关于S₄中4-循环与不交对换乘积的关系及n-循环推广的技术问询

Hey there! Let's unpack this question thoroughly, starting with what those $S_4$ relations tell us, then moving to generalizing to n-cycles.

First, let's break down the core takeaways from those equalities (using (abc...) notation for cycles):

• Double transpositions are squares of 4-cycles: Every element in the double transposition conjugacy class (the pairs of disjoint 2-cycles like (12)(34)) is the square of some 4-cycle. Since a 4-cycle has order 4, squaring it reduces its order to 2—exactly the order of a double transposition, which makes sense algebraically.
• 2-to-1 correspondence: Each double transposition is the square of two distinct 4-cycles. For example, (13)(24) = (1234)^2 = (1432)^2—notice that (1432) is the inverse of (1234), and squaring an element and its inverse always yields the same result (because $(\sigma^{-1})2 = (\sigma²⁾{-1}$, and since $\sigma^2$ has order 2, it’s its own inverse).
• Subgroup connections: This ties the cyclic subgroups generated by 4-cycles (each of order 4) to the Klein four-group $V_4$—the subgroup of $S_4$ made up of all double transpositions plus the identity. Every 4-cycle's square lands inside $V_4$, and every non-identity element of $V_4$ is hit by exactly two 4-cycles' squares.
• Conjugacy class mapping: Squaring acts as a 2-to-1 map from the 4-cycle conjugacy class (which has 6 elements: (1234), (1243), (1324), (1342), (1423), (1432)) to the double transposition conjugacy class (which has 3 elements: (12)(34), (13)(24), (14)(23)). This is a nice example of how power operations can bridge different conjugacy classes in symmetric groups.

To find n-cycles with similar properties, we need to use the rule for powers of cycles: For a k-cycle $\sigma$, $\sigma^m$ decomposes into $\gcd(k,m)$ disjoint cycles each of length $k/\gcd(k,m)$. For squaring ($m=2$), here's what happens based on n's parity:

Case 1: n is odd #

When n is odd, $\gcd(n,2)=1$, so squaring an n-cycle gives another n-cycle (the cycle length doesn’t split). Here, the squaring map is a bijection on the n-cycle conjugacy class (since 2 has an inverse modulo n when n is odd). This means every n-cycle is the square of exactly one other n-cycle—but this isn’t analogous to the $S_4$ case, because we’re staying within the same conjugacy class, not moving to a different one.

Case 2: n is even (n ≥ 4) #

Let $n=2k$. Then $\gcd(n,2)=2$, so squaring an n-cycle produces a product of two disjoint k-cycles. This is exactly the parallel to $S_4$ (where $k=2$, so two 2-cycles = double transposition).

For even n:

• Each product of two disjoint k-cycles can be written as the square of at least two distinct n-cycles. For example, take n=6 (k=3): the permutation (135)(246) is the square of the 6-cycle (123456) and also the square of its inverse (165432). You can even find other non-inverse n-cycles that square to the same permutation—like (153264), whose square is also (135)(246).
• Just like in $S_4$, this creates a many-to-one mapping from the n-cycle conjugacy class to the conjugacy class of "two disjoint k-cycles" (where $k=n/2$).

The key conclusion: Any even n ≥ 4 gives n-cycles with analogous properties to 4-cycles in $S_4$: squaring them results in a permutation that’s a product of two equal-length disjoint cycles, and each such permutation is the square of multiple distinct n-cycles.

内容的提问来源于stack exchange，提问作者1Spectre1