More about the stability of a molecule
You can test directly the stability of a molecule, you just have to choose Job style --> Stability. And then test with a singulet or a triplet.
But in this case what you are looking for is the stability of the wave function!
For the example of propene, we get the same, the too exist but at different energy.
For the singulet SCF Done: E(UHF) = -117.022407343 A.U. after 19 cycles Convg = 0.9626D-08 -V/T = 1.9986 S**2 = 0.1527
For the triplet SCF Done: E(UHF) = -116.910538298 A.U. after 15 cycles Convg = 0.1887D-08 -V/T = 1.9942 S**2 = 2.0203
So the results suggest that in term of energy the singulet is the more stable because it has got the lowest energy.
But with the example of O2 you get different results.
O2 HF 6-21G
Singulet
The wavefunction has an RHF -> UHF instability.
Triplet
The wavefunction is stable under the perturbations considered.
So here the singulet state does not exist.
Consider the energy of the two forms (singulet and triplet) of O2.
For the singulet SCF Done: E(RHF) = -148.686490556 A.U. after 8 cycles Convg = 0.2240D-08 -V/T = 2.0032 S**2 = 0.0000
For the triplet SCF Done: E(UHF) = -148.768525186 A.U. after 9 cycles Convg = 0.4772D-08 -V/T = 2.0031 S**2 = 2.0195
So if we just have a look at the energy we deduce that the triplet has a lower energy so it is the more stable... it is not in contradiction to our results concerning the stability of the wave function, but it is two different questions!
And if we consider the theory that we know about the orbital diagram of O2 the filling is : 1s2 *1s2 2s2 *2s2 2px2 2pz2 2py2 *2pz1 *2py1 *2px0
So it is quite logical that O2 is more stable as a triplet because of the repartition of the electrons. Back to Calculate the energy of a molecule