Brauer's induction 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).

Artin's theorem expresses any virtual character using induced characters of cyclic subgroups, but its coefficients may be rational. Brauer's theorem removes the denominators. It enlarges the subgroup family just enough: a cyclic group of order prime to \(p\), multiplied by a \(p\)-group, for various primes \(p\).

The proof has two different ingredients. A nilpotent group has enough abelian normal subgroups to produce every irreducible by induction from a line. For a general finite group, reducing character values modulo a prime shows that inductions from elementary subgroups generate the whole character ring locally. A determinant argument then returns an ordinary integer prime to that prime. Combining these integers gives the integral theorem.

Basic references are Kramár's Artin's and Brauer's Theorems on Induced Characters, Teleman's Representation Theory, and Gruson–Serganova's A Journey Through Representation Theory. We use Artin's induction theorem and rationality for cyclotomic fields and Artin induction, Mackey theory and Clifford's theorem for the inertia-group correspondence, and Induced representations and Frobenius reciprocity for induction, transitivity and the projection formula. Character orthogonality and algebraic-integer facts come from Characters and the orthogonality relations and Integrality of characters and Burnside's \(p^a q^b\) theorem.

All groups are finite and all representations are complex unless another ring is explicitly specified. Let \(R(G)\) be the ring of virtual characters, with pointwise multiplication. A representation is monomial if it is induced from a one-dimensional representation of a subgroup. A group is an M-group if all its irreducible representations are monomial.

1. Nilpotent groups can induce from lines

A group \(N\) is nilpotent if it has a central series

Here \([N,N_{i+1}]\) is generated by commutators \(xyx^{-1}y^{-1}\), with \(x\in N\), \(y\in N_{i+1}\). We may equivalently use the upper central series \(Z_0(N)=1\) and

Indeed (1) implies \(N_i\subseteq Z_i(N)\) by induction, and the upper central series is itself central. Intersecting a central series with a subgroup gives a central series there. Taking its images in a quotient gives a central series in that quotient. Thus subgroups and quotients of nilpotent groups are nilpotent.

Lemma 1.1 (an abelian normal subgroup beyond the centre). If \(N\) is nonabelian and nilpotent, it has an abelian normal subgroup \(A\) not contained in \(Z(N)\).

Proof. Its upper central series must have \(Z_2(N)>Z_1(N)\). Otherwise \(N/Z_1(N)\) would have zero centre; every subsequent term of the upper central series would equal \(Z_1(N)\), contradicting termination at the nonabelian group \(N\). Choose \(x\in Z_2(N)\setminus Z_1(N)\), and let \(A=\langle x,Z_1(N)\rangle\). It is abelian because the other generators are central. For every \(g\in N\), \(gxg^{-1}\in xZ_1(N)\), so \(A\) is normal. It contains the noncentral element \(x\). \(\square\)

Theorem 1.2 (nilpotent monomiality). Every irreducible representation of a finite nilpotent group is induced from a one-dimensional representation of a subgroup.

Proof. Induct on \(|N|\), the trivial group being immediate. Let \(V\) be irreducible. If its kernel \(K\) is nontrivial, it is an irreducible representation of the smaller nilpotent quotient \(N/K\). By induction it is induced there from a line on some \(\overline H\leq N/K\). Let \(H\) be the inverse image of \(\overline H\), and inflate that line to \(H\). The induced model for \(N\) is the inflation of the quotient-induced model: the cosets \(N/H\) and \((N/K)/\overline H\) are the same, with the same line operators on each coset. This proves the result in this case.

We may therefore suppose \(V\) faithful. If \(N\) is abelian, Schur's lemma makes \(V\) one-dimensional, and \(H=N\) works. Otherwise choose \(A\) as in Lemma 1.1. Its irreducibles are linear, so \(V|_A\) is a sum of weight spaces.

There must be at least two distinct weights. If there were just one, every element of \(A\) would act by a scalar. For \(a\in A\), \(g\in N\), the commutator \(gag^{-1}a^{-1}\) would then act trivially. Faithfulness would make \(a\) central in \(N\), contradicting the choice of \(A\).

