非不交2-循环（对换）是否不一定可交换？置换分解实例求证

Great question! Let's unpack this clearly—your conclusion is 100% correct.

First, let's recap a key rule in permutation groups: Two permutations commute (can be swapped without changing the result) if and only if their support sets (the set of elements they actually rearrange) are disjoint. For 2-cycles (transpositions), this means:

• If two transpositions share no common elements (like (12) and (34)), they will always commute.
• If they share at least one common element (like (12) and (13)), they do not commute—swapping their order will produce a different permutation.

Let's test this with a simple example to make it concrete:

• Compute (12)(13) (using right-to-left composition, standard in permutation group theory):
  1. Apply (13) first: 1→3, 3→1, all other elements stay the same.
  2. Apply (12) to the result: 3 remains unchanged, 1→2, 2→1.
  3. Final permutation: (1 3 2)
• Now compute (13)(12):
  1. Apply (12) first: 1→2, 2→1, all other elements stay the same.
  2. Apply (13) to the result: 2 remains unchanged, 1→3, 3→1.
  3. Final permutation: (1 2 3)

These two results are distinct, which directly proves non-disjoint transpositions don't commute.

Going back to your example:
$$(1632)(457)=(12)(13)(16)(47)(45)$$
The transpositions on the right-hand side are not pairwise disjoint—(12) shares element 1 with (13), which in turn shares element 1 with (16). If you were to reorder these non-disjoint transpositions (say, swap (12) and (13)), the resulting permutation would no longer equal (1632)(457).

A quick side note: While every permutation can be written as a product of transpositions, this decomposition is not unique. The order of transpositions only doesn't matter when they're disjoint—otherwise, the order is critical to getting the correct permutation.

内容的提问来源于stack exchange，提问作者user462561