theorem
  ord 1_G = 1
proof
A1: for n st (1_G) |^ n = 1_G & n <> 0 holds 1 <= n by NAT_1:14;
  (not 1_G is being_of_order_0) & (1_G) |^ 1 = 1_G by Lm4;
  hence thesis by A1,Def11;
end;
