Artin Induction and Rational Characters

By [F1], write $x={\sum }_{i=1}^r{\mathrm{Ind}}_{C_i}^G({\theta }_i)$ with each $C_i\le G$ cyclic and each ${\theta }_i\in R(C_i)$.

Multiplying by $\psi$ and applying [F2] termwise gives $x\psi ={\sum }_{i=1}^r({\mathrm{Ind}}_{C_i}^G({\theta }_i))\psi ={\sum }_{i=1}^r{\mathrm{Ind}}_{C_i}^G\text{\hspace{0.17em}⁣}\left({\theta }_i{\mathrm{Res}}_{C_i}^G\psi \right)$. Each factor ${\theta }_i{\mathrm{Res}}_{C_i}^G\psi$ lies in $R(C_i)$, so every summand on the right again belongs to $I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$ by [F1].

Therefore $x\psi \in I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$. Since $x\in I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$ and $\psi \in R(G)$ were arbitrary, $I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$ is an ideal of $R(G)$.

Let $E:=⟨c⟩$. By [F1], for every subgroup $D\le C$ one has $c\in D$ if and only if $E\le D$. Choose a generator $g$ of $C$. For each divisor $m$ of $\mathrm{\mid }C\mathrm{\mid }$, the subgroup $⟨g^{\mathrm{\mid }C\mathrm{\mid }\mathrm{/}m}⟩$ has order $m$, so cyclic subgroups of $C$ are uniquely indexed by the divisors of $\mathrm{\mid }C\mathrm{\mid }$. If $H,K\le C$ have orders $m,n$, then $H\le K$ holds exactly when $m\mid n$, because with the displayed indexing one has $⟨g^{\mathrm{\mid }C\mathrm{\mid }\mathrm{/}m}⟩\le ⟨g^{\mathrm{\mid }C\mathrm{\mid }\mathrm{/}n}⟩$ exactly when $\mathrm{\mid }C\mathrm{\mid }\mathrm{/}n$ divides $\mathrm{\mid }C\mathrm{\mid }\mathrm{/}m$.

The quotient $\mathrm{\mid }C:E\mathrm{\mid }$ is a positive integer. If $E\le D\le C$, then $\mathrm{\mid }C:E\mathrm{\mid }=\mathrm{\mid }C:D\mathrm{\mid }\mathrm{\mid }D:E\mathrm{\mid }$, so the index $\mathrm{\mid }C:D\mathrm{\mid }$ divides $\mathrm{\mid }C:E\mathrm{\mid }$. Conversely, if $d$ is a positive divisor of $\mathrm{\mid }C:E\mathrm{\mid }$, put $m:=\mathrm{\mid }C\mathrm{\mid }\mathrm{/}d$. Then $\mathrm{\mid }E\mathrm{\mid }$ divides $m$, so by step 1.1 the unique subgroup $D\le C$ of order $m$ contains $E$ and satisfies $\mathrm{\mid }C:D\mathrm{\mid }=d$. Therefore the positive divisors of $\mathrm{\mid }C:E\mathrm{\mid }$ are exactly the indices $\mathrm{\mid }C:D\mathrm{\mid }$ for subgroups $D$ with $E\le D\le C$. Hence [F3] gives ${\sum }_{E\le D\le C}\mu (\mathrm{\mid }C:D\mathrm{\mid })={\sum }_{d\mid \mathrm{\mid }C:E\mathrm{\mid }}\mu (d)=\{\begin{array}{ll}1,& \mathrm{\mid }C:E\mathrm{\mid }=1,\\ 0,& \mathrm{\mid }C:E\mathrm{\mid }>1.\end{array}$ Since $\mathrm{\mid }C:E\mathrm{\mid }=1$ holds exactly when $E=C$, this is $1_{\{⟨c⟩=C\}}$.

Because $C$ is abelian, $x^{-1}cx=c$ for every $x\in C$, so [F2] becomes ${\mathrm{Ind}}_D^C{1}_D(c)=\mathrm{\mid }C\mathrm{\mid }\mathrm{/}\mathrm{\mid }D\mathrm{\mid }$ when $c\in D$ and ${\mathrm{Ind}}_D^C{1}_D(c)=0$ when $c\mathrm{\notin }D$. Using step 1.1, the right-hand side of the claimed formula evaluates at $c$ to ${\sum }_{D\le C}\mu (\mathrm{\mid }C:D\mathrm{\mid })\mathrm{\mid }D\mathrm{\mid }{\mathrm{Ind}}_D^C{1}_D(c)=\mathrm{\mid }C\mathrm{\mid }{\sum }_{E\le D\le C}\mu (\mathrm{\mid }C:D\mathrm{\mid })=\mathrm{\mid }C\mathrm{\mid }{1}_{\{⟨c⟩=C\}}$, which is exactly ${\eta }_C(c)$ by step 2.1.

