# s and sr generators for D2n

• Oct 3rd 2011, 03:37 AM
dwsmith
s and sr generators for D2n
Use generators and relations to show that every element of \$\displaystyle D_{2n}\$ which is not a power of r has order 2. Deduce that \$\displaystyle D_{2n}\$ is generated by the two elements s and sr, both of which have order 2.

Little confused on what to do and how to start. The first part seems trivial but how to show it is another thing.
• Oct 3rd 2011, 03:52 AM
Deveno
Re: s and sr generators for D2n
we know that sr = r^-1s, right? this, in turn implies sr^k = (r^-1s)r^(k-1) = (r^-1)(sr)(r^(k-2)) = r^-2sr^(k-2)

=......= r^(-k)s (the dots are the same step, applied repeatedly to move r's to the left of s).

so (r^ks)(r^ks) = r^k(sr^k)s = (r^k)(r^(-k)s)s = es^2 = s^2 = e. so that's the easy part.

now all we have to do is show that D2n = <r,s> is contained in <sr,s>; or, what is the same thing, that r and s are in <sr,s>

clearly s is in <sr,s>, and r = s(sr) is in <sr,s> as well.

(of course, we haven't shown that sr has order 2, but by the first part, sr is not in <r>, so...)