How can I show that the map $\theta:S_4\rightarrow S_3$ defined by $θ(g) = g^∗$ is an homomorphism. I've been trying this problem and I can only prove for all the cases which is exhausting, how can I show directly that it's an homomorhpism.
$g^*$ is the the permutation induced on the partitions. That is with a permutation we substitute each element on the partition by its' next element on the permutation.
For example,if $$A = \{\{1, 2\}, \{3, 4\}\}; \\ B = \{\{1, 3\}, \{2, 4\}\}; \\ C = \{\{1, 4\}, \{2, 3\}\}.$$ and if g is the cyclic permutation (1, 2, 3, 4), then $g^∗ = (A, C)(B)$.
An action of a group $G$ on a set $E$ gives rise to a group morphism: $G\to S(E)$, sending $g\in G$ to the permutation $\sigma_g: x\in E\mapsto g\cdot x\in E$.
You can check that this map is indeed a morphism just using the group action axioms (the fact that $\sigma_g$ is bijective comes from the fact that an inverse is $\sigma_{g^{-1}}$, again using the definition on a group action)
Your map is exactly the morphism you obtain with the action you describe in your post...