Monomial Characters and M Groups - Examples

By [F1] the group $G=A⋊⟨s⟩$ has order $2n$, the subgroup $A=⟨r⟩\cong C_n$ is cyclic of index $[G:A]=2$ and normal, $s^2=1$, and $sa{s}^{-1}=a^{-1}$ for $a\in A$; in particular $s$ has order $2$ modulo $A$ and $G\mathrm{/}A\cong C_2$.

$A$ is cyclic, hence abelian, so by [F2] every irreducible complex character of $A$ is one-dimensional, i.e. a group homomorphism $A\to {\mathbb{C}}^{\times }$; moreover $\mathbb{C}$ is a splitting field for every finite group.

By [F7] a linear character of $G$ is monomial: taking $H=G$ and the one-dimensional module $L$ affording it, evaluation at $1$ is an isomorphism ${\mathrm{Ind}}_G^{G}L\to L$, so $\chi ={\mathrm{Ind}}_G^G\chi$ has the monomial form of [F8]; conversely, if $\chi ={\mathrm{Ind}}_H^G\lambda$ with $\lambda$ linear then $\chi (1)=[G:H]\lambda (1)=[G:H]$, so a monomial character of $G$ is linear exactly when it is induced from $G$ itself.

The homomorphisms ${\lambda }_k:A\to {\mathbb{C}}^{\times }$, $k\in \mathbb{Z}$, defined by ${\lambda }_k(r^j):={\zeta }^{jk}$, are well defined because ${\zeta }^n=1$, and ${\lambda }_k(r^{j+l})={\zeta }^{(j+l)k}={\lambda }_k(r^j){\lambda }_k(r^l)$; for $0\le k<n$ they are pairwise distinct, since ${\zeta }^k={\lambda }_k(r)\mathrm{\ne }{\lambda }_{k^{\mathrm{\prime }}}(r)={\zeta }^{k^{\mathrm{\prime }}}$ for $k\mathrm{\ne }{k}^{\mathrm{\prime }}$ by [F11]. Conversely, if $\lambda :A\to {\mathbb{C}}^{\times }$ is any homomorphism then $\lambda (r{)}^n=\lambda (r^n)=1$, so by [F11] there is $k$ with $\lambda (r)={\zeta }^k$, whence $\lambda (r^j)=\lambda (r{)}^j={\zeta }^{jk}={\lambda }_k(r^j)$ for all $j$ and $\lambda ={\lambda }_k$. Hence $\mathrm{Irr}(A)=\{{\lambda }_k:0\le k<n\}$ consists of exactly $n$ distinct linear characters by step 1.2, and ${\lambda }_k={\lambda }_{k^{\mathrm{\prime }}}$ exactly when $k\equiv k^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}n)$.

The claim that $D_n$ is supersolvable also uses a genuine normal series with prime-order factors. Write $n=p_1\cdots p_t$ with primes repeated according to multiplicity and let $A_i$ be the unique subgroup of the cyclic group $A$ of order $p_1\cdots p_i$, with $A_0=1$ and $A_t=A$. Each $A_i$ is characteristic in $A$ and therefore normal in $D_n$ because $A⊴D_n$ by step 1.1; each $A_i\mathrm{/}{A}_{i-1}$ has prime order $p_i$, and $D_n\mathrm{/}A$ has order two. Hence $1=A_0◃\cdots ◃A_t=A◃D_n$ is a normal prime-factor series, including $n=1$ when $t=0$, as required by Finite supersolvable groups are M-groups.

For each $k$ the conjugate ${}^s{\lambda }_k$ is ${\lambda }_{-k}$: by [F3] and the inversion action of step 1.1, $({}^s{\lambda }_k)(r^j)={\lambda }_k(s^{-1}{r}^{j}s)={\lambda }_k(r^{-j})={\zeta }^{-jk}={\lambda }_{-k}(r^j)$ for all $j$, so the two homomorphisms of $A$ agree on the generator $r$ of $A$. By step 2.1 the orbit of ${\lambda }_k$ under the action of $G$ on $\mathrm{Irr}(A)$ is therefore $\{{\lambda }_k,{\lambda }_{-k}\}$, of size $1$ exactly when $k\equiv -k(\mathrm{m}\mathrm{o}\mathrm{d}n)$.

