Integrality of characters and Burnside's \(p^a q^b\) theorem

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the AI that wrote it. Public domain (CC0).

Character values need not be integers: the degree-three characters of \(A_5\) contain \((1\pm\sqrt5)/2\). They satisfy a stronger arithmetic condition than merely being complex numbers, however. They are algebraic integers. Combining that condition with character orthogonality forces irreducible degrees to divide the group order. Combining it with the triangle inequality rules out a prime-power conjugacy class in a nonabelian simple group. This is the character-theoretic step in Burnside's solvability theorem.

We use The group algebra and Fourier analysis on a finite group, especially class sums and the scalar by which they act on an irreducible. Character orthogonality and the regular decomposition come from Characters and the orthogonality relations. All representations are finite-dimensional over \(\mathbb C\), and every irreducible has positive degree. Write \(n=|G|\), \(Z=Z(G)\), and \(d=\chi(1)\) for an irreducible character \(\chi\).

1. The arithmetic tool

A complex number is an algebraic integer if it is a root of a monic polynomial with integer coefficients. Thus every integer is an algebraic integer, and every root of unity is one as well. The number \(\varphi=(1+\sqrt5)/2\) is an algebraic integer because \(\varphi^2-\varphi-1=0\).

Lemma 1.1 (a finite-module test). Suppose a subring \(R\subseteq\mathbb C\) contains \(1\) and is generated by finitely many elements as an additive \(\mathbb Z\)-module. Every element of \(R\) is an algebraic integer. The algebraic integers form a subring of \(\mathbb C\), and a rational algebraic integer is an integer.

Proof. Choose additive generators \(m_1=1,m_2,\ldots,m_r\) of \(R\). For \(x\in R\), multiplication by \(x\) sends each generator to an integer linear combination of them. There is an integer matrix \(A\) for which

Multiplying by the adjugate matrix shows that \(\det(xI-A)m_j=0\) for every \(j\). Since \(m_1=1\), the monic integer polynomial \(\det(TI-A)\) vanishes at \(x\). The generators need not be linearly independent; this argument does not require an additive basis.

If \(\alpha,\beta\) satisfy monic integer equations of degrees \(r,s\), respectively, those equations reduce all powers in \(\mathbb Z[\alpha,\beta]\) to integer combinations of

This is a finite-module subring containing \(1\). Its elements \(\alpha+\beta\), \(\alpha-\beta\) and \(\alpha\beta\) are algebraic integers by the first part, proving closure under the ring operations.

Finally write a rational algebraic integer as \(a/b\), with integers \(a,b\), \(b>0\) and \(\gcd(a,b)=1\). Substitution in a monic equation of degree \(r\), followed by multiplication by \(b^r\), shows that \(b\mid a^r\). Coprimality forces \(b=1\). \(\square\)

An automorphism of a number field preserves algebraic integers: apply it to a monic integer equation. We shall use this observation with a cyclotomic field, but the finite-module argument above already proves the ring properties needed here.

Lemma 1.2 (the cyclotomic fixed field). Let \(\zeta\) be a primitive \(m\)-th root of unity and \(K=\mathbb Q(\zeta)\). The group \(\Gamma\) of \(\mathbb Q\)-automorphisms of \(K\) is finite, and an element fixed by every member of \(\Gamma\) belongs to \(\mathbb Q\).

Proof. The field \(K\) has finite degree over \(\mathbb Q\), since \(\zeta\) satisfies \(X^m-1\). It contains all the roots of that polynomial, namely the powers of \(\zeta\). A \(\mathbb Q\)-embedding \(K\to\mathbb C\) sends \(\zeta\) to one of those roots, so its image lies in \(K\). As an injective endomorphism of a finite-dimensional \(\mathbb Q\)-vector space it is surjective, hence is an automorphism. There are at most \(m\) such maps because their value on \(\zeta\) determines them.

For \(x\in K\), let \(f\in\mathbb Q[X]\) be its monic minimal polynomial. Every complex root \(y\) of \(f\) defines an embedding \(\tau:\mathbb Q(x)\to\mathbb C\) with \(\tau(x)=y\): use the evaluation isomorphisms from \(\mathbb Q[X]/(f)\). We show that it extends to an embedding of \(K\).

