# Internal Semidirect with Factors Isomorphic to “Outside” Groups

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?

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