Skip to main content

Timeline for A question on almost simple groups

Current License: CC BY-SA 3.0

8 events
when toggle format what by license comment
Dec 31, 2012 at 16:34 comment added majid arezoomand A very nice proof!
Dec 31, 2012 at 10:35 history edited Geoff Robinson CC BY-SA 3.0
added 1083 characters in body
Dec 31, 2012 at 7:39 history edited Geoff Robinson CC BY-SA 3.0
typos
Dec 31, 2012 at 3:34 comment added Mart @Geoff Robinson: Thank you very much, it was most helpful!
Dec 30, 2012 at 20:01 history edited Geoff Robinson CC BY-SA 3.0
Show $p$ does divide $|S|$ in response to Mart's question.
Dec 30, 2012 at 18:09 comment added Derek Holt Yes that is indeed a more direct argument!
Dec 30, 2012 at 17:38 comment added Mart @Geoff Robinson: Thanks. By Derek's answer if $S\unlhd G$ with $S$ simple and $G\leq Aut(S)$, then $p$ is prime divisor of $S$. Whether by your answer it implies that $p$ is prime divisor of $S$?
Dec 30, 2012 at 16:27 history answered Geoff Robinson CC BY-SA 3.0