# Can a factorisation be pulled out of a semidirect product in this case?

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.

• This question doesn't make sense. There is no relation given between $G$ and $H$, so there is no relation between $N$ and $H$. Hence I don't see how to interpret "$N \cap H$". Moreover, $N$ is a normal subgroup of $G$, and not a subgroup of $G \ltimes H$, so you can't take the quotient by $N$. You have to mention how $N$ "fits in" $G \ltimes H$ first. Also, writing an element of the semi-direct product as $gh$ is dangerous, it's much safer to write $(g,h)$. And the inverse of such element is not $h^{-1}g^{-1}$ but $(g^{-1},\varphi_{g^{-1}}(h^{-1}))$ where $\varphi$ is the action of $G$ on $H$. – TastyRomeo May 11 '17 at 9:13
• I assume there is a function $\varphi:G\times H\to h$ by which $G$ acts on $H$. Also, I identify $G$ with $(G,e_H)$, $H$ with $(e_G,H)$ and $N$ with $(N,e_H)$. Thanks for the pointer to the inverse of $(g,h)$. – Wauzl May 11 '17 at 12:26
• You should edit this in the question! +1 because the question now makes sense and shows a decent attempt. As for a wild guess: the statement will probably be true only for "nice enough" actions. – TastyRomeo May 11 '17 at 12:47
• I just realized that I only need this if $\varphi_n(h)=h$ and $\varphi_g(h_1h_2)=\varphi_g(h_1)\varphi_g(h_2)$ for all $n\in N$, $g\in G$ and $h,h_1,h_2\in H$. I think I got it now. I'm editing my question. – Wauzl May 11 '17 at 12:52
• It is not true in general that every element of $N$ commutes with every element of $H$. – Derek Holt May 11 '17 at 13:04