
theorem FAuthelp:
for p being Prime
for n being non zero Nat
for F being Field st card F = p|^n
for m,k being Nat st 0 <= m & m <= n-1 & 0 <= k & k <= n-1 & m <> k
holds (Frob F)`^m <> (Frob F)`^k
proof
let p be Prime, n be non zero Nat, F be Field;
assume AS1: card F = p|^n;
let m,k being Nat;
assume AS2: 0 <= m & m <= n-1 & 0 <= k & k <= n-1 & m <> k;
Char F = p by AS1,T5; then
reconsider F as p-characteristic finite Field by AS1,RING_3:def 6;
now
  assume K: (Frob F)`^m = (Frob F)`^k;
  per cases by AS2,XXREAL_0:1;
  suppose A: k > m; then
    k - m > k - k by XREAL_1:15; then
    reconsider mk = k - m as Element of NAT by INT_1:3;
    reconsider mk as non zero Nat by A;
    B: Frob F`^mk = (Frob F)`^0 by K,TT3 .= id F by T1;
    mk + m = k; then
    mk <= k by NAT_1:11; then
    0 < mk & mk <= n - 1 by AS2,XXREAL_0:2;
    hence contradiction by lemAuthelp,B,AS1;
    end;
  suppose A: m > k; then
    m - k > m - m by XREAL_1:15; then
    reconsider mk = m - k as Element of NAT by INT_1:3;
    reconsider mk as non zero Nat by A;
    B: Frob F`^mk = (Frob F)`^0 by K,TT3 .= id F by T1;
    mk + k = m; then
    mk <= m by NAT_1:11; then
    0 < mk & mk <= n - 1 by AS2,XXREAL_0:2;
    hence contradiction by lemAuthelp,B,AS1;
    end;
  end;
hence thesis;
end;
