Monomial Characters and M Groups

For $t\in T$ define $f_t:G\to \mathbb{C}$ by $f_t(t{h}^{\mathrm{\prime }}):=\lambda (h^{\mathrm{\prime }}{)}^{-1}$ for $h^{\mathrm{\prime }}\in H$, and $f_t(g):=0$ for $g\mathrm{\notin }tH$. This is well defined because a decomposition $g=t{h}^{\mathrm{\prime }}$ with $t\in T$, $h^{\mathrm{\prime }}\in H$ is unique ([F3]), and $f_t$ lies in ${\mathrm{Ind}}_H^G{\mathbb{C}}_{\lambda }$: for $g=t{h}^{\mathrm{\prime }}$ and $h\in H$ one has $gh=t(h^{\mathrm{\prime }}h)$ and $\lambda (h^{\mathrm{\prime }}h{)}^{-1}=\lambda (h{)}^{-1}\lambda (h^{\mathrm{\prime }}{)}^{-1}$, while for $g\mathrm{\notin }tH$ also $gh\mathrm{\notin }tH$ and both sides vanish.

For claim 2 put $L:=\mathbb{C}{v}_1$ and $H:=\{x\in G:x\cdot L=L\}$. This $H$ is a subgroup of $G$ containing $1$, and each $x\in H$ acts on the line $L$ by a nonzero scalar; writing $x\cdot z=\lambda (x)z$ for $z\in L$ defines a group homomorphism $\lambda :H\to {\mathbb{C}}^{\times }$, because the action of $G$ on $V$ is a group action. Thus $\lambda$ is a linear character of $H$ and $L$ is a one-dimensional $H$-module.

For claim 2 let $W$ be the span of all lines $x\cdot L=\mathbb{C}(x\cdot v_1)$ with $x\in G$. Each spanning line is one of the basis lines $\mathbb{C}{v}_i$, because the action permutes the basis lines, so $W\subseteq V$; and $W$ is $G$-invariant with $W\mathrm{\ne }0$, since $y\cdot (x\cdot L)=(yx)\cdot L$ and $L\subseteq W$. Irreducibility of $V$ forces $W=V$, so every basis line equals some $x\cdot L$ and the distinct lines $x\cdot L$, $x\in G$, are exactly the $r$ basis lines.

The function $f_t$ of step 1.1 takes the value $1$ at $t$ and the value $0$ at every other element of $T$; hence ${\mathrm{ev}}_T(f_{t_i})$ is the $i$-th standard basis vector of ${⨁}_{i=1}^n{\mathbb{C}}_{\lambda }$. By [F2] the map ${\mathrm{ev}}_T$ is an isomorphism, so $f_1,\dots ,f_n$ is a basis of ${\mathrm{Ind}}_H^G\lambda$ and $n=[G:H]$.

Fix $x\in G$ and $t\in T$. By [A1] there are unique $t^{\mathrm{\prime }}\in T$ and $h\in H$ with $xt=t^{\mathrm{\prime }}h$. For $s\in T$ the value $(x\cdot f_t)(s)=f_t(x^{-1}s)$ is nonzero exactly when $x^{-1}s\in tH$, that is $s\in xtH=t^{\mathrm{\prime }}H$, hence exactly when $s=t^{\mathrm{\prime }}$. At that point $x^{-1}{t}^{\mathrm{\prime }}=t{h}^{-1}$ by the relation $xt=t^{\mathrm{\prime }}h$, so $(x\cdot f_t)(t^{\mathrm{\prime }})=f_t(t{h}^{-1})=\lambda (h^{-1}{)}^{-1}=\lambda (h)\mathrm{\ne }0$.

