Brauer Induction and Elementary Subgroups

Put $C_H=H\cap C$. It is cyclic, normal in $H$, and $H\mathrm{/}{C}_H$ embeds in $G\mathrm{/}C\cong P$; hence it is a $p$-group. A Sylow $p$-subgroup $P_H$ of $H$ maps isomorphically onto $H\mathrm{/}{C}_H$, because its image has the full $p$-power order and $C_H$ has order prime to $p$.

Thus $H=C_H⋊P_H$. If $G=C\times P$, its Sylow $p$-subgroup is unique, so $P_H=H\cap P$ and it commutes with $C_H$; consequently $H=C_H\times P_H$. ∎

The projection formula gives ${\mathrm{Ind}}_H^G\theta \cdot \chi ={\mathrm{Ind}}_H^G(\theta \cdot {\mathrm{Res}}_H^G\chi )$.

Its right side is one of the defining summands of $I_{\mathcal{F}}(G)$; additivity handles finite sums and additive inverses, so multiplication by every $\chi$ preserves the subgroup. ∎

On the cyclic group $⟨g⟩$, every irreducible complex character is linear. For each such character $\psi$, the $p^n$-th powers of $\psi (g)$ and $\psi (g_{p^{\mathrm{\prime }}})$ agree, since $g^{p^n}=g_{p^{\mathrm{\prime }}}^{p^n}$.

Write the restriction as ${\sum }_i{a}_i{\psi }_i$ with $a_i\in A$ and the ${\psi }_i$ linear. In $A\mathrm{/}pA$, the freshman's dream and step 1.1 give the following congruence.

Both character values are integers. The power basis $1,{\zeta }_{\mathrm{\mid }G\mathrm{\mid }},\dots ,{\zeta }_{\mathrm{\mid }G\mathrm{\mid }}^{\phi (\mathrm{\mid }G\mathrm{\mid })-1}$ makes $\mathbb{Z}\cdot 1$ a direct summand of $A$, so $pA\cap \mathbb{Z}=p\mathbb{Z}$. Fermat's congruence then gives $\chi (g)\equiv \chi (g{)}^{p^n}\equiv \chi (g_{p^{\mathrm{\prime }}}{)}^{p^n}\equiv \chi (g_{p^{\mathrm{\prime }}})(\mathrm{m}\mathrm{o}\mathrm{d}p)$. ∎

By the projection formula, $({\mathrm{Ind}}_H^G{1}_H)({\mathrm{Ind}}_K^G{1}_K)={\mathrm{Ind}}_H^G({\mathrm{Res}}_H^G{\mathrm{Ind}}_K^G{1}_K)$. Mackey's formula expresses the restricted character as a finite sum of ${\mathrm{Ind}}_{H\cap xK{x}^{-1}}^H{1}_{H\cap xK{x}^{-1}}$. Each intersection is hyperelementary by subgroup closure; transitivity then makes every summand induced back to $G$ equal to ${\mathrm{Ind}}_{H\cap xK{x}^{-1}}^G{1}_{H\cap xK{x}^{-1}}$. Thus products of the displayed generators stay in $P({\mathcal{H}}^{\mathrm{\prime }})$, and bilinearity makes its additive span a subring.

If $1_H\in I_{\mathcal{E}}(H)$, express it by [F3] as an integral sum of characters induced from elementary subgroups of $H$. Inducing this sum to $G$ and using transitivity puts ${\mathrm{Ind}}_H^G{1}_H$ in $I_{\mathcal{E}}(G)$. Apply this to every summand of a relation for $1_G$ in $P({\mathcal{H}}^{\mathrm{\prime }})$. ∎

For $x\in X$, the value set $A_x=\{a(x):a\in A\}$ is an ideal $n_x\mathbb{Z}$. If every $A_x$ contained $1$, choose $a_x\in A$ with $a_x(x)=1$ and form ${\prod }_{x\in X}(1_X-a_x)=0$.

Expanding the finite product would put $1_X$ in $A$, a contradiction. Hence for some $x$, the ideal $A_x=n_x\mathbb{Z}$ does not contain $1$, so $n_x=0$ or $n_x>1$. If $n_x=0$, choose any prime $p$; otherwise choose a prime divisor of $n_x$. In either case $A_x\subseteq p\mathbb{Z}$. Permutation characters are integer-valued fixed-point counts by [F1], and their hyperelementary span is closed under products by [F2], so that span gives the stated instance of the subring. ∎

Put $A=\mathbb{Z}[{\zeta }_{\mathrm{\mid }G\mathrm{\mid }}]$. For $r\in R$, choose a Sylow $p$-subgroup $P_r$ of $C_G(r)$ and set $H_r=⟨r⟩\times P_r$. On $⟨r⟩$ the delta function ${\chi }_r(c)=\mathrm{\mid }⟨r⟩\mathrm{\mid }{\delta }_{c,r}$ lies in $AR(⟨r⟩)$ by Fourier inversion on this cyclic group. Inflate it across $P_r$ to ${\psi }_r\in AR(H_r)$ and form the following sum.