The restriction of $\chi$ to the normal subgroup $A$ is a nonzero finite-dimensional $A$-module, so it has an irreducible $A$-submodule, whose character is some ${\lambda }_k\in \mathrm{Irr}(A)$ by step 2.1; this ${\lambda }_k$ occurs in ${\mathrm{Res}}_A^G\chi$, that is, $\chi \in \mathrm{Irr}(G\mid {\lambda }_k)$ in the notation of [F3].

Fix $k$. By [F3] the inertia group $I:=I_G({\lambda }_k)$ contains $A$ and is contained in $G$; by [F6] applied to $A\le I\le G$ one has $2=[G:A]=[G:I][I:A]$, so $[I:A]\in \{1,2\}$ and $I=A$ or $I=G$. By steps 1.1 and 3.1 the equality $I=G$ holds exactly when $s\in I$, i.e. exactly when ${\lambda }_k={\lambda }_{-k}$, which by step 2.1 is exactly the condition $n\mid 2k$; equivalently ${\lambda }_k(r{)}^2=1$. Hence the number of $s$-fixed characters of $A$ is the number of solutions of $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$, which by [F12] (with $a=2$, $b=0$) is $f=\gcd (2,n)\mathrm{.}$

Case $I_G({\lambda }_k)=A$. By [F4] applied to $N=A$ and $\theta ={\lambda }_k$, the irreducible characters of $G$ lying over ${\lambda }_k$ are exactly the ${\mathrm{Ind}}_A^G\psi$ with $\psi \in \mathrm{Irr}(A\mid {\lambda }_k)$. Every $\psi \in \mathrm{Irr}(A)$ is linear by step 1.2, so ${\mathrm{Res}}_A^A\psi =\psi$ occurs in itself, and $\psi$ lies over ${\lambda }_k$ exactly when $\psi ={\lambda }_k$, by the distinctness in step 2.1. Hence $\chi ={\mathrm{Ind}}_A^G{\lambda }_k$ with ${\lambda }_k$ linear: $\chi$ is monomial by [F8], and $\chi (1)=[G:A]{\lambda }_k(1)=2$ by [F7] and step 1.1.

Case $I_G({\lambda }_k)=G$: then ${\lambda }_k$ is $s$-fixed. By [F5] with $I=G$, ${\mathrm{Res}}_A^G\chi =e{\lambda }_k$ for the positive integer $e=\chi (1)\mathrm{/}[G:G]{\lambda }_k(1)=\chi (1)$, that is, $\chi (a)=\chi (1){\lambda }_k(a)$ for all $a\in A$; and by step 4.1 the $s$-fixedness says ${\lambda }_k(r{)}^2=1$, hence ${\lambda }_k(r^2)=1$ and ${\lambda }_k(a{)}^2=1$ for every $a\in A$.

Put $K:=\ker {\lambda }_k\le A$. Then $K$ is normal in $G$: for $a\in K$ and $g\in G$ one has ${\lambda }_k(g^{-1}ag)={\lambda }_k(a)=1$, because $I_G({\lambda }_k)=G$ fixes ${\lambda }_k$, so $g^{-1}ag\in K$. Moreover $K\le \ker \chi$: for $a\in K$ the formula of step 5.2 gives $\chi (a)=\chi (1){\lambda }_k(a)=\chi (1)$, so $a\in \ker \chi$ by [F9].

For example, when $n=4$ the subgroup $H=⟨r^2,s⟩\cong C_2\times C_2$ is normal of index two in $D_4$ and differs from $A=⟨r⟩$. Define its linear character $\nu$ by $\nu (r^2)=-1$ and $\nu (s)=1$. Conjugation by $r$ sends $s$ to $r^{2}s$, so ${}^r\nu (s)=\nu (r^{-1}sr)=\nu (r^{2}s)=-1\mathrm{\ne }\nu (s)$ and $I_{D_4}(\nu )=H$. Since $H$ is abelian, its only irreducible character lying over $\nu$ is $\nu$ itself; [F4] therefore makes ${\mathrm{Ind}}_H^{D_4}\nu$ an irreducible character of degree $[D_4:H]=2$ by [F7]. Thus the degree-two character of $D_4$ has an inducing subgroup other than $A$.

