Counting transversals in Latin squares over groups of even order

The number of transversals in a Latin square $L(G)$ whose rows and columns are indexed by a finite group $G$ has been determined computationally for all $G$ of order at most $23$ and for $C_{25}$. However, computation of the number of transversals of $L(G)$ has until now been considered intractable for all non-cyclic groups $G$ of order $32$. We show, for a group $G$ containing an order $2$ normal subgroup $N$, that the image of a transversal in $L(G)$ under the quotient mapping from $G$ to $G/N$ is a $2$-plex in $L(G/N)$. We then derive linear algebraic conditions that determine the number of transversals in $L(G)$ whose image under the quotient mapping is a given $2$-plex in $L(G/N)$. This allows us to count transversals in $L(G)$ by examining all $2$-plexes in $L(G/N)$. Although we can simplify the counting of $2$-plexes in $L(G/N)$ by taking advantage of their symmetry group, the linear algebraic conditions can vary over a symmetry class of $2$-plexes. However, in the case that $G$ takes the splitting form $C_2 \times H$, the linear algebraic conditions become invariant over each symmetry class and so the counting of transversals in $L(C_2 \times H)$ reduces to counting the $2$-plexes in $L(H)$ under the action of their symmetry group. We illustrate the power of these techniques by computing the exact number of transversals in $L(C_2^5)$.