The equality of step 3.1 holds for every $c\in C$, so the two class functions are equal.

For each cyclic subgroup $C\le G$, let ${\eta }_C$ be the class function from [F1], and define $f:={\sum }_{\begin{array}{c}C\le G\\ C\text{ cyclic}\end{array}}{\mathrm{Ind}}_C^G{\eta }_C$. The sum is finite because a finite group has only finitely many subgroups.

By [F2], for each cyclic $C\le G$ one has ${\mathrm{Ind}}_C^G{\eta }_C(g)=\frac{1}{\mathrm{\mid }C\mathrm{\mid }}{\sum }_{\begin{array}{c}x\in G\\ x^{-1}gx\in C\end{array}}{\eta }_C(x^{-1}gx)$. Fix $x\in G$. Among all cyclic subgroups $C\le G$, exactly one of them can make the summand indexed by $x$ nonzero, namely $C=⟨x^{-1}gx⟩$; for that subgroup, the value of ${\eta }_C$ is $\mathrm{\mid }C\mathrm{\mid }$. Therefore the double sum defining $f(g)$ contributes exactly $1$ for each $x\in G$, so $f(g)=\mathrm{\mid }G\mathrm{\mid }$.

Step 2.1 holds for every $g\in G$, hence $f=\mathrm{\mid }G\mathrm{\mid }{1}_G$ as class functions. Expanding each ${\eta }_C$ by [F1] and then using [F3] to replace ${\mathrm{Ind}}_C^G({\mathrm{Ind}}_D^C{1}_D)$ by ${\mathrm{Ind}}_D^G{1}_D$ expresses $f$ as an integral linear combination of characters ${\mathrm{Ind}}_D^G{1}_D$ with $D$ cyclic. Thus $\mathrm{\mid }G\mathrm{\mid }{1}_G$ has the required form.

By [F2], choose cyclic subgroups $C_i\le G$ and integers $a_i$ with $\mathrm{\mid }G\mathrm{\mid }{1}_G={\sum }_i{a}_i{\mathrm{Ind}}_{C_i}^G{1}_{C_i}$. Multiplying by $x$ in $R(G)$ gives $\mathrm{\mid }G\mathrm{\mid }x={\sum }_i{a}_i\left({\mathrm{Ind}}_{C_i}^G{1}_{C_i}\right)x$. Because [F1] makes $I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$ an ideal containing each ${\mathrm{Ind}}_{C_i}^G{1}_{C_i}$, this already shows $\mathrm{\mid }G\mathrm{\mid }x\in I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$.

Applying [F3] to each summand of step 1.1 yields $\mathrm{\mid }G\mathrm{\mid }x={\sum }_i{a}_i{\mathrm{Ind}}_{C_i}^G({\mathrm{Res}}_{C_i}^{G}x)$. By [A1], each ${\mathrm{Res}}_{C_i}^{G}x$ again lies in the rational representation ring of the cyclic subgroup $C_i$. This identity is an explicit expression of $\mathrm{\mid }G\mathrm{\mid }x$ as a sum of characters induced from cyclic subgroups.

Dividing the identity of step 2.1 by $\mathrm{\mid }G\mathrm{\mid }$ expresses $x$ as a rational linear combination of characters induced from cyclic subgroups of $G$. This is claim 1.

Conversely, if $x$ is any rational linear combination of characters induced from cyclic subgroups, then multiplying by a common positive denominator places that multiple of $x$ in $I_{\mathrm{c}\mathrm{y}\mathrm{c}}(G)$. Step 1.1 shows that one may take the specific controlled denominator $\mathrm{\mid }G\mathrm{\mid }$, so claims 1 and 2 are equivalent.