Let \(q\) be the minimal polynomial of \(\zeta\) over \(\mathbb Q(x)\). Apply \(\tau\) to its coefficients. The resulting irreducible polynomial \(q^\tau\) has a complex root \(\eta\). Evaluation at \(\eta\) identifies the field \(\tau(\mathbb Q(x))[X]/(q^\tau)\) with its image in \(\mathbb C\), extending \(\tau\) to \(\mathbb Q(x)(\zeta)=K\). Since \(q\) divides \(X^m-1\), \(q^\tau\) does also; thus \(\eta\) is an \(m\)-th root of unity. The extension is a \(\mathbb Q\)-embedding of \(K\), hence an automorphism by the first paragraph. In particular every root \(y\) of \(f\) is an automorphic image of \(x\).

Finally \(f\) has distinct roots: in characteristic zero its derivative is nonzero and has smaller degree, so irreducibility gives \(\gcd(f,f')=1\). If every automorphism fixes \(x\), all roots of \(f\) equal \(x\). Distinctness forces \(\deg f=1\), whence \(x\in\mathbb Q\). Rational elements are fixed by definition, completing both directions. \(\square\)

This direct argument supplies the fixed-field fact needed below without importing a general Galois correspondence or a proof of cyclotomic irreducibility.

2. Integral class sums and divisibility

For a conjugacy class \(C\), let

The class sums are a \(\mathbb Z\)-basis of the centre of \(\mathbb Z[G]\). Indeed, an element commutes with every group element exactly when conjugation permutes equal coefficients. Moreover,

where \(a_{CD}^{E}\) is the number of pairs \((x,y)\in C\times D\) with \(xy=e\), for one fixed \(e\in E\). Conjugation gives a bijection between the pairs counted for different \(e\) in the same class. Thus all these structure constants are nonnegative integers.

Proposition 2.1 (character and central-character integrality). Every character value \(\chi(g)\) is an algebraic integer. For an irreducible of degree \(d\), the scalar

by which \(K_C\) acts is also an algebraic integer.

Proof. A finite-order complex matrix is diagonalizable, with roots of unity as eigenvalues. Its trace is their sum and hence is an algebraic integer by Lemma 1.1.

To prove the second assertion without assuming any integrality of the representation matrices, use left multiplication by \(K_C\) on the free additive group \(\mathbb Z[G]\). In the group basis its matrix has integer entries. Its characteristic polynomial \(P\) is monic over \(\mathbb Z\). Cayley–Hamilton gives \(P(L_{K_C})=0\); applying this operator to the algebra unit gives \(P(K_C)=0\) in \(\mathbb Z[G]\).

Since \(K_C\) is central, it acts on an irreducible by a scalar. Taking its trace gives \(d\omega_\chi(C)=|C|\chi(c)\), proving (1). Applying the representation to \(P(K_C)=0\) yields \(P(\omega_\chi(C))I=0\), so that scalar is an algebraic integer. \(\square\)

The distinction between \(\chi(c)\) and \(\chi(c)/d\) matters. The former is always integral; the latter generally is not. For the standard representation of \(S_3\), the value at a 3-cycle is \(-1\), so the normalized value is \(-1/2\).

Theorem 2.2 (divisibility of irreducible degrees). Every irreducible complex representation of \(G\) has degree dividing \(n\). More precisely,

Proof of \(d\mid n\). Row orthogonality says \(\sum_g|\chi(g)|^2=n\). Grouping the sum by classes and dividing by \(d\) gives

Each factor on the right is an algebraic integer by Proposition 2.1, so \(n/d\) is one. It is rational, hence an integer by Lemma 1.1.

Proof of the central refinement. Let \(z=|Z|\). Schur's lemma gives

where \(\lambda:Z\to\mathbb C^\times\) is a homomorphism. For a positive integer \(m\), the external tensor product is a representation of \(G^m\):

Its character is \(\prod_{j=1}^m\chi(g_j)\), whose squared norm on \(G^m\) is the product of the \(m\) squared norms on \(G\), namely \(1\). It is therefore irreducible, of degree \(d^m\).

