this is one of interesting properties of suppose then since is surjective, we have since is nontrivial, we have so what we need to prove is that for any

subgroup of with we have we have proved before that we must have for some now define by see that is a

well-defined homomorphism. it's also surjective. finally which is again obvious because: so we've proved that is a

surjective homomorphism and its kernel is thus by the first isomorphism theorem for groups we must have