Let $C\le G$ be cyclic, and let $y\in R_{\mathbb{Q}}(C)$. By The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions, $y$ is an integral linear combination of characters of finite-dimensional $\mathbb{Q}$-representations of $C$, so it is enough to prove the next claim for one such character and then extend by linearity. Let $\rho :C\to \mathrm{GL}(V)$ be a finite-dimensional $\mathbb{Q}$-representation with character $y$, let $D\le C$, and let $d,d^{\mathrm{\prime }}$ be generators of $D$. Then $d^{\mathrm{\prime }}=d^u$ for some integer $u$ coprime to $\mathrm{\mid }D\mathrm{\mid }$. Because $d^{\mathrm{\mid }D\mathrm{\mid }}=1$, the matrix $\rho (d)$ has rational entries and satisfies $\rho (d{)}^{\mathrm{\mid }D\mathrm{\mid }}=I$, so over $\mathbb{C}$ it is diagonalizable with eigenvalues among the $\mathrm{\mid }D\mathrm{\mid }$-th roots of unity. Because the characteristic polynomial of $\rho (d)$ lies in $\mathbb{Q}[t]$, those eigenvalues occur with multiplicities stable under the Galois automorphism $\zeta \mapsto {\zeta }^u$ of the cyclotomic field. Thus the multisets of eigenvalues of $\rho (d)$ and $\rho (d^{\mathrm{\prime }})=\rho (d{)}^u$ agree, so $y(d^{\mathrm{\prime }})=\mathrm{tr}\rho (d^{\mathrm{\prime }})=\mathrm{tr}\rho (d)=y(d)$. Therefore every $y\in R_{\mathbb{Q}}(C)$ is constant on the set of generators of each subgroup of $C$. The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions

For each subgroup $D\le C$, choose a generator $d_D$ of $D$ when $D\mathrm{\ne }1$ and put $d_1=e$. Writing $b_D:=y(d_D)\in \mathbb{Q}$, step 1.1 gives $\mathrm{\mid }C\mathrm{\mid }y={\sum }_{D\le C}{b}_D{\tilde{\eta }}_D,$ where ${\tilde{\eta }}_D(c)=\mathrm{\mid }C\mathrm{\mid }$ when $⟨c⟩=D$ and ${\tilde{\eta }}_D(c)=0$ otherwise.

For each $D\le C$, one has ${\tilde{\eta }}_D={\mathrm{Ind}}_D^C{\eta }_D$. Indeed, if $c\in C$ generates $D$, then $c\in D$ and $C$ is abelian, so [F6] gives ${\mathrm{Ind}}_D^C{\eta }_D(c)=\mathrm{\mid }C\mathrm{\mid }$; if $⟨c⟩\mathrm{\ne }D$, then either $c\mathrm{\notin }D$ or $c$ does not generate $D$, and the induced value is $0$. Using [F4] inside $D$ and [F5] to induce further to $C$, each ${\tilde{\eta }}_D$ is therefore an integral linear combination of the permutation characters $U_E^C:={\mathrm{Ind}}_E^C{1}_E$ with $E\le D$. Thus every $y\in R_{\mathbb{Q}}(C)$ is a rational linear combination of the $U_E^C$.

Suppose now that $m_D(x)=0$ for every cyclic subgroup $D\le G$. Let $C\le G$ be cyclic. For each subgroup $E\le C$, the class functions ${\mathrm{Res}}_C^{G}x$ and $U_E^C$ are rational-valued, so [F1] makes their inner product symmetric. Using that symmetry and then Frobenius reciprocity [F2], we get $⟨{\mathrm{Res}}_C^{G}x,U_E^C{⟩}_C=⟨U_E^C,{\mathrm{Res}}_C^{G}x{⟩}_C=⟨1_E,{\mathrm{Res}}_E^{G}x{⟩}_E=⟨{\mathrm{Res}}_E^{G}x,1_E{⟩}_E=m_E(x)=0.$

By step 3.1, every $y\in R_{\mathbb{Q}}(C)$ is a rational linear combination of the class functions $U_E^C$, and both ${\mathrm{Res}}_C^{G}x$ and $y$ are rational-valued by The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions. Thus the $\mathbb{Q}$-bilinearity of [F1] on rational-valued class functions combines with step 4.1 to give $⟨{\mathrm{Res}}_C^{G}x,y{⟩}_C=0$ for every $y\in R_{\mathbb{Q}}(C)$. Applying [F2] again, we get $⟨{\mathrm{Ind}}_C^{G}y,x{⟩}_G=0$ for every cyclic subgroup $C\le G$ and every $y\in R_{\mathbb{Q}}(C)$.