For claim 2 choose a left transversal $S$ of the stabilizer $H$ from step 1.2. Such an $S$ exists by choosing one representative from each of the finitely many nonempty cosets in the finite group $G$; no axiom of choice is needed. The coset argument in [F3] applies to this $H$ and $S$. The map $S\to \{x\cdot L:x\in G\}$, $t\mapsto t\cdot L$, is a well-defined bijection: it is well defined because $th\cdot L=t\cdot L$ for $h\in H$; it is injective because $t\cdot L=t^{\mathrm{\prime }}\cdot L$ gives $t^{\mathrm{\prime }-1}t\cdot L=L$, that is $t^{\mathrm{\prime }-1}t\in H$, hence $t^{\mathrm{\prime }}=t$ by [F3]; and it is surjective because every $x\in G$ can be written $x=th$ with $t\in S$, $h\in H$, giving $x\cdot L=t\cdot L$. By step 1.3 the translates $L_t:=t\cdot L$, $t\in S$, are exactly the basis lines, and $V={⨁}_{t\in S}{L}_t$, each $u\in V$ having a unique expression $u={\sum }_{t\in S}t\cdot u_t$ with $u_t\in L$.

Thus the matrix of $x$ in the basis $f_1,\dots ,f_n$ has its only possibly nonzero entry in the column indexed by $t$ at the row indexed by $t^{\mathrm{\prime }}$, where $xt\in t^{\mathrm{\prime }}H$, and that entry equals $\lambda (h)\mathrm{\ne }0$. The map $t\mapsto t^{\mathrm{\prime }}$ is a permutation of $T$ by [A1], so every row also receives exactly one nonzero entry, namely from the unique $t$ with $xt\in t^{\mathrm{\prime }}H$. Hence the matrix is monomial with nonzero entries among the values of $\lambda$, which proves claim 1.

Define $\mathrm{\Psi }:V\to {\mathrm{Ind}}_H^{G}L$ by $\mathrm{\Psi }(v)(th):=h^{-1}\cdot u_t$ for $t\in S$, $h\in H$, where $v={\sum }_{t\in S}t\cdot u_t$ is the unique expression of step 2.3. This is well defined by [F3], it satisfies the covariance law $\mathrm{\Psi }(v)(g{h}^{\mathrm{\prime }})=h^{\mathrm{\prime }-1}\cdot \mathrm{\Psi }(v)(g)$ for $g=th$, $h^{\mathrm{\prime }}\in H$, and so lies in ${\mathrm{Ind}}_H^{G}L$ as in [F1]; the assignment $\mathrm{\Psi }$ is $\mathbb{C}$-linear because the coordinates $u_t$ depend linearly on $v$.

For $f\in {\mathrm{Ind}}_H^{G}L$ put $\mathrm{\Phi }(f):={\sum }_{t\in S}t\cdot f(t)$, an element of $V$ by step 2.3, and $\mathrm{\Phi }$ is $\mathbb{C}$-linear. For $v={\sum }_{t\in S}t\cdot u_t$ step 3.2 gives $\mathrm{\Psi }(v)(t)=u_t$, so $\mathrm{\Phi }(\mathrm{\Psi }(v))=v$. Conversely, for $f\in {\mathrm{Ind}}_H^{G}L$ and $t\in S$ the definition of $\mathrm{\Phi }$ gives $\mathrm{\Psi }(\mathrm{\Phi }(f))(t)=f(t)$, and both $\mathrm{\Psi }(\mathrm{\Phi }(f))$ and $f$ satisfy the covariance law [F1], so they agree on $G$; hence $\mathrm{\Psi }$ is surjective and $\mathrm{\Phi }={\mathrm{\Psi }}^{-1}$.