Choose one weight \(\lambda\), with inertia group \(T\), the stabilizer of its weight space. Clifford's theorem makes the weights a single \(N\)-orbit. Since there is more than one, \(T<N\). Its correspondence gives

for an irreducible \(T\)-representation \(W\). The proper subgroup \(T\) is nilpotent. By induction \(W\cong\operatorname{Ind}_H^T\eta\) for a line \(\eta\) on \(H\leq T\). Induction in stages now gives \(V\cong\operatorname{Ind}_H^N\eta\). \(\square\)

A finite \(p\)-group is nilpotent. Its centre is nontrivial by the class equation: noncentral conjugacy classes have sizes divisible by \(p\), so the centre has size divisible by \(p\). Apply the same argument to successive nontrivial quotients in the upper central series; their orders decrease until the series reaches the whole group.

We will use one consequence of Sylow existence, proved in Integrality of characters and Burnside's \(p^a q^b\) theorem, Lemma 4.2. Every \(p\)-subgroup \(H\) lies in a conjugate of a fixed Sylow \(p\)-subgroup \(P\). Indeed \(H\) acts on \(G/P\); its nontrivial orbits have sizes divisible by \(p\). Since \([G:P]\) is prime to \(p\), there is a fixed coset \(xP\). Its stabilizer condition gives \(H\subseteq xPx^{-1}\).

For a prime \(p\), a group is \(p\)-elementary if

Either factor may be trivial. It is elementary if (2) holds for some prime. A direct product of an abelian group and a nilpotent group is nilpotent: combine their central series. Thus all elementary groups are M-groups.

We will also need that every subgroup \(H\leq C\times P\) is \(p\)-elementary. For \((c,u)\in H\), taking powers with exponents equal to \(1\) modulo \(|C|\) and \(0\) modulo \(|P|\), or conversely, shows that \((c,1),(1,u)\in H\). These exponents exist by Bézout's identity. Therefore

whose first factor is cyclic of order prime to \(p\) and whose second is a \(p\)-group.

2. What remains after induction

For fixed \(p\), define the additive subgroup

Let \(I\) be the analogous sum over all elementary subgroups.

Lemma 2.1 (induction gives ideals). Each \(I_p\), and \(I\), is an ideal of \(R(G)\).

Proof. For \(\alpha\in R(E)\), \(\beta\in R(G)\), the projection formula says

The expression inside induction belongs to \(R(E)\). Addition gives the assertion for every summand and then for their sum. \(\square\)

It suffices to put \(1_G\) in \(I\). The next result supplies enough integer multiples of it.

Theorem 2.2 (the local induction lemma). For every prime \(p\), there is an integer \(n_p\) with \(p\nmid n_p\) and

The proof occupies the next two sections. Its scalar coefficients are temporary: statement (5) is an assertion in the ordinary integral character ring.

3. Character values modulo one prime

Choose a primitive \(|G|\)-th root \(\zeta\), and set \(A=\mathbb Z[\zeta]\). Every character value belongs to \(A\). This ring is a domain, is finite as an additive \(\mathbb Z\)-module, and consists of algebraic integers. The ideal \(pA\) is proper: otherwise \(1/p\in A\), contradicting the rational algebraic-integer criterion.

Choose a maximal ideal \(\pi\) containing \(pA\), and form the localization

It embeds in \(\mathbb C\). It is a local ring with maximal ideal \(\pi D\): a fraction whose numerator is outside \(\pi\) is invertible; a fraction whose numerator is in \(\pi\) belongs to \(\pi D\) and is not a unit. Its residue field \(k=D/\pi D\) has characteristic \(p\). In particular every integer prime to \(p\) is a unit of \(D\), by Bézout.

We use existence of maximal ideals only in this elementary form: the union of an increasing chain of proper ideals is proper, since it cannot contain \(1\). Zorn's lemma then supplies a maximal proper ideal containing any proper ideal.

Write

Concretely its elements are finite \(D\)-linear combinations of irreducible characters, with pointwise multiplication. The irreducible characters are linearly independent over \(\mathbb C\), hence over \(D\). Evaluating at representatives \(g_1,\ldots,g_c\) of the conjugacy classes gives an injective unital ring map

The ring \(R_D\) contains the diagonal constants \(D\,1_G\). Therefore \(B\) is a finite faithful \(R_D\)-module: its coordinate vectors generate it already over \(D\), and an element annihilating \(B\) must annihilate \(1_B\) and hence be zero.

