A finite transitive permutation group $G$ can always be ``decomposed'' into primitive permutation groups, called its primitive components, although the decomposition is not unique. See Chapter 1 of *Finite Permutation Groups* by Wielandt. 


Suppose $H$ is a primitive component of $G$, and $T$ is a composition factor of $H$ that is also a classical group $X(d,q)$. Here $X$ stands for $\mathrm{PSL}$, $\mathrm{PSU}$, etc. Then $T$ is a subquotient of a composition factor $T'$ of $G$. If $T'$ is an alternating group 
$\mathrm{Alt}\_k$, is there any lower bound on its degree $k$, in terms of $d$ and $q$?


If $T$ is a subgroup of $T'$ then $T$ has a faithful permutation representation of degree $k$. In this case I think a lower bound of order $q^{\Theta(d)}$ is known. See:


Cooperstein, Bruce N. "Minimal degree for a permutation representation of a classical group." Israel J. Math. 30 (1978), no. 3, 213-235. MR 506701 DOI: 10.1007/BF02761072.


I wonder if a similar bound is possible if $T$ is only a subquotient of $T'$.


A related question: for $k\in\mathbb{N}^+$, denote by $\Gamma\_k$ the family of finite groups whose nonabelian composition factors are all isomorphic to subgroups of $\mathrm{Sym}\_k$. If $G$ is a group in $\Gamma\_k$, are its primitive components also in $\Gamma\_k$ (or $\Gamma\_{k'}$ for some $k'$ depending on $k$)?



composition factors of primitive components
Here is a proof of the claim I made in my comment: if $m(G)$ denotes the minimal degree of a faithful permutation representation of a finite group $G$ and $S$ is a composition factor of $G$, then $m(S) \le m(G)$. I think that answers your questions. (Note that, since $m(H) \le m(G)$ for any $H \le G$, this implies $d(S) \le d(G)$ for any composition factor $S$ of any subgroup of $G$.)


The claim is clear if $S$ is abelian, so assume that $S$ is nonabelian. As you observed yourself, if $N$ denotes the largest normal solvable subgroup of $G$, then $m(G/N) \le m(G)$. This also follows from Proposition 1.3 of


D. Easdown and C. E. Praeger, On minimal faithful permutation
representations of finite groups, Bull. Aust. Math. Soc. 38 (1988),
207-220.


So we may assume that $G$ has no nontrivial solvable normal subgroup. So the socle $K$ of $G$ (the group generated by its minimal normal subgroups) is a direct product $\times\_{i=1}^k S\_i$ of nonabelian simple groups $S\_i$, which are permuted under conjugation by $G$.


et $L$ be the kernel of this conjugation action i.e. $L = \cap\_{i=1}^k N\_G(S\_i)$. Then $L/K$ is isomorphic to a subgroup of $\times\_{i=1}^k {\rm Out}(S\_i)$. From the classification of fintie simple groups, we know that each ${\rm Out}(S\_i)$ is solvable, and hece so is $L/K$. Also $G/L$ is isomorphic to a subgroup of $S\_k$.


If $S$ is isomorphic to one of the $S\_i$, ithen $m(S) = m(S\_i) \le m(G)$ because $S\_i \le G$.


Otherwise, $S$ is a composition factor of $G/L$ and $m(G/L) \le k$, so by induction we have $m(S) \le k$.


Now by Theorem 3.1 of the same paper by Easdown and Praeger, we have $m(G) = \sum\_{i=1}^k m(S\_i) \ge k$, so $m(S) \le m(G)$ as claimed.


Note also that $m(S)$ is known for all finite simple groups $S$. Unfortunately, many of the papers on this topic contain minor errors, but there is a hopefully correct table giving the results in the paper:


S.Guest, J.Morris, C.E.Praeger and P.Spiga.
On the maximum orders of elements of finite almost
 simple groups and primitive permutation groups., which you can find [here](http://arxiv.org/abs/1301.5166)


