If $H \unlhd G$ then is $H \unlhd G \rtimes K$ Suppose we have a semidirect product $G \rtimes_{\varphi} K$ and $H \unlhd G$, then is it necessarily true that $H \unlhd G \rtimes_{\varphi} K$?
We know that $G \unlhd G \rtimes_{\varphi} K$ but this embeds in a different way so I think something might go wrong. Can't think of an example though.

As Berci commented, this is only true if $H$ is $\varphi$-invariant. Here are two explicit counterexamples:

*

*Let $\varphi\colon C_2\to\mathrm{Aut}(C_2\times C_2)$ be the action given by swapping the two coordinates. Then the diagonal subgroup of $C_2\times C_2$ is $\varphi$-invariant, but the other two subgroups of order $2$ are not $\varphi$-invariant. More concretely, if we write $D_4=\langle r,s\mid r^4=s^2=(rs)^2=1\rangle$, then $D_4=(\langle s\rangle\times\langle r^2s\rangle)\rtimes\langle rs\rangle$. Here $\langle r^2\rangle$ is normal, but $\langle s\rangle$ and $\langle r^2s\rangle$ are not normal.


*Let $\varphi\colon C_3\to\mathrm{Aut}(C_2\times C_2)$ be the action given by rotating the three non-identity elements. Then none of the three subgroups of order $2$ are $\varphi$-invariant. More concretely, $A_4\cong(C_2\times C_2)\rtimes_\varphi C_3$ does not have a normal subgroup of order 2.