$\mathrm{\Psi }$ is $G$-equivariant. Let $v={\sum }_{t\in S}t\cdot u_t$ and $x\in G$; for each $t\in S$ write uniquely $xt=t^{\mathrm{\prime }}h$ with $t^{\mathrm{\prime }}\in S$, $h\in H$, so that $x\cdot v={\sum }_t(xt)\cdot u_t={\sum }_t{t}^{\mathrm{\prime }}\cdot (\lambda (h)u_t)$ has $L_{t^{\mathrm{\prime }}}$-component $\lambda (h)u_t$. By step 3.2, $\mathrm{\Psi }(v)(x^{-1}{t}^{\mathrm{\prime }})=\mathrm{\Psi }(v)(t{h}^{-1})=\lambda (h)u_t$, while the function $x\cdot \mathrm{\Psi }(v)$ takes the value $\mathrm{\Psi }(v)(x^{-1}{t}^{\mathrm{\prime }})$ at $t^{\mathrm{\prime }}$ by [F1]; as $t$ runs over $S$ so does $t^{\mathrm{\prime }}$, so $\mathrm{\Psi }(x\cdot v)$ and $x\cdot \mathrm{\Psi }(v)$ agree on $S$ and both satisfy the covariance law, hence they agree on $G$.

Steps 4.1 and 4.2 exhibit $\mathrm{\Psi }:V\to {\mathrm{Ind}}_H^{G}L={\mathrm{Ind}}_H^G\lambda$ as a $G$-equivariant $\mathbb{C}$-linear bijection, where $\lambda$ is the linear character of step 1.2, so $V$ is monomial in the sense of [F5]. Together with step 3.1 this proves both assertions.

Restricting $\rho$ to the abelian normal subgroup $A$ gives a representation of $A$ whose irreducible constituents are one-dimensional by [F1], because $\mathbb{C}$ is a splitting field for $A$; thus the constituent $\lambda$ of the restriction is a group homomorphism $\lambda :A\to {\mathbb{C}}^{\times }$.

Apply [F2] with $N=A$ and $\chi ={\chi }_V$: the constituents of the restriction of $V$ to $A$ are exactly the distinct conjugates ${}^g\lambda$ with $g\in G$, each occurring with one positive multiplicity $e$, and by [F3] the conjugate ${}^g\lambda$ equals $\lambda$ precisely when $g\in I_G(\lambda )$. Hence the restriction of $V$ to $A$ is $\lambda$-isotypical, that is all its constituents equal $\lambda$, precisely when $I_G(\lambda )=G$.

Suppose now, for the sake of a contradiction, that $I_G(\lambda )=G$. By step 2.1 the only constituent of the restriction of $V$ to $A$ is $\lambda$, so its character is a positive multiple of $\lambda$; equivalently $\rho (a)=\lambda (a){\mathrm{id}}_V$ for every $a\in A$.

Let $a\in A$ and $g\in G$. Using step 3.1 and [A1], $\rho (g^{-1}ag)=\rho (g{)}^{-1}\rho (a)\rho (g)=\lambda (a){\mathrm{id}}_V=\rho (a)$. Since $\rho$ is faithful by [F5], this gives $g^{-1}ag=a$; as $g$ was arbitrary, $a$ commutes with every element of $G$, so $a\in Z(G)$ by [F6]. Hence $A\subseteq Z(G)$, contradicting the hypothesis $A⊈Z(G)$. Therefore $I_G(\lambda )\mathrm{\ne }G$, and since $A\le I_G(\lambda )\le G$ by [F3], the inertia group $I_G(\lambda )$ is a proper subgroup of $G$.

Since $V$ is irreducible and $\lambda$ is a constituent of its restriction to $A$, the module $V$ lies over $\lambda$. By [F4] the $\lambda$-isotypical component $W=V_{\lambda }$ is an irreducible $I_G(\lambda )$-module lying over $\lambda$ and induction gives $V\cong {\mathrm{Ind}}_{I_G(\lambda )}^{G}W$. As $\lambda$ was an arbitrary constituent of the restriction to $A$, both assertions hold for every such constituent.