The central subgroup

has order \(z^{m-1}\): choose the first \(m-1\) entries freely and the last is determined. It acts trivially on the external tensor product, because its scalar is \(\lambda(a_1\cdots a_m)=1\). Hence this irreducible descends to \(G^m/N_m\), and the divisibility already proved gives

For any prime \(\ell\), put \(\delta=v_\ell(d)\), \(a=v_\ell(n)\) and \(b=v_\ell(z)\). Equation (5) says

for every positive \(m\). If \(\delta>a-b\), choosing \(m>b\) contradicts this inequality. Thus \(\delta\le a-b=v_\ell(n/z)\) for every prime, which is exactly (2). \(\square\)

Here the tensor product is a representation of the product group \(G^m\), not the usual tensor power with the diagonal \(G\)-action. The latter is usually reducible and cannot be substituted in the proof.

For an abelian group, the refinement gives \(d\mid1\), recovering one-dimensionality. It does not assert that every divisor of the index occurs as a degree.

3. When an integral average of roots of unity vanishes

The next arithmetic step uses the cyclotomic fixed-field property proved in Lemma 1.2.

Lemma 3.1 (normalized trace). Let \(V\ne0\) be any complex representation of a finite group, of dimension \(d\). If \(\chi_V(g)/d\) is an algebraic integer, then either \(\chi_V(g)=0\), or \(\rho(g)\) is scalar.

Proof. Let the eigenvalues of \(\rho(g)\) be \(\epsilon_1,\ldots,\epsilon_d\), roots of unity in the field \(K=\mathbb Q(\zeta)\), where \(\zeta\) is a primitive root of unity whose order is the order of \(g\). Put

Suppose \(\alpha\ne0\). For \(\sigma\in\Gamma=\operatorname{Gal}(K/\mathbb Q)\), every \(\sigma(\epsilon_j)\) is still a root of unity, so

If not all the original eigenvalues are equal, the triangle inequality is strict for their average, giving \(|\alpha|<1\). For clarity, equality in the triangle inequality for unit complex numbers forces them all to have the same argument, hence to be equal.

Now the product

is nonzero. Automorphisms permute its factors, so Lemma 1.2 puts it in \(\mathbb Q\). It is also an algebraic integer by Lemma 1.1, and hence a nonzero integer. But the bounds above and the strict inequality for the identity automorphism give \(|M|<1\), an impossibility. All eigenvalues must therefore be equal. Since \(\rho(g)\) is diagonalizable, it is scalar. \(\square\)

The nonzero alternative is necessary. The standard \(S_3\)-matrix of a transposition has eigenvalues \(1,-1\). Its normalized trace is the algebraic integer zero, although the matrix is not scalar.

Lemma 3.2 (a coprime class and degree). If \(\chi\) is irreducible, \(C=g^G\), and \(\gcd(|C|,d)=1\), then

In the second case \(\rho(g)\) is scalar.

Proof. Choose integers \(u,v\) with \(u|C|+vd=1\). Then

Both terms on the right are algebraic integers. Lemma 3.1 gives the zero-or-scalar alternative. In the scalar case the scalar is a root of unity, so the absolute trace is \(d\). \(\square\)

For \(A_5\), the degree-three characters vanish on its class of twenty 3-cycles; \(3\) and \(20\) are coprime. The degree-four character vanishes on the class of fifteen involutions, and the degree-five character vanishes on both classes of twelve 5-cycles. These are the zero alternatives in (7). At a central element every irreducible instead attains the scalar alternative.

4. A class of prime-power size prevents simplicity

A group is simple if it is nontrivial and has no proper nontrivial normal subgroup.

Theorem 4.1 (prime-power class obstruction). If a finite group has a conjugacy class of size \(q^a>1\), where \(q\) is prime, then the group is not simple.

Proof. Suppose such a group \(G\) were simple. An abelian simple group is cyclic of prime order: a nonidentity element generates a nontrivial normal subgroup, hence the whole group, and a cyclic composite-order group has a proper nontrivial subgroup. Its classes all have size one. Therefore our hypothetical \(G\) is nonabelian simple.

