Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 (Character Theory of Finite Groups)

Use the defining property of the Frobenius complement: $H\cap H^x=\{1\}$ whenever $x\notin H$. The induced character formula, extended linearly to any class function, isFor $h\in H\setminus \{1\}$, a summand requires $h\in H\cap xH{x}^{-1}$. The Frobenius intersection property forces $x\in H$. All $|H|$ remaining terms equal $\theta (h)$ since $\theta$ is a class function on $H$, so ${\theta }^G(h)=\theta (h)$. At the identity,Together these giveThe zero identity value is essential: at nonidentity elements the intersection argument already gives the equality, whereas induction multiplies the identity value by the subgroup index.

We prove closure of $N$ through character kernels, rather than assuming that a conjugacy-invariant set is a subgroup. Let $\eta \in \mathrm{Irr}(H)\setminus \{1_H\}$, write $d=\eta (1)$, and define the generalised characterHere $1_H,1_G$ are the trivial characters. Since $(\eta -d{1}_H)(1)=0$, part (i) gives $({\Psi }_{\eta }{)}_H=\eta$ and ${\Psi }_{\eta }(1)=d$. Put $\vartheta =\eta -d{1}_H$. By Frobenius reciprocity and character orthogonality,ConsequentlyTo justify the resulting irreducibility, write any generalised character as an integer combination ${\sum }_{\chi }{a}_{\chi }\chi$ of irreducible characters. Its squared norm is $\sum a_{\chi }^2$, so norm one makes it $\chi$ or $-\chi$ for one irreducible $\chi$. Positive degree rules out the negative sign. Thus ${\Psi }_{\eta }$ is an irreducible character, an instance of the fact that positive-degree norm-one virtual characters are irreducible.

If $x\in N\setminus \{1\}$, no conjugate of $x$ lies in $H$, so the induction formula gives ${\vartheta }^G(x)=0$. At $1$ it is also zero. Hence ${\Psi }_{\eta }(x)=d$ for all $x\in N$. For a finite-group representation we may choose an invariant positive Hermitian form, making every representing matrix unitary. Its trace equals its dimension only when all its unit-modulus eigenvalues are $1$. Thus the kernel of a character is $\{g:{\Psi }_{\eta }(g)={\Psi }_{\eta }(1)\}$, and $N\subseteq \ker {\Psi }_{\eta }$.

Conversely, if $g\notin N$, then $g$ is conjugate to some $h\in H\setminus \{1\}$. The character of the regular representation of $H$ isIf every nontrivial $\eta$ satisfied $\eta (h)=\eta (1)$, the sum at $h$ would instead equal $\sum \eta (1{)}^2=|H|$, a contradiction. Choose an $\eta$ for which these values differ. Since ${\Psi }_{\eta }$ restricts to $\eta$, it has ${\Psi }_{\eta }(g)=\eta (h)\ne d$, so $g\notin \ker {\Psi }_{\eta }$. ThereforeThis proves the Frobenius kernel theorem's subgroup assertion. The definition also gives $N\cap H=\{1\}$. Using the permitted cardinality $|N|=[G:H]$, normality ensures $NH$ is a subgroup andThus $N$ is the Frobenius kernel, and $H$ supplies the complementary factor.

Let $V$ afford the irreducible character $\chi$. The positive character inner product $⟨{\chi }_H,1_H{⟩}_H$ is the multiplicity of the trivial representation in $V$ restricted to $H$, so there is a nonzero vector $v$ fixed by every element of $H$. Since $N\subseteq \ker \chi$, every element of $N$ acts as the identity on all of $V$. By $G=NH$, every element of $G$ therefore fixes $v$.

The line $\mathbb{C}v$ is a nonzero invariant subspace of the irreducible representation $V$, so it must equal $V$. The action on this line is trivial, provingThe hypothesis on the kernel of a character is what makes an $H$-fixed vector into a $G$-fixed vector.

First show $C_N(h)=\{1\}$. If $z\in N$ commutes with $h$ and $z\ne 1$, then $z\notin H$ because $N\cap H=\{1\}$. But $z^{-1}hz=h$, so the nonidentity element $h$ lies in $H\cap H^z$, contradicting the Frobenius complement property. Thus conjugation by $h$ has no nonidentity fixed element in the Frobenius kernel.

