= Solution
A <permutation representation> is a <group homomorphism> $\rho:G\to\operatorname{Sym}(\Omega)$, equivalently a <group action> on a set $\Omega$. The <left regular action> takes $\Omega=G$ and $\rho(g)(h)=gh$. Each such map is a <bijection>, and $\rho(g_1g_2)=\rho(g_1)\rho(g_2)$. If $\rho(g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\operatorname{Sym}(G)$ with the <symmetric group> $S_n$, proving <Cayley theorem>:
$$
\boxed{G\cong\rho(G)\leq S_n.}
$$
For a nonabelian <simple group>, compose this faithful representation with the <sign of a permutation> homomorphism $S_n\to\{1,-1\}$. Its <kernel of a group homomorphism>[kernel] is a <normal subgroup> of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and \b[the same representation embeds $G$ in the <alternating group> $A_n$].
Finally let $S_3$ act on the two <cosets> of its <normal subgroup> $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with \b[kernel $A_3$]; explicitly, send even permutations to the identity and odd permutations to $(12)$.
Back to article page
