From jaapsch@hotmail.com Sat Nov 20 00:42:34 CST 2004 Article: 339026 of sci.math Path: news!news.niu.edu!arclight.uoregon.edu!wns13feed!worldnet.att.net!199.218.7.141!news.glorb.com!postnews.google.com!not-for-mail From: jaapsch@hotmail.com (Jaap) Newsgroups: sci.math Subject: Re: Question on alternating groups Date: 15 Nov 2004 06:24:35 -0800 Organization: http://groups.google.com Lines: 26 Message-ID: <30fbb89.0411150624.14997ec8@posting.google.com> References: NNTP-Posting-Host: 134.146.0.12 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit X-Trace: posting.google.com 1100528675 6033 127.0.0.1 (15 Nov 2004 14:24:35 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Mon, 15 Nov 2004 14:24:35 +0000 (UTC) Xref: news sci.math:339026 Simon Guest wrote in message news:... > Is it possible to generate the alternating group A_p with a p-cycle and > one other element (p a prime >=5), perhaps a 3-cycle? Sure it is, and p need not even be prime (but must be odd so that the p-cycle is an even permutation). With the p-cycle you can conjugate the 3-cycle (1 2 3) to get any 3-cycle of consecutive elements (n-1 n n+2). WHen you start conjugating these, pretty soon you can get any 3-cycle, and hence the whole of A_p. I collect and play with permutation puzzles a lot. One of these is Topspin by Binary Arts: http://www.geocities.com/jaapsch/puzzles/topspin.htm This has a loop of pieces which can slide as a whole (a p-cycle, though p here is even), and a turntable which does another small permutation (in this case it does (14)(23) instead of a 3-cycle). If you imagine your question as such a puzzle, you will see that it is very easy to build up the row of pieces one by one, until 1..p-2 is solved. The last two pieces will automatically be correct by the parity. Jaap Jaap's Puzzle Page http://www.geocities.com/jaapsch/puzzles/