# 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?

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.

• 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! Commented Nov 5, 2022 at 21:22
• One last thing... I think in the second paragraph, don't we need $\theta\in \text{Irr}N$? Commented Nov 5, 2022 at 22:08
• Yes, thanks, corrected! Commented Nov 6, 2022 at 8:38
• Would you mind, if you have a moment, answering my first comment? :) It would be much appreciated! Commented Nov 6, 2022 at 22:18
• 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! Commented Nov 7, 2022 at 14:31