Does the class of soluble groups whose $p$-length is $\leq 1$ for all $p$ form a saturated Fitting formation?

Question

Let $\mathcal{F}$ be the class of all soluble groups $G$ such that the $p$-length $l_{p}(G) \leq 1$ for all primes $p$. How do I show ‎$\mathcal{F}$ is a saturated Fitting formation?

Answer 1

Overview

This question turns out to be very similar to showing that $\pi$-groups form a (subgroup closed) saturated Fitting formation. If you have not shown that, I highly recommend working on it first. The next step are the $p$-nilpotent groups (and/or the $p$-closed groups). After that are the $p$-length 1 groups, then $p$-nil by $p$-nil, then $p$-length 2, then …. I worked out the $p$-nilpotent case explicitly, since it can be used to handle arbitrary $p$-length easily. I also worked one prime at a time, as one sometimes treats the even and odd primes separately. I didn't see any particular short cuts for $p$-length 1 (either one prime at a time, or all primes together), that didn't require more background material (like local definitions of formations).

At any rate, there are no sneaky tricks in this one. Just standard verifications using fundamental properties (that might take years to learn, so I'm not saying this as easy as articles like to make it seem).

Proof

Definition: A finite group is said to be $p$-nilpotent if it has a normal Sylow $p$-complement (a normal subgroup whose order is relatively prime to $p$, but whose index is a power of $p$).

Every finite group has a unique maximum amongst the normal $p$-nilpotent subgroups: it is $O_{p',p}(G)$ defined by $O_{p'}(G)$ being the unique largest normal $p'$-subgroup of $G$, and $O_{p',p}(G)/O_{p'}(G)$ being the unique largest normal $p$-subgroup of $G/O_{p'}(G)$. Set $F_p^{n+1}(G)/F_p^n(G) = O_{p',p}(G/F_p^n(G))$ and $F_p^0(G) = 1$.

Definition: The $p$-length of $G$ is the minimum number $n$ such that $F_p^n(G)$ has index relatively prime to $p$ in $G$; if such an $n$ exists $G$ is called $p$-solvable. If no such $n$ exists, then the $p$-length is infinity.

I assume you know how to prove that $p$-groups and $p'$-groups are each closed under subgroups, quotients, normal products, and Frattini extensions. The next step is showing the $p$-nilpotents are similar, and then we can put these together to handle arbitrary $p$-length.

Lemma: The class of $p$-nilpotent groups is closed under subgroups, quotients, normal products, and Frattini extensions.

Proof: Let $G$ be a finite group, $M,N \unlhd G$, and $H \leq G$. If $G$ is $p$-nilpotent with normal Sylow $p$-complement $G_{p'}$, then $H \cap G_{p'}$ is a normal Sylow $p$-complement for $H$, and $H$ is $p$-nilpotent. Similarly $G_{p'}N/N$ is a normal Sylow $p$-complement for $G/N$. Return to general $G$, but assume $M,N$ are $p$-nilpotent with normal Sylow $p$-complements $M_{p'}$ and $N_{p'}$. $M_{p'}$ is characteristic in $M$, and since $N$ normalizes $M$, $N$ also normalizes $M_{p'}$. Hence $M_{p'} N_{p'}$ is a normal Sylow $p$-complement of $MN$, and $MN$ is $p$-nilpotent. [This next part is the Frattini argument.] Suppose $G/\Phi(G)$ is $p$-nilpotent with normal Sylow $p$-complement $N/\Phi(G)$. Set $G_{p'}$ to be a Sylow $p$-complement of $N$. If $g \in G$, then $G_{p'}^g \leq N$ and so is conjugate by some element $f \in \Phi(G)$. Hence $g = gf^{-1} \cdot f \in N_G(G_{p'}) \Phi(G)$ and $G = N_G(G_{p'}) \Phi(G)$ so $G=N_G(G_{p'})$ and $G_{p'}$ is a normal Sylow $p$-complement for $G$, and $G$ is $p$-nilpotent. $\square$

Lemma: If $H \leq G$ and $M,N \unlhd G$, then $F_p(H) \geq F_p(G) \cap H$, $F_p(G/N) \geq F_p(G)N/N$, $F_p(MN) \geq F_p(M) F_p(N)$, and $F_p(G/\Phi(G)) = F_p(G)/\Phi(G)$.

Proof: $F_p(G) \cap H$ is $p$-nilpotent by the previous lemma. It is normal in $H$ since both $F_p(G)$ and $H$ are normalized by $H$. By the previous lemma, $(F_p(G) \cap H)F_p(H)$ is a normal product of $p$-nilpotent groups, so is also $p$-nilpotent. However, $F_p(H)$ is the unique largest $p$-nilpotent normal subgroup of $H$, and so $(F_p(G) \cap H)F_p(H) = F_p(H)$ and $(F_p(G) \cap H) \leq F_p(H)$.

The others follow similarly:

$F_p(G)N/N \cong F_p(G)/(N \cap F_p(G))$ is a quotient of a $p$-nilpotent group, so $p$-nilpotent by the previous lemma. It normal in $G/N$, and so contained in the unique largest normal $p$-nilpotent normal subgroup $F_p(G/N)$.

$F_p(M) F_p(N)$ is $p$-nilpotent by the previous lemma, since $F_p(M)$ and $F_p(N)$ are characteristic in $M$ and $N$, and $M$ and $N$ are normal in $G$. For similar reasons, it is normal in $MN$ as well. Hence we get $F_p(M) F_p(N) \leq F_p(MN)$.

