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 $\chi (g)\equiv \chi (g{)}^{p^n}\equiv \chi (g_{p^{\mathrm{\prime }}}{)}^{p^n}\equiv \chi (g_{p^{\mathrm{\prime }}})(\mathrm{m}\mathrm{o}\mathrm{d}p)$. ∎

Mackey's formula expresses ${\mathrm{Res}}_H^G{\mathrm{Ind}}_K^G{1}_K$ as a sum of permutation characters induced from $H\cap xK{x}^{-1}$. Those intersections are hyperelementary by subgroup closure, and induction back to $G$ shows that products of the displayed generators stay in $P({\mathcal{H}}^{\mathrm{\prime }})$.

If $1_H\in I_{\mathcal{E}}(H)$, 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 some $A_x=n_x\mathbb{Z}$ has $n_x>1$; any prime divisor $p$ of $n_x$ has the stated property. Permutation characters are integer-valued fixed-point counts, so their span is a subring of this form. ∎

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 $\psi (g)\equiv \psi (r_0)\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}p)$.

Assume first that $n\ge 1$ 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 $g$. Hence the integer-valued class function $l({\psi }^e-1_G)$ is pointwise divisible by $\mathrm{\mid }G\mathrm{\mid }$. The cyclic-generator identity The generator-indicator class function of a cyclic group is obtained by Mobius inversion, followed by the projection formula, shows that $\mathrm{\mid }G\mathrm{\mid }$ times any integer-valued class function belongs to the $A$-span of inductions from cyclic subgroups. Those subgroups are $p$-elementary, so $l({\psi }^e-1_G)$ lies 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.

The cyclotomic ring $A$ is a finite free $\mathbb{Z}$-module and $A\mathrm{/}\mathbb{Z}$ is torsion-free, so choose a $\mathbb{Z}$-basis of $A$ containing $1$. Expand the relation from step 5.1 in this basis and take its coefficient of $1$. Since all inducing characters there lie in integral character rings, this yields $l{1}_G\in I_{{\mathcal{E}}_p}(G)$. If $n=0$, then $l=\mathrm{\mid }G\mathrm{\mid }$ and the same integral relation follows directly from the cyclic-generator identity; cyclic subgroups are $p$-elementary in this case. ∎

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 decomposes $V{\mathrm{\mid }}_A$ into 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 $V$; faithfulness would then make $A$ central, contrary to hypothesis. Thus $G_{\lambda }<G$, and the direct sum of its translates identifies $V$ with the induction of its $\lambda$-isotypical component. ∎

Let $i$ be maximal with $G_i$ abelian. If $i=r$, $G$ is abelian, so $i<r$. By maximality there is $g\in G$ not commuting with some $a\in G_i$.

Since $G_i◃G$, it is an abelian normal subgroup, while the chosen $g,a$ show that it is not central. ∎

Factor $\chi$ through an irreducible character $\bar{\chi }$ of $G\mathrm{/}N$. By hypothesis choose $\bar{H}\le G\mathrm{/}N$ and a linear $\bar{\lambda }$ with $\bar{\chi }={\mathrm{Ind}}_{\bar{H}}^{G\mathrm{/}N}\bar{\lambda }$.

Let $H$ be the inverse image of $\bar{H}$ and inflate $\bar{\lambda }$ to a linear $\lambda$ of $H$. Compatibility of induction with quotient inflation gives $\chi ={\mathrm{Ind}}_H^G\lambda$. ∎

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 the monomiality theorem 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. ∎