A student decomposes the permutation (12345) into transpositions two different ways: one decomposition uses 4 transpositions, another uses 6. What can you conclude?
AThe student made an error in at least one decomposition, since the number of transpositions must be fixed
BBoth decompositions are valid, and both show that (12345) is an even permutation
CThe sign is undefined for this permutation because different decompositions give different counts
DThe permutation is even by one decomposition and odd by the other, so its sign cannot be determined
(12345) is a 5-cycle, contributing 5 − 1 = 4 transpositions, so it is an even permutation. The theorem guarantees that every factorization into transpositions for a given permutation has the same parity — both counts here are even (4 and 6). The number of transpositions is not fixed, but the parity (even or odd) is invariant. Options C and D describe what would hold if parity were not well-defined — but the well-definedness theorem rules this out.
Question 2 Multiple Choice
What is the sign of the permutation (123)(45) in S₅?
A+1, because the permutation has two cycles and 2 is even
B+1, because (123) is an even cycle and it determines the sign
C−1, because the total transposition count is odd: (3−1) + (2−1) = 3
D−1, because (45) is a transposition with sign −1, which cancels the positive sign of (123)
Each k-cycle contributes k − 1 transpositions. The 3-cycle (123) contributes 2 transpositions (sign +1); the 2-cycle (45) contributes 1 transposition (sign −1). Total: 3 transpositions, odd, so sign = −1. Option A incorrectly counts cycles rather than transpositions. Option B is incomplete — all cycles must be accounted for. Option D states the right answer with a misleading explanation; the correct computation multiplies signs: (+1)(−1) = −1.
Question 3 True / False
The sign of a permutation depends on which specific decomposition into transpositions you use, so different decompositions can yield different signs for the same permutation.
TTrue
FFalse
Answer: False
The well-definedness of sign is the foundational theorem underlying this topic. While a permutation has infinitely many decompositions into transpositions, the parity of the count (even or odd) is an invariant — it never changes regardless of which decomposition is chosen. Different decompositions give different counts, but always all-even or all-odd for any given permutation. The sign is a property of the permutation, not of the decomposition.
Question 4 True / False
A k-cycle is an odd permutation when k is even, and an even permutation when k is odd.
TTrue
FFalse
Answer: True
A k-cycle decomposes into k − 1 transpositions. When k is even, k − 1 is odd, so the k-cycle has an odd number of transpositions — it is an odd permutation. When k is odd, k − 1 is even, so the k-cycle is even. Examples: a 2-cycle (transposition) is odd; a 3-cycle is even; a 4-cycle is odd; a 5-cycle is even. This rule is worth memorizing: even-length cycles are odd permutations, and odd-length cycles are even permutations.
Question 5 Short Answer
Why does the well-definedness of the sign function matter for the construction of the alternating group Aₙ?
Think about your answer, then reveal below.
Model answer: The alternating group Aₙ is defined as the kernel of the sign homomorphism — the set of all permutations with sign +1 (even permutations). For this definition to be mathematically meaningful, the sign must be a function: each permutation must have exactly one sign value, not different values depending on how it is decomposed. If parity were not well-defined, a permutation could be 'in Aₙ' by one decomposition and 'outside it' by another, making Aₙ an incoherent set. Well-definedness is what makes Aₙ a genuine subgroup with a definite membership criterion, and the sign a genuine group homomorphism from Sₙ to {±1}.
This connects to deeper structural facts: Aₙ has index 2 in Sₙ (exactly half of all permutations are even), making it the unique normal subgroup of index 2. For n ≥ 5, Aₙ is simple, which is the algebraic heart of the proof that there is no general radical formula for quintic polynomials.