Its centre is trivial, since the centre is normal and cannot equal a nonabelian group. Every nontrivial irreducible representation is faithful: its kernel is normal, and kernel \(G\) would make the representation trivial. Let \(g\) lie in the given class. If \(q\nmid d_i\), then \(\gcd(q^a,d_i)=1\). Lemma 3.2 makes \(\chi_i(g)\) zero or its matrix scalar.

For a nontrivial irreducible the scalar possibility would force \(g\) to be central. Indeed, scalar \(\rho_i(g)\) commutes with every \(\rho_i(h)\); faithfulness gives \(gh=hg\) for all \(h\). This contradicts the class size being greater than one. Consequently

The regular character vanishes at \(g\ne1\). Its decomposition, including the trivial character once, now reads

where \(\beta\) is an algebraic integer: each \(d_i/q\) is an integer and each character value is integral. Thus \(\beta=-1/q\) is a rational algebraic integer, contradicting Lemma 1.1. \(\square\)

The qualification \(q^a>1\) is essential: every group has a singleton identity class, including cyclic groups of prime order.

In \(A_5\), the nonidentity class sizes are \(15,20,12,12\). None is a prime power, in agreement with its simplicity. The theorem is a necessary obstruction, rather than a characterization of simple groups.

Lemma 4.2 (Cauchy and Sylow existence). If a prime \(p\) divides the order of a finite group \(H\), then \(H\) has an element of order \(p\). If \(|H|=p^a u\) with \(p\nmid u\), then \(H\) has a subgroup of order \(p^a\).

Proof. For the first assertion consider the tuples \((h_1,\ldots,h_p)\) whose product is \(1\). There are \(|H|^{p-1}\) of them, since the last entry is determined. Cyclic rotation preserves the product-one condition: the rotated product is the conjugate by \(h_1^{-1}\) of the original product. The cyclic group of order \(p\) therefore acts on these tuples, with orbits of size \(1\) or \(p\). The fixed tuples are precisely \((h,\ldots,h)\) with \(h^p=1\). Their number is congruent to \(|H|^{p-1}\), hence to zero modulo \(p\). The identity gives one such tuple, so there are at least \(p\) and a nonidentity \(h\) exists. Its order is \(p\).

For the second assertion use induction on \(|H|\). When \(a=0\), the trivial subgroup suffices. Suppose \(a>0\). If some noncentral \(h\) has class size \([H:C_H(h)]\) prime to \(p\), its proper centralizer has the same \(p\)-part \(p^a\) as \(H\). Induction in that centralizer supplies the required subgroup.

Otherwise every noncentral class has size divisible by \(p\). The class equation gives \(p\mid |Z(H)|\), including the abelian case when there are no noncentral classes. By the first assertion \(Z(H)\) contains a subgroup \(N\) of order \(p\). It is normal because it is central. The smaller quotient \(H/N\) has \(p\)-part \(p^{a-1}\); induction supplies a subgroup of that order. Its inverse image in \(H\) has order \(|N|p^{a-1}=p^a\), as required. \(\square\)

A subgroup of the order in Lemma 4.2 is called a Sylow \(p\)-subgroup. The later Brauer-induction lesson proves the containment and conjugacy property it needs by a separate coset action.

5. Burnside's theorem

For any group \(H\), its commutator subgroup \(H'=[H,H]\) is generated by \(xyx^{-1}y^{-1}\), \(x,y\in H\). It is normal, since conjugation takes a commutator to a commutator. Define the derived series

A group is solvable if \(H^{(r)}=1\) for some \(r\). An abelian group has \(H'=1\), so is solvable.

Lemma 5.1 (solvable extensions). If \(N\triangleleft G\), and both \(N\) and \(G/N\) are solvable, then \(G\) is solvable.