In the quotient $\bar{G}:=G\mathrm{/}K$ the images $\bar{r},\bar{s}$ generate $\bar{G}$ (they are the images of the generators $r,s$ of $G$), and they commute: ${\bar{r}}^2=\overline{r^2}=\bar{1}$ because $r^2\in K$ by step 5.2, likewise ${\bar{s}}^2=\bar{1}$, and $\bar{s}\bar{r}{\bar{s}}^{-1}=\overline{sr{s}^{-1}}=\overline{r^{-1}}={\bar{r}}^{-1}=\bar{r}$, so $\bar{s}\bar{r}=\bar{r}\bar{s}$. A group generated by two commuting elements is abelian, so $\bar{G}$ is a finite abelian group.

Let $\rho$ be a representation affording $\chi$, so $\ker \rho =\ker \chi$ by [F9] and $K\subseteq \ker \rho$ by step 6.1. By [F10] the representation $\rho$ factors through $\bar{G}=G\mathrm{/}K$ and $V$ stays irreducible over $\bar{G}$; so $\chi (1)=\dim V=\psi (1)$ for some $\psi \in \mathrm{Irr}(\bar{G})$. Since $\bar{G}$ is finite abelian by step 7.1 and $\mathbb{C}$ is a splitting field, [F2] gives $\psi (1)=1$, hence $\chi (1)=1$: in this case $\chi$ is a linear character of $G$. By step 1.3 $\chi$ is monomial, and since $\chi (1)=1$ and $\chi \in \mathrm{Irr}(G\mid {\lambda }_k)$ with ${\lambda }_k$ of degree $1$, its restriction is ${\mathrm{Res}}_A^G\chi ={\lambda }_k$ (a degree-$1$ character occurring in a degree-$1$ character), so $\chi$ extends ${\lambda }_k$.

The linear characters of $G$ are exactly the $2f$ extensions of the $f$ $s$-fixed characters of $A$ found in step 4.1. Indeed, if ${\lambda }_k$ is $s$-fixed, i.e. ${\lambda }_k(r{)}^2=1$ by step 4.1, then for $\epsilon \in \{1,-1\}$ the formula ${\chi }_{k,\epsilon }(r^j{s}^m):={\lambda }_k(r{)}^j{\epsilon }^m(0\le j<n,m\in \{0,1\})$ is well defined by the uniqueness of the normal form [F1], and it is multiplicative: by [F1] the product of $r^j{s}^m$ and $r^{j^{\mathrm{\prime }}}{s}^{m^{\mathrm{\prime }}}$ is $r^{j+(-1{)}^m{j}^{\mathrm{\prime }}}{s}^{m+m^{\mathrm{\prime }}}$, and ${\chi }_{k,\epsilon }$ of that product is ${\lambda }_k(r{)}^j({\lambda }_k(r{)}^{(-1{)}^m}{)}^{j^{\mathrm{\prime }}}{\epsilon }^{m+m^{\mathrm{\prime }}}$, which equals ${\lambda }_k(r{)}^j{\epsilon }^m\cdot {\lambda }_k(r{)}^{j^{\mathrm{\prime }}}{\epsilon }^{m^{\mathrm{\prime }}}$ because ${\lambda }_k(r{)}^{-j^{\mathrm{\prime }}}={\lambda }_k(r{)}^{j^{\mathrm{\prime }}}$ and ${\epsilon }^2=1$; so ${\chi }_{k,\epsilon }$ is a linear character with ${\mathrm{Res}}_A^G{\chi }_{k,\epsilon }={\lambda }_k$ and ${\chi }_{k,\epsilon }(s)=\epsilon$, and ${\chi }_{k,1}\mathrm{\ne }{\chi }_{k,-1}$. Conversely a linear character $\chi$ of $G$ restricts to an $s$-fixed ${\lambda }_k$ by step 2.1, since ${\lambda }_k(r)=\chi (r)=\chi (sr{s}^{-1})=\chi (r{)}^{-1}$, and $\chi$ is determined by ${\lambda }_k$ together with $\chi (s)\in \{1,-1\}$ (as $\chi (s{)}^2=\chi (s^2)=1$), hence equals ${\chi }_{k,\epsilon }$ for one $\epsilon$; by step 8.1 every linear character arises in this way from an $s$-fixed ${\lambda }_k$. Distinct pairs $(k,\epsilon )$ give distinct characters, because their restrictions to $A$ differ or their values at $s$ differ, so there are exactly $2f$ linear characters in total.

Every $\chi \in \mathrm{Irr}(G)$ is monomial: if $I_G({\lambda }_k)=A$ it is ${\mathrm{Ind}}_A^G{\lambda }_k$ by step 5.1, and if $I_G({\lambda }_k)=G$ it is linear by step 8.1, hence monomial by step 1.3. Therefore $D_n=\mathrm{Dih}(C_n)$ is an $M$-group by [F8].

