
theorem Frobcard:
for p being Prime
for n being non zero Nat
for F be Field st card F = p|^n
holds card{(Frob F)`^m where m is Nat : 0 <= m & m <= n-1} = n
proof
let p be Prime, n be non zero Nat, F be Field;
assume AS: card F = p|^n;
defpred P[object,object] means
  ex x being Element of Seg n,
     y being Element of NAT st $1 = x & y = x-1 & $2 = (Frob F)`^y;
B1: for x,y1,y2 being object st x in Seg n & P[x,y1] & P[x,y2] holds y1 = y2;
B2: now let x be object;
    assume x in Seg n;
    then reconsider m = x as Element of Seg n;
    1 <= m by FINSEQ_1:1; then
    reconsider z = m - 1 as Element of NAT by INT_1:3;
    thus ex y being object st P[x,y]
      proof
      take (Frob F)`^z;
      thus thesis;
      end;
    end;
consider f being Function such that
C: dom f = Seg n &
   for x being object st x in Seg n holds P[x,f.x] from FUNCT_1:sch 2(B1,B2);
A1: now let o be object;
    assume o in {(Frob F)`^m where m is Nat : 0 <= m & m <= n-1};
    then consider i being Nat such that
    A2: o = (Frob F)`^i & 0 <= i & i <= n-1;
    0 + 1 <= i + 1 & i + 1 <= (n-1)+1 by A2,XREAL_1:6; then
    reconsider x = i + 1 as Element of Seg n by FINSEQ_1:1;
    P[x,f.x] by C;
    hence o in rng f by C,A2,FUNCT_1:def 3;
    end;
now let o be object;
    assume o in rng f;
    then consider u being object such that
    A2: u in dom f & o = f.u by FUNCT_1:def 3;
    reconsider j = u as Element of NAT by A2,C;
    P[u,f.u] by C,A2; then
    consider x being Element of (Seg n), y being Element of NAT such that
    A3: u = x & y = x-1 & f.x = (Frob F)`^y;
    1 <= x & x <= n by FINSEQ_1:1; then
    0 <= y & y <= n-1 by A3,XREAL_1:9;
    hence o in {(Frob F)`^m where m is Nat : 0 <= m & m <= n-1} by A2,A3;
    end; then
A: rng f = {(Frob F)`^m where m is Nat : 0 <= m & m <= n-1} by A1,TARSKI:2;
now assume not f is one-to-one;
  then consider x1,x2 being object such that
  A1: x1 in dom f & x2 in dom f & f.x1 = f.x2 & x1 <> x2;
  reconsider j1 = x1, j2 = x2 as Element of NAT by A1,C;
  consider n1 being Element of Seg n,
           y1 being Element of NAT such that
  A2: x1 = n1 & y1 = n1-1 & f.x1 = (Frob F)`^y1 by A1,C;
  consider n2 being Element of Seg n,
           y2 being Element of NAT such that
  A3: x2 = n2 & y2 = n2-1 & f.x2 = (Frob F)`^y2 by A1,C;
  1 <= j1 & j1 <= n & 1 <= j2 & j2 <= n by A1,C,FINSEQ_1:1; then
  0 <= y1 & y1 <= n-1 & 0 <= y2 & y2 <= n-1 by A2,A3,XREAL_1:9; then
  y1 = y2 by A3,A2,A1,AS,FAuthelp;
  hence contradiction by A1,A2,A3;
  end;
then card {(Frob F)`^m where m is Nat : 0 <= m & m <= n-1}
   = card Seg n by A,C,CARD_1:70 .= n by FINSEQ_1:57;
hence thesis;
end;
