Overlapping latin subsquares and full products

We derive necessary and sufficient conditions for there to exist a latin square of order n containing two subsquares of order a and b that intersect in a subsquare of order c. We also solve the case of two disjoint subsquares. We use these results to show that: (a) A latin square of order n cannot have more than n m n h / m h subsquares of order m, where h = ⌈(m + 1)/2⌉. Indeed, the number of subsquares of order m is bounded by a polynomial of degree at most √ 2m + 2 in n. (b) For all n ≥ 5 there exists a loop of order n in which every element can be obtained as a product of all n elements in some order and with some bracketing.