reserve i for Element of NAT;

theorem Th15:
  for G,H being strict Group, h being Homomorphism of G,H for A,B
being strict Subgroup of G holds A is strict normal Subgroup of B implies h.:A
  is strict normal Subgroup of h.:B
proof
  let G,H be strict Group;
  let h be Homomorphism of G,H;
  let A,B be strict Subgroup of G;
  assume A is strict normal Subgroup of B;
  then reconsider A1=A as strict normal Subgroup of B;
  reconsider C=h.:A1 as strict Subgroup of h.:B by Th12;
  for b2 being Element of h.:B holds b2* C c= C * b2
  proof
    let b2 be Element of h.:B;
A1: b2 in h.:B by STRUCT_0:def 5;
    now
      consider b1 being Element of B such that
A2:   b2=(h|B).b1 by A1,GROUP_6:45;
      reconsider b1 as Element of G by GROUP_2:42;
A3:   b2=h.b1 by A2,FUNCT_1:49;
      reconsider b=b1 as Element of B;
      let x be object;
      assume x in b2* C;
      then consider g being Element of h.:B such that
A4:   x = b2 * g and
A5:   g in C by GROUP_2:103;
      consider g1 being Element of A1 such that
A6:   g=(h|A1).g1 by A5,GROUP_6:45;
      reconsider g1 as Element of G by GROUP_2:42;
      g=h.g1 by A6,FUNCT_1:49;
      then
A7:   x =h.b1 * h.g1 by A3,A4,GROUP_2:43
        .=h.(b1 *g1) by GROUP_6:def 6;
      g1 in A1 by STRUCT_0:def 5;
      then
A8:   b1 * g1 in b1 * A1 by GROUP_2:103;
A9:   h.:(A1 * b1) = h.:A1 * h.b1 by Th13;
      b1 * A1=b * A1 by GROUP_6:2
        .=A1 * b by GROUP_3:117
        .=A1 * b1 by GROUP_6:2;
      then x in h.:(A1 *b1) by A7,A8,FUNCT_2:35;
      hence x in C* b2 by A3,A9,GROUP_6:2;
    end;
    hence thesis;
  end;
  hence thesis by GROUP_3:118;
end;
