If a subgroup H of a semidirect product intersects the kernel trivially, must H lie in a complement?

Question

Let $G = N \rtimes K$ be a semidirect product (with $N$ normal in $G$). If $H < G$ satisfies $H\cap N = \{1\}$, must there exist a complement $M < G$ to $N$ such that $H < M$?

By a complement to $N$, I just mean a subgroup $M < G$ such that $M\cap N = \{1\}$ and $NM=G$.

Note that the projection $G\to G/N \cong K$ embeds $H$ as a subgroup of $K$, so $H$ is certainly isomorphic to a subgroup of a complement to $N$, but I'm asking for more.

I am particularly interested in the case where all complements to $N$ are conjugate to one another inside $G$.

Answer 1

No, not necessarily. There is a counterexample with $N = C_2$ and $K=C_4$ with trivial action of $K$ on $N$. So $G = N \times K$.

To get an example in which all complements are conjugate replace $K$ by a simple group, such as $A_5$. Then the complement is unique, but there are still diagonal elements of order $2$ that are not contained in that complement. In fact $K=A_4$ would also work.

Answer 2

Let $G=NK$ then,

$NH=NH\cap NK=N(NH\cap K)$ by Dedekind rule. Now assume that $(|N|,|K|)=1$

By Schur-Zassenhaus theorem $H$ and $NH\cap K$ are conjugate in $NH$. $$H^n=NH\cap K$$ $$H^n\subseteq K$$ $$H\subseteq K^{n^{-1}}=M$$ and clearly $M$ is also a complement.