This lies in the $A$-scalar extension of $I_{{\mathcal{E}}_p}(G)$. Each ${\psi }_r$ is integer-valued, so the induction formula makes every value of $\psi$ rational. On the other hand $\psi$ is an $A$-linear combination of characters, hence all its values are algebraic integers. A rational algebraic integer is an integer, so $\psi$ is integer-valued.

If $r_0\in R$, a conjugate of $r_0$ lying in $H_r$ lies in $⟨r⟩$. The definition of ${\chi }_r$ and [F1] therefore give the following value.

Thus $\psi (r_0)$ is an integer prime to $p$. For arbitrary $g$, its $p^{\mathrm{\prime }}$-part is conjugate to some $r_0\in R$, so [F2] shows that $\psi (g)\equiv \psi (r_0)\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}p)$.

To use the integer-valued function from step 4.1, first let $\beta$ be any integer-valued class function on $G$. For each conjugacy-class representative $g$, put $C_g=⟨g⟩$ and define ${\theta }_g(c)=\mathrm{\mid }{C}_g\mathrm{\mid }{\delta }_{c,g}$ on $C_g$. The same cyclic Fourier inversion used in step 1.1 makes ${\theta }_g\in AR(C_g)$. By [F1], ${\mathrm{Ind}}_{C_g}^G{\theta }_g$ is $\mathrm{\mid }{C}_G(g)\mathrm{\mid }$ on the conjugacy class of $g$ and zero elsewhere. Consequently $\mathrm{\mid }G\mathrm{\mid }\beta ={\sum }_{[g]}\beta (g)[G:C_G(g)]{\mathrm{Ind}}_{C_g}^G{\theta }_g\mathrm{.}$ Every cyclic subgroup is $p$-elementary by splitting its order into coprime $p$-power and $p^{\mathrm{\prime }}$ factors, so this identity puts $\mathrm{\mid }G\mathrm{\mid }\beta$ in the $A$-scalar extension of $I_{{\mathcal{E}}_p}(G)$.

Assume $n\ge 1$ and put $e=p^{n-1}(p-1)$. Euler's congruence gives $\psi (g{)}^e\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{p}^n)$ for every $g$. Thus $\beta =l({\psi }^e-1_G)\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$ is an integer-valued class function, and step 5.1 puts $l({\psi }^e-1_G)$ in the $A$-scalar extension of $I_{{\mathcal{E}}_p}(G)$. The same is true of $l{\psi }^e$ by [F3] and step 1.1. Subtraction puts $l{1}_G$ in that scalar extension. If $n=0$, then $l=\mathrm{\mid }G\mathrm{\mid }$, so step 5.1 applied to $\beta =1_G$ gives the same conclusion directly.

The cyclotomic ring $A$ is a finite free $\mathbb{Z}$-module with the power basis $1,{\zeta }_{\mathrm{\mid }G\mathrm{\mid }},\dots ,{\zeta }_{\mathrm{\mid }G\mathrm{\mid }}^{\phi (\mathrm{\mid }G\mathrm{\mid })-1}$. Expand the relation from step 6.1 in this basis and take its coefficient of $1$. Since all inducing characters there lie in integral character rings, this yields $l{1}_G\in I_{{\mathcal{E}}_p}(G)$, including the case $n=0$.

The fibre is $P$-stable by invariance and has cardinality $\mathrm{\mid }\ker \lambda \mathrm{\mid }$, a divisor of $\mathrm{\mid }N\mathrm{\mid }$ and hence prime to $p$. The fixed-point congruence supplies a $P$-fixed element in that fibre.

A fixed element lies in $C_N(P)\subseteq \ker \lambda$, so its character value is $1$. Since the fibre has that same value, $\lambda (\nu )=1$; as $\nu$ was arbitrary, $\lambda$ is trivial. ∎

A cyclic group has a normal series with prime-order factors. The case $P=1$ is immediate. If $P\mathrm{\ne }1$, its centre contains a subgroup $Z$ of order $p$, and $P\mathrm{/}Z$ has smaller order; by the induction hypothesis it has a normal prime-factor series, whose inverse images preceded by $1◃Z$ give one for $P$.

