theorem ThB119:
  for H being Subgroup of G holds H is normal Subgroup of G iff
  for a holds H + a c= a + H
proof
  let H be Subgroup of G;
  thus H is normal Subgroup of G implies for a holds H + a c= a + H by Th117;
  assume
A1: for a holds H + a c= a + H;
  now
    let a;
    H + (-a) c= (-a) + H by A1;
    then a + (H + (-a)) c= a + ((-a) + H) by Th4;
    then a + (H + (-a)) c= a + (-a) + H by ThB105;
    then a + (H + (-a)) c= 0_G + H by Def5;
    then a + (H + (-a)) c= carr H by ThB109;
    then a + H + (-a) c= carr H by ThB106;
    then a + H + (-a) + a c= carr H + a by Th4;
    then a + H + ((-a) + a) c= H + a by ThB34;
    then a + H + 0_G c= H + a by Def5;
    hence a + H c= H + a by Th37;
  end;
  then for a holds a + H = H + a by A1;
  hence thesis by Th117;
end;
