I have talked a bit about the Frattini subgroup. Time for its big brother.
The definition of the Fitting subgroup F(G) of a finite group G is the unique maximal normal nilpotent subgroup of G. As such, of course, it contains the Frattini subgroup Φ(G). (I showed here the Frattini argument used to prove its nilpotence.)
But wait a minute, that definition makes no sense until you have proved that the object in question exists. I learned this proof as a student; I can remember that it seemed to me to be more complicated than you would expect, but I cannot remember how it went.
So here is a simple proof, using nothing more than Sylow’s theorem. Since Sylow’s theorem is the most important theorem about finite groups, this is entirely fitting.
We need also the fact that a finite group is nilpotent if and only if it has a unique Sylow p-subgroup for each prime p. This is the same property I used for the Frattini subgroup. You can, if you wish, use it as a definition of nilpotence for finite groups. In the end you don’t escape the work, since this won’t do for infinite groups, so you have to come to terms with central series. But it will do as a working definition here.
The key lemma is the following:
If A and B are normal nilpotent subgroups of G, then AB is a normal nilpotent subgroup of G.
Given this, it is clear that the product of all the normal nilpotent subgroups is the required unique maximal one.
So let A and B be as given. It is elementary that AB is a normal subgroup of G. Also, |AB| = |A|.|B|/|A∩B|.
Choose a prime p, and let R be a Sylow p-subgroup of A∩B. By Sylow’s theorem, R is contained in a Sylow p-subgroup P of A, and in a Sylow p-subgroup Q of B. Moreover, P∩Q = R, since there is no larger p-subgroup in A∩B.
Now P is the unique Sylow p-subgroup of A, and so is characteristic in A, and thus normal in G. Similarly Q is normal in G. So PQ is a normal subgroup of G, and its order is |P|.|Q|/|P∩Q|. Comparing the equations for |AB| and |PQ|, we see that the latter is the p-part of the former; so PQ is a normal (and hence unique) Sylow p-subgroup of AB.
Since this holds for all primes p, AB is nilpotent.
Peter Cameron's Blog 
“…this is entirely fitting”
I can never resist a good pun.
Did you know that, if wearing a face mask makes your glasses steam up, you may be entitled to condensation?
A very nice proof Peter! I also like a slight variation of this which goes as follows.
We use that fact that a finite group is nilpotent if and only if it is the direct product of its Sylow subgroups. With A and B as in your proof, write A = P_1 x … x P_r and B = Q_1 x … x Q_r, where (for each i) P_i and Q_i are Sylow p_i-subgroups of A and B respectively for the same prime p_i (some of these groups may be trivial). As you showed, each P_i and Q_i is normal in G and P_i Q_i is a p_i-group. Also, if i \not= j, then [P_i, Q,j] is a subgroup of P_i \cap Q_j = {1}; so P_i Q_j = P_i x Q_j. Now
AB = (P_1 x … x P_r)(Q_1 x … x Q_r) = P_1 Q_1 x … x P_r Q_r
is a direct product of its Sylow subgroups and we’re done.