Normality of $N$ makeswell-defined. If $f_h(a)=f_h(b)$, cancellation gives $a^{-1}ha=b^{-1}hb$, and hence $b{a}^{-1}$ commutes with $h$. Since $b{a}^{-1}\in N$ and $C_N(h)=\{1\}$, we obtain $a=b$. The map is injective, and an injection of a finite set into itself is surjective. ConsequentlyThis is the commutator bijection from a fixed-point-free automorphism, applied to $\alpha (y)=h^{-1}yh$.

Because $N$ is an abelian group, every $\phi \in \mathrm{Irr}(N)$ is a linear character, hence a homomorphism to ${\mathbb{C}}^{\times }$. Its inertia group of a character contains $N$, whose inner conjugations act trivially on $N$. Write an arbitrary $g\in G$ as $g=nh$ with $n\in N$ and $h\in H$, using $G=NH$. Conjugation by $n$ does not affect $\phi$, so $g$ stabilizes $\phi$ if and only if $h$ does.

Suppose $h\ne 1$ stabilizes $\phi$. Then $\phi (h^{-1}yh)=\phi (y)$ for every $y\in N$, and its multiplicativity givesPart (iv) makes $y\mapsto [h,y]$ surjective onto $N$, so $\phi (x)=1$ for every $x\in N$. This would be the trivial character, contrary to the hypothesis. Therefore no nonidentity $h\in H$ stabilizes $\phi$, andThe argument uses both abelianness, to make $\phi$ multiplicative, and the fixed-point-free commutator bijection.

We use complex representations. By Maschke's theorem, restriction to any finite subgroup is a semisimple module, so it has a canonical isotypic decomposition. In particular, the $\theta$-isotypic component $W_{\theta }$ is the sum of all copies of the irreducible $N$-module with character $\theta$. The central character idempotentprojects onto this component. These projections preserve every $N$-submodule, which consequently decomposes as the direct sum of its intersections with the isotypic components.

Let $W$ afford $\xi \in \mathrm{Irr}(T\mid \theta )$. The component $W_{\theta }$ is nonzero. Since $T=I_G(\theta )$ fixes $\theta$, it preserves $W_{\theta }$. Irreducibility of $W$ as a $T$-module therefore gives $W=W_{\theta }$: its restriction to $N$ consists entirely of copies of $\theta$.

Form the induced representationwhere $\mathcal{R}$ represents the left cosets $G/T$. Because $N◃G$, for $n\in N$ we haveThus $g\otimes W$ is $N$-isotypic of type ${\theta }^g$, where ${\theta }^g(n)=\theta (g^{-1}ng)$. These types are distinct for distinct cosets $gT$, precisely by the definition of the inertia group of a character $T$.

Let $Y$ be a nonzero $G$-submodule of $V$. Its isotypic decomposition as an $N$-module shows that it meets some $g\otimes W$ nontrivially. Acting by $g^{-1}$ gives $Y\cap (1\otimes W)\ne 0$. This intersection is a $T$-submodule of the irreducible $W$, hence equals $1\otimes W$. The $G$-translates of that component span $V$, so $Y=V$. The induced representation is therefore irreducible, and its restriction contains $\theta$. Its induced character satisfiesThis proves the irreducibility assertion directly from isotypic components; no form of the correspondence being proved has been assumed.

Let $V$ be an irreducible $G$-module with character $\chi \in \mathrm{Irr}(G\mid \theta )$, and let $W=V_{\theta }\ne 0$ be its $\theta$-isotypic component. It is invariant under the inertia group of a character $T$. The sum of its $G$-translates is a nonzero $G$-submodule, so irreducibility makes it all of $V$. Translates associated with different cosets $gT$ have distinct $N$-types and hence form a direct sum:To see that $W$ is irreducible over $T$, let $0\ne W_0\subseteq W$ be a $T$-submodule. Then ${\sum }_{g\in \mathcal{R}}g{W}_0$ is a nonzero $G$-submodule of $V$, hence equals $V$. Its $\theta$-isotypic part is exactly $W_0$, whereas the corresponding part of $V$ is $W$. Thus $W_0=W$.

