= Solution
{c}
A left <group action> is a map $a:G\times X\to X$ satisfying
$$
a(e,x)=x,\qquad a(gh,x)=a\bigl(g,a(h,x)\bigr)
\quad(g,h\in G,\ x\in X).
$$
Define $\rho_a(g):X\to X$ by $\rho_a(g)(x)=a(g,x)$. The action axioms give
$$
\rho_a(g)\circ\rho_a(g^{-1})=\rho_a(e)=\operatorname{id}_X
=\rho_a(g^{-1})\circ\rho_a(g).
$$
Hence each $\rho_a(g)$ is a <bijection>, with inverse $\rho_a(g^{-1})$, and belongs to the <symmetric group> $\operatorname{Sym}(X)$. Finally,
$$
\rho_a(gh)(x)=a(gh,x)=a\bigl(g,a(h,x)\bigr)
=\bigl(\rho_a(g)\circ\rho_a(h)\bigr)(x),
$$
so $\rho_a:G\to\operatorname{Sym}(X)$ is a <group homomorphism>.
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