian tower in ${P}$, as desired. $\Box$

We fix some notation. If ${\rho : G \rightarrow \text{GL}(V)}$ is a complex representation of ${G}$, then we omit the homomorphism ${\rho}$ and simply write ${V}$ instead. We denote the character of the representation ${V}$ by ${\chi_V}$. If ${g \in G}$, then ${\text{Cent}_G(g)}$, is the centralizer of ${g}$ in ${G}$ and ${C(g)}$ is the conjugacy class of ${g}$ in ${G}$. Note that ${(G : \text{Cent}_G(g)) = |C(g)|}$, by the orbit-stabilizer formula. We write ${\text{Stab}(g)}$ for the stabilizer of ${g}$ in ${G}$.

2. The Proof of Burnside’s Theorem

The proof will proceed by a series of claims. We assume that ${G}$ is non-abelian and ${a > 0}$, otherwise ${G}$ is solvable by section one. We wish to derive a contradiction.

Claim 1 We can, without loss of generality, assume that ${G}$ is simple.

Proof: If ${N}$ is a normal subgroup of ${G}$, then ${N}$ and ${G/N}$ are solvable, by induction, therefore, ${G}$ is solvable. $\Box$

Assuming this, we note that ${Z(G) = 1}$ because ${Z(G)}$ is normal, and ${G}$ is non-abelian. This also implies that every non-trivial representation of ${G}$ is faithful.

Claim 2 Let ${V}$ be an irreducible representation of ${G}$. Then, for all ${g \in G}$

$\displaystyle \begin{array}{rcl} \frac{|C(g)|\chi_V(g)}{\chi_V(1)} \end{array}$

is an algebraic integer.

Proof: Let ${C(g_1), \cdots C(g_k)}$ be the conjugacy classes of ${G}$. Then

$\displaystyle \begin{array}{rcl} \varphi_{g_i} &=& \sum_{g \in C(g_i)} g \end{array}$

is in the center of ${\mathbf{Z}[G]}$ because it is invariant under conjugation. Further, the center of ${\mathbf{Z}[G]}$ is generated by these elements. Indeed, suppose ${ x= \sum_{g\in G} a(g)g}$ is in the center of ${G}$. Write ${x = \sum_{i=1}^k \sum_{g \in C(g_i)} a(g)g}$. Suppose that ${h,h' \in C(g_i)}$, and choose ${y}$ such that ${h' = yh'y^{-1}}$. Then conjugation by ${y}$ fixes ${\sum_{g \in C(g_i)} a(g)g}$, so ${a(h) = a(h')}$. It follows that ${a(g)}$ is a class function and so ${\sum_{g \in C(g_i)} a(g)g = a(C(g_i)) \varphi_{g_i}}$. We have shown that the center of ${\mathbf{Z}[G]}$ is finitely generated over ${G}$, and hence it is integral over ${\mathbf{Z}}$. Note that ${\varphi_{g} : V \rightarrow V}$ is a ${G}$ invariant map. Thus, by Schur’s lemma there exists ${\lambda \in \mathbf{C}}$ such that ${\varphi_{g} = \lambda\text{id}_V}$. Because ${\varphi_{g}}$ is integral over ${G}$, and ${\lambda}$ satisfies the same polynomial over ${\mathbf{Z}}$ that ${\varphi_{g}}$ does, it follows that ${\lambda}$ is integral over ${\mathbf{Z}}$. Thus, ${\lambda \text{dim} V = \text{Tr}(\varphi_{g}) = |C(g)|\chi_V(g)}$ and so ${\lambda = \frac{|C(g)|\chi_{V}(g)}{\chi_V(1)}}$ is integral over ${\mathbf{Z}}$. $\Box$

Claim 3 Let ${V}$ be an irreducible representation of ${G}$. Then ${\chi_V(1) \mid |G|}$.

