Timeline for On J.G.Thompson's conjecture about conjugacy classes of finite simple groups
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 19, 2016 at 2:23 | vote | accept | C. Simon | ||
| Oct 19, 2016 at 1:58 | vote | accept | C. Simon | ||
| Oct 19, 2016 at 2:21 | |||||
| Oct 18, 2016 at 11:34 | vote | accept | C. Simon | ||
| Oct 18, 2016 at 11:37 | |||||
| Oct 17, 2016 at 21:22 | answer | added | Igor Rivin | timeline score: 5 | |
| Oct 17, 2016 at 12:19 | comment | added | Nick Gill | Shalev has a stronger conjecture than Thompson's in this paper -- annals.math.princeton.edu/wp-content/uploads/… (Conj 10.3). Shalev proposes sufficient conditions for a conjugacy class $C$ in a finite simple group $G$ to satisfy $G=CC$. (Basically the conjecture proposes that any real class that is "big enough" should do the trick.) Proving this stronger conjecture is, presumably, harder, so I don't suggest you try and attack it... But it at least gives you an idea of what classes might work. Proving Thompson's conjecture would be pretty amazing. | |
| Oct 17, 2016 at 9:36 | comment | added | Geoff Robinson | As I mentioned, the class $C$ must be closed under taking inverses. It follows easily from this that any product of the form $ab$ with $a,b \in C$ is a commutator ( for $a = c^{-1}b^{-1}c$ for some $c \in G$). Hence Thompson's conjecture implies that every element of $G$ is a commutator when $G$ is simple, which is a (now proved) conjecture of Ore. | |
| Oct 17, 2016 at 8:49 | comment | added | Derek Holt | Before you can prove it, you will need to identify the class $C$ for which this might work. What I would do would be to start by doing some computer experiments for $n=5,6,7,8,9,\ldots$ in order to find out for which $C$ the conjecture is true and, based on that, try to formulate a conjecture, | |
| Oct 17, 2016 at 6:29 | comment | added | verret | @GerryMyerson For $n=5$, the answer to your question is yes. On the other hand, note that $n$-cycles fall in two distinct conjugacy classes in $A_n$ when $n$ is odd. For $n=5$, we don't have $A_5=CC$ where $C$ is one of these classes. | |
| Oct 17, 2016 at 5:51 | comment | added | Gerry Myerson | For $n\ge5$ odd, is every element of $A_n$ a product of two $n$-cycles? | |
| Oct 17, 2016 at 4:59 | comment | added | C. Simon | "The requirement is stronger than asking that evey element of $G$ is a commutator."? | |
| Oct 17, 2016 at 4:50 | history | edited | C. Simon | CC BY-SA 3.0 |
added 38 characters in body
|
| Oct 17, 2016 at 2:56 | history | edited | Myshkin | CC BY-SA 3.0 |
+ top level tag (gr.)
|
| Oct 17, 2016 at 1:08 | comment | added | Geoff Robinson | What do you mean $A_{n}$ in $S_{n}$? The simple group is $A_{n}$. The problem is to prove that if $G$ is a finite simple group, there is a conjugacy class $C$ such that every element of $G$ can be expressed in the form $ab$ with $a,b \in C.$ The class $C$ has to be closed under taking inverses. The conjecture has been checked for many simple groups, The requirement is stronger than asking that every element of $G$ is a commutator. | |
| Oct 17, 2016 at 0:44 | review | First posts | |||
| Oct 17, 2016 at 1:30 | |||||
| Oct 17, 2016 at 0:42 | review | Suggested edits | |||
| Oct 17, 2016 at 0:43 | |||||
| Oct 17, 2016 at 0:39 | history | asked | C. Simon | CC BY-SA 3.0 |