= Solution
{c}
For each $g\in G$, left <translation in a group> $L_g:x\mapsto gx$ is a <bijection>, with <inverse function> $L_{g^{-1}}$. Thus $\theta(g)=L_g$ belongs to $\operatorname{Sym}(G)$. For $g,h,x\in G$, <associativity> of the <group operation> gives
$$
\theta(gh)(x)=(gh)x=g(hx)=(\theta(g)\circ\theta(h))(x).
$$
Hence $\theta$ is a <group homomorphism>. If $\theta(g)=\theta(h)$, evaluation at the <identity element> $e$ gives $g=ge=he=h$. Therefore
$$
\boxed{\theta:G\hookrightarrow\operatorname{Sym}(G)\text{ is an injective group homomorphism}.}
$$
This is the left regular representation used in <Cayley's theorem>.
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