A solvable group that is not an M-group

The element $u$ satisfies $u^2=\bar{u}$ and $u^3=1$. Indeed $u=\frac{1}{2}(-1+(i+j+k))$ with $(i+j+k{)}^2=i^2+j^2+k^2+(ij+ji)+(ik+ki)+(jk+kj)=-3$ by [F1], so $u^2=\frac{1}{4}(1-2(i+j+k)+(i+j+k{)}^2)=\frac{1}{2}(-1-i-j-k)=\bar{u}$; by [A1] the product $u\bar{u}$ has scalar part $\frac{1}{4}-\frac{1}{4}((1)(-1)+(1)(-1)+(1)(-1))=\frac{1}{4}+\frac{3}{4}=1$ and imaginary part $\frac{1}{4}(-1,-1,-1)-\frac{1}{4}(1,1,1)+0=0$, the cross product vanishing because the imaginary parts of $u$ and $\bar{u}$ are antiparallel, so $N(u)=u\bar{u}=1$ and $u^{-1}=\bar{u}=u^2$ by [F1], whence $u^3=u{u}^2=1$. Moreover $u\mathrm{\notin }{Q}_8$ and $u^2\mathrm{\notin }{Q}_8$: the coordinates of $u$ and of $u^2$ are half-integers, whereas all coordinates of elements of $Q_8$ lie in $\{0,1,-1\}$ by [F2]. Hence $1,u,u^2$ are three distinct elements with $u\cdot u^2=u^3=1$ and $u^a{u}^b=u^{a+b}$ (exponent reduced modulo $3$), so $⟨u⟩=\{1,u,u^2\}$ is a subgroup of order $3$ and $Q_8\cap ⟨u⟩=\{1\}$.

The derived subgroup of $Q_8$ is $Q_8^{\mathrm{\prime }}=\{\pm 1\}$. First $-1\in Q_8^{\mathrm{\prime }}$, since $[i,j]=ij{i}^{-1}{j}^{-1}=(ij)(ji{)}^{-1}=k\cdot (-k{)}^{-1}=k\cdot k=k^2=-1$ by [F1] and [F5]. Second, $Q_8\mathrm{/}\{\pm 1\}$ has order $8\mathrm{/}2=4$ by [F8] and is therefore abelian by [F8], so [F6] gives $Q_8^{\mathrm{\prime }}\subseteq \{\pm 1\}$.

The subring ${\mathbb{C}}^{\mathrm{\prime }}:=\mathbb{R}+\mathbb{R}i$ of $\mathbb{H}$ is a field isomorphic to $\mathbb{C}$ via $a+bi\mapsto a+bi$, and one obtains a complex structure on $\mathbb{H}$ by $a\cdot x:=xa$ for $a\in {\mathbb{C}}^{\mathrm{\prime }}$ and $x\in \mathbb{H}$. This is a ${\mathbb{C}}^{\mathrm{\prime }}$-vector space structure: $a\cdot (x+y)=(x+y)a=xa+ya=a\cdot x+a\cdot y$, $(a+a^{\mathrm{\prime }})\cdot x=x(a+a^{\mathrm{\prime }})=xa+x{a}^{\mathrm{\prime }}$, $(a{a}^{\mathrm{\prime }})\cdot x=x(a{a}^{\mathrm{\prime }})=(xa)a^{\mathrm{\prime }}=a\cdot (a^{\mathrm{\prime }}\cdot x)$ and $1\cdot x=x$, all by associativity and distributivity in the ring $\mathbb{H}$ of [F1]. The elements $1,j$ form a ${\mathbb{C}}^{\mathrm{\prime }}$-basis: for $a=a_0+a_{1}i$ and $b=b_0+b_{1}i$ one computes $a\cdot 1+b\cdot j=a+jb$ with coordinates $(a_0,a_1,b_0,-b_1)$ by [F1], which vanishes only for $a=b=0$; and every $x\in \mathbb{H}$ with coordinates $(x_0,x_1,x_2,x_3)$ is $x=(x_0+x_{1}i)\cdot 1+(x_2-x_{3}i)\cdot j$, since $(x_2-x_{3}i)\cdot j=j(x_2-x_{3}i)=x_{2}j-x_{3}ji=x_{2}j+x_{3}k$. Hence $\mathbb{H}$ with this structure is a $2$-dimensional complex vector space.

