# Generators and relations

• Oct 2nd 2011, 07:47 PM
dwsmith
Generators and relations
Use generators and relations to show that if $\displaystyle x\in\text{D}_{2n}$ which is not a power of r, then $\displaystyle rx=xr^{-1}$

$\displaystyle \text{D}_{2n}=<r,s:r^n=s^2=1, \ rs=sr^{-1}>$

Basically, x in this case is a flip. I am not sure what needs to proved. x is just a flip so by definition it satisfy $\displaystyle rx=xr^{-1}$
• Oct 2nd 2011, 08:20 PM
Amer
Re: Generators and relations
since x is not a power of r it will be a multiplication of power of r and s say

$\displaystyle x = r^k s$

we have to show that $\displaystyle r r^k s = r^k s r^{-1}$
• Oct 2nd 2011, 08:26 PM
dwsmith
Re: Generators and relations
Quote:

Originally Posted by Amer
since x is not a power of r it will be a multiplication of power of r and s say

$\displaystyle x = r^k s$

we have to show that $\displaystyle r r^k s = r^k s r^{-1}$

Can I do this:

Since $\displaystyle rs=sr^{-1}$, $\displaystyle r^ksr^{-1}=r^krs=r^{k+1}s\equiv rr^ks=r^{k+1}s$
• Oct 3rd 2011, 12:16 AM
Swlabr
Re: Generators and relations
Quote:

Originally Posted by dwsmith
Can I do this:

Since $\displaystyle rs=sr^{-1}$, $\displaystyle r^ksr^{-1}=r^krs=r^{k+1}s\equiv rr^ks=r^{k+1}s$

Yup, that works.