Timeline for What groups have a second maximal subgroup below exactly four maximal subgroups?
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| when toggle format | what | by | license | comment | |
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| S Mar 30, 2014 at 14:28 | history | suggested | Sebastien Palcoux | CC BY-SA 3.0 |
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| Mar 30, 2014 at 14:20 | review | Suggested edits | |||
| S Mar 30, 2014 at 14:28 | |||||
| May 7, 2011 at 0:43 | comment | added | William DeMeo | I think you are right about what Basile means re. "examples of Feit and Palfy," and his Theorem D does imply there is no M_4 interval with A_n or S_n at the top, for n>4. As I'm sure you know, Feit was the first to give a (rather terse) description of an M_7 upper interval, in Sub[A31]. Palfy fleshed this out in "On Feit's Examples of Intervals in Subgroup Lattices", J. Algebra 116 (1988), where he gives examples of M_n (n=2,3,5,7,11) in Sub[A_p], for prime p>3. Thanks again for the suggestions. | |
| May 7, 2011 at 0:20 | comment | added | William DeMeo | The congruence lattice L of a finite algebra (A;F) is hereditary if every 0-1 sublattice of L is the congruence lattice of an algebra with the same universe A. In other words, L is a congruence lattice of (A; F) and, given any 0-1 sublattice L' of L, you can add operations G to the algebra so that L' is the congruence lattice of (A; F, G). To say it another way, all the sublattices are representable "where they sit" in the partition lattice of A. Besides Hegedus-Palfy, another reference is "OPC Lattices and Congruence Heredity" by John Snow, Algebra Universalis, 58 (2008). | |
| May 6, 2011 at 15:56 | comment | added | John Shareshian | It seems very likely that to characterize all [H,G]=M_4, one would need to use the CFSG. For a start, in his thesis Alberto Basile showed that one cannot have [H,G]=M_4 if G=S_n or A_n, n \geq 5. (I think - see Theorem D. If I understand what is meant by "examples of Feit and Palfy", then we never see M_4.) The thesis is on the arxiv. | |
| May 6, 2011 at 15:49 | comment | added | John Shareshian | 1) Jack, right, V is absolutely irreducible. It seems if V is not faithful then H does not have trivial core in G. Also, in case 2, the condition that no element of M have eigenvalue -1 on W need not hold if W=V, as in G=S_3. 2) William, I am trying to understand the definition of hereditary in the paper of Hegedus-Palfy. Is there a better place? Also, I would recommend Aschbacher's solo papers over our joint one. | |
| May 6, 2011 at 9:37 | comment | added | William DeMeo | Thank you very much for these interesting suggestions. I hadn't been thinking about semidirect products this way, but it looks like I should be. As for Baddeley, Borner, Lucchini, yes, I'm familiar with their work, as well as your recent work with Aschbacher. However, I'm particularly interested in M_4 because, afaik, no one has found a "hereditary" M_4. It seems that, if such an M_4 exists, we can assume it's an interval in a subgroup lattice. If we could then characterize such groups, perhaps we could prove there's no hereditary M_4... a longshot, but your suggestions are a great start! | |
| May 5, 2011 at 16:10 | comment | added | Jack Schmidt | I'd be interested in a few examples of type 2. They seemed rare, but perhaps I stopped looking too early. | |
| May 5, 2011 at 16:08 | comment | added | Jack Schmidt | In part 1, you just mean V is absolutely irreducible, right? Must V be faithful as well? | |
| May 5, 2011 at 14:29 | history | answered | John Shareshian | CC BY-SA 3.0 |