Let $K⊴G$ and put $\bar{G}:=G\mathrm{/}K$, ${\bar{G}}_j:=G_{j}K\mathrm{/}K$ for $j=0,\dots ,r$. Each ${\bar{G}}_j$ is a subgroup of $\bar{G}$ containing ${\bar{G}}_{j-1}$ and is normal in $\bar{G}$, because $G_j⊴G$ and $K⊴G$ ([F2], [F3]); moreover ${\bar{G}}_0=K\mathrm{/}K=1$ and ${\bar{G}}_r=G\mathrm{/}K=\bar{G}$. For each $j$ the map $\phi :{\bar{G}}_j\to G_{j}K\mathrm{/}{G}_{j-1}K$, $gK\mapsto g{G}_{j-1}K$, is well defined and a surjective homomorphism with kernel ${\bar{G}}_{j-1}=G_{j-1}K\mathrm{/}K$, so by [F5] the quotient ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ is isomorphic to $G_{j}K\mathrm{/}{G}_{j-1}K$; that group is in turn a quotient of the prime-order group $G_j\mathrm{/}{G}_{j-1}$, since $G_j\to G_{j}K\mathrm{/}{G}_{j-1}K$ is surjective with $G_{j-1}$ in its kernel. Hence every factor ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ has order $1$ or prime.

Now assume that $G$ is nonabelian, so $G=G_r⊈Z(G)$ by [F4], while $G_0=1\subseteq Z(G)$. Let $i\in \{1,\dots ,r\}$ be the least index with $G_i⊈Z(G)$; such an index exists because $G_r⊈Z(G)$, and $i\ge 1$ because $G_0=1$ is central. By minimality $G_{i-1}\subseteq Z(G)$.

By step 1.2 the subgroup $G_{i-1}$ is contained in $Z(G)$, hence in $Z(G_i)$; by [F1] the factor $G_i\mathrm{/}{G}_{i-1}$ has prime order, so $G_i=G_{i-1}⟨x⟩$ for any $x\in G_i\setminus G_{i-1}$ by [A1], and every element of $G_i$ is of the form $g{x}^k$ with $g\in G_{i-1}$. Two such elements $g{x}^k$ and $g^{\mathrm{\prime }}{x}^{k^{\mathrm{\prime }}}$ commute, because $g,g^{\mathrm{\prime }}\in G_{i-1}\subseteq Z(G_i)$ and powers of $x$ commute with each other. Hence $G_i$ is abelian.

Thus $A:=G_i$ is abelian by step 2.1, it is normal in $G$ by hypothesis, and $A=G_i⊈Z(G)$ by the choice of $i$ in step 1.2. This is the asserted abelian normal subgroup of $G$ and proves the first assertion.

Finally assume that $\bar{G}=G\mathrm{/}K$ is nonabelian, so $\bar{G}={\bar{G}}_r⊈Z(\bar{G})$ by [F4] while ${\bar{G}}_0=1\subseteq Z(\bar{G})$. Let $j$ be the least index with ${\bar{G}}_j⊈Z(\bar{G})$; then $j\ge 1$, ${\bar{G}}_{j-1}\subseteq Z(\bar{G})$, and ${\bar{G}}_j\mathrm{\ne }{\bar{G}}_{j-1}$, so by step 1.1 the factor ${\bar{G}}_j\mathrm{/}{\bar{G}}_{j-1}$ has prime order. The computation of step 2.1, applied with the pair $(G_i,G_{i-1})$ replaced by the pair $({\bar{G}}_j,{\bar{G}}_{j-1})$ inside $\bar{G}$, shows in the same way that ${\bar{G}}_j$ is abelian, while ${\bar{G}}_j⊈Z(\bar{G})$ by the choice of $j$. Hence the nonabelian quotient $\bar{G}$ also contains an abelian normal subgroup that is not contained in its center, which is the second assertion.