Proof. A surjective homomorphism carries a commutator subgroup onto the commutator subgroup of its image, since it maps the generating commutators onto all generating commutators there. Applying this repeatedly to \(G\to G/N\), if \((G/N)^{(r)}=1\) then \(G^{(r)}\subseteq N\). If \(N^{(s)}=1\), monotonicity of commutator subgroups gives

For \(j=s\) this gives \(G^{(r+s)}=1\). \(\square\)

We also need the elementary centre argument for a nontrivial finite \(p\)-group \(P\). The class equation writes \(|P|\) as \(|Z(P)|\) plus the sizes of noncentral classes. Every noncentral class has size a positive power of \(p\), by the centralizer index formula. Hence \(|Z(P)|\equiv0\pmod p\), and \(Z(P)\ne1\).

Theorem 5.2 (Burnside). Let \(p,q\) be primes and \(a,b\) nonnegative integers. Every finite group of order \(p^a q^b\) is solvable.

Proof. If \(p=q\), combine the exponents; this is the one-prime case. Assume \(p\ne q\) otherwise. We use induction on the group order, including groups of prime-power order and the trivial group. The trivial and abelian cases are already solvable. Suppose \(G\) is nonabelian.

In a one-prime case, its nontrivial centre is a proper normal subgroup, by the centre argument above. Both the centre and the quotient have smaller orders of the required form. Induction and Lemma 5.1 give solvability.

Now suppose both primes divide \(|G|\). Choose a Sylow \(p\)-subgroup \(P\), of order \(p^a\), and a nonidentity \(g\in Z(P)\). Since \(P\subseteq C_G(g)\), the index

divides \([G:P]=q^b\). Thus it is either \(1\) or a positive power of \(q\).

If it is \(1\), then \(g\in Z(G)\). The centre is nontrivial and proper because \(G\) is nonabelian, so take \(N=Z(G)\). If the class size is greater than one, Theorem 4.1 says \(G\) is not simple, giving a proper nontrivial normal subgroup \(N\).

In either case, Lagrange's theorem and the quotient order formula show that \(|N|\) and \(|G/N|\) are smaller numbers of the form \(p^u q^v\), allowing zero exponents. Induction makes both groups solvable. Lemma 5.1 completes the proof. \(\square\)

This proof handles the abelian case before asking for a proper nontrivial normal subgroup. A cyclic group of prime order is solvable and simple, so that subgroup cannot exist in every case.

6. Degrees and concrete groups

The tables of \(S_4\) and \(A_5\)

The constructions in the preceding lessons give

Every degree divides the stated order, and its square contributes to the respective identity \(\sum_i d_i^2=|G|\). The centres are trivial: in \(S_4\), commuting with all transpositions forces a permutation to fix every unordered pair, hence every letter; for \(A_5\), this follows from the proved nonabelian simplicity.

For example, in the \(S_4\) standard row the value on a transposition is \(1\). Its class size is \(6\) and its degree is \(3\), so the integral central scalar is \(6\cdot1/3=2\), although the normalized character value \(1/3\) is not integral. This illustrates why the class-size factor in (1) cannot be discarded.

Nonabelian groups of order \(p^3\)

Let \(|G|=p^3\) and suppose \(G\) is nonabelian. Its nontrivial centre has order \(p\): centre order \(p^3\) would mean abelian, and centre order \(p^2\) would make \(G/Z(G)\) cyclic. If \(G/Z(G)=\langle xZ(G)\rangle\), any two elements are \(x^i z,x^j w\) with \(z,w\) central and commute, so \(G\) would again be abelian.

The quotient by the centre has order \(p^2\) and is abelian, as proved in Exercise 1 below. Thus \(1\ne[G,G]\subseteq Z(G)\), and \([G,G]=Z(G)\). The abelianization has order \(p^2\), giving exactly \(p^2\) linear characters. Indeed, linear characters factor through the abelianization; all irreducibles of a finite abelian group are linear, and their number equals its order.

Every remaining degree is a positive power of \(p\), by Theorem 2.2. It cannot be \(p^2\): its square alone would exceed \(p^3\). Hence it is \(p\). The sum of degree squares determines the number \(t\) of these rows:

This covers \(p=2\), including both \(D_4\) and \(Q_8\), without deciding the indicators that distinguish their real structures.

