# Problem 6.10, I. Martin Isaacs' Character Theory

Let $$N \triangleleft G$$ with $$(|G : N|, |N|) = 1$$. Suppose that every subgroup of $$G/N$$ is an $$M$$-group. Show that $$G$$ is a relative $$M$$-group with respect to $$N$$.

Here are definitions of $$M$$-group and relative $$M$$-group in the book:

Let $$\chi$$ be a character of $$G$$. Then $$\chi$$ is monomial if $$\chi = \lambda^G$$, where $$\lambda$$ is a linear character of some (not necessarily proper) subgroup of $$G$$. The group $$G$$ is an $$M$$-group if every $$\chi \in \operatorname{Irr}(G)$$ is monomial.

Let $$N \triangleleft G$$ and let $$\chi \in \operatorname{Irr}(G)$$. Then $$\chi$$ is a relative $$M$$-character with respect to $$N$$ if there exists $$H$$ with $$N \subseteq H \subseteq G$$ and $$\psi \in \operatorname{Irr}(H)$$ such that $$\psi^G = \chi$$ and $$\psi_N \in \operatorname{Irr}(N)$$. Iff every $$\chi \operatorname{Irr}(G)$$ is a relative $$M$$-character with respect to $$N$$, then $$G$$ is a relative $$M$$-group with respect to $$N$$.

Since $$G/N$$ is an $$M$$-group, it is solvable. I want to use Theorem 6.22, but I don't know how to prove that every chief factor of every subgroup of $$G/N$$ has nonsquare order. And I don't know how to use the fact that $$(|G : N|, |N|) = 1$$.

Can anyone help me?

Let $\psi \in {\rm Irr}(G)$. If $\psi_N$ is not homogeneous, then $\psi$ is induced from a character of a proper subgroup $H$ of $G$ (i.e. the inertia group of an irreducible component of $\psi_N$). So we can assume that $G=H$ and hence that $\psi_N$ is homogeneous.
So $\psi_N = e \theta$ for some $\theta \in {\rm Irr}(H)$ and $e \ge 1$. Since $(|N|,|G:N|) = 1$, Corollary (6.28) of Isaacs says that $\theta$ extends to some $\chi \in {\rm Irr}(G)$. Then Corollary (6.17) of Isaacs implies that $\psi = \chi \beta$ for some $\beta \in {\rm Irr}(G/N)$.
Now use the fact that $G/N$ is an $M$-group to deduce that $\beta = \lambda^{H/N}$ for some subgroup $H/N$ of $G/N$ and a linear character $\lambda$ of $H/N$, and so (by Frobenius Reciprocity) $\psi = (\chi_H \lambda)^G$ with $(\chi_H\lambda)_N = \chi_N$ irreducible.