Since $K⊴G$ by [F6] and $K\le H\le G$, the quotient $H\mathrm{/}K$ is a group and $h\mapsto \bar{h}$ is a homomorphism $H\to \bar{H}$; hence the $H$-action on the inflated module, which is the $\bar{H}$-action through $h\mapsto \bar{h}$ by [F2], is well defined. For $\bar{f}\in {\mathrm{Infl}}_{\bar{G}}^G({\mathrm{Ind}}_{\bar{H}}^{\bar{G}}W)$ define $\mathrm{\Phi }(\bar{f}):G\to W$ by $\mathrm{\Phi }(\bar{f})(g):=\bar{f}(\bar{g})=\bar{f}(\pi (g))$. If $h\in H$, then $\pi (gh)=\bar{g}\bar{h}$ by [F3], so $\mathrm{\Phi }(\bar{f})(gh)=\bar{f}(\bar{g}\bar{h})={\bar{h}}^{-1}\cdot \bar{f}(\bar{g})=h^{-1}\cdot \mathrm{\Phi }(\bar{f})(g)$; hence $\mathrm{\Phi }(\bar{f})\in {\mathrm{Ind}}_H^G({\mathrm{Infl}}_{\bar{H}}^{H}W)$ by [F1].

$\mathrm{\Phi }$ is $\mathbb{C}$-linear, and it is injective: if $\mathrm{\Phi }(\bar{f})=0$, then $\bar{f}(\bar{g})=\mathrm{\Phi }(\bar{f})(g)=0$ for every $g\in G$, and every element of $\bar{G}$ is some $\bar{g}$ by [F3], so $\bar{f}=0$.

$\mathrm{\Phi }$ is surjective. Given $F\in {\mathrm{Ind}}_H^G({\mathrm{Infl}}_{\bar{H}}^{H}W)$, define $\bar{f}:\bar{G}\to W$ by $\bar{f}(\bar{g}):=F(g)$. This is well defined: if $\bar{g}={\bar{g}}^{\mathrm{\prime }}$, then $g^{\mathrm{\prime }}=gk$ with $k\in K\le H$ by [F3], so $F(g^{\mathrm{\prime }})=F(gk)=k^{-1}\cdot F(g)=F(g)$ by [A1] and [F2]. Moreover $\bar{f}$ is $\bar{H}$-covariant, since for $\bar{h}\in \bar{H}$ one has $\bar{f}(\bar{g}\bar{h})=F(gh)=h^{-1}\cdot F(g)={\bar{h}}^{-1}\cdot \bar{f}(\bar{g})$ for any $h\in H$ with image $\bar{h}$, so $\bar{f}\in {\mathrm{Ind}}_{\bar{H}}^{\bar{G}}W$ and $\mathrm{\Phi }(\bar{f})=F$.

$\mathrm{\Phi }$ is $G$-equivariant: for $x,g\in G$ and $\bar{f}$ as in step 1.1, $\mathrm{\Phi }(x\cdot \bar{f})(g)=(x\cdot \bar{f})(\bar{g})=\bar{f}({\bar{x}}^{-1}\bar{g})=\bar{f}(\overline{x^{-1}g})=\mathrm{\Phi }(\bar{f})(x^{-1}g)=(x\cdot \mathrm{\Phi }(\bar{f}))(g)$, using [F3] and the action rules of [F1].

Steps 2.1, 2.2 and 2.3 show that $\mathrm{\Phi }$ is a $G$-equivariant $\mathbb{C}$-linear bijection, which is the asserted isomorphism. For the final clause let $\bar{H}\le \bar{G}$, put $H={\pi }^{-1}(\bar{H})$, let $W$ be one-dimensional with ${\mathrm{Ind}}_{\bar{H}}^{\bar{G}}W$ irreducible, and note $K\le H\le G$ and $H\mathrm{/}K=\bar{H}$. By the isomorphism just proved the inflation of ${\mathrm{Ind}}_{\bar{H}}^{\bar{G}}W$ is isomorphic to ${\mathrm{Ind}}_H^G({\mathrm{Infl}}_{\bar{H}}^{H}W)$, and ${\mathrm{Infl}}_{\bar{H}}^{H}W$ is one-dimensional because it has the same underlying space as $W$ by [F2]; the inflation is irreducible because inflation preserves irreducibility in both directions by [F4] and ${\mathrm{Ind}}_{\bar{H}}^{\bar{G}}W$ is irreducible. Hence it is a monomial irreducible $G$-module induced from the one-dimensional $H$-module ${\mathrm{Infl}}_{\bar{H}}^{H}W$ by [F5].