Combinatorics of symmetric group:S5

This page discusses some of the combinatorics associated with symmetric group:S5, that relies specifically on viewing it as a symmetric group on a finite set.

Partitions, subset partitions, and cycle decompositions
Denote by $$\mathcal{B}(n)$$ the set of all unordered set partitions of $$\{ 1,2,\dots,n\}$$ into subsets and by $$P(n)$$ the set of unordered integer partitions of $$n$$. There are natural combinatorial maps:

$$S_n \to \mathcal{B}(n) \to P(n)$$

where the first map sends a permutation to the subset partition induced by its cycle decomposition, which is equivalently the decomposition into orbits for the action of the cyclic subgroup generated by that permutation on $$\{ 1,2,\dots,n\}$$. The second map sends a subset partition to the partition of $$n$$ given by the sizes of the parts. The composite of the two maps is termed the cycle type, and classifies conjugacy classes in $$S_n$$, because cycle type determines conjugacy class.

Further, if we define actions as follows:


 * $$S_n$$ acts on itself by conjugation
 * $$S_n$$ acts on $$\mathcal{B}(n)$$ by moving around the elements and hence changing the subsets
 * $$S_n$$ acts on $$P(n)$$ trivially

then the maps above are $$S_n$$-equivariant, i.e., they commute with the $$S_n$$-action. Moreover, the action on $$\mathcal{B}(n)$$ is transitive on each fiber above $$P(n)$$ and the action on $$S_n$$ is transitive on each fiber above the composite map to $$P(n)$$. In particular, for two elements of $$\mathcal{B}(n)$$ that map to the same element of $$P(n)$$, the fibers above them in $$S_n$$ have the same size.

There are formulas for calculating the sizes of the fibers at each level.

In our case $$n = 5$$: