Let $G$ be a permutation group on $\Omega$. The orbits of $G$ partition $\Omega$, and I would like to prove that $G$ is a subdirect product of the transitive groups $G_i$ induced by the action of $G$ on the corresponding orbits $O_i$.
So It is clear to me that the transitive constituents are the quotient groups $G/S_i$ where $S_i$ is the intersection of the stabilisers of the points in $\Omega_i$. I cannot make the next step, so I would appreciate some help finishing this.
We have $\Omega=\bigcup_i\Omega_i$ where the $\Omega_i$ are disjoint and $G$ acts transitively on each $\Omega_i$. The latter means that we have homomorphisms $f_i\colon G\to \operatorname{Sym}(\Omega_i)$, and the $G_i:=\operatorname{im}(f_i)$ are the transitive constituents. Combining the $f_i$, we obtain a homomorphism $\prod_i f_i\colon G\to\prod_i G_i$. If $g\in \ker \prod_i f_i$, then $g\in \ker f_i$ for all $i$, which means that $g$ acts trivially on each $\Omega_i$, hence acts trivially on $\Omega$. As $G$ is a permutation group of omega, this implies $g=1$. Therefore $\prod_i f_i$ lets us view $G$ as a subgroup of $\prod_i f_i$. Finaly, $\pi_i\circ \prod_i f_i$ is just $f_i$, hence is onto $G_i$. In other words, $G$ is a subproduct of the $G_i$.