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 $k$)?

  • $\begingroup$ Could you please define the primitive components of a permutation group or provide a reference for a definition? $\endgroup$
    – Derek Holt
    Mar 22 '17 at 8:27
  • $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    Mar 22 '17 at 8:28
  • $\begingroup$ @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. $\endgroup$
    – Zeyu
    Mar 22 '17 at 8:42
  • $\begingroup$ @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. $\endgroup$
    – Zeyu
    Mar 22 '17 at 8:47
  • $\begingroup$ Wielandt only defines primitive components for transitive permutation groups. I'll sketch a proof of my claim about $m(S) \le m(G)$ later. $\endgroup$
    – 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.