Lemma 3.1 (maximal ideals are reduced evaluations). Every maximal ideal of \(R_D\) is the kernel of evaluation at some \(g_i\), followed by reduction to \(k\).

Proof. Let \(\mathfrak m\) be maximal. First show \(\mathfrak mB\ne B\). If equality held, choose module generators \(b_1,\ldots,b_t\). We could write \(b_i=\sum_j a_{ij}b_j\) with all \(a_{ij}\in\mathfrak m\). Multiplication by the adjugate of \(I-(a_{ij})\) would show that

annihilates each generator, hence all of \(B\). Faithfulness would give \(d=0\). But \(d\equiv1\pmod{\mathfrak m}\), a contradiction.

Choose a maximal ideal \(\mathfrak q\) of \(B\) containing the proper ideal \(\mathfrak mB\). Its contraction to \(R_D\) is proper and contains \(\mathfrak m\), so equals \(\mathfrak m\).

The coordinate idempotents of \(D^c\) are orthogonal and sum to \(1\). In the field quotient by \(\mathfrak q\), exactly one becomes \(1\) and all the others become \(0\). On that surviving coordinate the quotient map is a field quotient of \(D\), whose kernel must be its unique maximal ideal \(\pi D\). Thus \(\mathfrak q\) is reduction modulo \(\pi D\) in one coordinate. Contracting gives the claimed reduced evaluation. It is onto \(k\) because \(R_D\) contains all constants from \(D\). \(\square\)

For \(g\in G\), write \(g=ab=ba\) with \(a\) of order prime to \(p\) and \(b\) of \(p\)-power order. These are the \(p'\)-part and \(p\)-part. If \(|g|=dp^r\), choose \(u\equiv1\pmod d\), \(u\equiv0\pmod{p^r}\); then \(a=g^u\) and \(b=g^{1-u}\) have the required properties.

Lemma 3.2 (the \(p\)-part disappears in residue). For every \(f\in R_D\),

Proof. First let \(f\) be an ordinary character. The commuting finite-order matrices of \(a,b\) can be simultaneously diagonalized: each eigenspace of the first is stable under the second, which is diagonalizable there. Thus the eigenvalues of \(g\) are products \(\alpha_j\beta_j\), where \(\beta_j^{p^r}=1\). In the characteristic-\(p\) field \(k\),

A field has no nonzero nilpotents, so \(\overline{\beta_j}=1\). Summing \(\alpha_j\beta_j\) and \(\alpha_j\) proves the congruence. It extends to virtual characters and then \(D\)-linear combinations. \(\square\)

4. An elementary subgroup detects every evaluation

Fix \(a\) of order \(d\) prime to \(p\). Let \(P\) be a Sylow \(p\)-subgroup of its centralizer \(C_G(a)\). The factors \(C=\langle a\rangle\) and \(P\) commute and intersect trivially, so

is \(p\)-elementary. Sylow existence is the only group-theoretic existence theorem used at this step.

The point indicator \(\delta_a\) on \(C\) has the cyclic Fourier expansion

To check it, choose a generator and sum a geometric progression: the sum is \(d\) at \(a\) and zero elsewhere. Its coefficients belong to \(D\), since \(d\) is a unit there and the character values belong to \(A\). Inflate it to \(E\), trivially on \(P\), giving \(\varphi\in D\otimes R(E)\).

Let \(J_p\) be the \(D\)-linear span in \(R_D\) of all inductions from \(p\)-elementary subgroups. By (4) it is an ideal and contains

Regrouping the conjugators in the induction formula by their target element gives

Indeed a conjugate of \(a\) has order prime to \(p\). In \(C\times P\), such an element has trivial \(P\)-coordinate. It therefore lies in \(C\), where \(\varphi\) is the point indicator at \(a\). The sum leaves just that one element. The final quotient is an integer prime to \(p\): \(E\leq C_G(a)\), and \(|P|\) is its full \(p\)-part.

Now take any maximal ideal of \(R_D\). By Lemma 3.1 it corresponds to a reduced evaluation at some \(g\). Let \(a\) be the \(p'\)-part of \(g\). Equations (7) and (9) show that \(F_a(g)\) is nonzero modulo \(\pi D\). Thus this maximal ideal does not contain \(J_p\). No maximal ideal contains \(J_p\), so \(J_p=R_D\).