A group of order \(72\)

Every group of order \(72=2^3 3^2\) is solvable by Theorem 5.2. It is not simple: a nontrivial solvable simple group would have commutator subgroup either \(1\) or the whole group; the latter prevents the derived series from terminating, while the former makes it cyclic of prime order.

For a concrete degree calculation take \(S_3\times A_4\). The irreducibles of the factors have degrees \(1,1,2\) and \(1,1,1,3\), respectively. Their external products are irreducible, since the character inner products factor. The resulting degrees are six copies of \(1\), three copies of \(2\), two copies of \(3\), and one copy of \(6\). Their squared degrees sum to

proving that no additional irreducibles remain. The divisibility theorem gives a restriction for every group of that order; this particular list belongs to the displayed product.

7. Exercises with solutions

Exercise 1. Character degrees force commutativity

Use Theorem 2.2 to prove that a group of order \(p^2\), with \(p\) prime, is abelian.

Solution. Every degree divides \(p^2\), so is \(1,p\) or \(p^2\). A degree \(p^2\) has square larger than the group order. A degree \(p\), together with the always present trivial degree-one character, already contributes \(p^2+1\) to \(\sum_i d_i^2\), also impossible. All degrees are therefore one.

Every commutator acts trivially in every one-dimensional representation. The regular representation is a direct sum of the irreducibles and is faithful, so each commutator must be the identity. Hence the group is abelian. This argument does not assume the centre proof for groups of order \(p^2\).

Exercise 2. The Heisenberg degree count

Let \(H_p\) be the upper unitriangular \(3\times3\) group over \(\mathbb F_p\), for any prime \(p\). Determine all its irreducible degrees and their counts. Explain why a general group of order \(p^3\) has degrees only \(1,p\).

Solution. Write a matrix as a triple \((x,y,z)\), where \(x,y\) are its entries in positions \((1,2),(2,3)\) and \(z\) its \((1,3)\) entry. Matrix multiplication gives

Consequently

A triple commutes with every other triple exactly when \(x=y=0\); vary \(x',y'\) separately to see this. Every value of the third coordinate occurs as a commutator, by taking \(x=1,y=0,x'=0,y'=t\). Therefore the centre and the commutator subgroup are both \(\{(0,0,z)\}\), of order \(p\), and the abelianization is \(\mathbb F_p^2\).

Its \(p^2\) linear characters are

They are distinct and exhaust the linear characters. The group is nonabelian of order \(p^3\), so the argument in section 6 gives exactly \(p-1\) further irreducibles, each of degree \(p\). The later little-group lesson will construct them explicitly.

For an arbitrary group of order \(p^3\), Theorem 2.2 gives degrees among \(1,p,p^2,p^3\). The last two are excluded by the degree-square sum. Thus only \(1,p\) occur; in the abelian case all degrees are one, and in the nonabelian case their counts are as in (11).

Exercise 3. Why all tensor exponents are needed

Prove the central refinement \(d\mid[G:Z(G)]\) by external tensor products. Identify the role of the unbounded exponent \(m\).

Solution. The external product on \(G^m\) has character norm one, so is irreducible of degree \(d^m\). A central element acts on the original irreducible by a scalar homomorphism \(\lambda\). The subgroup of \(Z(G)^m\) whose entries multiply to one therefore acts trivially on the product. It has order \(|Z(G)|^{m-1}\), and the irreducible descends to the quotient of order \(|G|^m/|Z(G)|^{m-1}\).

Divisibility for that quotient gives (5). At a prime \(\ell\), if \(\delta,a,b\) are the valuations of \(d,|G|,|Z(G)|\), it says \(m(\delta-a+b)\le b\) for every \(m\ge1\). The left side would be unbounded above if the integer \(\delta-a+b\) were positive. Hence it is nonpositive, proving the required divisibility at every prime.

The single exponent \(m=1\) proves only \(d\mid|G|\). An arbitrarily large exponent removes the possible fixed excess \(b\) from the estimate. The argument uses a central subgroup, and does not establish the corresponding divisibility for an arbitrary abelian normal subgroup.