Conjugation by $u$ permutes the generators of $Q_8$: one computes $ui=\frac{1}{2}(-1+j-i-k)$ and $u^{-1}=\bar{u}=\frac{1}{2}(-1-i-j-k)$, and expanding the four products $(-1+j-i-k)(-1)$, $(-1+j-i-k)(-i)$, $(-1+j-i-k)(-j)$, $(-1+j-i-k)(-k)$ with the multiplication table of [F1] gives $(-1+j-i-k)(-1-i-j-k)=4k$, so $ui{u}^{-1}=\frac{1}{4}\cdot 4k=k$; the same expansion with $uj=\frac{1}{2}(-1-i-j+k)$ and $uk=\frac{1}{2}(-1+i-j-k)$ gives $uj{u}^{-1}=i$ and $uk{u}^{-1}=j$. Since $-1$ is central by [F1] and every element of $Q_8$ is $\pm 1,\pm i,\pm j,\pm k$ by [F2], conjugation by $u$ maps the set $Q_8$ bijectively onto itself; the same holds for $u^2=u^{-1}$, whose conjugation is the inverse permutation.

For $q\in {\mathbb{H}}^{\times }$ the left multiplication $L_q(x):=qx$ is ${\mathbb{C}}^{\mathrm{\prime }}$-linear, because $L_q(a\cdot x)=q(xa)=(qx)a=a\cdot L_q(x)$ by associativity [F1]; moreover $L_1={\mathrm{id}}_{\mathbb{H}}$ and $L_{q{q}^{\mathrm{\prime }}}=L_q\circ L_{q^{\mathrm{\prime }}}$, so $L:{\mathbb{H}}^{\times }\to {\mathrm{GL}}_{{\mathbb{C}}^{\mathrm{\prime }}}(\mathbb{H})$ is a group homomorphism. Restricting $L$ to the subgroup $T$ gives a $2$-dimensional complex representation $V:=\mathbb{H}$ of $T$ in the sense of [F9], since $T$ is a group and the restriction of a homomorphism is a homomorphism.

In the ${\mathbb{C}}^{\mathrm{\prime }}$-basis $(1,j)$ the matrices are $L_i=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$ and $L_j=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$: by step 1.3, $L_i(1)=i=i\cdot 1$ and $L_i(j)=ij=k=(-i)\cdot j$ because $(-i)\cdot j=j(-i)=-ji=k$ by [F1], while $L_j(1)=j=1\cdot j$ and $L_j(j)=j^2=-1=(-1)\cdot 1$.

$Q_8$ is normal in $T$. Let $N=\{g\in {\mathbb{H}}^{\times }:g{Q}_8{g}^{-1}=Q_8\}$; conjugation by any $g$ is a bijection of ${\mathbb{H}}^{\times }$, so $g\in N$ if and only if $g^{-1}\in N$, and if $g,h\in N$ then $(gh)Q_8(gh{)}^{-1}=g(h{Q}_8{h}^{-1})g^{-1}=g{Q}_8{g}^{-1}=Q_8$, so $gh\in N$. Thus $N$ is a subgroup of ${\mathbb{H}}^{\times }$ by [F4], and it contains $Q_8$ (conjugation by an element of the subgroup $Q_8$ preserves $Q_8$) and $u,u^2$ by step 2.1. As $T=⟨Q_8\cup \{u\}⟩$ is the smallest subgroup containing $Q_8\cup \{u\}$ by [F4], one has $T\subseteq N$, that is $Q_8⊴T$.

