Index of maximal subgroups of $p$-solvable groups

Question

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?).

Answer 1

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.

Answer 2

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.