Let $\xi$ be the irreducible character of $W$ as a $T$-module. It lies over $\theta$, and the natural map of induced representationsis an isomorphism: it identifies each summand with the corresponding $gW$, and these summands are direct and exhaust $V$. Consequently $\chi ={\xi }^G$, proving surjectivity of induction on the stated sets.

For injectivity, the construction in (a) shows that the $\theta$-isotypic component of ${\mathrm{Ind}}_T^{G}W$ is exactly $1\otimes W$, with its original $T$-action. Isomorphic induced $G$-modules therefore have isomorphic intrinsic $\theta$-components as $T$-modules. Since complex representations with the same character are isomorphic, ${\xi }_1^G={\xi }_2^G$ forces ${\xi }_1={\xi }_2$. Hence the Clifford correspondence is the bijectionwhose inverse takes the character of the $\theta$-isotypic component.

An M-group is a finite group for which every irreducible character is induced from a linear character of some subgroup. Let $L◃K$ have both $L$ and $K/L$ abelian, and fix an irreducible $K$-module $V$ with character $\chi$. Since $L$ is abelian, $V_L$ has a linear constituent. Choose, among all pairs $(A,\phi )$ with $L\subseteq A\subseteq K$ and a linear character $\phi$ occurring in ${\chi }_A$, one for which $|A|$ is maximal.

Every subgroup containing $L$ is normal in $K$: its image in the abelian quotient $K/L$ is normal, and it is the full preimage of that image. Thus $A◃K$, so the inertia group of a character $I_K(\phi )$ is defined and contains $A$.

Suppose $x\in I_K(\phi )\setminus A$. The nonzero $\phi$-isotypic component $V_{\phi }$ is preserved by $x$, and every $a\in A$ acts there as the scalar $\phi (a)$ because $\phi (1)=1$. The finite-order operator representing $x$ is diagonalizable over $\mathbb{C}$; choose an eigenvector $0\ne v\in V_{\phi }$. Then $\mathbb{C}v$ is invariant under both $A$ and $x$, hence under $D=⟨A,x⟩$. This line affords a linear character $\psi$ of $D$ with ${\psi }_A=\phi$. Since it is a subrepresentation of $V_D$, it is a constituent of ${\chi }_D$. But $D$ strictly contains $A$ and still contains $L$, contradicting maximality.

Therefore $I_K(\phi )=A$. Apply the Clifford correspondence with normal subgroup $A$. Its inertia group is $A$ itself, and $\mathrm{Irr}(A\mid \phi )=\{\phi \}$, so the unique irreducible $K$-character lying over $\phi$ is ${\phi }^K$. In particular,Since $\chi$ was arbitrary, every irreducible character is a monomial character, proving that finite metabelian groups are monomial:

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

Suppose first that $U$ is not contained in $(S^{\lambda }{)}^{\perp }$. Since polytabloids span $S^{\lambda }$, there exist $u\in U$ and a tableau $t$ with $⟨u,e_t⟩\ne 0$. The column antisymmetrizer identity from the preliminaries givesTherefore $e_t\in U$. Because $U$ is an $S_n$-submodule, it contains all translates $g{e}_t=e_{gt}$, and these span the Specht module. Hence $S^{\lambda }\subseteq U$. If the initial supposition fails, $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the James submodule theoremNow let $W$ be a submodule of $S^{\lambda }$. Apply the dichotomy to $W\subseteq M^{\lambda }$. Either $W=S^{\lambda }$, or $W\subseteq S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Positivity of the Hermitian form gives $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }=\{0\}$: a vector in this intersection has squared norm zero. The same intersection is zero for the bilinear form by the agreement of complements explained above. Since $S^{\lambda }\ne 0$, the only submodules are $0$ and itself. ThereforeThis use of characteristic zero and nondegeneracy is why the conclusion does not automatically extend to every modular Specht module.

