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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?

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I think an argument along the following lines will work, but I haven't filled in all of the details.

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 $\theta \in {\rm Irr}(N)$ 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.

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  • $\begingroup$ Hi Derek! If you have a moment, would you mind elaborating on the first paragraph? In particular, I am reading this as if... if $\psi_N$ is not homogenous, then the $e$ in definition 6.10 is equal to $1$, and since $e$ divides $|G:H|$, then $|G:H|$=1, which would violate $H$ being a proper subgroup of $G$. is this correct? Also, I know that if $\psi_N$ is not homogenous, then $psi_N$ is not a multiple of an irreducible character. So that implies that we are in case c of theorem 6.11? Or am I not thinking of this quite correctly? Thank you! $\endgroup$
    – User7238
    Commented Nov 5, 2022 at 21:22
  • $\begingroup$ One last thing... I think in the second paragraph, don't we need $\theta\in \text{Irr}N$? $\endgroup$
    – User7238
    Commented Nov 5, 2022 at 22:08
  • $\begingroup$ Yes, thanks, corrected! $\endgroup$
    – Derek Holt
    Commented Nov 6, 2022 at 8:38
  • $\begingroup$ Would you mind, if you have a moment, answering my first comment? :) It would be much appreciated! $\endgroup$
    – User7238
    Commented Nov 6, 2022 at 22:18
  • $\begingroup$ I can't make much sense of what you wrote. Why should $e$ in Definition 6.10 be equal to $1$? And why would that imply that $|G:H|=1$? When I wrote "we can assume that $G=H$", I meant that it is sufficient to prove the result with $H$ in place of $G$. But unfortunately, at first sight, that seems to depend on $H/N$ being and $M$-group, and I can't for the moment see how to justify that, because AFAIK being an $M$-group is not closed under taking subgroups. So I am afraid that I can't help you for the moment! $\endgroup$
    – Derek Holt
    Commented Nov 7, 2022 at 14:31

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