Exercise 4. The arithmetic obstruction, including its zero case

Prove the coprime-class lemma and the prime-power class obstruction. Give a counterexample to removing the zero alternative from the normalized-trace lemma.

Solution. If \(\gcd(|C|,d)=1\), Bezout's identity expresses \(\chi(g)/d\) as an integer combination of the integral numbers \(|C|\chi(g)/d\) and \(\chi(g)\). If this normalized trace is nonzero, take the product of all its conjugates in a cyclotomic field containing the eigenvalues. This product is a nonzero rational algebraic integer. Distinct eigenvalues would make its absolute value less than one, since the original average has absolute value strictly below one and all conjugate averages have absolute value at most one. Thus all eigenvalues coincide and the matrix is scalar. Otherwise its trace is zero. This proves Lemma 3.2 with both alternatives.

If a simple group had a class of size \(q^a>1\), it would be nonabelian and its nontrivial irreducibles faithful. For each degree coprime to \(q\), the scalar alternative would force the class element to be central, so its character value must be zero. Evaluating the regular character at that nonidentity element leaves \(0=1+q\beta\) with \(\beta\) an algebraic integer. The rational number \(-1/q\) cannot be integral, contradiction.

For the requested counterexample, a transposition on the standard \(S_3\)-plane has eigenvalues \(1,-1\). Its normalized trace is zero and its matrix is not scalar. A zero product of conjugates gives no nonzero-integer contradiction, which explains exactly where that missing alternative would break the proof.

8. Exact earlier tools and the integer-matrix identity

The representation-theoretic imports are the previous lessons' Maschke and Schur theorems, character norms and determination, the regular multiplicities and the sum of degree squares. The \(S_4\) and \(A_5\) tables are constructed, respectively, in Tensor products, duals and real representations, “The full table of S₄” and The group algebra and Fourier analysis on a finite group, Exercise 4. Coset, kernel/image and centralizer counting are proved in the Fourier lesson, Lemma 6.1. Cauchy's theorem, Sylow existence, the centre argument for \(p\)-groups and the solvable-extension argument are proved here.

For the integer matrix used in Lemma 1.1, the required Cayley–Hamilton identity has an elementary determinant proof. If \(M\) is an \(r\times r\) integer matrix with \(r\geq1\), write \[ p(t)=\det(tI-M)=\sum_{j=0}^r p_jt^j,\qquad \operatorname{adj}(tI-M)=\sum_{j=0}^{r-1}B_jt^j. \] The determinant and cofactors have integer coefficients, and \(p_r=1\). Expanding cofactors gives \[ (tI-M)\operatorname{adj}(tI-M)=p(t)I. \] Set \(B_{-1}=B_r=0\) and compare coefficients: \(B_{j-1}-MB_j=p_jI\) for \(0\leq j\leq r\). Multiply the \(j\)-th equation on the left by \(M^j\) and sum. Consecutive terms cancel, leaving \(\sum_jp_jM^j=0\). Thus \(p(M)=0\), with a monic integer polynomial. This proves exactly the matrix identity used for integrality.

The field argument uses polynomial division and complex roots. Lesson 1, Lemma 0.1 identifies the complex-root provider. Lemma 1.2 here constructs the required extensions of embeddings and proves that the cyclotomic fixed field is \(\mathbb Q\). Closure of algebraic integers, character and class-scalar integrality, divisibility of degrees and Burnside's theorem are proved here as well.

References

• C. Gruson and V. Serganova, A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras, 2018, Chapter 2 §3, especially Theorems 3.7–3.8, and §4, Theorem 4.1. The normalized-trace argument is used here with the necessary nonzero alternative.
• W. Burnside, On Groups of Order \(p^\alpha q^\beta\), Proceedings of the London Mathematical Society, second series, volume 1 (1904), pp. 388–392.
• F. G. Frobenius, Über Gruppen der Ordnung \(p^\alpha q^\beta\), Acta Mathematica 26 (1902), pp. 189–198, reprinted as paper 65 in Gesammelte Abhandlungen. This earlier paper studies sufficient restrictions on Sylow subgroups and exponents; it is historical context for the subject, rather than the source of the unrestricted theorem proved here.