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Two disjoint synthemes form a six-cycle, whose complement is a triangular prism. The prism has a unique partition into three perfect matchings, so the pair extends uniquely to a total of synthemes. Eight synthemes are disjoint from a fixed syntheme; grouping them four per total yields two totals through it, and double counting gives six totals in all.

// Target: combinatorics.bigb

A one-factorization partitions every edge of a graph into perfect matchings. Each matching uses every vertex exactly once. A total of synthemes is precisely a one-factorization of the complete graph on six points.

A duad is an unordered pair of letters, a syntheme is a partition of the six letters into three duads, and a total of synthemes is a set of five synthemes partitioning the 15 duads. There are $6!/(2^3\cdot 3!)=15$ synthemes. Fix one, $M$. Each of its three duads lies in three synthemes, and two distinct duads of $M$ can occur together only in $M$ itself. Thus precisely $1+3\cdot 2=7$ synthemes meet $M$ in a duad, leaving eight disjoint from it.

For any syntheme $M^{\prime }$ disjoint from $M$, their union as edges in $K_6$ is a six-cycle: the two perfect matchings alternate, and no two-cycle is possible in a simple graph. Its complement is a triangular prism, namely two triangles joined by three corresponding edges. The prism has four perfect matchings. One consists of all three joining edges; each of the other three consists of one joining edge and the remaining edge in each triangle. To partition the prism into three perfect matchings, one cannot use the all-joining matching, since what remains is two odd triangles. The other three do partition it. Hence two disjoint synthemes extend to a unique synthematic total.

There are eight choices for $M^{\prime }$ and four choices of $M^{\prime }$ within each synthematic total containing $M$. Thus $M$ lies in exactly two synthematic totals. Counting incidences of synthemes and synthematic totals now gives the counting synthematic totals formulaTwo distinct synthematic totals share at most one syntheme, since sharing two would force them to be the unique synthematic total extending that pair. Each syntheme gives a pair of synthematic totals containing it, and the 15 synthemes account for all $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ pairs. Therefore every pair of synthematic totals shares exactly one syntheme.

This also explains why the action on the six synthematic totals defines an automorphism of $S_6$. A permutation fixing all synthematic totals fixes their pairwise intersections, hence every syntheme. Each duad is the intersection of two distinct synthemes containing it, so every duad is then fixed. Intersecting two duads through a point shows that every point is fixed. The action is faithful and, by equality of orders, maps $S_6$ onto the full symmetric group of the synthematic totals. Part iii shows that a transposition acts as a triple transposition, establishing that this is an outer automorphism.

A point is an element of $H$. A duad is an unordered two-element subset. A syntheme is a partition of $H$ into three duads, and a total of synthemes is a collection of five synthemes whose duads partition all fifteen duads. In graph terms these are vertices, edges, perfect matchings and one-factorization of $K_6$.

The first counts are $6$, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$, andsynthemes. To count totals, first observe that two edge-disjoint synthemes have union a six-cycle. Its complement in $K_6$ is a triangular prism: two triangles on alternate cycle vertices, joined by the three remaining cross edges. Its perfect matchings are the matching using all three cross edges and three matchings using one cross edge each. The all-cross matching cannot be used in a factorization, because the remaining two odd triangles cannot be matched. The other three matchings partition the prism edges. Therefore every pair of disjoint synthemes extends to a unique total.

Fix a syntheme $s$. Each of its three duads belongs to three synthemes. Inclusion-exclusion shows that $3\cdot 3-3+1=7$ synthemes share a duad with $s$, including $s$ itself. Thus eight are disjoint from $s$. A total containing $s$ uses four of these, and each disjoint syntheme determines exactly one such total. Hence $s$ belongs to $8/4=2$ totals. Counting incidences givesTwo different totals share at most one syntheme, by the unique-completion assertion. There are fifteen pairs of totals and fifteen synthemes each belonging to two totals; consequently each pair of totals has exactly one common syntheme. This incidence property drives the next construction.

A syntheme partitions six points into three duads, equivalently a perfect matching of the complete graph. There are $6!/(2^{3}3!)=15$ synthemes. Every duad belongs to three. Two edge-disjoint synthemes extend to a unique total of synthemes: their union is a six-cycle, whose complement has a unique factorization into three matchings.