Variations to Cayley's Embedding Theorem for Groups

Question

Early in a course in Algebra the result that every group can be embedded as a subgroup
of a symmetric group is introduced. One can further work on it to embed it as a subgroup of a suitable (higher degree) alternating group.

Inverting the view point we can say that the family of simple groups $A_n, n\geq 5$, contains all finite groups as their subgroups.

My question now is, is the same true for each of the other infinite families listed in the Classification of Finite Simple Groups?

In case the answer to this question is negative it might lead to some categorization. Cayley's embedding theorem is often considered a 'useless theorem', as no result about that group can be proved using that embedding. (Is that correct?) Other simple groups being somewhat more special (structure preserving maps of some non-trivial structure), we can categorize groups according to which infinite family(ies) they fall into. And groups embeddable in a particular family, but not embeddable in another may exhibit some special property.

Hope this provides a motivation for the question.

Answer 1

In general, for the groups in a family of fixed Lie rank, there will be a bound on the degree of an alternating group that can occur as a subgroup, so the answer to your question is no. This is easily seen from the fact that they have representations of a fixed degree. For example $E_8(q)$ has a representation of degree $248$ over ${\mathbb F}_q$, so it cannot possibly contain $A_n$ for $n > 250$, and I would guess that there is a much lower bound than that.

Of course, if by a family you mean one of the doubly infinite families like $A_n(q)$ for arbitrary $n$ then the answer is yes, because, for each such family, by making $n$ sufficient large, the groups will contain alternating groups of arbitrarily large degrees as subgroup of their Weyl groups.

To be more specific, the image of the natural permutation representation of $A_n$ over ${\mathbb F}_q$ preserves a unitary form and an orthogonal form with matrix $I_n$, so $A_n < L_n(q)$, $A_n < U_n(q)$ and $A_n$ lies in one of the types of orthogonal groups. I am not sure if it lies in the orthogonal type that preserves a diagonal form with non-square determinant (but I think it does), but that type certainly contains $A_{n-1}$. It is also easy to see that $A_n < {\rm Sp}_{2n}(q)$ for all $q$. That deals with all of the doubly infinite families.

Answer 2

There are plenty of ways of characterizing bounded rank for finite simple groups. Let $G$ be a finite group. Define

• $r_n(G)$ as the largest $k$ such that $(\mathbf{Z}/n\mathbf{Z})^k$ embeds into $G$
• $\mathrm{nc}(G)$ the largest $k$ such that there exist non-abelian subgroups $H_1\dots,H_k$ such that $[H_i,H_j]=1$ for all $i\neq j$, and $\mathrm{ns}(G)$ the same with non-solvable subgroups.
• $\rho(G)$ the smallest dimension of a faithful representation for $G$ (over any field of any characteristic, finite fields being enough).
• $\mathrm{lni}(G)$ is the largest nilpotency length of a nilpotent subgroup of $G$, and $\mathrm{lso}(G)$ is the largest solvability length of a solvable subgroup of $G$.
• $a(G)$ the largest $n$ such that $\mathrm{Alt}_n$ embeds into $G$

Then if $(S_n)$ is a sequence of finite simple groups, the conditions $r_6(S_n)\to\infty$, $\mathrm{nc}(S_n)\to\infty$, $\mathrm{ns}(S_n)\to\infty$, $\rho(S_n)\to\infty$, $r_m(S_n)\to\infty$ where $m$ is any number with at least 2 distinct prime divisors, $\mathrm{lni}(S_n)\to\infty$, $\mathrm{lso}(S_n)\to\infty$, $a(S_n)\to\infty$, are all equivalent. The same holds with "$\to\infty"$ replaced by "is bounded". This is just a sample: many variants are possible, and follow from the classification with not much efforts. Of course this dramatically fails for arbitrary finite groups, for which "being of bounded rank" is not univocally defined.

Answer 3

Another bit of information available from Cayley's Theorem:

It is possible to prove, without using the transfer homomorphism, that a finite group $G$ with a cyclic (and nontrivial) Sylow 2-subgroup has a normal 2-complement.
First we show that having a cyclic, nontrivial Sylow 2-subgroup implies the group has a subgroup of index 2: a generator of a Sylow 2-subgroup gets sent, via Cayley's embedding, to an odd permutation of the elements of $G$.
Next, note that if that subgroup of index 2 again has even order, it again has a subgroup of index 2.

Answer 4

Another use of regarding a group (there called $H$) as a subgroup of the symmetric group $S_{|H|}$ is given by Marty Isaacs in Subgroup property stronger than being characteristic