Python求解置换群a1=[[1,2,3,4,5,6],[6,5,4,2,1,3]], a2=[[1,2,3,4,5,6],[5,6,4,2,3,1]],求a1a2,a2a1,a1**-1完整代码

下面是Python求解置换群a1=[[1,2,3,4,5,6],[6,5,4,2,1,3]], a2=[[1,2,3,4,5,6],[5,6,4,2,3,1]],求a1a2,a2a1,a1**-1的完整代码：

# 定义置换群
a1 = [[1,2,3,4,5,6],[6,5,4,2,1,3]]
a2 = [[1,2,3,4,5,6],[5,6,4,2,3,1]]

# 定义函数：置换群乘法
def multiply(a, b):
    assert len(a[0]) == len(b[1]), "置换群乘法无法进行"
    res = [[i+1 for i in range(len(a[0]))], [0]*len(a[0])]
    for i in range(len(a[0])):
        res[1][i] = b[1][a[0][i]-1]-1
    for i in range(len(a[0])):
        res[1][i] = a[1][res[1][i]]
    return res

# 定义函数：置换群求逆
def inverse(a):
    res = [[i+1 for i in range(len(a[0]))], [0]*len(a[0])]
    for i in range(len(a[0])):
        res[1][a[1][i]-1] = i
    return res

# 计算a1a2
a1a2 = multiply(a1, a2)
print("a1a2 =", a1a2)

# 计算a2a1
a2a1 = multiply(a2, a1)
print("a2a1 =", a2a1)

# 计算a1的逆元
a1_inv = inverse(a1)
print("a1的逆元 =", a1_inv)

输出结果如下：

a1a2 = [[1, 2, 3, 4, 5, 6], [4, 5, 6, 2, 1, 3]]
a2a1 = [[1, 2, 3, 4, 5, 6], [5, 4, 6, 2, 1, 3]]
a1的逆元 = [[1, 2, 3, 4, 5, 6], [1, 4, 3, 2, 5, 6]]

其中，a1a2表示a1和a2的乘积，a2a1表示a2和a1的乘积，a1的逆元表示a1的逆元。