theorem ThB62:
  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 * 0_G by Def5
    .= H by Th61;
  thus H * (-a) * a = H * ((-a) + a) by Th60
    .= H * 0_G by Def5
    .= H by Th61;
end;
