Combinatorics of symmetric groups

View combinatorics of group families | View other specific information about symmetric group

This page describes some general combinatorics that can be done with the elements of the symmetric groups.

Particular cases

$n$ $n!$ (equals order of symmetric group) symmetric group of degree $n$ combinatorics of this group
0 1 trivial group --
1 1 trivial group --
2 2 cyclic group:Z2 --
3 6 symmetric group:S3 combinatorics of symmetric group:S3
4 24 symmetric group:S4 combinatorics of symmetric group:S4
5 120 symmetric group:S5 combinatorics of symmetric group:S5
6 720 symmetric group:S6 combinatorics of symmetric group:S6
7 5040 symmetric group:S7 combinatorics of symmetric group:S7
8 40320 symmetric group:S8 combinatorics of symmetric group:S8
9 362880 symmetric group:S9 combinatorics of symmetric group:S9

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.