 Solution

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and defineSince a transposition exchanges two columns, it has determinant $-1$. Thus any expression of $\sigma$ as $r$ transpositions has sign $(-1{)}^r$, proving that the parity is independent of the expression. Moreover, $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, so the determinant provesHence the sign homomorphism is a surjective homomorphism for $n\ge 2$.

For any homomorphism $\phi :S_n\to \{\pm 1\}$, all transpositions have the same image because they are conjugate. Since transpositions generate $S_n$, image $+1$ would make $\phi$ trivial. Surjectivity therefore forces every transposition to map to $-1$, and hence $\phi =\mathrm{sgn}$.