I am looking for a way to describe permutation groups, say $S_n$, where the permuted elements are not simple numbers but symmetric pairs of numbers. So, in some sense this is the "$S_n$ power of $S_2$". I know the order of these groups, but I will need them in terms of cycles.
The context of this question is tensor calculation in physics. We express the order parameter of nematic material by a symmetric rank-2 tensor, and then take higher powers of this to describe all possible deformations of the material. I want to automatise this step in a computer program. For this I need the different conjugacy classes of the permutation group of the indices, their number of elements, and how they decompose into cycles.
If anyone could point me to a reference or drop me some hints where to find such a "power" of two permutation groups, please let me know.
From the first comment I learn that I must be more specific -- Here is an example of $S_2$ and $S_4$: The combined group (for which I do not know the proper name) can be generated by understanding (1,2,3,4,5,7,8) as four successive pairs, and then permute within each pair, and permute the pairs. So, the group can be generated by the following exchanges:
(12), (34), (56), (78) : switch within one of the four pairs
(13)(24), (15)(26), (17)(28) : exchange pairs
The resulting group has 20 conjugacy classes (already this I find nontrivial). Each of them has a "pattern" of cycle lengths. By this I mean that the identity element can be expressed as (1)(2)(3)(4)(5)(6)(7)(8), thus 8 cycle lengths of one. The element (12)(34)(56)(78) has 4 cycle lengths of two. Elements such as (12) have 6 cycle lengths of one and 1 cycle of length two. When we continue like this, we get the following table of cycle lengths:
class(1 2 3 4 5 6 7 8): 1 elements, cycle lengths: [8, 0, 0, 0, 0, 0, 0, 0]
class(2 1 4 3 6 5 8 7): 1 elements, cycle lengths: [0, 4, 0, 0, 0, 0, 0, 0]
class(2 1 3 4 5 6 7 8): 4 elements, cycle lengths: [6, 1, 0, 0, 0, 0, 0, 0]
class(1 2 4 3 6 5 8 7): 4 elements, cycle lengths: [2, 3, 0, 0, 0, 0, 0, 0]
class(2 1 3 4 5 6 8 7): 6 elements, cycle lengths: [4, 2, 0, 0, 0, 0, 0, 0]
class(8 7 3 4 5 6 2 1): 12 elements, cycle lengths: [4, 2, 0, 0, 0, 0, 0, 0]
class(1 2 5 6 4 3 7 8): 12 elements, cycle lengths: [4, 0, 0, 1, 0, 0, 0, 0]
class(4 3 2 1 6 5 8 7): 12 elements, cycle lengths: [0, 4, 0, 0, 0, 0, 0, 0]
class(2 1 6 5 3 4 8 7): 12 elements, cycle lengths: [0, 2, 0, 1, 0, 0, 0, 0]
class(3 4 1 2 8 7 6 5): 12 elements, cycle lengths: [0, 4, 0, 0, 0, 0, 0, 0]
class(6 5 7 8 1 2 4 3): 12 elements, cycle lengths: [0, 0, 0, 2, 0, 0, 0, 0]
class(2 1 3 4 7 8 5 6): 24 elements, cycle lengths: [2, 3, 0, 0, 0, 0, 0, 0]
class(1 2 4 3 8 7 5 6): 24 elements, cycle lengths: [2, 1, 0, 1, 0, 0, 0, 0]
class(7 8 5 6 4 3 1 2): 24 elements, cycle lengths: [0, 2, 0, 1, 0, 0, 0, 0]
class(5 6 2 1 4 3 7 8): 32 elements, cycle lengths: [2, 0, 2, 0, 0, 0, 0, 0]
class(3 4 6 5 1 2 7 8): 32 elements, cycle lengths: [2, 0, 0, 0, 0, 1, 0, 0]
class(3 4 8 7 6 5 2 1): 32 elements, cycle lengths: [0, 1, 2, 0, 0, 0, 0, 0]
class(3 4 7 8 6 5 2 1): 32 elements, cycle lengths: [0, 1, 0, 0, 0, 1, 0, 0]
class(5 6 8 7 3 4 2 1): 48 elements, cycle lengths: [0, 0, 0, 2, 0, 0, 0, 0]
class(7 8 2 1 3 4 5 6): 48 elements, cycle lengths: [0, 0, 0, 0, 0, 0, 0, 1]
In some sense, this is the "product" of $S_2$ and $S_4$, which have the tables of cycle lengths as follows:
class(1 2): 1 element, cycle lengths: [2, 0]
class(2 1): 1 element, cycle lengths: [0, 1]
class(1 2 3 4): 1 element, cycle lengths: [4, 0, 0, 0]
class(4 3 2 1): 3 element, cycle lengths: [0, 2, 0, 0]
class(4 2 3 1): 6 element, cycle lengths: [2, 1, 0, 0]
class(2 4 1 3): 6 element, cycle lengths: [0, 0, 0, 1]
class(2 3 1 4): 8 element, cycle lengths: [1, 0, 1, 0]
The above tables are taken from a computer code that generates the whole group, identifies the classes, etc. My question is: Can I deduce the long table directly from the small ones, without generating the whole group?