= Solution
A <group homomorphism> $h:G\to F$ preserves multiplication: $h(g_1g_2)=h(g_1)h(g_2)$. It sends the <identity element> to the identity and <inverse elements> to <inverse elements>. Its <group kernel> is $K=\{g\in G:h(g)=e_F\}$. This contains $e_G$ and is closed under multiplication and <inverse elements>, so it is a <subgroup>. For $k\in K$ and $g\in G$,
$$
h(gkg^{-1})=h(g)e_Fh(g)^{-1}=e_F.
$$
Thus $gKg^{-1}\subseteq K$; applying the same argument with $g^{-1}$ gives equality. Hence \b[the <group kernel> is a <normal subgroup>].
Let $S_2=\{e,\tau\}$ with $\tau=(12)$. The <parity of a permutation> defines the desired map from $S_3$:
$$
\boxed{h(e)=h((123))=h((132))=e,\qquad h((12))=h((13))=h((23))=\tau.}
$$
The <sign of a permutation> is multiplicative, so parity adds modulo two under composition, exactly matching $\tau^2=e$. Thus this is a <group homomorphism>; it is surjective because a transposition maps to $\tau$. Its <group kernel> is
$$
\boxed{K=A_3=\{e,(123),(132)\}.}
$$
This is the <alternating group> on three objects, a <normal subgroup> of index two in $S_3$.
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