We must return to integer coefficients. The following descent is the last part of the local lemma.

Put \(M=R(G)/I_p\), a finitely generated abelian group. The equality \(J_p=R_D\) gives \(D\otimes M=0\): tensoring a quotient imposes exactly the \(D\)-linear relations from \(I_p\). Reducing further to the residue field gives

Here the identification follows directly from the tensor relations: multiplication by \(p\) is zero in \(k\), so the second factor is \(M/pM\); its remaining scalar action is over \(\mathbb F_p\). A nonzero \(\mathbb F_p\)-vector space cannot become zero after extension to the nonzero field \(k\), because any basis remains a basis after scalar extension. Hence \(M=pM\).

Choose abelian-group generators \(u_1,\ldots,u_t\) of \(M\). There are integers \(a_{ij}\) such that \(u_i=p\sum_j a_{ij}u_j\). The adjugate argument now shows that

annihilates every generator and thus all of \(M\). It satisfies \(n_p\equiv1\pmod p\). In particular \(n_p\,1_G\in I_p\). If \(M=0\), choose \(n_p=1\). This proves Theorem 2.2, including primes not dividing \(|G|\).

5. Brauer's integral theorem

Theorem 5.1 (Brauer induction). Every virtual character of \(G\) is an integer linear combination of

Proof. By Theorem 2.2, \(I\) contains \(n_p1_G\) for every prime \(p\), with \(p\nmid n_p\). These integers generate the unit ideal in \(\mathbb Z\). To see this without an infinite Bézout sum, start with any nonzero \(n_{p_0}\). If its absolute value exceeds \(1\), choose one \(n_q\) for each prime divisor \(q\) of \(n_{p_0}\). The gcd of this finite list is \(1\), since no prime dividing the first integer divides every entry. The Euclidean algorithm expresses \(1\) as an integer combination of that finite list. Therefore \(1_G\in I\).

Lemma 2.1 gives \(I=R(G)\). Each character induced from an elementary subgroup \(E\) is an integer sum of inductions of its irreducibles. By Theorem 1.2, each such irreducible is \(\operatorname{Ind}_H^E\lambda\) for a line on a subgroup \(H\leq E\). That subgroup is itself elementary, by the subgroup argument after (2). Induction in stages gives (12). \(\square\)

The theorem permits negative coefficients. It does not assert that an arbitrary irreducible is itself induced from a line, or even from a proper elementary subgroup.

6. Lines, sums and differences in small groups

For \(S_3\), the degree-two standard character is \(\operatorname{Ind}_{C_3}^{S_3}\omega\), with \(\omega\) nontrivial. The two remaining characters satisfy

These are the explicit integral decompositions established in the preceding lesson. Both cyclic subgroups are elementary.

For \(Q_8\), the group itself is \(2\)-elementary. Its degree-two irreducible is already monomial from the proper subgroup \(\langle i\rangle\cong C_4\): induce the line \(\eta(i)=i\). The induced trace is \(2,-2,0,0,0\) on its five classes. In particular this induction does not make the representation realizable over \(\mathbb Q\); the line has nonrational values.

A more informative example is

This is isomorphic to \(\mathrm{SL}_2(\mathbb F_3)\). For a direct verification, put

over \(\mathbb F_3\). They generate \(Q_8\). The matrix \(v=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) has order three and permutes the three pairs \(\{\pm i\},\{\pm j\},\{\pm k\}\) cyclically by conjugation. Choose signs of the generators and \(t=v\) or \(v^{-1}\) to give (13). The subgroup generated has \(8\cdot3=24\) elements. There are exactly \(24\) determinant-one matrices: eight nonzero first columns and three second columns of determinant one for each. This proves the isomorphism.

The quaternion representation has a two-dimensional extension \(\rho\) to (13). View unit quaternions as complex \(2\times2\) matrices by

The unit quaternion \(u=(-1+i+j+k)/2\) has order three: with \(w=(i+j+k)/\sqrt3\), it equals \(-1/2+(\sqrt3/2)w\), where \(w^2=-1\). Conjugation by \(u\) cyclically permutes \(i,j,k\) in one direction; choose \(u\) or \(u^{-1}\) so it agrees with (13). Its complex eigenvalues are \(\omega,\omega^2\), where \(\omega=e^{2\pi i/3}\), since its trace is \(-1\), determinant is \(1\), and it has order three. Sending \(t\) to this matrix defines the extension. It is irreducible since its restriction to \(Q_8\) is.

