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 $\frac{n}{m}\left(\genfrac{}{}{0px}{}{n}{h}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{m}{h}\right)$ subsquares of order $m$, where $h=\lceil (m+1)\mathrm{/}2\rceil$. Indeed, the number of subsquares of order $m$ is bounded by a polynomial of degree at most $\sqrt{2m}+2$ in $n$. (b) For all $n\ge 5$ 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.