theorem Th25:
  for N being normal Subgroup of G, S being Element of G./.N st
    S = a * N holds S" = a" * N
proof
  let N be normal Subgroup of G, S be Element of G./.N;
  reconsider g = a" * N as Element of G./.N by Th14;
  assume
A1: S = a * N;
A2: g * S = @g * @S by Def3
    .= a" * N * a * N by A1,GROUP_3:9
    .= a" * (N * a) * N by GROUP_2:106
    .= a" * (a * N) * N by GROUP_3:117
    .= a" * a * N * N by GROUP_2:105
    .= 1_G * N * N by GROUP_1:def 5
    .= 1_G * (N * N) by GROUP_4:45
    .= 1_G * carr N by GROUP_2:76
    .= carr N by GROUP_2:37;
A3: 1_(G./.N) = carr N by Th24;
  S * g = @S * @g by Def3
    .= N * a * (a" * N) by A1,GROUP_3:117
    .= N * a * a" * N by GROUP_3:9
    .= N * (a * a") * N by GROUP_2:107
    .= N * 1_G * N by GROUP_1:def 5
    .= carr N * carr N by GROUP_2:109
    .= carr N by GROUP_2:76;
  hence thesis by A2,A3,GROUP_1:def 5;
end;
