Recipe

Determine the parity of a permutation

Use when: you have a permutation, given as one or more cycles (disjoint or not), and need to classify it as even or odd

Goal: a definitive even-or-odd classification, reached from cycle lengths alone

Taught in Parity -- even and odd permutations

1. 1

   If the permutation is already written as disjoint cycles, skip ahead to Step 2. If it's a product of possibly-overlapping cycles instead, multiply them out into disjoint cycle notation first, applying the rightmost cycle first, then drop any resulting fixed points.

   Example

   (1\ 2\ 3)(3\ 4)(5\ 6) on \{1,\ldots,6\}: applying (5\ 6) first, then (3\ 4), then (1\ 2\ 3), sends 1\to2, 2\to3, 3\to4, 4\to1, 5\to6, 6\to5. That's the disjoint decomposition (1\ 2\ 3\ 4)(5\ 6).

   Why: the cycle-length shortcut in Step 3 only works once every cycle in the list is genuinely disjoint, since it counts each cycle's own length exactly once.

2. 2

   For each disjoint cycle, count its elements and note whether that count is even or odd; ignore any leftover fixed points entirely.

   Example

   (1\ 2\ 3\ 4)(5\ 6) has a length-4 cycle (even) and a length-2 cycle (even).

   Why: a k-cycle always needs exactly k-1 transpositions, so the cycle's own length already fixes whether that count is even or odd.

3. 3

   Count how many of the cycles from Step 2 have even length.

   Example

   (1\ 2\ 3\ 4)(5\ 6) has two even-length cycles, the 4-cycle and the 2-cycle.

   • the count of even-length cycles is even — the whole permutation is even.
   • the count of even-length cycles is odd — the whole permutation is odd.

   ⚡ Shortcut

   When the permutation is a single k-cycle, this collapses to one check. The cycle is even exactly when k is odd, and odd exactly when k is even, since one even-length cycle alone is an odd count.

   Why: each even-length cycle contributes an odd transposition count (k-1 is odd when k is even), so it takes an even number of those odd contributions to add up to an overall even total.