Make parity independent of a chosen factorization

Try

Write (1 2 3)=(1 3)(1 2), composing right to left. Now insert the identity pair (1 2)(1 2) into a factorization. The number of transpositions changes by two, but its parity stays the same. This suggests a sign, yet inserting pairs alone does not prove that every possible factorization has the same parity.

We need a quantity attached directly to the permutation, independently of the chosen sequence of swaps. A determinant provides one such definition.

Predict

For a permutation σ, let its permutation matrix Pσ send the standard basis vector eᵢ to eσ(i). Predict how PσPτ acts on a basis vector and why this matches right-to-left composition. Then recall what happens to a determinant when two columns are exchanged.

Experiment

Define sign(σ)=det(Pσ). Permutation matrices have determinant either one or negative one. Composition multiplies determinants, and a transposition has determinant negative one.

Permutation form	Number of swaps in displayed construction	Sign
Transposition	1	−1
3-cycle	2	1
4-cycle	3	−1
Two disjoint transpositions	2	1

Check the cycle factorization (a₁ … aₖ)=(a₁ aₖ)…(a₁ a₃)(a₁ a₂) on each moved label. It uses k−1 transpositions, so a k-cycle has sign (−1)^(k−1). Count the parity of a disjoint cycle decomposition by multiplying those signs.

Understand

If a permutation has a factorization into r transpositions, determinant multiplicativity gives sign(σ)=(−1)^r. Because the determinant was defined without choosing that factorization, every other factorization must produce the same value. This proves parity is well-defined. The argument relies on the standard determinant facts; if those are unfamiliar, review row/column swaps and multiplicativity before treating this as a complete proof.

An even permutation has sign one. The set of even permutations is the alternating group Aₙ. The sign map is a homomorphism Sₙ→｛1,−1｝, where the target operation is multiplication. Its identity fiber is closed under products and inverses, so Aₙ is a subgroup. Conjugation preserves sign because sign(gσg⁻¹)=sign(g)sign(σ)sign(g)⁻¹=sign(σ); hence Aₙ is normal.

For n≥2, choose a fixed transposition τ. The map σ↦τσ is a bijection that reverses sign, pairing every even permutation with exactly one odd permutation. Therefore |Aₙ|=n!/2. The condition n≥2 matters: there is no transposition in S₁, and its one element is even.

Parity does not determine a permutation uniquely. In S₄, both the identity and a 3-cycle are even, but their element orders differ. Likewise an even permutation need not be a product of disjoint transpositions; a 3-cycle is an immediate counterexample. “Even” refers to the parity of any transposition factorization, not the appearance of the final cycle notation.

Apply

Compute the sign of a product containing a 5-cycle and a disjoint 2-cycle. Explain why you need not expand the entire product into a mapping table. Then use a table to verify one example as an independent check.

Repair the circular argument “parity is well-defined because every factorization has the same parity.” Identify the invariant that makes that statement a theorem. The practice asks for a proof route and its hypotheses, not only correct signs for a few permutations.

Continue

Polygon symmetries next provide a compact noncommutative family. Their relation between rotation and reflection will make factor order visible without a large permutation table.