= Solution
A <normal subgroup> $H$ satisfies $gHg^{-1}=H$ for every $g\in G$. A <group homomorphism> preserves multiplication, $\Phi(ab)=\Phi(a)\Phi(b)$. It follows that $\Phi(e_1)=e_2$ and $\Phi(g^{-1})=\Phi(g)^{-1}$.
The <kernel of a group homomorphism> $K=\{g:\Phi(g)=e_2\}$ contains the identity and satisfies $ab^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi(gkg^{-1})=\Phi(g)e_2\Phi(g)^{-1}=e_2$. Thus $gKg^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a <normal subgroup>.
The product $f=\prod_{i<j}(x_i-x_j)$ is a <Vandermonde determinant> up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. Therefore
$$
\boxed{f\longmapsto-f\quad\text{under every transposition}.}
$$
If a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1)^rf$ and $(-1)^sf$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus \b[all decompositions have the same parity], and an even decomposition forces every decomposition to be even. This defines the <sign of a permutation>; its kernel is the <alternating group>, linking the parity argument to the normal-kernel result above.
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