Student A writes σ = (1 2 3 4) as a product of 3 transpositions and concludes σ is odd. Student B writes the same permutation as a product of 5 transpositions and also concludes σ is odd. Student C insists there must exist a decomposition into 4 transpositions, making σ even. Who is correct?
AStudent C — four-cycles must be even because 4 is even
BStudents A and B — σ is odd, and the well-definedness theorem guarantees every decomposition has odd parity
CAll three could be right, depending on which transpositions are chosen
DStudent C — every permutation has a decomposition into an even number of transpositions
The well-definedness theorem is the heart of this topic: while permutations have infinitely many decompositions into transpositions, the parity of the count is always the same. For a 4-cycle (a b c d), the decomposition (a b)(a c)(a d) uses 3 transpositions — odd parity. Any other decomposition must also use an odd number. The proof goes through inversions: each transposition changes the inversion count by an odd amount, so the parity of the total change is determined by the number of transpositions applied, independent of which ones. Student C's claim is simply false — there is no even-length decomposition of a 4-cycle.
Question 2 Multiple Choice
What is the sign of the permutation σ = (1 2 3)(4 5) in S₅?
A+1 (even) — the two cycles together involve 5 elements, and 5 is odd
B−1 (odd) — the 3-cycle decomposes into 2 transpositions (even) and the 2-cycle is 1 transposition (odd), giving 3 total transpositions (odd parity)
C+1 (even) — the 3-cycle and the 2-cycle have opposite parities that cancel
D−1 (odd) — all non-trivial permutations in S₅ are odd
A k-cycle decomposes into exactly k−1 transpositions. The 3-cycle (1 2 3) needs 3−1 = 2 transpositions (even); the 2-cycle (4 5) needs 2−1 = 1 transposition (odd). Total: 2 + 1 = 3 transpositions. Since 3 is odd, σ is an odd permutation with sign −1. The sign function is a homomorphism, so sgn(σ) = sgn((1 2 3)) · sgn((4 5)) = (+1)(−1) = −1. The common mistake is thinking that a 3-cycle is 'odd' because 3 is odd — but a k-cycle has parity (k−1), so a 3-cycle is *even*.
Question 3 True / False
A 5-cycle such as (1 2 3 4 5) is an even permutation.
TTrue
FFalse
Answer: True
A k-cycle decomposes into k−1 transpositions. For k = 5, that is 4 transpositions. Since 4 is even, a 5-cycle is an even permutation. The general rule: a k-cycle is even when k is odd, and odd when k is even. Mnemonically, 'odd-length cycles are even' — which is counterintuitive but follows directly from the k−1 transposition formula. This is why 3-cycles are even (they are in A₃ and A₄) while transpositions (2-cycles) are odd.
Question 4 True / False
The sign of a permutation depends on which specific decomposition into transpositions you choose — different decompositions may yield different parities.
TTrue
FFalse
Answer: False
This is the key well-definedness theorem: parity is an intrinsic property of the permutation, not of any particular decomposition. The proof uses inversions: an inversion is a pair (i, j) with i < j but σ(i) > σ(j). Each transposition changes the inversion count by an odd number, so the parity of the inversion count changes with each transposition. The total parity of the inversion count after all transpositions have been applied depends only on whether an even or odd number of transpositions was used. Since the inversion count of the identity is 0 (even), any product of transpositions equal to the identity must have even length — from which well-definedness follows.
Question 5 Short Answer
Why does a k-cycle have sign (−1)^(k−1), and what does this tell you about whether a 3-cycle is even or odd?
Think about your answer, then reveal below.
Model answer: A k-cycle (a₁ a₂ ··· aₖ) can be decomposed into exactly k−1 transpositions: (a₁ a₂)(a₁ a₃)···(a₁ aₖ). Since sign is a homomorphism and each transposition has sign −1, the sign of the k-cycle is (−1)^(k−1). For k = 3: sign = (−1)² = +1, so a 3-cycle is even. This means 3-cycles belong to the alternating group A₃ (and A₄, Aₙ for n ≥ 3).
The formula (−1)^(k−1) captures the counterintuitive pattern: odd-length cycles are even permutations, and even-length cycles are odd permutations. Transpositions (k=2) have sign −1 (odd); 3-cycles have sign +1 (even); 4-cycles have sign −1 (odd). For a product of disjoint cycles, sum the (k−1) values across all cycles — if the total is even, the permutation is even.