Generators and relations

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October 2nd 2011, 07:47 PM
dwsmith

Generators and relations

Use generators and relations to show that if $x\in\text{D}_{2n}$ which is not a power of r, then $rx=xr^{-1}$

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

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 $rx=xr^{-1}$
October 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

$x = r^k s$

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

$x = r^k s$

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

Can I do this:

Since $rs=sr^{-1}$ , $r^ksr^{-1}=r^krs=r^{k+1}s\equiv rr^ks=r^{k+1}s$
October 3rd 2011, 12:16 AM
Swlabr

Re: Generators and relations

Quote:

Originally Posted by dwsmith

Can I do this:

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

Yup, that works.

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