By [F7], the rational virtual character $x$ is itself a rational linear combination of the induced characters ${\mathrm{Ind}}_C^{G}y$ from step 5.1. Linearity of [F1] in the first argument and step 5.1 therefore give $⟨x,x{⟩}_G=0$. The positive definiteness in [F1] forces $x=0$. Therefore the map $x\mapsto (m_C(x){)}_{(C)}$ is injective.

For an honest $\mathbb{Q}$-representation $V$, the equality $m_C({\chi }_V)=\dim V^C$ is exactly [F3].

Choose representatives $C_1,\dots ,C_r$ of the conjugacy classes of cyclic subgroups of $G$. By [F1], the group $R_{\mathbb{Q}}(G)$ injects into ${\mathbb{Q}}^r$, so its rank is at most $r$.

For each $i$, let $U_i:={\mathrm{Ind}}_{C_i}^G{1}_{C_i}$, viewed as an element of $R_{\mathbb{Q}}(G)$. Suppose that $a_1{U}_1+\cdots +a_r{U}_r=0$ with $a_i\in \mathbb{Q}$. Reorder the representatives so that $\mathrm{\mid }{C}_1\mathrm{\mid }\ge \cdots \ge \mathrm{\mid }{C}_r\mathrm{\mid }$, and choose $i$ minimal with $a_i\mathrm{\ne }0$. Let $g_i$ be a generator of $C_i$.

If $U_j(g_i)\mathrm{\ne }0$, then [F2] and [F3] show that $g_i$ fixes some coset $x{C}_j$, so $x^{-1}{g}_{i}x\in C_j$. Therefore a conjugate of the cyclic subgroup $C_i=⟨g_i⟩$ lies in $C_j$, which implies $\mathrm{\mid }{C}_i\mathrm{\mid }\le \mathrm{\mid }{C}_j\mathrm{\mid }$. In the relation from step 1.2, every index $j<i$ has $a_j=0$ by minimality of $i$, so only $j\ge i$ can contribute. For such $j$, the ordering gives $\mathrm{\mid }{C}_j\mathrm{\mid }\le \mathrm{\mid }{C}_i\mathrm{\mid }$, hence $\mathrm{\mid }{C}_j\mathrm{\mid }=\mathrm{\mid }{C}_i\mathrm{\mid }$. A subgroup of $C_j$ with the same finite order as $C_j$ must equal $C_j$, so the conjugate of $C_i$ lying in $C_j$ is all of $C_j$. Thus $C_j$ is conjugate to $C_i$, and because $C_1,\dots ,C_r$ were chosen as distinct conjugacy-class representatives, this forces $j=i$. On the other hand, $g_i$ fixes the coset $C_i$ itself, so [F2] and [F3] give $U_i(g_i)>0$. Evaluating the relation from step 1.2 at $g_i$ therefore yields $a_i{U}_i(g_i)=0$, a contradiction. Thus the $U_i$ are linearly independent.

Step 2.1 gives $r$ linearly independent elements of $R_{\mathbb{Q}}(G)$, while step 1.1 shows that the rank is at most $r$. Hence $\mathrm{rank}{R}_{\mathbb{Q}}(G)=r$, the number of cyclic conjugacy classes.

Choose cyclic subgroups $C_i\le G$ and integers $a_i$ with $\mathrm{\mid }G\mathrm{\mid }{1}_G={\sum }_i{a}_i{\mathrm{Ind}}_{C_i}^G{1}_{C_i}$ as in [F1]. Multiplying by $x$ in $R(G){\otimes }_{\mathbb{Z}}\mathbb{Q}$ gives $\mathrm{\mid }G\mathrm{\mid }x={\sum }_i{a}_i\left({\mathrm{Ind}}_{C_i}^G{1}_{C_i}\right)x$.

Applying [F2] termwise to the identity of step 1.1 yields $\mathrm{\mid }G\mathrm{\mid }x={\sum }_i{a}_i{\mathrm{Ind}}_{C_i}^G({\mathrm{Res}}_{C_i}^{G}x)$. By hypothesis each ${\mathrm{Res}}_{C_i}^{G}x$ lies in $R(C_i)$, so [A1] places every induced summand in $R(G)$. Since $R(G)$ is closed under integral linear combinations, the whole right-hand side lies in $R(G)$.

Therefore $\mathrm{\mid }G\mathrm{\mid }x\in R(G)$, as claimed.