reserve G for strict Group,
  a,b,x,y,z for Element of G,
  H,K for strict Subgroup of G,
  p for Element of NAT,
  A for Subset of G;

theorem
  for G being Group, H being Subgroup of G, a being Element of H
  for b being Element of G st a = b & G is finite holds ord a = ord b
proof
  let G be Group, H be Subgroup of G;
  let a be Element of H;
  let b be Element of G such that
A1: a =b and
A2: G is finite;
A3: not a is being_of_order_0 by A2,GR_CY_1:6;
A4: not b is being_of_order_0 by A2,GR_CY_1:6;
A5: a |^ (ord a) = 1_H by A3,GROUP_1:def 11;
A6: b |^ ord b = 1_G by A4,GROUP_1:def 11;
  a |^ (ord a) = b |^ (ord a) by A1,GROUP_4:1;
  then
A7: ord b divides ord a by A5,A6,GROUP_1:44,GROUP_2:44;
  a |^ (ord b) = 1_G by A1,A6,GROUP_4:1;
  then a |^ (ord b) = 1_H by GROUP_2:44;
  then ord a divides ord b by GROUP_1:44;
  hence thesis by A7,NAT_D:5;
end;
