Exercise 8.32: Conjugation Table

from itertools import permutations, product
from pprint import pformat

from numpy import zeros
from sympy.combinatorics.generators import symmetric
from IPython.display import Markdown


def display_symmetric(n: int):
    perms = list(symmetric(n))
    perm_dict = dict()
    for i, perm in enumerate(perms):
        perm_dict[i] = perm
    inv_perms = {v: k for k, v in perm_dict.items()}

    display(Markdown(f'**n = {n}**'))
    print(pformat(perm_dict))

    exp_table        = zeros([len(perms), len(perms)], dtype=int)
    conjugacy_table2 = zeros([len(perms), len(perms)], dtype=int)
    for i, j in product(range(len(perms)), repeat=2):
        a = perms[i]
        b = perms[j]
        exp_table[i, j]        = inv_perms[a ^ b]
        conjugacy_table2[i, j] = inv_perms[b * a * ~b]

    display(Markdown('<br/>exp_table'))
    print(exp_table)

    display(Markdown('<br/>conjugacy_table2'))
    print(conjugacy_table2)

[display_symmetric(n) for n in range(3,4)];

n = 3

{0: Permutation(2),
 1: Permutation(1, 2),
 2: Permutation(2)(0, 1),
 3: Permutation(0, 1, 2),
 4: Permutation(0, 2, 1),
 5: Permutation(0, 2)}

exp_table

[[0 0 0 0 0 0]
 [1 1 5 5 2 2]
 [2 5 2 1 5 1]
 [3 4 4 3 3 4]
 [4 3 3 4 4 3]
 [5 2 1 2 1 5]]

conjugacy_table2

[[0 0 0 0 0 0]
 [1 1 5 2 5 2]
 [2 5 2 5 1 1]
 [3 4 4 3 3 4]
 [4 3 3 4 4 3]
 [5 2 1 1 2 5]]