reserve n for Element of NAT;
reserve i for Integer;
reserve G,H,I for Group;
reserve A,B for Subgroup of G;
reserve N for normal Subgroup of G;
reserve a,a1,a2,a3,b,b1 for Element of G;
reserve c,d for Element of H;
reserve f for Function of the carrier of G, the carrier of H;
reserve x,y,y1,y2,z for set;
reserve A1,A2 for Subset of G;
reserve N for normal Subgroup of G;
reserve S,T1,T2 for Element of G./.N;

theorem
  for G being strict Group, N be strict normal Subgroup of G holds
  G./.N is commutative Group iff G` is Subgroup of N
proof
  let G be strict Group, N be strict normal Subgroup of G;
  thus G./.N is commutative Group implies G` is Subgroup of N
  proof
    assume
A1: G./.N is commutative Group;
    now
      let a,b be Element of G;
      reconsider S = a * N,T = b * N as Element of G./.N by Th14;
A2:   [.S,T.] = 1_(G./.N) & 1_(G./.N) = carr N by A1,Th24,GROUP_5:37;
      S" = a" * N & T" = b" * N by Th25;
      then
A3:   S" * T" = (a" * N) * (b" * N) by Def3;
      S * T = (a * N) * (b * N) & [.S,T.] = (S" * T") * (S * T)
        by Def3,GROUP_5:16;
      then [.S,T.] = (a" * N) * (b" * N) * ((a * N) * (b * N)) by A3,Def3;
      then carr N = (a" * N) * (b" * N) * (a * (N * (b * N))) by A2,GROUP_3:10
        .= (a" * N) * (b" * N) * (a * (N * b * N)) by GROUP_3:13
        .= (a" * N) * (b" * N) * (a * (b * N * N)) by GROUP_3:117
        .= (a" * N) * (b" * N) * (a * (b * N)) by Th5
        .= (a" * N) * (b" * N) * (a * b * N) by GROUP_2:105
        .= a" * (N * (b" * N)) * (a * b * N) by GROUP_3:10
        .= a" * (N * b" * N) * (a * b * N) by GROUP_3:13
        .= a" * (b" * N * N) * (a * b * N) by GROUP_3:117
        .= a" * (b" * N) * (a * b * N) by Th5
        .= (a" * b" * N) * (a * b * N) by GROUP_2:105
        .= (a" * b") * (N * (a * b * N)) by GROUP_3:10
        .= (a" * b") * (N * (a * b) * N) by GROUP_3:13
        .= (a" * b") * ((a * b) * N * N) by GROUP_3:117
        .= (a" * b") * ((a * b) * N) by Th5
        .= (a" * b") * (a * b) * N by GROUP_2:105
        .= [.a,b.] * N by GROUP_5:16;
      hence [.a,b.] in N by GROUP_2:113;
    end;
    hence thesis by Th7;
  end;
  assume
A4: G` is Subgroup of N;
  now
    let S,T be Element of G./.N;
A5: [.S,T.] = (S" * T") * (S * T) by GROUP_5:16;
    consider b being Element of G such that
A6: T = b * N and
    T = N * b by Th13;
A7: T" = b" * N by A6,Th25;
    consider a being Element of G such that
A8: S = a * N and
    S = N * a by Th13;
    S" = a" * N by A8,Th25;
    then
A9: S" * T" = (a" * N) * (b" * N) by A7,Def3;
    [.a,b.] in N by A4,Th7;
    then
A10: carr N = [.a,b.] * N by GROUP_2:113
      .= (a" * b") * (a * b) * N by GROUP_5:16
      .= (a" * b") * ((a * b) * N) by GROUP_2:105
      .= (a" * b") * ((a * b) * N * N) by Th5
      .= (a" * b") * (N * (a * b) * N) by GROUP_3:117
      .= (a" * b") * (N * (a * b * N)) by GROUP_3:13
      .= (a" * b" * N) * (a * b * N) by GROUP_3:10
      .= a" * (b" * N) * (a * b * N) by GROUP_2:105
      .= a" * (b" * N * N) * (a * b * N) by Th5
      .= a" * (N * b" * N) * (a * b * N) by GROUP_3:117
      .= a" * (N * (b" * N)) * (a * b * N) by GROUP_3:13
      .= (a" * N) * (b" * N) * (a * b * N) by GROUP_3:10
      .= (a" * N) * (b" * N) * (a * (b * N)) by GROUP_2:105
      .= (a" * N) * (b" * N) * (a * (b * N * N)) by Th5
      .= (a" * N) * (b" * N) * (a * (N * b * N)) by GROUP_3:117
      .= (a" * N) * (b" * N) * (a * (N * (b * N))) by GROUP_3:13
      .= (a" * N) * (b" * N) * (a * N * (b * N)) by GROUP_3:10;
    S * T = (a * N) * (b * N) by A8,A6,Def3;
    then [.S,T.] = (a" * N) * (b" * N) * ((a * N) * (b * N)) by A9,A5,Def3;
    hence [.S,T.] = 1_(G./.N) by A10,Th24;
  end;
  hence thesis by GROUP_5:37;
end;
