Finite supersolvable groups are M-groups

We prove the assertion by induction on $\mathrm{\mid }G\mathrm{\mid }$. If $\mathrm{\mid }G\mathrm{\mid }=1$ then $V$ is the one-dimensional trivial representation, and [F9] with $H=G=1$ shows that $V$ is induced from a linear character of the subgroup $G$ itself, so $V$ is monomial; this is the base case. In the remaining cases $G$ is nontrivial, $V$ is a fixed irreducible complex $G$-representation, and $K=\ker \rho$ is its kernel.

Induction hypothesis: every finite group $H$ with $\mathrm{\mid }H\mathrm{\mid }<\mathrm{\mid }G\mathrm{\mid }$ that is supersolvable in the sense of [F2], that is, possesses a series with terms normal in $H$ and prime-order factors, has the property that each of its irreducible complex representations is induced from a linear character of a subgroup.

Suppose next that $K=1$ and that $G$ is abelian. By [F7] the irreducible representation $V$ over the splitting field $\mathbb{C}$ is one-dimensional, so its representing map is a linear character of $G$; by [F9] with $H=G$ the module $V$ is isomorphic to ${\mathrm{Ind}}_G^{G}V$, hence is induced from a linear character of the subgroup $G$. So $V$ is monomial in this case as well.

Suppose first that $K\mathrm{\ne }1$. Then $K$ acts trivially on $V$, so by [F8] the representation descends to an irreducible representation $\bar{\rho }$ of $\bar{G}=G\mathrm{/}K$ on $V$, that is $V={\mathrm{Infl}}_{\bar{G}}^G\bar{V}$ for the irreducible $\bar{G}$-module $\bar{V}$ affording $\bar{\rho }$. The images ${\bar{G}}_i:=G_{i}K\mathrm{/}K$ form a chain of subgroups normal in $\bar{G}$ with ${\bar{G}}_0=1$ and ${\bar{G}}_r=\bar{G}$, and each ${\bar{G}}_i\mathrm{/}{\bar{G}}_{i-1}$ is a quotient of the prime-order group $G_i\mathrm{/}{G}_{i-1}$: the map $gK\mapsto g{G}_{i-1}K$ from $G_{i}K\mathrm{/}K$ onto $G_{i}K\mathrm{/}{G}_{i-1}K$ has kernel $G_{i-1}K\mathrm{/}K$ and is surjective, so by [F11] ${\bar{G}}_i\mathrm{/}{\bar{G}}_{i-1}\cong G_{i}K\mathrm{/}{G}_{i-1}K$, and the latter is a quotient of $G_i\mathrm{/}{G}_{i-1}$ since the natural map $G_i\to G_{i}K\mathrm{/}{G}_{i-1}K$ is onto with $G_{i-1}$ in its kernel. Hence $\bar{G}$ satisfies the hypothesis of the theorem with $\mathrm{\mid }\bar{G}\mathrm{\mid }<\mathrm{\mid }G\mathrm{\mid }$, and by step 1.2 there are $\bar{H}\le \bar{G}$ and a one-dimensional $\bar{H}$-module $\bar{L}$ with $\bar{V}\cong {\mathrm{Ind}}_{\bar{H}}^{\bar{G}}\bar{L}$. Writing $H={\pi }^{-1}(\bar{H})$ for the quotient map $\pi :G\to \bar{G}$, the final clause of [F5] gives $V={\mathrm{Infl}}_{\bar{G}}^G\bar{V}\cong {\mathrm{Ind}}_H^G({\mathrm{Infl}}_{\bar{H}}^H\bar{L})$ with ${\mathrm{Infl}}_{\bar{H}}^H\bar{L}$ one-dimensional. So $V$ is monomial in this case.

There remains the case $K=1$ with $G$ nonabelian, which by step 1.3 exhausts the remaining possibilities. Then $\rho$ is faithful by [F10], so [F3] applied to $G$ gives an abelian normal subgroup $A⊴G$ with $A⊈Z(G)$. Restricting $V$ to the abelian group $A$ and using [F7], the module $V{\mathrm{\mid }}_A$ has a constituent $\lambda$, which is a linear character of $A$; applying [F4] to $A$ and $V$ shows that $I:=I_G(\lambda )$ is a proper subgroup of $G$ and that $V\cong {\mathrm{Ind}}_I^{G}W$ for an irreducible $I$-module $W$ lying over $\lambda$. In particular $\mathrm{\mid }I\mathrm{\mid }<\mathrm{\mid }G\mathrm{\mid }$.

The intersection $I_j:=I\cap G_j$ is normal in $I$ for each $j$, since $G_j⊴G$, and the map $I\cap G_j\to G_j\mathrm{/}{G}_{j-1}$, $x\mapsto x{G}_{j-1}$, has kernel $I_{j-1}$, so $I_j\mathrm{/}{I}_{j-1}$ is isomorphic to a subgroup of the prime-order group $G_j\mathrm{/}{G}_{j-1}$ and therefore has order $1$ or prime; deleting repeated terms gives a series for $I$ whose terms are normal in $I$ and whose factors have prime order. Since $\mathrm{\mid }I\mathrm{\mid }<\mathrm{\mid }G\mathrm{\mid }$ by step 2.2, the induction hypothesis of step 1.2 applied to $I$ gives $W\cong {\mathrm{Ind}}_L^I\mu$ for some subgroup $L\le I$ and some linear character $\mu$ of $L$.

Combining steps 2.2 and 3.1 with transitivity of induction [F6], $V\cong {\mathrm{Ind}}_I^G{\mathrm{Ind}}_L^I\mu \cong {\mathrm{Ind}}_L^G\mu$ with $\mu$ a linear character of $L$, so $V$ is monomial. The cases $\mathrm{\mid }G\mathrm{\mid }=1$ (step 1.1), $K\mathrm{\ne }1$ (step 2.1), $K=1$ with $G$ abelian (step 1.3) and $K=1$ with $G$ nonabelian (steps 2.2 and 3.1) are exhaustive, so every irreducible complex representation of $G$ is induced from a linear character of a subgroup and $G$ is an $M$-group by [F1].