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Here is a conjecture of mine:

If $G$ is the internal semidirect product of $N \unlhd G$ and $Q \le G$, and $\phi_1 : N' \to N$ and $\phi_2 : Q' \to Q$ are isomorphisms, then there is some $\theta : Q' \to \text{Aut } N'$ such that $G \cong N' \rtimes_\theta Q'$

My thought was to take $\theta (x) = i_x$ with $i_x(y) = \phi_1( \phi_2(x) \phi_1(y) \phi_2(x)^{-1})$; and then show that $\phi : G \to N' \rtimes_\theta Q'$ given by $\phi(nq) = (\phi_1^{-1}(n),\phi_2^{-1}(q))$ . Assuming that $\theta$ is a homomorphism, I was able to show that $\phi$ is an isomorphism. However, when I went back to verify that $\theta$ is in fact a homomorphism, I ran into seemingly insuperable difficulties. Is $\theta$ as I have defined it a homomorphism? Is it the "right" homomorphism?

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    $\begingroup$ Do you mean $i_x(y) = \phi_1^{-1}( \phi_2(x) \phi_1(y) \phi_2(x)^{-1})$? $\endgroup$ – Arnaud D. Mar 20 at 11:48
  • $\begingroup$ @ArnaudD. Dang it! That was my problem. Without that inverse I wasn't getting the right cancellation. $\endgroup$ – user193319 Mar 20 at 11:50

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