
theorem Th11:
  for G being Group, A,B being normal Subgroup of G st
  (the carrier of A) /\ (the carrier of B) = {1_G} holds
  for a,b be Element of G st a in A & b in B holds a*b = b*a
  proof
    let G be Group, A,B be normal Subgroup of G;
    assume A1: (the carrier of A) /\ (the carrier of B) = {1_G};
    let a,b be Element of G;
    assume A2: a in A & b in B;
    A3: a*b*(b*a)" = a*b*(a"*b") by GROUP_1:17
    .=(a*b*a") * b" by GROUP_1:def 3;
    A4: b" in B by A2,GROUP_2:51;
    a*b in a*B by GROUP_2:27,A2;
    then a*b in B*a by GROUP_3:117;
    then consider s be Element of G
    such that A5: a*b = s*a & s in the carrier of B by GROUP_2:28;
    a*b*a" in B by GROUP_3:1,A5;
    then
    A6: a*b*(b*a)" in the carrier of B by STRUCT_0:def 5,A3,A4,GROUP_2:50;
    A7: a*b*(b*a)" = a*b*(a"*b") by GROUP_1:17
    .=a*(b*(a"* b")) by GROUP_1:def 3
    .=a*(b*a"* b") by GROUP_1:def 3;
    a" in A by A2,GROUP_2:51;
    then
    b*a" in b*A by GROUP_2:27;
    then b*a" in A*b by GROUP_3:117;
    then consider t be Element of G
    such that A8: b*a" = t*b & t in the carrier of A by GROUP_2:28;
    b*a"* b" in A by GROUP_3:1,A8;
    then a*b*(b*a)" in the carrier of A by STRUCT_0:def 5,A7,A2,GROUP_2:50;
    then a*b*(b*a)" in (the carrier of A) /\ (the carrier of B)
    by XBOOLE_0:def 4,A6;
    then a*b*(b*a)" = 1_G by A1,TARSKI:def 1;
    then 1_G * (b*a) = a*b by GROUP_1:14;
    hence thesis by GROUP_1:def 4;
  end;