The degree-$2$ irreducible characters are exactly the characters of step 5.1, that is, the ${\mathrm{Ind}}_A^G{\lambda }_k$ for ${\lambda }_k$ that is not $s$-fixed. Each such ${\lambda }_k$ has orbit $\{{\lambda }_k,{\lambda }_{-k}\}$ of size two by step 3.1, distinct orbits give disjoint sets $\mathrm{Irr}(G\mid {\lambda }_k)$ by [F4], and by step 4.1 every orbit of size two arises from a character that is not $s$-fixed. Since by step 4.1 exactly $f$ of the $n$ characters of $A$ are $s$-fixed and the remaining $n-f$ split into two-element orbits, there are exactly $\frac{n-f}{2}$ irreducible characters of degree $2$, each induced from the cyclic index-two subgroup $A$; the remaining irreducible characters are the $2f$ linear ones of step 9.1. This constructs an inducing subgroup for every irreducible character without asserting that the subgroup is unique.

The degenerate cases are covered by the same statements. For $n=1$ the group is $G\cong C_2$ by [F1], $f=\gcd (2,1)=1$ and $\frac{n-f}{2}=0$, and step 9.1 returns the two linear characters of $C_2$ and no character of degree $2$; for $n=2$ one has $G\cong C_2\times C_2$ by [F1], $f=\gcd (2,2)=2$ and $\frac{n-f}{2}=0$, and step 9.1 returns all four linear characters of $C_2\times C_2$, again with no character of degree $2$. Both agree with steps 9.2 and 10.1, since in these cases every irreducible character is linear and hence monomial.

The example is verified: for every $n\ge 1$ each irreducible complex character of $D_n=\mathrm{Dih}(C_n)$ is either a linear character extending an $s$-fixed linear character of the cyclic subgroup $A=⟨r⟩$ (the $2f$ characters of step 9.1) or the monomial character ${\mathrm{Ind}}_A^{D_n}\lambda$ of a linear character $\lambda$ of $A$ that is not $s$-fixed, of degree $2$ (the $\frac{n-f}{2}$ characters of step 10.1); in either case it is monomial, so $D_n$ is an $M$-group, with $f=\gcd (2,n)$ and including the degenerate cases $n=1,2$.

By [F1] the set $U$ with the displayed multiplication is a group of order $p^3$, its identity is $(0,0,0)$, its inverses are $(x,y,z{)}^{-1}=(-x,-y,-z+xy)$, and $e_1,e_2,e_3$ generate $U$ with $e_1^p=e_2^p=e_3^p=1$.

$H=\{(0,y,z)\}$ is a subgroup: for $(0,y,z),(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }})\in H$ the product is $(0,y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }}+0\cdot y^{\mathrm{\prime }})=(0,y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }})\in H$, and for $h=(0,y,z)$ the inverse computed from the formula of step 1.1 is $(0,-y,-z)$ (since $(-0,-y,-z+0\cdot y)=(0,-y,-z)$), which lies in $H$; so $H$ is closed under products and inverses, and $\mathrm{\mid }H\mathrm{\mid }=p^2$ because $y,z$ range over $\mathbb{Z}\mathrm{/}p$. The subset $Z=\{(0,0,z)\}$ is contained in $H$ and has $\mathrm{\mid }Z\mathrm{\mid }=p$.

For $(a,b)\in (\mathbb{Z}\mathrm{/}p{)}^2$ define ${\chi }_{a,b}(x,y,z):={\omega }^{ax+by}$. This is well defined on the triple $(x,y,z)\in U$, and it is a homomorphism: by step 1.1 the first two coordinates of a product add, so ${\chi }_{a,b}((x,y,z)(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }}))={\omega }^{a(x+x^{\mathrm{\prime }})+b(y+y^{\mathrm{\prime }})}={\chi }_{a,b}(x,y,z){\chi }_{a,b}(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})$. Different pairs give different characters, because ${\chi }_{a,b}(e_1)={\omega }^a$ and ${\chi }_{a,b}(e_2)={\omega }^b$ determine $a,b$ by the distinctness of the powers of $\omega$ in [F3]. Hence $U$ has at least $p^2$ characters of degree $1$; each of them is an irreducible character of $U$ and, being a linear character of the subgroup $U$ itself, is monomial in the sense of [F7]: in the covariant-function model, evaluation at $1$ identifies ${\mathrm{Ind}}_U^U{\chi }_{a,b}$ with its one-dimensional space, with inverse $v\mapsto (g\mapsto {\chi }_{a,b}(g{)}^{-1}v)$.

