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Recipe

List every element of A_n

Use when: you need every element of the alternating group A_n for a small n, not just a verdict on one permutation

Goal: a complete, non-redundant list of A_n's n!/2 elements, organized by cycle type

Taught in The alternating group Aₙ

Quick recipe

  1. List every permutation of \{1,\ldots,n\}, grouped by cycle type
  2. Classify each cycle type's parity using the cycle-length shortcut
  3. Keep only the even cycle types, and list every permutation of each
  4. Check the total count against n!/2
  1. 1

    Organize S_n's n! permutations by cycle type: the identity, single transpositions, 3-cycles, pairs of disjoint transpositions, and so on. That way none are missed or double-counted.

    Example

    For n=3: identity e; transpositions (1\ 2), (1\ 3), (2\ 3); 3-cycles (1\ 2\ 3), (1\ 3\ 2). That's 1+3+2=6=3! total.

    Why: grouping by cycle type gives a systematic, exhaustive count instead of guessing at which permutations exist.

  2. 2

    Apply the disjoint even-length-cycle count from the previous section to each cycle type as a whole. Every permutation sharing that cycle type shares the same parity.

    Example

    e has zero even-length cycles (even count, so even). A transposition has one even-length cycle (odd count, so odd). A 3-cycle has only one length-3 cycle, so zero even-length cycles (even count, so even).

    Why: cycle type alone, not which specific labels appear, determines parity. So a whole type is classified in one pass.

  3. 3

    Discard every odd cycle type entirely; write out every individual permutation belonging to each remaining even cycle type.

    Example

    For n=3, the even cycle types are the identity and the 3-cycles, giving e, (1\ 2\ 3), (1\ 3\ 2).

    • n is small enough to enumerate by hand (roughly n \le 4) — list every permutation directly.
    • n is too large to list by hand — report only the total count |A_n| = n!/2 instead of writing out every element.

    Why: fully listing only the even cycle types guarantees no odd permutation slips in and no even one is left out.

  4. 4

    Count the elements just listed and confirm the total matches n!/2, the formula this section's theorem proves.

    Example

    Three elements were found for A_3, and 3!/2=3. The count matches.

    Why: matching n!/2 catches a missed or double-counted cycle type before it propagates into later work with A_n.

Common mistakes

  • Testing only a couple of specific permutations and assuming that settles which cycle types are even.
  • Re-checking each individual permutation's parity from scratch instead of classifying a whole cycle type at once.
  • Trying to hand-list A_n for an n too large to enumerate practically.
  • Assuming the leftover odd permutations form their own subgroup the same way A_n does.