theorem Th62:
  for H being strict Subgroup of G holds H |^ a |^ a" = H & H |^ a " |^ a = H
proof
  let H be strict Subgroup of G;
  thus H |^ a |^ a" = H |^ (a * a") by Th60
    .= H |^ 1_G by GROUP_1:def 5
    .= H by Th61;
  thus H |^ a" |^ a = H |^ (a" * a) by Th60
    .= H |^ 1_G by GROUP_1:def 5
    .= H by Th61;
end;