Concatenate the series for $C$ and the series $C\times P_i$ from the series of $P$. Each term is normal in $C\times P$ and every factor has prime order, including the cases $C=1$ or $P=1$. ∎

Complete reducibility [F1] decomposes $V{\mathrm{\mid }}_A$ into irreducible summands, and [F2] makes every such summand one-dimensional. Thus $V{\mathrm{\mid }}_A$ is the direct sum of its linear weight spaces. The translates of the $\lambda$-weight space are the weight spaces in its $G$-orbit, and their direct sum is $V$ by irreducibility.

If the inertia group $G_{\lambda }$ were $G$, every $a\in A$ would act by a scalar on $V$. Its commutator with every $g\in G$ would then act trivially, so faithfulness would make $A$ central, contrary to hypothesis. Thus $G_{\lambda }<G$.

Put $W=V_{\lambda }$. Distinct left cosets of $G_{\lambda }$ carry $W$ to distinct $A$-weight spaces, whose direct sum is $V$ by step 1.1; the usual transversal model therefore identifies ${\mathrm{Ind}}_{G_{\lambda }}^{G}W$ with $V$. If $0\mathrm{\ne }U\subseteq W$ is a $G_{\lambda }$-submodule, the direct sum of its translates over the left cosets is a nonzero $G$-submodule of $V$. Irreducibility makes this sum all of $V$, and its intersection with $W$ is $U$, so $U=W$. Hence the inducing representation $W$ is irreducible, completing the claim.

Let $i$ be maximal with $G_i$ abelian; such an $i$ exists since $G_0=1$. If $i=r$, then $G$ is abelian, contrary to the hypothesis, so $i<r$. Suppose $G_i\le Z(G)$. By [F1] and [F2], $G_{i+1}\mathrm{/}{G}_i$ is cyclic of prime order; choose $x\in G_{i+1}$ whose coset generates it. Then every element of $G_{i+1}$ has the form $a{x}^k$ with $a\in G_i$. All such elements commute because $G_i\le Z(G)$ and powers of $x$ commute, so $G_{i+1}$ is abelian, contradicting maximality of $i$. Thus $G_i\nleqq Z(G)$, and there are $a\in G_i$ and $g\in G$ with $ag\mathrm{\ne }ga$.

By [F3], $G_i⊴G$. It is abelian by construction and not central by step 1.1, so it is the required abelian normal subgroup.

Choose a representation $\rho$ affording $\chi$. By [F1], $N\subseteq \ker \rho$, so [F2] factors $\rho$ through an irreducible representation $\bar{\rho }$ of $Q:=G\mathrm{/}N$. Its character $\bar{\chi }$ satisfies $\chi (g)=\bar{\chi }(gN)$. By the hypothesis and [F3], choose $\bar{H}\le Q$ and a one-dimensional $\bar{H}$-module $W$ with character $\bar{\lambda }$ such that $\bar{\chi }={\mathrm{Ind}}_{\bar{H}}^Q\bar{\lambda }$.

Let $\pi :G\to Q$ be the quotient map and $H={\pi }^{-1}(\bar{H})$. Then $N\le H\le G$ and $H\mathrm{/}N=\bar{H}$. Make $W$ an $H$-module by $h\cdot w:=(hN)\cdot w$; its linear character is $\lambda (h)=\bar{\lambda }(hN)$. In particular, $N$ acts trivially on $W$.

For $\bar{f}\in {\mathrm{Ind}}_{\bar{H}}^{Q}W$, define $\mathrm{\Phi }(\bar{f})(g):=\bar{f}(gN)$. If $h\in H$, then [F4] and [F5] give $\mathrm{\Phi }(\bar{f})(gh)=\bar{f}((gN)(hN))=(hN{)}^{-1}\cdot \bar{f}(gN)=h^{-1}\cdot \mathrm{\Phi }(\bar{f})(g)$. Thus $\mathrm{\Phi }(\bar{f})\in {\mathrm{Ind}}_H^{G}W$, and $\mathrm{\Phi }$ is linear.

Conversely, for $F\in {\mathrm{Ind}}_H^{G}W$ and $n\in N$, [F4] gives $F(gn)=n^{-1}\cdot F(g)=F(g)$ by step 2.1. Hence $\bar{f}(gN):=F(g)$ is well defined; the same covariance calculation shows that $\bar{f}\in {\mathrm{Ind}}_{\bar{H}}^{Q}W$. This construction is inverse to $\mathrm{\Phi }$, so $\mathrm{\Phi }$ is a linear bijection.

For $x,g\in G$, the action in [F4] gives $\mathrm{\Phi }((xN)\cdot \bar{f})(g)=\bar{f}((x^{-1}g)N)=(x\cdot \mathrm{\Phi }(\bar{f}))(g)$. Thus $\mathrm{\Phi }$ is $G$-equivariant when the $Q$-module on the left is inflated along $\pi$.

