Prove that the sign function on permutations is unique in the following sense. If \(f\) is any function which assigns to each permutation of degree \(n\) an integer, and if \(f(\sigma\tau) = f(\sigma)f(\tau)\), then \(f\) is identically 0, or \(f\) is identically 1, or \(f\) is the sign function.
Solution
1. \(f\) is a group homomorphism \(S_n \to \ZZ^\times = \{\pm 1\}\) (or to \(\ZZ\) under multiplication).
the condition \(f(\sigma\tau) = f(\sigma)f(\tau)\) is exactly the homomorphism property.
2. For any transposition \(\tau\), \(f(\tau)^2 = f(\tau^2) = f(e) = 1\), so \(f(\tau) = \pm 1\).
\(\tau^2 = e\) and \(f(e) = f(e)^2\) forces \(f(e) = 1\) (or \(f \equiv 0\)).
3. All transpositions are conjugate in \(S_n\), so \(f\) takes the same value on every transposition.
conjugate elements have equal image under a homomorphism (since \(f(\sigma\tau\sigma^{-1}) = f(\sigma)f(\tau)f(\sigma)^{-1} = f(\tau)\)).
4. Case 1: \(f(\tau) = 1\) for every transposition \(\tau\).
4.1. Then \(f(\sigma) = 1\) for every \(\sigma \in S_n\).
every permutation is a product of transpositions (<1>3 of the transpositions-generate result), and \(f\) is multiplicative.
5. Case 2: \(f(\tau) = -1\) for every transposition \(\tau\).
5.1. Then \(f(\sigma) = (-1)^{\text{number of transpositions in a decomposition of } \sigma} = \operatorname{sgn}(\sigma)\).
\(f\) is multiplicative and equals \(-1\) on each transposition, so it equals the sign function.
6. Case 3: \(f \equiv 0\).
if \(f(e) = 0\) (or \(f\) is the zero map), then \(f(\sigma) = f(\sigma e) = f(\sigma) f(e) = 0\) for all \(\sigma\).
7. Hence \(f\) is identically \(0\), identically \(1\), or the sign function.
4, 5, and 6 exhaust the possibilities.
8. Q.E.D.
7.