$\Phi(G) \leq F_p(G)$ since $\Phi(G)$ is a nilpotent and thus $p$-nilpotent normal subgroup. Let $N/\Phi(G)$ be a $p$-nilpotent normal subgroup of $G$. Then $N$ is a normal subgroup of $G$, and $N$ is $p$-nilpotent by the lemma. Hence $N \leq F_p(G)$. Taking $N/\Phi(G) = F_p(G/\Phi(G))$ we are done. $\square$

Lemma: If $H \leq G$ and $M,N \unlhd G$, then $F_p^n(H) \geq F_p^n(G) \cap H$, $F_p^n(G/N) \geq F_p^n(G)N/N$, $F_p^n(MN) \geq F_p^n(M) F_p^n(N)$, and $F_p^n(G/\Phi(G)) = F_p^n(G)/\Phi(G)$.

Proof: Just an induction. Both of the first two use both of $F_p(H) \geq F_p(G) \cap H$ and $F_p(G/N) \geq F_p(G)N/N$, which is neat. The last two are just straightforward. $\square$

Proposition: If $H \leq G$ and $M,N \unlhd G$ then the $p$-length of $H$ and $G/N$ are less than or equal to the $p$-length of $G$. The $p$-length of $G/\Phi(G)$ is exactly equal to the $p$-length of $G$, and the $p$-length of $MN$ is the maximum of the $p$-lengths of $M$ and $N$.

Proof: By the previous lemma, and the definition.

Proposition: For each prime $p$, let $v_p \in \{0,1,\dots,\infty\}$. The set of finite groups whose $p$-length is less than or equal to $v_p$ is closed under subgroups, quotients, normal products, residual products, and Frattini extensions. Hence the set is a (subgroup closed), saturated, Fitting formation.

Proof: The only step we haven't said explicitly is closure under residual products: If $M,N \unlhd G$ and $G/M, G/N$ have $p$-length at most $v_p$, then $G/(M \cap N) \leq G/M \times G/N$ has $p$-length at most $v_p$, since a direct product is a very special sort of normal product. $\square$

Background

Definition: A finite group $G$ has $p$-length 1 if and only if $G=O_{p',p,p'}(G)$. Here $O_{p'}(G)$ is the largest normal $p'$-subgroup of $G$, $O_{p',p}(G)/O_{p'}(G)$ is the largest normal $p$-subgroup of $G/O_{p'}(G)$, and $O_{p',p,p'}(G)/O_{p',p}(G)$ is the largest normal $p'$-subgroup of $G/O_{p',p}(G)$.

$O_{p',p}(G)$ is also the largest normal $p$-nilpotent subgroup of $G$, and is the intersection of the centralizers of the $p$-chief factors of $G$. It is very similar to a $p$-local version of the Fitting subgroup. For instance $A_4$ is $O_{3',3}(S_4)$ and $S_4$ has 3-length 1. However, $S_4$ has 2-length 2, as $O_{2',2}(S_4) = K_4$ does not have odd index.

Definition: A formation $\mathfrak{F}$ is a class of finite groups such that $G \in \mathfrak{F}$ and $N \unlhd G$, then $G/N \in \mathfrak{F}$, and if $G$ is a finite group with $N,M \unlhd G$ and $G/N, G/M \in \mathfrak{F}$, then $G/(N\cap M) \in \mathfrak{F}$. A formation is saturated if $G/\Phi(G) \in \mathfrak{F}$ implies $G \in \mathfrak{F}$. A Fitting class is a class of finite groups $\mathfrak{F}$ such that if $H \leq G$ is subnormal and $G \in \mathfrak{F}$, then $H \in \mathfrak{F}$, and if $M,N \unlhd G$ with $M,N \in \mathfrak{F}$, then $MN \in \mathfrak{F}$.

Formations are used to form residuals. The formation of abelian groups has residual $[G,G]$, the intersection of all $N$ such that $G/N$ is abelian. If $N \unlhd G$ then $G/N$ is abelian iff $[G,G] \leq N$. That is the gist of a formation. A saturated formation has much better properties, but I won't use them in this answer. A Fitting class lets you form a radical. The Fitting class of nilpotent groups has radical the Fitting subgroup: the normal product of all normal nilpotent subgroups of a group. A Fitting class is defined exactly so that the radical is the unique largest normal $\mathfrak{F}$-subgroup of $G$, and a subgroup is a subnormal $\mathfrak{F}$-subgroup of $G$ if and only if it is a subnormal subgroup of the radical. For instance, my view on this problem is that we are localizing the idea of the Fitting length of a solvable group to the $p$-length of a $p$-solvable group. We take the Fitting class of $p$-nilpotent groups (instead of all nilpotent groups), and define $F_p(G)$ to be the radical of this class (instead of the standard Fitting subgroup), and then define $F_p^n(G)$ inductively (just like the Fitting series). Since we don't require $G=F_p^n(G)$, just that it have $p'$-index, the groups of $p$-length 1 are more like the nilpotent-by-abelian than the strictly nilpotent groups.

Gaschütz (1962) introduced formations, and is a fairly important paper to read; for instance your question is Beispiel 4.6. He does not prove it^{nor does he give the associated projector :~(}, but the claim was considered standard even in Hall–Higman's (very important) paper from 1956. I include both these papers because they are fundamental steps in finite group theory.

Bibliography

• Hall, P.; Higman, Graham. “On the $p$-length of $p$-soluble groups and reduction theorems for Burnside's problem.” Proc. London Math. Soc. (3) 6 (1956), 1–42. MR72872 DOI:10.1112/plms/s3-6.1.1

• Gaschütz, Wolfgang. “Zur Theorie der endlichen auflösbaren Gruppen.” Math. Z. 80 (1962/1963) 300–305. MR179257 DOI:10.1007/BF01162386