Let $0\ne f:S^{\lambda }\to M^{\mu }$ be a module homomorphism. Since $S^{\lambda }$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero polytabloid $e_t$ then has $f(e_t)\ne 0$. The sign sum in the column group satisfiesbecause every coefficient in the square receives $|C_t|$ equal contributions. Thus ${\kappa }_t{e}_t=|C_t|e_t$, and equivariance givesThe column antisymmetrizer therefore acts nontrivially on $M^{\mu }$, so some basis tabloid $[s]$ has ${\kappa }_t[s]\ne 0$. The stated dominance from a nonzero column antisymmetrizer yieldsFor $\mu =\lambda$, the same calculation forces $f(e_t)=|C_t{|}^{-1}{\kappa }_{t}f(e_t)$ into the one-dimensional space $\mathbb{C}{e}_t$. Write $f(e_t)=c{e}_t$. Since $e_t$ generates the Specht module, equivariance gives $f(g{e}_t)=cg{e}_t$ for every $g$, so $f$ is $c$ times the inclusion $S^{\lambda }\hookrightarrow M^{\lambda }$. That inclusion is nonzero, henceAs a useful consequence, if $S^{\lambda }\cong S^{\mu }$, their inclusions into $M^{\lambda },M^{\mu }$ give dominance both ways. Antisymmetry of the dominance order on partitions then gives $\lambda =\mu$, so different partitions produce nonisomorphic simple modules.

The preceding parts give a simple Specht module for each partition and show that different partitions give nonisomorphic simples. The conjugacy classes of $S_n$ are indexed by partitions of $n$, namely cycle types. We use the standard finite-group character theorem: the number of irreducible complex characters equals the number of conjugacy classes, and they form an orthonormal basis of complex class functions. Consequently the ${\chi }^{\lambda }$ are the complete list of irreducible characters of $S_n$.

By Maschke's theorem, the Young permutation module $M^{\mu }$ decomposes as a direct sum of these simple modules. For a semisimple complex module, Schur lemma makes the multiplicity of a simple module equal to the dimension of the homomorphism space from that simple; character orthogonality identifies the same number with the character inner product. HencePart (ii) makes $a_{\lambda \mu }=0$ unless $\lambda ⊵\mu$, and gives $a_{\mu \mu }=1$. Taking characters provesThus the permutation-character decomposition is triangular in the dominance order on partitions, with diagonal multiplicity one. These are the dominance and diagonal-multiplicity consequences of Young's rule, obtained here without assuming its full multiplicity formula.

The permutation character ${\xi }^{\mu }(g)$ is the number of $\mu$-tabloids fixed by $g$, since the trace of a permutation matrix counts fixed basis elements. It is therefore an integer. Arrange the partitions in decreasing lexicographic order. If $\lambda$ strictly dominates $\mu$, the first part at which they differ is larger for $\lambda$, so this ordering places $\lambda$ before $\mu$.

For the first partition $(n)$, there is just one tabloid, and ${\chi }^{(n)}={\xi }^{(n)}=1_{S_n}$ is integer-valued. At a general partition $\mu$, the decomposition in (iii) can be rearranged asBy induction, all characters in the sum are already integer-valued; the permutation character and all coefficients are also integral. Thus ${\chi }^{\mu }$ is integer-valued. Finite induction proves that symmetric-group characters are integer-valued:The diagonal coefficient one is essential: it lets us solve for ${\chi }^{\mu }$ without division, preserving integrality.

Take $g=(12\cdots n)$. Since $n\equiv 3(\mathrm{mod}4)$ is odd, its sign is $(-1{)}^{n-1}=1$, so $g\in A_n$. Define $s$ to fix $1$ and reverse the other cyclic labels: it interchanges $2$ with $n$, $3$ with $n-1$, and so on. Then $sg{s}^{-1}=g^{-1}$, whileThe centralizer of an $n$-cycle in $S_n$ is $⟨g⟩$: a commuting permutation is determined by the image of one point, and that image fixes the corresponding power of the cycle. All its elements are even, because $g$ is even. Any other conjugator taking $g$ to $g^{-1}$ differs from $s$ by a centralizer element and therefore remains odd. Thus $g$ and $g^{-1}$ are not conjugate in $A_n$. This is the parity obstruction described by inversion of an odd cycle in an alternating group.