The representation $V$ is faithful: if $L_q={\mathrm{id}}_{\mathbb{H}}$ for some $q\in T\subseteq {\mathbb{H}}^{\times }$, then $q=q\cdot 1=L_q(1)=1$.

The $T$-module $V$ is irreducible. Let $W\subseteq V$ be a $T$-invariant ${\mathbb{C}}^{\mathrm{\prime }}$-subspace with $W\mathrm{\ne }0$. If $W\mathrm{\ne }V$ then ${\dim }_{{\mathbb{C}}^{\mathrm{\prime }}}W=1$ by step 1.3, so $W={\mathbb{C}}^{\mathrm{\prime }}w$ is a line, and $W$ is invariant under $L_i=\mathrm{diag}(i,-i)$ of step 2.3, hence equals one of the eigenspaces of the two distinct eigenvalues $i\mathrm{\ne }-i$, namely ${\mathbb{C}}^{\mathrm{\prime }}\cdot 1$ or ${\mathbb{C}}^{\mathrm{\prime }}\cdot j$. But $W$ is also invariant under $L_j$, whereas $L_j(1)=j\mathrm{\notin }{\mathbb{C}}^{\mathrm{\prime }}\cdot 1$ and $L_j(j)=-1\mathrm{\notin }{\mathbb{C}}^{\mathrm{\prime }}\cdot j$, since ${\mathbb{C}}^{\mathrm{\prime }}\cdot 1\cap {\mathbb{C}}^{\mathrm{\prime }}\cdot j=\{0\}$ and $1,j\mathrm{\ne }0$ by steps 1.3 and 2.3. This contradiction shows $W=V$, so $V$ has no nonzero proper $T$-invariant subspace and is irreducible by [F9].

Every element of $T$ has the form $q{u}^m$ with $q\in Q_8$ and $m\in \{0,1,2\}$. For $S=\{q{u}^m:q\in Q_8,m\in \{0,1,2\}\}$ one has $(q{u}^m)(q^{\mathrm{\prime }}{u}^{m^{\mathrm{\prime }}})=q(u^m{q}^{\mathrm{\prime }}{u}^{-m})u^{m+m^{\mathrm{\prime }}}$ with $u^m{q}^{\mathrm{\prime }}{u}^{-m}\in Q_8$ by step 3.1 (for $m=0,1,2$, using $u^2=u^{-1}$) and $u^{m+m^{\mathrm{\prime }}}=u^{m^{\mathrm{\prime }\mathrm{\prime }}}$ for the residue $m^{\mathrm{\prime }\mathrm{\prime }}$ of $m+m^{\mathrm{\prime }}$ modulo $3$ by step 1.1, so the product lies in $S$; and $(q{u}^m{)}^{-1}=u^{-m}{q}^{-1}=(u^{-m}{q}^{-1}{u}^m)u^{-m}\in S$. Hence $S$ is a subgroup by [F4], and it contains $Q_8$ and $u$, so $T\subseteq S$ and therefore $T=S$. The expression is unique: $q{u}^m=q^{\mathrm{\prime }}{u}^{m^{\mathrm{\prime }}}$ gives $q^{-1}{q}^{\mathrm{\prime }}=u^{m-m^{\mathrm{\prime }}}\in Q_8\cap ⟨u⟩=\{1\}$ by step 1.1, hence $q=q^{\mathrm{\prime }}$ and $m=m^{\mathrm{\prime }}$ because $1,u,u^2$ are distinct. Consequently $\mathrm{\mid }T\mathrm{\mid }=\mathrm{\mid }{Q}_8\mathrm{\mid }\cdot \mathrm{\mid }⟨u⟩\mathrm{\mid }=8\cdot 3=24$, the group $T$ is the internal semidirect product of $Q_8$ by $⟨u⟩\cong C_3$ in the sense of [F3], and $u^m\mapsto u^m{Q}_8$ is an isomorphism $⟨u⟩\to T\mathrm{/}{Q}_8$, so $T\mathrm{/}{Q}_8\cong C_3$ is cyclic of order $3$.

