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?