I have to prove the following statement:
If $G$ is a finite soluble group and $N \trianglelefteq G$, then any $\pi$-Hall subgroup of $G/N$ is of the type $HN/N$ for some $H$ a $\pi$-Hall subgroup of $G$.
My work: I know that if $H$ is a $\pi$-Hall subgroup of $G$ then $HN/N$ is $\pi$-Hall subgroup of $G/N$ with $N \trianglelefteq G$.
My problem is the other direction. As $N \trianglelefteq G$ , $G/N$ is solvable, then by The first theorem of Hall for finite soluble groups, $G/N$ has a $K/N$ $\pi$-Hall subgroup. It is also known that $N\leq K \leq G$.
But, how do I prove that $K/N$ can be write as $HN/N$ for some $H$ a $\pi$-Hall subgroup of $G$?
For convenience, given a natural number $n=\prod_pp^{a_p}$, let $n_\pi:=\prod_{p\in\pi}p^{a_p}$ and $n_\pi':=\prod_{p\notin\pi}p^{a_p}$. Now Hall's theorem is the assertion that for each solvable group $G$ of order $n$, it contains a subgroup of order $n_\pi$.
Expanding on Derek's comment: Let $\overline K\subset G/N$ be a $\pi$-Hall subgroup, i.e., of order $[G:N]_\pi$. Now let $K$ be the pre-image of $\overline K$ under the surjection $G\to G/N$, which has order $|N|\cdot[G:N]_\pi$.
Then now $K$ contains a $\pi$-Hall subgroup $H$, which must have order $|N|_\pi\cdot[G:N]_\pi=|G|_\pi$, i.e., $H$ is a $\pi$-Hall subgroup of $G$. Now we hope to check $HN/N=\overline K$. But $HN/N$ is a $\pi$-Hall subgroup of $G/N$, and $HN/N\subseteq \overline K$, so they must be equal (their order is both $[G:N]_\pi$).