$M<K\rtimes H$ is a semidirect product?

Question

Let $H,K$ be two finite groups, $K$ abelian, and let $M$ be a subgroup of $K\rtimes H$.

Consider the projection $\pi:K\rtimes H \rightarrow H$ on the Second factor.

Let us suppose that $\pi(M)=H$. Is true in this case that $M$ is isomorphic to $K'\rtimes H$ for a suitable $K' <K$?

My claim is that it is true and that it is possible to construct an injective group omomorphism

$j:Ker \pi_{|M} \rtimes H \rightarrow M$ and then conclude using some classical arguments about group extensions.

Any help (or remark about the incompleteness of the question) is well accepted. Thank you.

Answer 1

No, there are many counter-examples.

For example, $K\rtimes H$ could be the dihedral group of order $8$, with $K$ a Klein group, $H$ generated by an involution outside of $K$ and $M$ cyclic of order $4$. We have $\pi(M)=H$ but $M$ cannot be expressed as a non-trivial semidirect product.

You can also take $K\rtimes H=C_4\times C_2$, with $K$ cyclic of order $4$, $H$ cyclic of order $2$ and $M$ a different cyclic subgroup of order $4$.