theorem
  h1 * 1:(G,H) = 1:(G,I) & 1:(H,I) * h = 1:(G,I)
proof
  thus h1 * 1:(G,H) = 1:(G,I)
  proof
    let a be Element of G;
    thus (h1 * 1:(G,H)).a = h1.(1:(G,H).a) by FUNCT_2:15
    .= 1_I by Th31
    .= 1:(G,I).a;
  end;
  let a;
  thus (1:(H,I) * h).a = 1:(H,I).(h.a) by FUNCT_2:15
  .= 1:(G,I).a;
end;