Let \(\alpha(t)=\omega\), \(\alpha|_{Q_8}=1\), and write \(\chi_r\) for the character of \(\rho\otimes\alpha^r\), with indices modulo three. They are distinct: their values at \(t\) are \(-\omega^r\). The central element \(z=-1\) acts as \(-I\) in all three.

Let \(K=\langle z,t\rangle\cong C_6\), and let \(\lambda_r(z)=-1,\ \lambda_r(t)=\omega^r\). Reciprocity gives

For the first equality each multiplicity is one by restriction to \(Q_8\); the displayed constituents already fill degree six. For the second, the two eigencharacters of \(K\) on \(\rho\otimes\alpha^s\) are \(\lambda_{s+1},\lambda_{s+2}\); the two constituents indicated by reciprocity fill degree four. Subtract (14), and use \(\psi=\operatorname{Ind}_{C_4}^{Q_8}\eta\):

This is a concrete Brauer expression using proper cyclic subgroups.

Yet \(G\) is not an M-group. A monomial irreducible of degree two would require a subgroup of index two. Such a subgroup would give a nontrivial homomorphism \(G\to C_2\). In any abelian quotient of (13), conjugation gives \(i=j=k\), and \(ij=k\) then forces their common image to be the identity. The quotient is generated by the image of \(t\), of order dividing three. It has no nontrivial \(C_2\) quotient. Thus these degree-two irreducibles are not monomial. Nilpotent monomiality cannot be extended to all solvable groups: the series \(1<\{\pm1\}<Q_8<G\) has abelian factors, so this particular counterexample is solvable.

7. Exercises with complete solutions

Exercise 1. List the elementary subgroups of \(S_4\) up to conjugacy, and justify completeness.

Solution. The list has eight entries:

Every \(2\)-subgroup lies in a conjugate of this Sylow \(D_8\). Its subgroups have orders \(1,2,4,8\). The order-four possibilities are \(C_4\) and \(V_4\); a \(V_4\) with a transposition must also contain its disjoint commuting transposition, whereas one without transpositions is the displayed normal \(V_4\). This proves the classification of \(2\)-subgroups.

For \(p=2\), a nontrivial odd cyclic factor must be \(C_3\), since the odd element orders in \(S_4\) are \(1,3\). Its centralizer is \(C_3\), so it cannot commute with a nontrivial \(2\)-group. This adds just \(C_3\). For \(p=3\), a nontrivial \(3\)-group is \(C_3\), whose centralizer again permits no nontrivial cyclic factor prime to three. With trivial \(3\)-part the subgroup is cyclic, already on the list. For all other primes the \(p\)-part is trivial, so once again the subgroup is cyclic. All eight displayed subgroups are elementary, proving both directions.

Exercise 2. Prove monomiality for groups of order \(p^3\). In the nonabelian case give a subgroup of index \(p\) from which every degree-\(p\) irreducible is induced.

Solution. An abelian group has only linear complex irreducibles. Suppose \(N\) is nonabelian of order \(p^3\). Its centre has order \(p\): it is nontrivial, and a centre of index \(p\) would make \(N/Z(N)\) cyclic, forcing \(N\) abelian. To verify that last assertion, write every element as \(x^a z\) for one lift \(x\) of a quotient generator and central \(z\); all such elements commute.

Choose \(x\notin Z(N)\). The subgroup \(A=\langle x,Z(N)\rangle\) is abelian and has order \(p^2\). It cannot have order \(p\), since it contains the centre and an element outside it, and cannot have order \(p^3\), since \(N\) is nonabelian. A subgroup of index \(p\) in a \(p\)-group is normal. Here is a direct proof: its coset action has transitive \(p\)-group image in \(S_p\); the \(p\)-part of \(p!\) is \(p\), so the image has order \(p\) and a point stabilizer is trivial. The original subgroup is therefore the action kernel.

