# Thread: abstract algebra - dihedral groups

1. Originally Posted by dori1123
So for $x = r^0s = s$, $xr = sr$ does not equal to $rx = rs = sr^{-1}$ (unless $r = r^{-1}$?). For $x = r^ks$, $xr = r^ksr = sr^{-k}r = sr^{-k+1}$ and $rx = rr^ks = r^{k+1}s$, so $r^ksr$ does not equal to $rr^ks$. Right?
Of course $r\not = r^{-1}$ because that would imply $\text{ord}(r) = 1 \text{ or }2$ which we know is impossible since $\text{ord}(r) = k>2$ (Here we are assuming the dihedral group is sufficiently large). Thus, this shows that neither of these two elements commutes.

2. Thank you very much for helping me and thank you for your patience.

Page 2 of 2 First 12