
theorem GRCY28:
  for G being finite Group, a be Element of G,
  n, p, s being Element of NAT
  st card gr{a} = n & n = p * s
  holds ord(a |^ p) = s
  proof
    let G be finite Group, a be Element of G, n, p, s be Element of NAT;
    assume
    AS1: card gr{a} = n & n = p * s;
    reconsider a0 = a as Element of gr{a} by GR_CY_2:2, STRUCT_0:def 5;
    A0: gr{a0} = gr{a} by GR_CY_2:3;
    ord(a0 |^ p) = card(gr{a0 |^ p}) by GR_CY_1:7
    .= card(gr{a |^ p}) by GROUP_4:2, GR_CY_2:3
    .= ord(a |^ p) by GR_CY_1:7;
    hence ord(a |^ p) = s by A0, AS1, GR_CY_2:8;
  end;
