Timeline for Suzuki group order
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2013 at 17:39 | vote | accept | Tom | ||
| Jan 4, 2013 at 18:42 | answer | added | Jim Humphreys | timeline score: 2 | |
| Jan 3, 2013 at 22:16 | answer | added | Geoff Robinson | timeline score: 3 | |
| Dec 25, 2012 at 20:19 | comment | added | Simon Thomas | @Jim: Rutgers doesn't have it online either. My guess/memory is that it also treats the Suzuki and Ree groups as the corresponding result for these groups is also needed for the main application: the classification of the simple periodic linear groups. In any case, these cases were done earlier by Kegal and Stingl. | |
| Dec 25, 2012 at 16:48 | comment | added | Jim Humphreys | @Simon: Thanks for calling this article to my attention. Probably our library stores it somewhere in print form, but it isn't available to us online. Judging from Carter's review, it deals with Chevalley and Steinberg (split and quasi-split) types. Does it also treat Suzuki and Ree groups? | |
| Dec 25, 2012 at 15:37 | comment | added | Simon Thomas | @Jim: Assuming the Suzuki groups are "sufficiently large", any inclusion is natural. In fact, this is true for any (possibly twisted) Lie type. For example, see: MR0734665 (85k:20094) Reviewed Hartley, B.(4-MANC); Shute, G.(1-WIP) Monomorphisms and direct limits of finite groups of Lie type. Quart. J. Math. Oxford Ser. (2) 35 (1984), no. 137, 49–71. | |
| Dec 25, 2012 at 13:58 | comment | added | Jim Humphreys |
What is the motivation? If it's just a numerical question, why the tag 'group-theory'? Perhaps the more interesting group-theoretic question is when or if one Suzuki group occurs as a subgroup of another. By construction they all occur in certain $B_2$ Chevalley groups, which in turn sometimes lie in each other (depending on field inclusions). But is there any reason to expect natural inclusions among Suzuki groups?
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| Dec 25, 2012 at 11:55 | comment | added | Peter Mueller | I doubt that this is research level, in particular because the OP doesn't even provide the order formula for the Suzuki groups, which is $q^2(q^2+1)(q-1)$ for $q=2^m$, $m\ge3$ odd. Anyway, for instance if $q_1=2^3$ and $q^2=2^m$ with $m\equiv3\pmod{6}$, then the former Suzuki group order divides the latter one. By easy number theory, one can work out all such cases. | |
| Dec 25, 2012 at 9:58 | history | asked | Tom | CC BY-SA 3.0 |