This proof sketch of the third isomorphism theorem uses the symbols in my writeup there without repetition of the formulation, so you may want to read that first.

It's easy to see L is normal in H.

Define the natural epimorphism f: G -> G/N by f(g) = gN. Consider its restriction h = f|H to H. Clearly, Ker h = L (the intersection of H and N). But since G=HN, h is an epimorphism (indeed, for any coset gN with g in G, g=hn for h in H, n in N, so gN=hN).

The theorem follows by applying the first isomorphism theorem to h.

Y'know, if you log in, you can write something here, or contact authors directly on the site. Create a New User if you don't already have an account.