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Oct 30, 2013 at 4:11 history edited paul Monsky CC BY-SA 3.0
Notification of publication added.
Apr 19, 2011 at 18:02 answer added Jim Humphreys timeline score: 1
Apr 19, 2011 at 15:26 answer added paul Monsky timeline score: 6
Apr 17, 2011 at 8:32 answer added Geoff Robinson timeline score: 3
Apr 14, 2011 at 15:59 history edited paul Monsky CC BY-SA 3.0
new material (an outline of my argument) added
Apr 12, 2011 at 16:56 answer added Jim Humphreys timeline score: 1
Apr 12, 2011 at 12:32 comment added Frieder Ladisch Isaacs' proof takes more than a page, and before stating the theorem he says: "As is often the case when trying to identify a particular group up to isomorphism, there is a great deal of tedious and detailed work involved. Since the result hardly seems to be worth that much effort, the proof which follows is somewhat more sketchy than most in this book." However, for $PSL_2$ over prime fields, I guess the situation is simpler, since the prime divisor $p$ is big relatively to the order of the group.
Apr 12, 2011 at 2:00 comment added Steve D @Junkie: Yes, but no simple group of order 360 has 6 Sylow 5-groups!
Apr 12, 2011 at 1:07 comment added Junkie @Steve D: All Rotman's exercise (8.12) says is "Prove that any two simple groups of order 360 are isomorphic, and conclude that $PSL(2,9)\cong A_6$. (Hint: Show that a Sylow 5-subgroup has 6 conjugates.)" My recollection is that the proof of Isaacs took nearly a page of rather dense character theory calculations.
Apr 12, 2011 at 1:03 comment added Junkie The 1912 census of Siceloff jstor.org/stable/2370496 says there is only one such group for $p=17,19$, citing Frobenius and Burnside's "Theory of Groups", and the latter appears on page 456 (p476 of the PDF) I think latexnical.com/library/Burnside/William/… : "In fact, if a simple group $h$ is a self-conjugate sub-group of $G$ it must be contained in $H$. Also, since $h$ is a doubly transitive group of degree $p^n + 1$, it must contain every operation of order $p$ that occurs in $G$. Now we may shew that these operations generate $H$..."
Apr 12, 2011 at 0:50 comment added Steve D That said, I would love to see a general argument using Burnside's Transfer Theorem!!
Apr 12, 2011 at 0:45 comment added Steve D I think it is well-known the exercise in Rotman is incorrect.
Apr 12, 2011 at 0:45 comment added Junkie "If simple, [$G$ of order 1092] must contain 14 subgroups of order 13, each being self-conjugate in a group of order $6.13$. Since a group of degree 14 cannot contain operations of order 26 or 39, this latter subgroup must be metacyclical in type. Again there must be 78 subgroups of order 7, each self-conjugate in a subgroup order $2.7$; and this must be dihedral in type, as otherwise the 78 subgroups would contain $78.12$ distinct operations. Hence the distribution of the operations of the group in conjugate sets is necessarily identical with that of the known simple group of this order."
Apr 12, 2011 at 0:43 comment added Junkie Burnside does $p=13$ in his 1895 paper (on the last page). plms.oxfordjournals.org/content/s1-26/1/325.full.pdf The proof is only a paragraph long, relying on known transitive groups of degree 14 it seems.
Apr 12, 2011 at 0:33 history edited Frieder Ladisch CC BY-SA 3.0
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Apr 12, 2011 at 0:31 comment added Junkie The Warwick wiki looks broken in rendering LaTeX, but it does $p=7$, as part of the standard "all simple groups less than order 500" exercise. wiki.dcs.warwick.ac.uk/ma3e2tidywiki/index.php/… $ $ The case of $p=9$ (not prime, but) is done a few places like Isaacs and an exercise in Rotman, usually in the guise of $A_6$. $ $ Cole does $p=11$ in his 1893 census. jstor.org/stable/2369516?seq=12
Apr 11, 2011 at 23:44 history asked paul Monsky CC BY-SA 3.0