Taking characters of the isomorphic $G$-modules in steps 4.1–4.2, [F4] gives $\chi (g)=\bar{\chi }(gN)={\mathrm{Ind}}_H^G\lambda (g)$ for every $g\in G$. By [F3], $\chi$ is monomial.

Induct on $\mathrm{\mid }G\mathrm{\mid }$. The trivial group is the base case. If $G$ is abelian, $\chi$ is linear. If $\ker \chi \mathrm{\ne }1$, then $G\mathrm{/}\ker \chi$ is supersolvable of smaller order and the quotient lemma makes $\chi$ monomial.

Otherwise $\chi$ is faithful. A nonabelian $G$ has a noncentral normal abelian subgroup, so the proper-inertia proposition writes $\chi ={\mathrm{Ind}}_H^G\theta$ with $H<G$ and $\theta \in \mathrm{Irr}(H)$. Intersecting a supersolvable series with $H$ and deleting repeated terms makes $H$ supersolvable; by the induction hypothesis, $\theta$ is induced from a linear character, and transitivity makes $\chi$ so induced. ∎

The elementary-group lemma makes $E$ supersolvable, and [F2] writes every irreducible constituent of $\xi$ as an induction of a linear character.

Add the resulting expressions with the integral multiplicities defining the virtual character $\xi$.

If $G$ is trivial, the assertion is immediate. Otherwise, for every prime $p\mid \mathrm{\mid }G\mathrm{\mid }$, write $\mathrm{\mid }G\mathrm{\mid }=p^{n_p}{l}_p$ with $p\nmid l_p$. By [F1], $l_p{1}_G\in I_{{\mathcal{E}}_p}(G)\subseteq I_{\mathcal{E}}(G)$. The integers $l_p$ have greatest common divisor $1$: for each prime divisor $q$ of $\mathrm{\mid }G\mathrm{\mid }$, the particular integer $l_q$ is prime to $q$. Bézout therefore gives $1_G\in I_{\mathcal{E}}(G)$.

By [F2], every $\chi \in R(G)$ satisfies $\chi =\chi 1_G\in I_{\mathcal{E}}(G)$. Thus it is an integral sum of characters ${\mathrm{Ind}}_H^G\theta$ with $H$ elementary and $\theta \in R(H)$.

By [F3], each such $\theta$ is an integral combination of characters induced from linear characters of elementary subgroups $K\le H$. Transitivity [F4] changes ${\mathrm{Ind}}_H^G{\mathrm{Ind}}_K^H\lambda$ into ${\mathrm{Ind}}_K^G\lambda$, which is exactly the claimed form. ∎

Brauer induction writes $1_G={\sum }_i{n}_i{\mathrm{Ind}}_{H_i}^G{\lambda }_i$ with $H_i$ elementary and ${\lambda }_i$ linear.

The projection formula yields $\chi ={\sum }_i{n}_i{\mathrm{Ind}}_{H_i}^G({\lambda }_i\cdot {\mathrm{Res}}_{H_i}^G\chi )=0$, as every restriction is zero. ∎

The forward implication is restriction stability. For the converse, choose the Brauer relation $1_G={\sum }_i{n}_i{\mathrm{Ind}}_{H_i}^G{\lambda }_i$.

Pointwise multiplication and the projection formula give $f={\sum }_i{n}_i{\mathrm{Ind}}_{H_i}^G({\lambda }_{i}f{\mathrm{\mid }}_{H_i})$. Each inner product is in $R(H_i)$, so the right side is in $R(G)$. ∎

Brauer induction expresses $[V]$ integrally as inductions of linear characters of elementary subgroups. Every such linear character $\lambda$ satisfies $\lambda (h{)}^m=1$, so it takes values in $F$ and its induced representation has an $F$-model. Thus $[V]$ is the scalar extension of a virtual $F$-representation.

By [F2], decompose that virtual $F$-representation as ${\sum }_i{c}_i[W_i]$ with the $W_i$ distinct irreducible $F$-representations. Base change preserves intertwiner spaces, so distinct $W_i{\otimes }_F\mathbb{C}$ have disjoint irreducible complex constituents. Each is semisimple, with positive constituent multiplicities. Since their signed sum is the single irreducible basis element $[V]$, exactly one summand occurs, its coefficient and the multiplicity of $V$ are both $1$, and it has no other constituent. Hence $V\cong W_i{\otimes }_F\mathbb{C}$ for that index. Every irreducible complex representation is therefore defined over $F$, so $F$ is a splitting field.

The exponent $m$ divides $\mathrm{\mid }G\mathrm{\mid }$, so $\mathbb{Q}({\zeta }_{\mathrm{\mid }G\mathrm{\mid }})$ contains every $m$-th root of unity. Applying the proved assertion to this field gives the final statement. ∎