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Suppose I am forming a semidirect product of finite groups $G \rtimes_\theta A$. Here, $\theta : A \to \mathrm{Aut}(G)$ is a homomorphism. Now, suppose I notice that the homomorphism $\theta$ factors through a quotient of $A$. That is, suppose $\pi : A \to Q$ is a surjective group homomorphism, and $\phi: Q \to \mathrm{Aut}(G)$ is such that $\theta = \phi \circ \pi$, so that $\theta$ "really comes from an action of $Q$ on $G$".

Question: In this situation, what can I say about the semidirect product $G \rtimes_\theta A$? Is it, in some way, built out of the semidirect product $G \rtimes_\phi Q$?

For instance, one could guess that $G \rtimes_\theta A \cong (G \rtimes_\theta Q) \times K$, where $K$ is the kernel of $\pi$. But that is obviously wrong... for instance, in the case where $\theta$ is trivial, this says that $G \times A \cong G \times Q \times K$, which should usually be false, since we likely do not have $A \cong Q \times K$.

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What you can say is that there is a surjective homomorphism $G \rtimes_\theta A\to G \rtimes_\phi Q$ given by $(g,a)\mapsto (g,\pi(a))$. Indeed, it is immediate from the definition of the semidirect product and the fact that $\theta=\phi\circ\pi$ that this map is a homomorphism, and it is obviously surjective. Furthermore, its kernel is the set of pairs $(1,a)$ with $a\in K$, which is isomorphic to $K$. So $G \rtimes_\theta A$ can canonically be written as an extension of $G \rtimes_\phi Q$ by $K$.

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