Proof: We maintain the notation from the previous claim. Note that by the second orthogonality relations:

$\displaystyle \begin{array}{rcl} \sum_{i=1}^k |C(g_i)|\overline{\chi_{V}(C(g_i))}\chi_V(C(g_i)) &=& |G| \end{array}$

If we divide both sides by ${\text{dim} V}$, we get a sum of algebraic integers on the left, and a rational number on the right. Thus, ${\text{dim} V \mid |G|}$. $\Box$

Claim 4 G has no non-trivial 1-dimensional representations.

Proof: If ${V}$ is a non-trivial one 1-dimensional representation, then ${G \hookrightarrow \mathbf{C}^\ast}$, so ${G}$ is abelian. This is a contradiction. $\Box$

Claim 5 For every ${g \in G}$, ${g \neq 1}$, there exists an irreducible representation ${V}$ such that ${\chi_{V}(g) \neq 0}$ and ${\text{dim} V = p^r}$ for some ${r > 0}$.

Proof: Let ${V_1, \cdots, V_k}$ be the irreducible representations of ${G}$, where ${V_1}$ is the trivial representation. By claim 2.3, ${\text{dim}(V_i) \mid |G|}$ for all ${i}$, so ${\chi_{V_i}(1) = p^{c}q^d}$ for some ${c \leq a, b \leq d}$. If ${r_G}$ is the character of the regular representation, then ${r_G(g) = \sum_{i} \chi_{V_i}(1) \chi_{V_i}(g)}$ is ${0}$ if ${g \neq 1}$, and ${|G|}$ otherwise. Now, suppose ${q}$ divides ${\chi_{V_i}(1)}$ for ${i = 2, \cdots, n}$. Then

$\displaystyle \begin{array}{rcl} \sum_{i=2}^k \chi_{V_i}(1) \chi_{V_i}(g) &=& -1 \end{array}$

because ${V_1}$ is the only 1-dimensional representation of ${G}$. This is a contradiction: if we divide both sides by ${q}$, we get an algebraic integer on the left, and a rational number on the right. In particular, it follows that there is an ${i}$ such that ${\chi_{V_i}(1) = p^r}$ for some ${ r > 0}$ and ${\chi_{V_i}(g) \neq 0}$. $\Box$

The following claim holds for any finite group.

Claim 6 If there exists ${g \in G}$ and a representation ${V}$ such that ${|\chi_V(g)| = \chi_V(1)}$, then ${g \in Z(G)}$.

Proof: Because ${g}$ is unitary, it is diagonalizable: ${g = \text{diag}(\lambda_1, \cdots, \lambda_{\chi_V(1)})}$. Note that ${|\lambda_1 + \lambda_2| \leq |\lambda_1| + |\lambda_2|}$ with equality if, and only if, ${\lambda_1 = \lambda_2}$. Thus, because

$\displaystyle \begin{array}{rcl} \chi_V(1) = \left|\sum_{i=1}^{\chi_V(1)}\lambda_i \right| \leq \sum_{i=1}^{\chi_{V}(1)} |\lambda_i|= \chi_{V}(1), \end{array}$

it follows by induction that ${\lambda_i = \lambda_1}$ for all ${i}$. Thus, ${g = \lambda_{1}\text{id}}$, and so ${g}$ is in the center of the image of ${G}$, but ${G}$ is simple so it is isomorphic to its image and hence ${g \in Z(G) }$. $\Box$

By the Sylow theorems, ${G}$ has a ${p}$-Sylow subgroup ${P}$. By proposition 1.2, ${Z(P)}$ is non-trivial. Thus, we can choose ${g \in Z(P)}$. By taking an appropriate power of ${g}$, we can assume that ${\text{ord} (g) = p}$. Note that ${P \subseteq \text{Cent}_G(g)}$, so ${(G : \text{Cent}_G(g)) = |C(g)|}$ is not divisible by ${p}$, i.e. ${|C(g)| = q^k}$ for some ${k \geq 0}$. Let ${V}$ be a representation such that ${\text{dim} V = p^r}$ and ${\chi_{V}(g) \neq 0}$. Note that the the map

