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Math Help - abstract algebra - dihedral groups

  1. #16
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    Quote Originally Posted by dori1123 View Post
    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.
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  2. #17
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    Thank you very much for helping me and thank you for your patience.
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