In a latin square of order n, a k-plex is a selection of kn entries in which each row, column and symbol occurs k times. A 1-plex is also called a transversal. An indivisible k-plex is one that contains no c-plex for 0<c<k0<c<k. For orders n∉{2,6}n∉{2,6}, existence of latin squares with a partition into 1-plexes was famously shown in 1960 by Bose, Shrikhande and Parker. A main result of this paper is that, if k divides n and 1<k<n1<k<n then there exists a latin square of order n with a partition into indivisible k-plexes.Define κ(n)κ(n) to be the largest integer k such that some latin square of order n contains an indivisible k-plex. We report on extensive computations of indivisible plexes and partitions in latin squares of order at most 9. We determine κ(n)κ(n) exactly for n≤8n≤8 and find that κ(9)∈{6,7}κ(9)∈{6,7}. Up to order 8 we count all indivisible partitions in each species.For each group table of order n≤8n≤8 we report the number of indivisible plexes and indivisible partitions. For group tables of order 9 we give the number of indivisible plexes and identify which types of indivisible partitions occur. We will also report on computations which show that the latin squares of order 9 satisfy a conjecture that every latin square of order n has a set of ⌊n/2⌋⌊n/2⌋ disjoint 2-plexes.By extending an argument used by Mann, we show that for all n≥5n≥5 there is some k∈{1,2,3,4}k∈{1,2,3,4} for which there exists a latin square of order n that has k disjoint transversals and a disjoint (n−k)-plex that contains no c-plex for any odd c.

We show that for any Latin square $L$ of order $2m$, we can partition the rows and columns of $L$ into pairs so that at most $(m+3)/2$ of the $2\times 2$ subarrays induced contain a repeated symbol. We conjecture that any Latin square of order $2m$ (where $m\geq 2$, with exactly five transposition class exceptions of order $6$) has such a partition so that every $2\times 2$ subarray induced contains no repeated symbol. We verify this conjecture by computer when $m\leq 4$.