$\displaystyle \begin{array}{rcl} G \hookrightarrow \text{GL}(V) \stackrel{\det}{\rightarrow} \mathbf{C}^\ast \end{array}$

is a 1-dimensional representation of ${G}$, thus, it is trivial, so ${\det(h) = 1}$, for all ${h \in G}$. Let ${\lambda_1, \cdots, \lambda_{p^r}}$ be the eigenvalues of ${g}$ listed with multiplicity. Note that ${\lambda_i}$ is a ${p}$-th root of unity for all ${i}$, because ${g^{p} = 1}$ but ${g^i \neq 1}$ for all ${ 1\leq i < p}$. Let ${\lambda_1, \cdots, \lambda_{j_0} = 1}$ and ${\lambda_{j_0+1}, \cdots, \lambda_{p^r} \neq 1}$. By the formula

$\displaystyle \begin{array}{rcl} \prod_{i=j_0}^{p^r} \lambda_i = \det (g) = 1, \end{array}$

each conjugate of ${\lambda_i}$, ${i \geq j_0}$ appear with the constant multiplicity because the product is stable under the action of ${\text{Gal}(\overline{\mathbf{Q}}/ \mathbf{Q})}$, in other words, if ${\lambda_i}$ has multiplicity ${l}$, then each ${\lambda_{i'}}$ has multiplicity ${l}$ for ${i' \geq j_0}$. Therefore, because each ${\lambda_i}$, ${i \geq j_0}$ has ${p-1}$ conjugates, we have ${l(p-1) = p^r - j_0}$. In particular, ${l \equiv j_0 \mod p}$ Recall that for any primitive ${p}$-th root of unity ${\zeta_{p}}$, ${\sum_{i=1}^{p-1} \zeta_{p}^i = -1}$, thus,

$\displaystyle \begin{array}{rcl} \chi_V(g) &=& \sum_{i=1}^{p^r} \lambda_i \\ &=& \sum_{i=1}^{j_0} \lambda_i + \sum_{i=j_0+1}^{p^r} \lambda_{i} \\ &=& j_0 - l \\ &\equiv& 0 \mod p. \end{array}$

By claim 2.2, it follows that ${\frac{|C(g)| \chi_V(g)}{\chi_V(1)}}$ is an algebraic integer, and hence an integer. However, ${|C(g)| = q^k}$, is relatively prime to ${p}$, so ${\frac{\chi_{V}(g)}{\chi_V(1)} \in \mathbf{Z}}$, and ${\chi_V(g) = p^rb}$ for some ${b \in \mathbf{Z}}$, because ${\chi_V(g) \neq 0}$. Now, because

$\displaystyle \begin{array}{rcl} |\chi_V(g)| \leq \sum_{i=1}^{p^r} |\lambda_i| = \chi_V(1) \end{array}$

it follows that ${b = \pm 1}$ and ${|\chi_V(g)| = \chi_V(1)}$. By claim 2.6, it follows that ${g \in Z(G) = \{1\}}$, which is a contradiction. Thus, ${G}$ is solvable.

3. Consequences

There are several key steps in this argument. The most important assumptions are that ${G}$ is simple, and for any ${g \in G}$, there is a prime power dimensional representation ${V}$ such that ${\chi_V(g) \neq 0}$. Retracing the steps of the proof yields the following consequence:

Corollary 1 Suppose ${G}$ is a simple group and ${V}$ is an irreducible representation of ${G}$ of dimension ${p^r}$, for some ${ r > 0}$. Let ${P}$ be a ${p}$-Sylow subgroup of ${G}$. If ${g \in Z(P)}$ of order ${p}$, and ${\chi_V(g) \neq 0}$, then ${G}$ is abelian.

${\Box}$