
theorem Th15:
  for G being finite commutative Group,
  m be Nat,
  H being finite Subgroup of G
  st the carrier of H = {x where x is Element of G: x|^m = 1_G} holds
  for q being Prime st q in support prime_factorization card H
  holds not q,m are_coprime
  proof
    let G be finite commutative Group,
    m be Nat,
    H be finite Subgroup of G;
    assume A1: the carrier of H = {x where x is Element of G: x|^m = 1_G};
    let q be Prime;
    assume A2: q in support prime_factorization card H;
    assume A3: q,m are_coprime;
    consider a be Element of H such that
    A4: a <> 1_H & ord a = q by A2,Lm5;
    a in {x where x is Element of G: x|^m = 1_G} by A1; then
    consider x be Element of G such that
    A5: x=a & x|^m = 1_G;
    A6:a|^m =1_G by A5,GROUP_4:2;
    q gcd m = 1 by A3,INT_2:def 3;
    then
    consider x,y be Integer such that
    A7: x*q + y*m = 1 by NAT_D:68;
    a = a|^1 by GROUP_1:26
    .=(a|^(q*x)) * (a|^(m*y)) by GROUP_1:33,A7
    .= ((a|^q) |^x) *(a|^(m*y)) by GROUP_1:35
    .= ((a|^q) |^x) * ((a|^m) |^y) by GROUP_1:35
    .= ((1_H) |^ x) * ((a|^m) |^y) by A4, GROUP_1:41
    .= ((1_H) |^ x) * ((1_H) |^y) by A6,GROUP_2:44
    .= 1_H * ((1_H) |^y) by GROUP_1:31
    .= 1_H * 1_H by GROUP_1:31
    .= 1_H by GROUP_1:def 4
    .= 1_G by GROUP_2:44;
    hence contradiction by GROUP_2:44,A4;
  end;
