I want to know if my proof of the following is true, since I am not proficient enough in group theory to be sure by myself.
Proposition
Let $G,H$ be groups, $N\triangleleft G$. Let there be a function $\varphi\colon G\times H\to H$ by which $G$ acts on $H$ that is trivial on $N$ (i.e. $\varphi_n(h)=h$) and is an homomorphism for every $g\in G$ (i.e. $\varphi_g(h_1h_2)=\varphi_g(h_1)\varphi_g(h_2)$). We identify $G=(G,e_H)$ and $H=(e_G,H)$, as well as $N=(N,e_H)$, where $e_G\in G$ and $e_H\in H$ are the unit elements of their respective groups. Then $N\triangleleft G\ltimes H$ and $$ (G/N)\ltimes H = (G\ltimes H)/N.$$
Proof
Let $(g,h)\in G\ltimes H$. Now consider for $n\in N$ \begin{align*} n^{(g,h)} &= (n,e_H)^{(g,h)} = (g,h)^{-1}(n,e_H)(g,h) \\ &= \bigl(g^{-1},\varphi_{g^{-1}}(h^{-1})\bigr)(n,e_H)(g,h) \\ &= \bigl(g^{-1}n,\varphi_{g^{-1}}(h^{-1})\varphi_{g^{-1}}(e_H)\bigr)(g,h)\\ &= \bigl(g^{-1}n,\varphi_{g^{-1}}(h^{-1}e_H)\bigr)(g,h) \\ &= \bigl(g^{-1}n,\varphi_{g^{-1}}(h^{-1})\bigr)(g,h) \\ &= \bigl(g^{-1}ng,\varphi_{g^{-1}}(h^{-1})\varphi_{g^{-1}n}(h)\bigr) \\ &= \bigl(n^g,\varphi_{g^{-1}}(h^{-1})\varphi_{g^{-1}}(\varphi_n(h))\bigr) \\ &= \bigl(n^g,\varphi_{g^{-1}}(h^{-1})\varphi_{g^{-1}}(h)\bigr) \\ &= \bigl(n^g,\varphi_{g^{-1}}(h^{-1}h)\bigr) \\ &= \bigl(n^g,\varphi_{g^{-1}}(h^{-1}h)\bigr) =(n^g,e_H) \in N. \end{align*} Thus, we have $N\triangleleft G\ltimes H$.
Now let $g\in G$ and $h\in H$ and consider $$ (G\ltimes H)/N\ni(g,h)(N,e_H) = \bigl(gN, h\varphi_g(e_H)\bigr) = (gN,h) \in (G/N)\ltimes H. $$ Therefore each element of one group has to be element of the other and vice versa and thus they are equal.