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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


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