Combinatorics of symmetric group:S5

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This article gives specific information, namely, combinatorics, about a particular group, namely: symmetric group:S5.
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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:

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 1 (5 times) \frac{5!}{(1!)^55!} 1 (1 - 1)!^5 1 \frac{5!}{1^55!} 1
2 + 1 + 1 + 1 2 (1 time), 1 (3 times) \frac{5!}{(2!)^11!(1!)^3(3!)} 10 (2 - 1)!(1 - 1)!^3 1 \frac{5!}{2^11!(1)^33!} 10
3 + 1 + 1 3 (1 time), 1 (2 times) \frac{5!}{(3!)^11!(1!)^22!} 10 (3 - 1)!(1 - 1)!^2 2 \frac{5!}{3^11!(1)^22!} 20
2 + 2 + 1 2 (2 times), 1 (1 time) \frac{5!}{(2!)^22!(1!)^11!} 15 (2 - 1)!^2(1 - 1)! 1 \frac{5!}{2^22!1^11!} 15
4 + 1 4 (1 time), 1 (1 time) \frac{5!}{(4!)^11!(1!)^11!} 5 (4 - 1)!(1 - 1)! 6 \frac{5!}{4^11!1^11!} 30
3 + 2 3 (1 time), 2 (1 time) \frac{5!}{(3!)^11!(2!)^11!} 10 (3 - 1)!(2 - 1)! 2 \frac{5!}{3^11!2^11!} 20
5 5 (1 time) \frac{5!}{(5!)^11!} 1 (5 - 1)! 24 \frac{5!}{5^11!} 24
Total (7 rows -- number of rows equals number of unordered integer partitions of 5) -- -- 52 (equals the size of \mathcal{B}(n), termed the Bell number, at n = 5) -- 37 (see Oeis:A107107) -- 120 (equals 5!, the size of the symmetric group)