
theorem
  for F be commutative Skew-Field, G be finite Subgroup of MultGroup(F)
  holds G is cyclic Group
proof
  let F be commutative Skew-Field, G be finite Subgroup of MultGroup(F);
  set a = the Element of G;
  defpred P[Nat,Element of G,Element of G] means ord $2 < ord $3;
  assume not G is cyclic Group;
  then
A1: not ex x be Element of G st ord x = card (G) by GR_CY_1:19;
A2: for x be Element of G holds ord x < card (G)
  proof
    let x be Element of G;
    ord x <= card (G) by GR_CY_1:8,NAT_D:7;
    hence thesis by A1,XXREAL_0:1;
  end;
A3: for n being Nat for x being Element of G ex y being Element
  of G st P[n,x,y]
  proof
    let n be Nat, x be Element of G;
    set n = ord x;
    n < card G by A2;
    then
A4: card gr {x} <> card G by GR_CY_1:7;
    the carrier of (gr {x}) c= the carrier of G by GROUP_2:def 5;
    then the carrier of (gr {x}) c< the carrier of G by A4,XBOOLE_0:def 8;
    then (the carrier of G) \ (the carrier of (gr {x})) <> {} by XBOOLE_1:105;
    then consider z be object such that
A5: z in (the carrier of G) \ (the carrier of (gr {x})) by XBOOLE_0:def 1;
    reconsider z as Element of G by A5;
    set m = ord z;
    set l = m lcm n;
    n divides (m lcm n) by INT_2:def 1;
    then consider j be Integer such that
A6: l = n*j by INT_1:def 3;
A7: 1 <= card gr {x} by GROUP_1:45;
    then
A8: 1 <= n by GR_CY_1:7;
    then l/n = j by A6,XCMPLX_1:89;
    then
A9: j is Element of NAT by INT_1:3;
    not m divides n
    proof
      assume m divides n;
      then consider j be Integer such that
A10:  n = m*j by INT_1:def 3;
A11:  0 < n by A7,GR_CY_1:7;
      z|^n = (z|^m)|^j by A10,GROUP_1:35
        .= (1_(G))|^j by GROUP_1:41
        .= 1_(G) by GROUP_1:31;
      then z is Element of gr {x} by A11,Lm11;
      hence contradiction by A5,XBOOLE_0:def 5;
    end;
    then
A12: n <> l by INT_2:def 1;
A13: 1 <= card gr {z} by GROUP_1:45;
    then
A14: m <> 0 by GR_CY_1:7;
    1 <= m by A13,GR_CY_1:7;
    then consider m0,n0 be Element of NAT such that
A15: l = n0*m0 and
A16: n0 gcd m0 = 1 and
A17: n0 divides n and
A18: m0 divides m and
A19: n0 <> 0 and
A20: m0 <> 0 by A8,INT_7:17;
    ex u be Integer st m = m0* u by A18,INT_1:def 3;
    then m/m0 is Integer by A20,XCMPLX_1:89;
    then reconsider d2 = m/m0 as Element of NAT by INT_1:3;
    ex j be Integer st n = n0*j by A17,INT_1:def 3;
    then n/n0 is Integer by A19,XCMPLX_1:89;
    then reconsider d1 = n/n0 as Element of NAT by INT_1:3;
    set y = (x|^d1)*(z|^d2);
    m = d2*m0 by A20,XCMPLX_1:87;
    then
A21: ord (z|^d2) = m0 by INT_7:30;
    n <> 0 by A7,GR_CY_1:7;
    then j <> 0 by A14,A6,INT_2:4;
    then n*1 <= n*j by A9,NAT_1:14,XREAL_1:64;
    then
A22: n < l by A12,A6,XXREAL_0:1;
    n = d1*n0 by A19,XCMPLX_1:87;
    then ord (x|^d1) = n0 by INT_7:30;
    then ord y = m0*n0 by A16,A21,INT_7:25;
    hence ex y be Element of G st n < ord y by A15,A22;
  end;
  consider f being sequence of  the carrier of G such that
A23: f.0 = a & for n being Nat holds P[n,f.n qua Element of G
,f.(n+1)
   qua Element of G] from
  RECDEF_1:sch 2(A3);
  deffunc F(Nat) = ord (f.$1);
  consider g be sequence of  NAT such that
A24: for k be Element of NAT holds g.k=F(k) from FUNCT_2:sch 4;
A25: for k be Nat holds g.k=F(k)
   proof let k be Nat;
     k in NAT by ORDINAL1:def 12;
    hence thesis by A24;
   end;
A26: now
    let k be Nat;
A27: g.(k+1) = ord (f.(k+1)) by A25;
    g.k = ord (f.k) by A25;
    hence g.k < g.(k+1) by A23,A27;
  end;
A28: for k,m be Element of NAT holds g.k < g.(k+1+m)
  proof
    let k be Element of NAT;
    defpred P[Nat] means g.k < g.(k+1+$1);
A29: now
      let m be Nat;
      assume
A30:  P[m];
      g.(k+1+m) < g.((k+1+m) + 1) by A26;
      hence P[m+1] by A30,XXREAL_0:2;
    end;
A31: P[0] by A26;
    for m be Nat holds P[m] from NAT_1:sch 2(A31,A29);
    hence thesis;
  end;
A32: for k,m be Element of NAT st k < m holds g.k < g.m
  proof
    let k, m be Element of NAT;
    assume
A33: k < m;
    then m-k is Element of NAT by INT_1:5;
    then reconsider mk = m-k as Nat;
    m-k <> 0 by A33;
    then consider n0 be Nat such that
A34: mk = n0+1 by NAT_1:6;
    reconsider n = n0 as Element of NAT by ORDINAL1:def 12;
    m = k+1+n by A34;
    hence thesis by A28;
  end;
  now
    let x1,x2 be object;
    assume that
A35: x1 in NAT and
A36: x2 in NAT and
A37: g.x1 = g.x2;
    reconsider xx1 = x1,xx2 = x2 as Element of NAT by A35,A36;
A38: xx2 <= xx1 by A32,A37;
    xx1 <= xx2 by A32,A37;
    hence x2 = x1 by A38,XXREAL_0:1;
  end;
  then g is one-to-one by FUNCT_2:19;
  then dom g, rng g are_equipotent by WELLORD2:def 4;
  then card dom g = card rng g by CARD_1:5;
  then
A39: card rng g = card NAT by FUNCT_2:def 1;
  rng g c= Segm card G
  proof
    let y be object;
    assume y in rng g;
    then consider x be object such that
A40: x in NAT and
A41: y=g.x by FUNCT_2:11;
    reconsider x as Element of NAT by A40;
    reconsider gg = g.x as Element of NAT;
    g.x = ord (f.x) by A25;
    then gg < card G by A2;
    hence y in Segm(card G) by A41,NAT_1:44;
  end;
  hence contradiction by A39;
end;