The derived subgroup of $T$ is contained in $Q_8$: by step 4.1 the quotient $T\mathrm{/}{Q}_8$ is isomorphic to the abelian group $C_3$, so [F6] gives $[T,T]\subseteq Q_8$.

Conversely $Q_8\subseteq T^{\mathrm{\prime }}=[T,T]$. By [F5] and step 2.1, $[u,i]=ui{u}^{-1}{i}^{-1}=k(-i)=-j$, $[u,j]=uj{u}^{-1}{j}^{-1}=i(-j)=-k$ and $[u,k]=uk{u}^{-1}{k}^{-1}=j(-k)=-i$ all lie in $T^{\mathrm{\prime }}$, where $i^{-1}=-i$, $j^{-1}=-j$ and $k^{-1}=-k$ by [F2]; as $T^{\mathrm{\prime }}$ is a subgroup containing $i=-(-i)$, $j=-(-j)$, $k=-(-k)$ and $-1=i\cdot i$, it contains all eight elements of $Q_8$. Together with step 5.1 this gives $T^{\mathrm{\prime }}=Q_8$.

The derived series of $T$ terminates: by steps 6.1 and 1.2, $T^{\mathrm{\prime }\mathrm{\prime }}=[T^{\mathrm{\prime }},T^{\mathrm{\prime }}]=[Q_8,Q_8]=Q_8^{\mathrm{\prime }}=\{\pm 1\}$, and $T^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=[\{\pm 1\},\{\pm 1\}]=1$ because $\{\pm 1\}$ is abelian (its commutators are trivial by [F5]). Hence $T^{(3)}=1$ and $T$ is solvable by [F7].

The group $T$ has no subgroup of index $2$. Suppose $H\le T$ satisfies $[T:H]=2$. Then $H⊴T$ and $T\mathrm{/}H$ has order $2$ by [F8], so the quotient map $\pi :T\to T\mathrm{/}H$ is a homomorphism onto an abelian group; by [F6] it factors as $\pi =\psi \circ q$ with $q:T\to T\mathrm{/}{T}^{\mathrm{\prime }}$ the quotient map and $\psi :T\mathrm{/}{T}^{\mathrm{\prime }}\to T\mathrm{/}H$. By steps 6.1 and 4.1, $T\mathrm{/}{T}^{\mathrm{\prime }}=T\mathrm{/}{Q}_8\cong C_3$; writing $\bar{x}$ for a generator of $T\mathrm{/}{T}^{\mathrm{\prime }}$ one has $\psi (\bar{x}{)}^3=\psi ({\bar{x}}^3)=1$ and $\psi (\bar{x}{)}^2=1$ because $T\mathrm{/}H$ has order $2$, so the order of $\psi (\bar{x})$ divides both $3$ and $2$ by [F8] and is therefore $1$; hence $\psi$ is trivial and so is $\pi$, contradicting that $\pi$ is surjective onto a group of order $2$.

The character ${\chi }_V$ of the irreducible $2$-dimensional $T$-module $V$ is an irreducible complex character of $T$ by [F9] and [F10], and it is not monomial. Suppose ${\chi }_V={\mathrm{Ind}}_H^T\lambda$ for some $H\le T$ and linear character $\lambda$; by [F10] the underlying representation is then $V\cong {\mathrm{Ind}}_H^{T}L$ for the one-dimensional $H$-module $L$ affording $\lambda$, so ${\dim }_{{\mathbb{C}}^{\mathrm{\prime }}}V=[T:H]{\dim }_{{\mathbb{C}}^{\mathrm{\prime }}}L=[T:H]$ by [F10], that is $[T:H]=2$, contradicting step 7.2. Hence $T$ is a solvable group, by step 7.1, with an irreducible complex character that is not monomial, so $T$ is not an $M$-group by [F10]; the statement that every finite solvable group is an $M$-group is therefore false.