关于ℤ₄₂中3的倍数集合在模42乘法下的单位元与可逆元素求解的技术问询

Hey there! Let's break this down step by step, skipping the tedious table approach and leaning on number theory instead. First, let's confirm the identity element you found, then tackle the invertible elements with a clean mathematical method.

1. 验证单位元（Identity Element） #

You mentioned you found the identity is 15—great call! Let's formalize that to be sure. The set ( S = {0, 3, 6, ..., 39} ) consists of all elements of the form ( 3k ) where ( k \in \mathbb{Z}_{14} ) (since ( 3 \times 14 = 42 \equiv 0 \pmod{42} )).

We need an element ( e = 3m \in S ) such that for every ( 3k \in S ):
[ (3k)(3m) \equiv 3k \pmod{42} ]
Simplify this congruence:
[ 9km \equiv 3k \pmod{42} ]
Divide the entire equation by 3 (since gcd(3,42)=3, we can reduce both the coefficients and modulus by 3):
[ 3km \equiv k \pmod{14} ]
Rearrange to get:
[ k(3m - 1) \equiv 0 \pmod{14} ]
This must hold for all ( k \in \mathbb{Z}_{14} ), so ( 3m - 1 \equiv 0 \pmod{14} ), or ( 3m \equiv 1 \pmod{14} ). The inverse of 3 modulo 14 is 5 (since ( 3 \times 5 = 15 \equiv 1 \pmod{14} )), so ( m=5 ), hence ( e=3 \times 5 = 15 ). Perfect match for your result—no tables required!

2. 寻找可逆元素（Invertible Elements） #

An element ( a = 3k \in S ) is invertible if there exists some ( b = 3m \in S ) such that ( (3k)(3m) \equiv 15 \pmod{42} ) (our identity). Let's translate this into a solvable congruence:
[ 9km \equiv 15 \pmod{42} ]
Again, divide through by gcd(9,15,42)=3:
[ 3km \equiv 5 \pmod{14} ]
For this linear congruence to have a solution ( m \in \mathbb{Z}_{14} ), the gcd of ( 3k ) (the coefficient of m) and 14 must divide the constant term 5. Since 5 is prime, gcd(3k,14) must equal 1 (2 and 7 don't divide 5).

Since 3 and 14 are coprime, ( \gcd(3k,14) = 1 ) is equivalent to ( \gcd(k,14) = 1 ). Now, list all ( k \in \mathbb{Z}_{14} ) (0 to 13) that are coprime to 14:

• ( k = 1, 3, 5, 9, 11, 13 )

Map these back to elements of S by multiplying by 3:

• ( 3 \times 1 = 3 )
• ( 3 \times 3 = 9 )
• ( 3 \times 5 = 15 ) (the identity, which is always invertible)
• ( 3 \times 9 = 27 )
• ( 3 \times 11 = 33 )
• ( 3 \times 13 = 39 )

Let's verify one pair to confirm: take 3 and 33. ( 3 \times 33 = 99 ), and ( 99 - 2 \times 42 = 15 ), which is our identity modulo 42. Perfect! Another example: ( 9 \times 39 = 351 ), ( 351 - 8 \times 42 = 15 )—that works too.

Note that 0 is not invertible, since 0 multiplied by any element in S is 0, which can never equal 15 modulo 42.

关键思路总结 #

• Instead of brute-force tables, we leverage the structure of S: all elements are multiples of 3, so we can reduce the modulus from 42 to 14 by dividing through by the common factor.
• For invertibility, we use the linear congruence solvability rule: ( ax \equiv b \pmod{m} ) has a solution iff ( \gcd(a,m) \mid b ). This lets us narrow down valid elements without checking every pair.

备注：内容来源于stack exchange，提问作者pasha