Combinatorics of symmetric group:S4

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 $n = 4$ :

Partition	Partition in grouped form	Formula calculating size of fiber for $\mathcal{B}(n) \to P(n)$	Size of fiber for $\mathcal{B}(n) \to P(n)$	Formula calculating size of fiber for $S_n \to \mathcal{B}(n)$	Size of fiber for $S_n \to \mathcal{B}(n)$	Formula calculating size of fiber for $S_n \to P(n)$ , which is precisely the conjugacy class size for that cycle type (product of previous two formulas)	Size of fiber for $S_n \to P(n)$ , which is precisely the conjugacy class size for that cycle type (product of previous two sizes)
1 + 1 + 1 + 1	1 (4 times)	$\frac{4!}{(1!)^44!}$	1	$(1 - 1)!^4$	1	$\frac{4!}{1^44!}$	1
2 + 1 + 1	2 (1 time), 1 (2 times)	$\frac{4!}{(2!)(1!)^2(2!)}$	6	$(2 - 1)!(1 - 1)!^2$	1	$\frac{4!}{(2)(1)^2(2!)}$	6
2 + 2	2 (2 times)	$\frac{4!}{(2!)^2(2!)}$	3	$(2 - 1)!^2$	1	$\frac{4!}{(2)^2(2!)}$	3
3 + 1	3 (1 time), 1 (1 time)	$\frac{4!}{(3!)^1(1!)(1!)^1(1!)}$	4	$(3 - 1)!(1 - 1)!$	2	$\frac{4!}{(3)^1(1!)(1^1)(1!)}$	8
4	4 (1 time)	$\frac{4!}{(4!)^1(1!)}$	1	$(4 - 1)!$	6	$\frac{4!}{(4^1(1!)}$	6
Total (5 rows -- number of rows equals number of unordered integer partitions of 4)	--	--	15 (equals the size of $\mathcal{B}(n)$ , termed the Bell number, for $n = 4$ )	--	11 (see Oeis:A107107)	--	24 (equals 4!, the size of the symmetric group)