Character degrees divide \(p^3\), and their squares do not exceed \(|N|=p^3\). Thus a nonlinear irreducible has degree \(p\). It is faithful: a nontrivial kernel would give a quotient of order at most \(p^2\), which is abelian. A group of order \(p^2\) is abelian by the same centre and cyclic-quotient argument.

Restriction to \(A\) has at least two distinct linear weights, since faithfulness and a single scalar weight would put \(A\) in the centre. Clifford's theorem makes their orbit size a divisor of \([N:A]=p\), hence exactly \(p\). The inertia group is \(A\). Its irreducible weight space is a line, and the correspondence gives \(V=\operatorname{Ind}_A^N\lambda\). Linear irreducibles require no induction from a proper subgroup. This proves the claim, also for \(p=2\).

Exercise 3. Prove directly from the projection formula that inductions from any chosen family of subgroups generate an ideal in the character ring.

Solution. For a family \(\mathcal F\), let \(L=\sum_{H\in\mathcal F}\operatorname{Ind}_H^G R(H)\). It is an additive subgroup by construction. For \(\beta\in R(G)\), a generator \(\operatorname{Ind}_H^G\gamma\) satisfies

Distributing multiplication across finite sums gives \(\beta L\subseteq L\). No closure property of the subgroup family is needed. The same calculation after extending coefficients to \(D\) proves that the temporary ideal \(J_p\) in the local proof is an ideal.

Exercise 4. Express the trivial character of \(A_4\) as an explicit integer combination of characters induced from lines on elementary subgroups.

Solution. Let \(V=\{1,(12)(34),(13)(24),(14)(23)\}\), choose any nontrivial linear character \(\lambda\) of \(V\), and let \(C=\langle(123)\rangle\). We claim

Check the four conjugacy classes. The coset permutation character from \(C\) has values \(4,0,1,1\) at the identity, the double transpositions, and the two three-cycle classes. At a three-cycle, its centralizer has size three and exactly one of the two nonidentity elements of \(C\) belongs to the relevant \(A_4\) class; the induction formula therefore gives \(3/3=1\). A double transposition has no conjugate in \(C\).

The normal subgroup \(V\) has index three. On its identity the second induced character is \(3\). Its three nonidentity elements are permuted transitively by \(A_4/V\), so at a double transposition the value is the sum of \(\lambda\) at those three elements, namely \(-1\); the sum over all of \(V\) is zero for a nontrivial line. Outside \(V\) it is zero. Its values are consequently \(3,-1,0,0\). Subtracting gives \(1,1,1,1\), proving (16). The group \(V\) is \(2\)-elementary and \(C\) is cyclic elementary. The negative term is essential in this expression.

8. Earlier results used without proof

Maschke's theorem over characteristic-zero fields and Schur's lemma are Theorems 2.3 and 3.1 of Representations and complete reducibility.

The character basis, degree-square identity, and Hom multiplicities are proved in §§2–4 of Characters and the orthogonality relations. The centralizer Sylow subgroup used in (9) exists by Lemma 4.2 of Integrality of characters and Burnside's \(p^a q^b\) theorem, with its proof included there. We also use that lesson's Lemma 1.1 on algebraic integers and Theorem 2.2 on divisibility of character degrees; Lemma 1.2 supplies the earlier normalized-trace argument's cyclotomic fixed-field step.

Induction, reciprocity, transitivity and the projection formula are proved in §§1–4 of Induced representations and Frobenius reciprocity. The inertia-group correspondence is Theorem 4.1 of Mackey theory and Clifford's theorem. General cyclotomic irreducibility is Lemma 1.1 of Artin's induction theorem and rationality. Its Artin theorem gives rational cyclic induction; the proof here does not use Brauer's later splitting-field theorem. The localization, finite-module, residue-field and integer-descent steps needed for the local lemma were all proved above.

References

• J. Kramár, Artin's and Brauer's Theorems on Induced Characters, 2005, §3, for a proof using integer-valued class functions and elementary centralizers.
• C. Teleman, Representation Theory, §18, for the elementary-subgroup induction statement; that treatment refers to Serre for its proof.
• C. Gruson and V. Serganova, A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras, Universitext, Springer, 2018, Chapter 2 §§7 and 12, for Clifford theory and the Artin precursor.