关于Aₙ中偶置换与奇置换可交换的两类证明及验证疑问

Hey there! Let's work through these permutation group questions clearly, since you already have a solid grasp of $A_n$—that makes this much easier. First, let's clear up your confusion with $A_4$, then dive into the proofs, and finally connect the dots between the conclusions.

First: Your $A_4$ Verification Confusion #

You mentioned that when testing the first conclusion in $A_4$, you thought elements didn't commute. Let's take a concrete example to fix this:
Let $f = (1\ 2)(3\ 4) \in A_4$—this is two disjoint 2-cycles, so it fits the first conclusion's condition. We need a single odd permutation that commutes with $f$ (not all odd permutations!). Try the transposition $g = (1\ 2)$ (which is odd):

• $g \circ f = (1\ 2) \circ (1\ 2)(3\ 4) = (3\ 4)$
• $f \circ g = (1\ 2)(3\ 4) \circ (1\ 2) = (3\ 4)$

They commute perfectly! The mistake was probably picking an odd permutation that doesn't interact nicely with $f$ (like $(1\ 3)$), but the conclusion only requires existence, not universality.

Proof 1: If $f \in A_n$ has two disjoint cycles of equal length, then $f$ commutes with some odd permutation #

Let's split this into two cases based on the length of the cycles:

Case 1: The two disjoint cycles are of odd length $k$

Let $f = \sigma \tau$, where $\sigma = (a_1\ a_2\ \dots\ a_k)$ and $\tau = (b_1\ b_2\ \dots\ b_k)$ are disjoint $k$-cycles (odd $k$). Define the permutation:
$$g = (a_1\ b_1)(a_2\ b_2)\dots(a_k\ b_k)$$
This is a product of $k$ transpositions. Since $k$ is odd, the total parity is odd (each transposition is odd, odd number of odd permutations multiply to odd). Now check commutativity:

• Conjugating $\sigma$ by $g$ gives $\tau$, and conjugating $\tau$ by $g$ gives $\sigma$
• Since $\sigma$ and $\tau$ are disjoint, $\sigma\tau = \tau\sigma$
• So $gfg^{-1} = g\sigma\tau g^{-1} = (g\sigma g^{-1})(g\tau g^{-1}) = \tau\sigma = \sigma\tau = f$
  Thus $g$ is an odd permutation that commutes with $f$.

Case 2: The two disjoint cycles are of even length $k$

Here, $\sigma$ (one of the even-length cycles) is itself an odd permutation (even-length cycles are odd). Since $\sigma$ and $\tau$ are disjoint, we have:
$$\sigma \circ f = \sigma \circ (\sigma\tau) = \sigma^2\tau$$
$$f \circ \sigma = (\sigma\tau) \circ \sigma = \sigma\tau\sigma = \sigma^2\tau$$
They're equal! So $\sigma$ (an odd permutation) commutes with $f$.

Proof 2: If $f \in A_n$ contains an even-length disjoint cycle, then $f$ commutes with some odd permutation #

This one is straightforward. Let $\sigma$ be the even-length cycle in $f$'s cycle decomposition. Since $\sigma$ is even-length, it's an odd permutation. Write $f = \sigma\pi$, where $\pi$ is the product of the remaining disjoint cycles in $f$.

Because $\sigma$ and $\pi$ are disjoint, they commute:
$$\sigma \circ f = \sigma \circ (\sigma\pi) = \sigma^2\pi$$
$$f \circ \sigma = (\sigma\pi) \circ \sigma = \sigma\pi\sigma = \sigma^2\pi$$
So $\sigma$ (an odd permutation) commutes with $f$, which is exactly what we needed to show.

Connection Between the Conclusions #

The two conclusions overlap but aren't identical:

• The first conclusion covers scenarios where $f$ has two equal-length cycles (even or odd), including cases where all cycles are odd-length (like two disjoint 3-cycles in $A_6$).
• The second conclusion only covers $f$ with at least one even-length cycle, which is a subset of the first conclusion's even-length case, but doesn't apply to $f$ with only odd-length cycles (unless there are two equal ones).

So they're related, but the first is more general for equal-length cycle pairs, while the second is a simpler, focused result for even-length cycles.

内容的提问来源于stack exchange，提问作者user536082