For any complex character of a finite group, a unitary realization givesIf every $\theta \in \mathrm{Irr}(A_n)$ had real value at $g$, all these characters would take equal values on $g$ and $g^{-1}$. But irreducible characters separate conjugacy classes: they form a basis of class functions, so equality of every irreducible value would also give equality of the class indicator functions. The two distinct conjugacy classes cannot have that property. ConsequentlyIn other words, these alternating groups are not ambivalent groups, even though all symmetric-group irreducible characters are integer-valued.

In character theory, a p-elementary group is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An elementary group is p-elementary for at least one prime $p$; it need not be an elementary abelian group.

A generalised character means an integer linear combination of irreducible complex characters. Brauer's characterisation of characters states that a complex class function $f$ on a finite group $G$ is a generalised character if and only if $f_E$ is a generalised character for every elementary subgroup $E\le G$. Equivalently,An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.

For the extension construction, set $d=\theta (1)$. The determinant character $\det \theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a linear character. For any $g\in G$, the subgroup $J_g=N⟨g⟩$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and ${\mu }_{J_g}$ extends its determinant. The permitted coprime-degree determinant extension theorem therefore gives a unique character ${\chi }_g\in \mathrm{Irr}(J_g)$ withDefine a function on $G$ by $f(g)={\chi }_g(g)$. We next verify that these individually defined values form a compatible class function.

If $x\in G$, conjugation carries $J_g$ to $J_{xg{x}^{-1}}$. Transporting ${\chi }_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a class function. Uniqueness identifies the transported character with ${\chi }_{xg{x}^{-1}}$. Consequently $f(xg{x}^{-1})=f(g)$.

Now take an elementary subgroup $E\le G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of solvable groups is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives ${\chi }_J\in \mathrm{Irr}(J)$ extending $\theta$ with determinant ${\mu }_J$.

For $g\in E$, we have $J_g\le J$. The restriction $({\chi }_J{)}_{J_g}$ is irreducible: an invariant subspace for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is ${\mu }_{J_g}$. By uniqueness on $J_g$, it equals ${\chi }_g$. Thereforean ordinary character and hence a generalised character of $E$. Brauer's characterisation of characters now makes $f$ a generalised character of $G$. If $n\in N$, then $J_n=N$ and ${\chi }_n=\theta$, so $f(n)=\theta (n)$. We have proved, by gluing determinant-normalized character extensions,Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.

For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_{\pi }\times C_{{\pi }^{\prime }}$, where each factor has the indicated prime divisors in its order. If $p\in \pi$, put $E_{\pi }=P\times C_{\pi }$ and $E_{{\pi }^{\prime }}=C_{{\pi }^{\prime }}$. If $p\notin \pi$, put $E_{\pi }=C_{\pi }$ and $E_{{\pi }^{\prime }}=P\times C_{{\pi }^{\prime }}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the prime-set decomposition of elementary groupsHere a pi-group has order divisible only by primes in $\pi$, and a pi-element has such an element order; the identity qualifies for both complementary prime sets.

Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of conjugacy classes and are disjoint, because membership in both would force every prime divisor of the element order into $\pi \cap {\pi }^{\prime }=\varnothing$, hence force order $1$. Let $E\le K$ be elementary and use $E=E_{\pi }\times E_{{\pi }^{\prime }}$. If both factors were nontrivial, choose $a\ne 1$ in the first and $b\ne 1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and ${\pi }^{\prime }$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a pi-group or entirely a complementary-prime group.

Write $m=|K{|}_{\pi }$ and $r=|K{|}_{{\pi }^{\prime }}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd (m,r)=1$. The Chinese remainder theorem supplies an integer $D$ satisfyingDefine the class functionFor an elementary $\pi$-subgroup $E$, its order divides $m$. Let ${\rho }_E$ be the character of its regular representation, equal to $|E|$ at $1$ and zero elsewhere. Thenwith integral coefficient because $D\equiv 1(\mathrm{mod}|E|)$. For an elementary ${\pi }^{\prime }$-subgroup, its order divides $r$, andagain has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a generalised character. A second application of Brauer's characterisation of characters provesThe freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi =\varnothing$, ${\pi }^{\prime }=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.