Let \(n \ge 2\), and let \(G\) be a finite group of odd order \(|G| = m\). Let \(\phi: S_n \to G\) be a group homomorphism. (1) Prove that if \(\tau \in S_n\) is a transposition, then \(\tau \in \ker\phi\). (2) Prove that \(\phi\) is the trivial homomorphism (\(\ker\phi = S_n\)). (3) Does this result depend on whether \(n\) is even or odd?
Solution
Let \(\tau\in S_n\) be a transposition. Since \(\tau^2=1\), \[ \phi(\tau)^2=1. \] Thus the order of \(\phi(\tau)\) divides \(2\). It also divides \(|G|\), which is odd. Hence \[ \operatorname{ord}(\phi(\tau))=1, \] so \(\phi(\tau)=1\) and every transposition lies in \(\ker\phi\).
Transpositions generate \(S_n\). Therefore \[ \ker\phi=S_n, \] and \(\phi\) is trivial.
The parity of \(n\) plays no role; the argument works for every \(n\ge2\).