theorem Th17:
  g * h = h * g iff (g * h)" = g" * h"
proof
  thus g * h = h * g implies (g * h)" = g" * h" by Th16;
  assume (g * h)" = g" * h";
  then
A1: (h * g) * (g * h)" = h * g * g" * h" by Def3
    .= h * (g * g") * h" by Def3
    .= h * 1_G * h" by Def5
    .= h * h" by Def4
    .= 1_G by Def5;
  (g * h) * (g * h)" = 1_G by Def5;
  hence thesis by A1,Th6;
end;
