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This question concerns finite groups.

It is a well-known fact that every subgroup of a solvable group must again be solvable; this is easily proven by looking at the derived series of a given subgroup.

What I have been thinking about for awhile now is if/how this generalizes to arbitrary finite groups? Specifically, given some set $S$ of finite simple groups, consider the set $\Gamma_S$ of all finite groups whose composition series only includes groups in $S$. Then the above statement on solvable groups can be rephrased as:

THEOREM: Let $S = \{ \mathbb{Z}/p\mathbb{Z}\}_{p\in P}$. If $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in\Gamma_S$.

Trying to generalize to arbitrary $S$, the obvious generalization to try is:

CONJECTURE (1): For $S$ an arbitrary set of finite simple groups, if $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in\Gamma_S$.

Unfortunately, this is easily seen to be false; let $S = \{A_6\}$. Then $A_6\in \Gamma_S$ but $A_5\leq A_6$ and $A_5\notin \Gamma_S$. The failure in this example then leads me to the following second attempt at a generalization:

CONJECTURE (2): Let $S$ be an arbitrary set of finite simple groups, and $c(S)$ denote the set of all finite simple groups which appear as composition factors of some subgroup of some element of $S$. If $G\in \Gamma_S$ then $\forall H \leq G$ one has $H\in \Gamma_{c(S)}$.

I have been trying to figure out how to prove Conjecture (2). One thought is to use some appropriate analogue of the derived series for the general case, although coming up with the right analogue seems elusive. I have also thought about using the characters of elements of $S$, but this too does not lead to any immediate insights.

So, does anyone know whether Conjecture (2) is indeed true, and if so have either a (short enough for a post) proof or reference to where this question or similar ones might have been considered before?

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I don't know whether conjecture 2 is true. I feel it is false because (I suspect) one can engineer an (arbitrarily long but finite) sequence of groups H_1 <= H_2, ... such that H_i is not in the composition series of any decomposition of H_{i+1}. Gerhard "Ask Me About Tame Congruences" Paseman, 2011.12.15 –  Gerhard Paseman Dec 15 '11 at 22:52
    
@Gerhard: That's pretty much why I'm worried about Conjecture (2)'s validity; in the simplest case I'm wondering if something like taking $S = \{A_5 ,A_6 ,A_7\}$ then having an $A_8$ show up as a subgroup of some $G\in\Gamma_S$; there are no obvious obstructions that I can see. –  ARupinski Dec 15 '11 at 23:07
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I am confused why this would be false. If $1=N_n\lhd N_{n-1}\lhd\cdots\lhd N_1\lhd G$ is a composition series, then induction on $n$ should be enough. Let $H$ be a subgroup of $G$. If $n=1$, then clearly we're OK. Otherwise, $H/(H\cap N_1)$ is a subgroup of $G/N_1$, so that's OK, and by induction so is $H\cap N_1\le N_1$. –  Steve D Dec 15 '11 at 23:29
    
In more generality, let $\mathbf{P}$ be the closure operator of "poly-" and $\mathbf{S}$ the closure operator of "subgroups". Then $\Gamma_S=\mathbf{P}S$, and you're asking if $\mathbf{S}\mathbf{P}S\le \Gamma_{c(S)}$. But the Schreier refinement theorem guarantees that $\mathbf{P}\mathbf{S}S\le \Gamma_{c(S)}$, and it is always true, on the level of closure operators, that $\mathbf{S}\mathbf{P}\le \mathbf{P}\mathbf{S}$. –  Steve D Dec 15 '11 at 23:58
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What Steve said. You don't even need induction: $H\cap N_1$ is a subgroup of $N_1$, which in turn is in $S$, so all terms in the composition series of $H\cap N_1$ are in $c(S)$. I actually don't think that this question is at "MO-level": it would be a nice exercise for an introductory group theory course. –  Alex B. Dec 16 '11 at 0:34

1 Answer 1

up vote 3 down vote accepted

Say that a group H is a divisor or factor of a group G if H is a quotient group of a subgroup of G. Let C be a family of finite simple groups and let C' be the smallest class of finite groups containing C that is closed under finite direct products and divisors.

The following are equivalent for a finite group G:

  1. Each simple group divisor of G belongs to C.
  2. G has a subnormal series ending at 1 with successive quotients in C'
  3. Same as 2 but with normal series.
  4. G embeds in an iterated wreath product of simple groups divisors of a group in C.
  5. Each composition factor of G is a divisor of a group in C.

The class of groups above are closed under extensions and sub.

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The book of Ribes and Zalesskii on profinite groups has information about formations and generalizations, which seems what the question is about. –  Benjamin Steinberg Dec 16 '11 at 0:33
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I think another common term for "divisor" is "section". –  Steve D Dec 16 '11 at 0:44
    
Steve, thanks! I knew group theorists had another name for it, but I couldn't recall it. –  Benjamin Steinberg Dec 16 '11 at 1:02
    
"Factor group" is an old-fashioned synonym for "quotient group". Subgroups of quotient groups are the same as quotient groups of subgroups. I like to call them subquotients. –  Tom Goodwillie Dec 16 '11 at 1:33
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I had meant section where I wrote factor. I remembered the term chief factors in Marshall Hall for composition factors and thought that he must use factor for what I would call divisor. A subgroup of a quotient is of course a quotient of a subgroup but the converse is false. If G is a simple group, then it has no subquotients because it has no quotients. But it can have many divisors. –  Benjamin Steinberg Dec 16 '11 at 1:44

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