       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

```

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