Characters of $H$: for $(c,d)\in (\mathbb{Z}\mathrm{/}p{)}^2$ define ${\mu }_{c,d}(0,y,z):={\omega }^{cy+dz}$. Each ${\mu }_{c,d}$ is a homomorphism, because the coordinates of $H$ multiply by adding by step 2.1: ${\mu }_{c,d}((0,y,z)(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }}))={\omega }^{c(y+y^{\mathrm{\prime }})+d(z+z^{\mathrm{\prime }})}={\mu }_{c,d}(0,y,z){\mu }_{c,d}(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }})$. The $p^2$ characters ${\mu }_{c,d}$ are pairwise distinct, since ${\mu }_{c,d}(0,1,0)={\omega }^c$ and ${\mu }_{c,d}(0,0,1)={\omega }^d$ are determined by $(c,d)$ by [F3]. Since the abelian group $H$ has only one-dimensional irreducible complex characters by [F2], and every such character is a homomorphism determined by its two values on $(0,1,0)$ and $(0,0,1)$, each of which is a $p$-th root of unity by [F3], there are exactly $p^2$ of them, so $\mathrm{Irr}(H)=\{{\mu }_{c,d}:(c,d)\in (\mathbb{Z}\mathrm{/}p{)}^2\}\mathrm{.}$ Moreover ${\mu }_{c,d}(Z)=1$ exactly when $d=0$.

Conjugation formula: for $u=(x,y,z)\in U$ and $h=(0,s,t)\in H$ one has $uh{u}^{-1}=(0,s,t+xs)$. Indeed $uh=(x,y+s,z+t+xs)$ by the multiplication law, and multiplying by the inverse $u^{-1}=(-x,-y,-z+xy)$ from step 1.1 gives first coordinate $x-x=0$, second coordinate $(y+s)-y=s$, and third coordinate $(z+t+xs)+(-z+xy)+x(-y)=t+xs$. This formula gives $uH{u}^{-1}=H$, so $H$ is normal. An element $(x,y,z)$ commuting with $e_2$ must have $x=0$ by this formula; comparison of its products with $e_1$ then forces $y=0$. Conversely every $(0,0,z)$ commutes with all triples by the multiplication law. Thus $Z$ is exactly the centre, and $(x,y,z)Z\mapsto (x,y)$ identifies $U\mathrm{/}Z$ with the additive group $(\mathbb{Z}\mathrm{/}p{)}^2$.

The action of $U$ on $\mathrm{Irr}(H)$: by the definition [F4] and step 3.2, $({}^u{\mu }_{c,d})(0,s,t)={\mu }_{c,d}(u^{-1}(0,s,t)u)={\mu }_{c,d}(0,s,t-xs)={\omega }^{cs+d(t-xs)}={\mu }_{c-dx,d}(0,s,t)$ for every $(0,s,t)\in H$, where $u=(x,y,z)$. Since characters are determined by their values, ${}^u{\mu }_{c,d}={\mu }_{c-dx,d}\mathrm{.}$

Orbits and inertia groups. Fix $(c,d)$ and let $u$ run over $U$, so that $x$ runs over $\mathbb{Z}\mathrm{/}p$ while the remaining coordinates are arbitrary. If $d\mathrm{\ne }0$, then $x\mapsto c-dx$ is injective by [F8] on the $p$-element set $\mathbb{Z}\mathrm{/}p$, hence bijective, and the orbit of ${\mu }_{c,d}$ is $\{{\mu }_{c^{\mathrm{\prime }},d}:c^{\mathrm{\prime }}\in \mathbb{Z}\mathrm{/}p\}$, of size $p$; the stabilizer is $\{u:x=0\}=H$ by step 3.2, so $I_U({\mu }_{c,d})=H$. If $d=0$, then ${}^u{\mu }_{c,0}={\mu }_{c,0}$ for all $u\in U$ by step 4.1, so $I_U({\mu }_{c,0})=U$. Hence the $p(p-1)$ characters ${\mu }_{c,d}$ with $d\mathrm{\ne }0$ split into $p-1$ orbits of size $p$ (the sets with a fixed $d\mathrm{\ne }0$), and the $p$ characters with $d=0$ are fixed points.