Re: Re: representing the dihedral group

*To*: mathgroup at smc.vnet.net*Subject*: [mg98989] Re: [mg98950] Re: representing the dihedral group*From*: Mark McClure <mcmcclur at unca.edu>*Date*: Thu, 23 Apr 2009 06:44:06 -0400 (EDT)*References*: <gsh100$gr$1@smc.vnet.net> <gsivel$9c7$1@smc.vnet.net>

On Wed, Apr 22, 2009 at 5:14 AM, <obott0 at gmail.com> wrote: >> The dihedral group of order 2n may be represented as >> <a,b|(ab)^n = a^2 = b^2 = 1>. > > 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? First, a does equal a^(-1), right? There is a fairly obvious homomorphism from your representation to mine generated by r -> ab s -> a We can show this homomorphism is surjective using the canonical forms in my representation. There was, however, a typo in my previous message. The four canonical forms should be (a*b)^m (a*b)^m * a (b*a)^m (b*a)^m * b Note that I had the trailing a and b switched before. Now we simply have to show that some element in your representation maps to each possible form. It's not too hard to check that r^m -> (a*b)^m r^m * s -> (a*b)^m * a r^(-m) -> (b*a)^m r^(-m) * s*r -> (b*a)^-m * b Mark