 Solution

For a permutation $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix, defined by $P_{\sigma }{e}_j=e_{\sigma (j)}$. Define the signature of a permutation by $ϵ(\sigma )=\det P_{\sigma }$. A permutation matrix is orthogonal, so $(\det P_{\sigma }{)}^2=1$ and its permutation sign lies in $\{1,-1\}$. A transposition exchanges two columns of the identity and therefore has determinant $-1$. Every permutation is a product of transpositions, by decomposing each permutation cycle asHence $ϵ(\sigma )=(-1{)}^m$ for any expression as $m$ transpositions; the determinant definition proves that this parity of a permutation is independent of the expression. Equivalently the permutation sign is positive for an even permutation and negative for an odd permutation.

Composition is taken rightmost first. Since $P_{\sigma \tau }{e}_j=e_{\sigma (\tau (j))}=P_{\sigma }{P}_{\tau }{e}_j$, we have $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. Multiplicativity of the determinant therefore proves the sign homomorphism identityThe alternating group consists of the even permutations:It contains the identity. If $\sigma ,\tau \in A_n$, their product has permutation sign $1$, and $ϵ({\sigma }^{-1})=ϵ(\sigma {)}^{-1}=1$, so their inverses also belong to $A_n$. This proves it is a subgroup. Finally, for every $g\in S_n$ and $a\in A_n$,Thus $A_n$ is a normal subgroup of $S_n$. This also exhibits directly why a kernel of a group homomorphism is normal.