Let $G \le S^\Omega$, and let $\Delta \dot{\cup} \Gamma$ be a nontrivial partition of $\Omega$. Suppose $\Delta$ (and hence $\Gamma$) is $G$-invariant. Let $G^\Delta$ denote the restriction of $G$ to $\Delta$ (while $G^\Delta$ is a subgroup of $S^\Delta$, it can also be identified with a subgroup of $S^\Omega$). I have question in the proof of the following result: if $G^\Delta$ and $G^\Gamma$ have no nontrivial homomorphic image in common, then $G=G^\Delta \times G^\Gamma$. (This result and its proof are from [Dixon and Mortimer, Permutation Groups, p. 23].)
I can see that the pointwise stabilizers $H_1=G_{(\Delta)}$ and $K_1=G_{(\Gamma)}$ are the kernels of the homomorphisms from $G$ onto $G^\Delta$ and $G^\Gamma$, respectively. Hence $G/H_1 \cong G^\Delta$ and $G/K_1 \cong G^\Gamma$. The proof says the common homomorphic image of $G^\Delta$ and $G^\Gamma$ is $G/H_1 K_1$ - why is that (and what is the definition of the homomorphic image of $G^\Delta$ in the first place)?
The proof continues that if their common homomorphic image is trivial, then $G=H_1 K_1$. Also, $G^\Delta = (H_1 K_1)^\Delta = K_1$, and similarly, $G^\Gamma=H_1$, whence $G=G^\Delta G^\Gamma$ (which can I see equals $G^\Delta \times G^\Gamma$).
I would appreciate an elaboration of the proof.
A homomorphic image of a group $G$ is defined to be a group that is isomorphic to a quotient group $G/N$ of $G$ for some $N$.
By one of the isomorphism theorems, $G/H_1K_1 \cong (G/H_1)/(H_1K_1/H_1)$, so $G/H_1K_1$ is a homomorphic image of $G/H_1 \cong G^\Delta$. Similarly, $G/H_1K_1$ is a homomorphic image of $G^\Gamma$. So $G^\Gamma$ and $G^\Delta$ have $G/H_1K_1$ as a common homomorphic image, and the assumption that they have no such nontrivial common image implies that $G/H_1K_1$ must be trivial i.e. $G=H_1K_1$.