If $G$ is finite $p$-solvable group (every chief factor has order either a power $p$ or relatively prime to $p$) must every maximal subgroup have index either a power of $p$ or relatively prime to $p$?

I've shown the maximal subgroups of $p$-power index behave like the maximal subgroups of solvable groups. A maximal subgroup of a finite solvable has prime power index, so it seems reasonable that the only other kind of maximal subgroup in a $p$-solvable group is one with prime-to-$p$ index. Surely this is true for simple $p$-solvable groups.

I suspect a maximal subgroup of a $p$-solvable group either (a) have index a power of $p$ and cover all chief factors except exactly one $p$-chief factor which it avoids or (b) have index relatively prime to $p$ and cover all $p$-chief factors (but not necessarily cover or avoid the $p'$-chief factors).

In case (b), I am interested if a maximal subgroup covers all but one $p'$-chief factor. I don't have much intuition here (what do maximal subgroups of $A_5 \wr A_5$ look like?).


If $N$ is a minimal normal subgroup of $G$, then a maximal subgroup $M$ must either contain $N$, in which case $M/N$ is maximal in $G/N$ and you can use induction, or it supplements $N$ (since otherwise $M < NM < G$), in which case it covers all chief factors except $N$.

The maximal subgroups of $A_5 \wr A_5$ are unsurprising. There are those contain the base group, those of form $M \wr A_5$ for $M$ maximal in $A_5$, and one class isomorphic to $A_5 \times A_5$ that intersects the base group in a diagonal subgroup.

$A_5 \wr A_6$ is more interesting: it has a maximal subgroup isomorphic to $A_6$ that complements the base group and has a twisted action on the base group.

  • $\begingroup$ Thanks! The first paragraph gives nice results for all finite groups. Do the second and third paragraphs follow more or less from the O'Nan Scott theorem? We consider the possible cores of the maximal subgroups, and then which ONan-Scott class the quotient could fall in, and then look for maximal subgroups that satisfy both conditions? $\endgroup$ – Jack Schmidt Jun 10 '13 at 17:21
  • $\begingroup$ Yes, you can use the O'Nan Scott theorem to describe the maximal subgroups of these and similar examples. It's done that way in a paper of Eick and Hulpke on computing maximal subgroups. There is another paper on the same topic by John Cannon and myself. We based the theory more on a paper by Kovacs from the 1980s - more details on request, but I expect you can find the papers. $\endgroup$ – Derek Holt Jun 10 '13 at 21:06
  • $\begingroup$ Thanks I'll read them again! It's been a few years, and I'm just now understanding how easy it is in the solvable case, so I expect to get more out of them now. $\endgroup$ – Jack Schmidt Jun 10 '13 at 21:09

Derek's answer was enough for me, but here are a few less obvious details:

Definition: A subgroup $H \leq G$ is said to cover a normal section $K/L$ (for $L,K \unlhd G$, $L \leq K$) if any of the following equivalent statements are true: (a) $HL \geq K$, (b) $[H \cap K: H \cap L]=[K:L]$, (c) $[HK:HL]=1$.

Let $G$ be a finite group with a chosen chief series.

Lemma: A maximal subgroup of $G$ covers all but exactly one chief factor.

Proof: This is the first paragraph of Derek Holt's answer. $\square$

Lemma: If $H \leq G$, then $|H|$ is a multiple of the orders of the chief factors it covers.

Proof: Use the (b) version of the definition.

Lemma: If $M$ is a maximal subgroup of $G$, then $[G:M]$ divides the order of the single chief factor it does not avoid.

Proof: Just the two lemmas combined.

Proposition: The index of a maximal subgroup of a finite $p$-solvable group is either a power of $p$ or relatively prime to $p$ based on the single chief factor (in each chief series) that it does not cover.


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