Re: representing the dihedral group {1,r,r^2...s,sr^2...}

*To*: mathgroup at smc.vnet.net*Subject*: [mg98950] Re: representing the dihedral group {1,r,r^2...s,sr^2...}*From*: obott0 at gmail.com*Date*: Wed, 22 Apr 2009 05:14:11 -0400 (EDT)*References*: <gsh100$gr$1@smc.vnet.net> <gsivel$9c7$1@smc.vnet.net>

> The dihedral group of order 2n may be represented as > <a,b|(ab)^n = a^2 = b^2 = 1>. If you start with an > arbitrary finite string of as and bs, simply remove > each consecutive pair of identical symbols. You will > be left with a product in one of the following four > forms: > (a*b)^m > (a*b)^m * b > (b*a)^m > (b*a)^m * a > > The exponent m may then be reduced modulo n. > Now, my ab is your r, my ba is your r^(n-1), my b is your s, > and (I think) my a is your r^(n-2)s. Thus, you should be > able to use this code to reduce in the alternative presentation > <r,s|r^n = s^2 = 1>. > Hope that helps, > Mark McClure Hmm.. I don't see how <a,b|(ab)^n = a^2 = b^2 = 1> is isomorphic to the dihedral group of order 2n. Using generators {r,s}, the relationship rs=s(r^-1) is satisfied. But abb = a which does not equal b(ab)^-1 = b*b^-1*a^-1 = a^-1. Can you explain? Otto