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Let $G$ be a finite M-group (ie, where all irreducible complex characters are induced from linear characters of subgroups). What subgroups of $G$ are necessarily themselves M-groups? For instance, clearly any nilpotent subgroup will be an M-group (so the Fitting subgroup works), and it has been shown that all normal Hall-subgroups are themselves M-groups. On the other hand, not all normal subgroups need to be M-groups, as has been shown by Dade, though it seems like most of them will be.

For instance, is it known whether the derived subgroup of an M-group is itself an M-group?

My reason for asking is that I am looking at a class of groups that behaves to some extent like M-groups, and it looks like all M-groups might indeed be in this class, but to show this it would be very helpful to know some subgroups (not necessarily normal) that are guaranteed to be M-groups themselves.

Edit: Given an irreducible character $\chi$, we can look at the set of subgroups of $G$ from which $\chi$ is induced from a linear character. Does anyone know an example of an M-group $G$ and an irreducible character $\chi$ of $G$ such that this set of subgroups does not contain any M-groups?

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Without having an expert opinion, I do know that the question of which subgroups of M-groups are themselves M-groups is extremely difficult. The basic problem with the subject seems to be that there is no internal group-theoretic characterization of M-groups: it all depends on the character theory of the group. There are hundreds of papers and many books to consult, but I'd be especially surprised if the derived group of an M-group were always an M-group (if true, that would be plainly stated in books by Isaacs, Huppert, etc.). – Jim Humphreys Feb 11 '11 at 13:31
Maybe you know this book already, but in case not: have a look at Michael Weinstein, "Between nilpotent and solvable". There is a whole chapter devoted to properties of M-groups. – Alex B. Feb 11 '11 at 14:14
FWIW: Dade's M-group of order 3584 has two subgroups of index 2 that are not M-groups, but the derived subgroup (their intersection) is an M-group. In fact those 2 subgroups of index 2 are the only subgroups that are not M-groups. Do you know if Sylow normalizers of M-groups are always M-groups? I haven't found counterexamples to the derived subgroup or the Sylow normalizer, but I haven't searched far enough to have much confidence. – Jack Schmidt Feb 12 '11 at 3:15
@Jack Schmidt: To the best of my knowledge, this is unknown and one of the open questions in the field (another open question is whether a normal subgroup $N$ of an M-group $G$ such that $|N|$ or $|G:N|$ is odd is always monomial). Isaacs (Hall subgroup normalizers...) has shown that $\mathbf{N}_G(H)/H'$ is an M-group, when $H$ is a Hall subgroup of the M-group $G$. – Frieder Ladisch Feb 12 '11 at 12:47
To reinforce other comments, I regret that there is no comprehensive up-to-date survey of M-groups: what is known/unknown/conjectured. For instance, the chapter in the book Alex mentions seems to cover just older material found in books by Isaacs and others. In any case, I'm aware that some people who have worked on M-groups became discouraged about the future prospects. – Jim Humphreys Feb 12 '11 at 13:31

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