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, A being non empty Subset of G holds
  for x being Element of G holds card A = card (x" * A * x)
proof
  let G be Group, A be non empty Subset of G;
  let x be Element of G;
  set B = x" * A * x;
  A is non empty & B is non empty proof set a = the Element of A;
    x" * a * x in B by Th15;
    hence thesis;
  end;
  then reconsider B as non empty Subset of G;
  deffunc F(Element of G) = x"*$1*x;
  consider f being Function of A, the carrier of G such that
A1: for a being Element of A holds f.a = F(a) from FUNCT_2:sch 4;
A2: dom f = A by FUNCT_2:def 1;
A3: rng f c= B
  proof
    let s be object;
    assume s in rng f;
    then consider a being object such that
A4: a in A and
A5: s = f.a by A2,FUNCT_1:def 3;
    reconsider a as Element of A by A4;
    s = F(a) by A1,A5;
    hence thesis by Th15;
  end;
A6: B c= rng f
  proof
    let s be object;
    assume s in B;
    then consider a being Element of G such that
A7: s = x" * a * x and
A8: a in A by Th15;
    ex x being Element of G st x in dom f & s = f.x
    proof
      take a;
      thus thesis by A1,A7,A8,FUNCT_2:def 1;
    end;
    hence thesis by FUNCT_1:def 3;
  end;
  reconsider f as Function of A,B by A2,A3,FUNCT_2:2;
  deffunc FF(Element of G) = x*$1*x";
  consider g being Function of B, the carrier of G such that
A9: for b being Element of B holds g.b = FF(b) from FUNCT_2:sch 4;
A10: dom g = B by FUNCT_2:def 1;
  rng g c= A
  proof
    let s be object;
    assume s in rng g;
    then consider a being object such that
A11: a in B and
A12: s = g.a by A10,FUNCT_1:def 3;
    reconsider a as Element of B by A11;
A13: s = FF(a) by A9,A12;
    consider c being Element of G such that
A14: a = x" * c * x and
A15: c in A by Th15;
    s = x * (x" * c) * x * x" by A13,A14,GROUP_1:def 3
      .= (x * x") * c * x * x" by GROUP_1:def 3
      .= (x * x") * c * (x * x") by GROUP_1:def 3
      .= 1_G * c * (x * x") by GROUP_1:def 5
      .= 1_G * c * 1_G by GROUP_1:def 5
      .= 1_G * c by GROUP_1:def 4
      .= c by GROUP_1:def 4;
    hence thesis by A15;
  end;
  then reconsider g as Function of B,A by A10,FUNCT_2:2;
A16: for a,b being Element of A st f.a = f.b holds a = b
  proof
    let a,b be Element of A such that
A17: f.a = f.b;
A18: x" * a * x in B by Th15;
A19: x" * b * x in B by Th15;
A20: g.(f.a) = g.(x" * a * x) by A1
      .= x * (x" * a * x) * x" by A9,A18
      .= x * (x" * a) * x * x" by GROUP_1:def 3
      .= (x * x") * a * x * x" by GROUP_1:def 3
      .= (x * x") * a * (x * x") by GROUP_1:def 3
      .= 1_G * a * (x * x") by GROUP_1:def 5
      .= a * (x * x") by GROUP_1:def 4
      .= a * 1_G by GROUP_1:def 5
      .= a by GROUP_1:def 4;
    g.(f.b) = g.(x" * b * x) by A1
      .= x * (x" * b * x) * x" by A9,A19
      .= x * (x" * b) * x * x" by GROUP_1:def 3
      .= (x * x") * b * x * x" by GROUP_1:def 3
      .= (x * x") * b * (x * x") by GROUP_1:def 3
      .= 1_G * b * (x * x") by GROUP_1:def 5
      .= b * (x * x") by GROUP_1:def 4
      .= b * 1_G by GROUP_1:def 5
      .= b by GROUP_1:def 4;
    hence thesis by A17,A20;
  end;
  A,B are_equipotent
  proof
    take f;
    thus thesis by A3,A6,A16,FUNCT_2:def 1,GROUP_6:1,XBOOLE_0:def 10;
  end;
  hence thesis by CARD_1:5;
end;
