= 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 $\epsilon(\sigma)=\det P_\sigma$. 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> as
$$
(a_1\,a_2\,\ldots,a_r)=(a_1\,a_r)(a_1\,a_{r-1})\cdots(a_1\,a_2).
$$
Hence $\epsilon(\sigma)=(-1)^m$ 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> identity
$$
\boxed{\epsilon(\sigma\tau)=\epsilon(\sigma)\epsilon(\tau).}
$$
The <alternating group> consists of the <even permutations>:
$$
\boxed{A_n=\{\sigma\in S_n:\epsilon(\sigma)=1\}=\ker\epsilon.}
$$
It contains the identity. If $\sigma,\tau\in A_n$, their product has <permutation sign> $1$, and $\epsilon(\sigma^{-1})=\epsilon(\sigma)^{-1}=1$, so their inverses also belong to $A_n$. This proves it is a <subgroup>. Finally, for every $g\in S_n$ and $a\in A_n$,
$$
\epsilon(gag^{-1})=\epsilon(g)\epsilon(a)\epsilon(g)^{-1}=1.
$$
Thus \b[$A_n$ is a <normal subgroup> of $S_n$]. This also exhibits directly why a <kernel of a group homomorphism> is normal.
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