Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it possible to determine the structure of maximal subgroups of finite simple groups?(Even if in special cases such as minimal simple groups, alternating groups,...)

share|improve this question
    
What do you mean by "the structure of" them? $\:$ –  Ricky Demer Nov 17 '12 at 9:23
    
If you take the minimal permutation representation of a finite simple group, the stabilizer of a point will be a maximal size subgroup. See mathoverflow.net/questions/16858/… –  Ian Agol Nov 29 '12 at 23:51

7 Answers 7

up vote 11 down vote accepted

The maximal subgroups of $A_n$ are given by the O'Nan-Scott Theorem. They lie in one of the following classes:

1) $A_n \cap (S_{n-k} \times S_k)$, that is the stabiliser of a $k$-set.

2) $A_n \cap (S_a wr S_b)$ where $n=ab$, that is the stabiliser of a partition.

3) $A_n\cap AGL(d,p)$ where $n=p^d$ for some prime $p$.

4) $A_n \cap (S_m wr S_k)$ where $n=m^k$, that is the stabiliser of a cartesian power.

5) $A_n \cap (T^{k+1}.(Out(T) \times S_{k+1}))$ where $n=|T|^k$ and $T$ is a finite nonabelian simple group

6) an almost simple group acting primitively on $n$ points.

For classical groups the main result is Aschbacher's Theorem in the paper pointed out by Rivin. More details of the structure of the subgroups is given in the book by Kleidman and Liebeck.

For exceptional groups of Lie type there are papers by Liebeck and Seitz as noted by Barnea.

Both these results and Aschbacher's Theorem have the same philosophy as the O'Nan-Scott Theorem, namely that a maximal subgroup is either one of a small number of natural families that are usually stabilisers of some geometric structure, or is almost simple.

For sporadic simple groups, all the information in in the online Atlas. They are all known except for the monster. In this case there are a couple of possibilities for maximal subgroups where it is not known if they are actually subgroups.

share|improve this answer
    
Thanks a lot for your comprehensive answer. –  sebastian Jan 13 '13 at 21:23

In general the question is too ambitious, but more can be said than might be expected. For example, it follows from results of J.G. Thompson that if $M$ is a maximal subgroup of a non-Abelian finite simple group $G$ and $M$ is nilpotent, then $M$ is a Sylow $2$-subgroup of $G$ and is non-Abelian ( this does occur "in nature", for example for $G = {\rm PSL}(2,17)).$ The proof by Feit and Thompson of the solvability of finite groups of odd order proceeds by analyzing the structure of maximal subgroups of a minimal non-Abelian simple group of odd order, and the interplay between them. Perhaps the person who has exploited the structure of maximal subgroups of finite simple groups most is H. Bender, and this has given rise to the term "the Bender method". A remarkable general result of Bender, which has had many extensions and appications is that if $G$ is a non-Abelian finite simple groups with distinct maximal subgroups $A$ and $B$ such that the generalized Fitting subgroups of $A$ and $B$ satisfy $F^{*}(A) \leq B$ and $F^{*}(B) \leq A$, then there is a prime $p$ such that $F^{*}(A)$ and $F^{*}(B)$are both $p$-groups (again, the exceptional situation does occur in nature, for example if $G$ is a simple group of Lie type of characteristic $p$ and rank greater than $1$). However, if $A$ and $B$ are both solvable, then $p = 2$ or $3$. It does, however, appear that there are situations in the study of finite simple groups where the Bender method is not as easily applicable as the method of signalizer functors.

share|improve this answer
    
Thank you so much! –  sebastian Nov 17 '12 at 14:15

I am not an expert, but I believe lots of work was done on the case of groups of Lie type by Seitz, Liebeck and others. I think the basic paper is "The maximal subgroups of classical algebraic groups" by Seitz. These results are of great importance in the development of the many of the random generation results that were found in the last 20 years. But I will leave it to the experts to add more information.

share|improve this answer

I think the basic paper is actually this one:

Aschbacher, Michael. "On the maximal subgroups of the finite classical groups." Inventiones mathematicae 76.3 (1984): 469-514.

share|improve this answer

The Atlas of Finite Group Representations has information on the maximal subgroups of some particular finite simple groups. This is at least a handy place to start. See

http://brauer.maths.qmul.ac.uk/Atlas/v3/lin/L253/#maxes

for an example. The front page is at

http://brauer.maths.qmul.ac.uk/Atlas/v3/

share|improve this answer

The recent book of Malle and Testerman, Linear algebraic groups and finite groups of Lie type, has several chapters on the subject. From the MR review:

some important recent developments are treated here for the first time. For instance, the authors describe the classification of the maximal subgroups of simple algebraic groups, and this is used in their subsequent analysis of the subgroup structure of finite groups of Lie type.

(etc. The review goes on to describe the Aschbacher/Liebeck-Seitz results mentioned in other replies.)

Also Robert A. Wilson's The finite simple groups addresses the question, for the classical groups in a devoted section (3.10), for the others by systematically giving references to original papers classifying the maximal subgroups.

share|improve this answer

Every finite group can be broken down into elements consisting of prime cycle permutations. Just go up the lattice. Every nontrivial subgroup has to contain at least one of these prime cycle permutations: http://oeis.org/A186202

share|improve this answer
1  
This OEIS link is essentially about the number of minimal subgroups of the symmetric groups. The question, however, is about the structure (not just the number) of maximal (not minimal) subgroups of finite simple groups (not the symmetric groups). –  Andreas Blass Nov 30 '12 at 0:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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