composition factors of primitive components

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 $k$)?

• Could you please define the primitive components of a permutation group or provide a reference for a definition? – Derek Holt Mar 22 '17 at 8:27
• I don't know whether this will help to answer your question, but if we denote the minimal degree of a faithful permutation representation of a finite group $G$ by $m(G)$ then $m(S) \le m(G)$ for any simple composition factor $S$ of $G$. – Derek Holt Mar 22 '17 at 8:28
• @DerekHolt A definition can be found at the end of Chapter 1 of Finite Permutation Groups by Wielandt. Basically a primitive component is obtained by repeatedly restricting the action to a nontrivial block, or taking the induced action on the the set of blocks, until it becomes primitive. – Zeyu Mar 22 '17 at 8:42
• @DerekHolt Yes I think that answers my question. I found a paper by Kovacs and Praeger that proves $m(G/N)\leq m(G)$ if $G/N$ has no nontrivial normal abelian subgroups. – Zeyu Mar 22 '17 at 8:47
• Wielandt only defines primitive components for transitive permutation groups. I'll sketch a proof of my claim about $m(S) \le m(G)$ later. – Derek Holt Mar 22 '17 at 8:50

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