All groups $G$ are finite, for $N \unlhd G$ we set $\overline{G} := G/N$ and for some $U\le G$ we write $\overline U = UN/N$. Also for a set $\pi$ of primes we denote by $O_{\pi}(G)$ the largets normal $\pi$-subgroup in $G$.
Lemma: Fix a set $\pi$ of primes and a group $N$, and consider a collection of subgroups $N_q \unlhd N$, where $q$ runs over some set of primes not in $\pi$ and $N = \prod_q N_q$. Assume for each prime $q$ that $N_q$ is a $(\pi \cup \{ q \})$-group. Then $N$ contains a nilpotent $\pi$-complement and all $\pi$-complements in $N$ are conjugate. Also, if $|\pi| = 1$, then $N$ is solvable.
Proof: First, note that $|\prod_{r\ne q} N_r|$ divides $\prod_{r\ne q} |N_r|$, which is not divisble by $q$. Then $N_q \cap \prod_{q \ne r} N_r$ is a normal $\pi$-subgroup of $N$, and it follows that the group $\overline N = N / O_{\pi}(N)$ is the direct product of the subgroups $\overline N_q$. Since each direct factor $\overline N_q$ of $\overline N$ has a nilpotent $\pi$-complement (its Sylow $q$-subgroup), it follows that the direct product $\overline N$ also has a nilpotent $\pi$-complement, and furthermore, since the $\pi$-complements in each factor are conjugate, it follows easily that the same is true for $\overline N$. Now let $\overline H$ be a nilpotent $\pi$-complement of $\overline N$, where $H \subseteq O_{\pi}(N)$. Applying the Schur-Zassenhaus theorem to $H$, we see that $H$ has a unique conjugacy class of $\pi$-complements, and these are nilpotent $\pi$-complements in $N$. Also, since all $\pi$-complements in $\overline N$ are conjugate to $\overline H$ and all $\pi$-complements in $H$ are conjugate, it follows that all $\pi$-complements in $N$ are conjugate, as required. Finally, if $\pi = \{ p \}$, then $N_q$ is a $\{p,q\}$-group, so $N_q$ is solvable by Burnside's $p^a q^b$-theorem. It follows that $N = \prod N_q$ is solvable, and the proof is complete. $\square$
I have some questions regarding this proof:
1) $\overline N = N / O_{\pi}(N)$ is the direct product of the subgroups $\overline N_q$.
I know the result that if $G$ is a product of normal subgroups $N_1, \ldots, N_n$, and $$ Z_i := N_i \cap \prod_{i\ne j} N_j $$ then $$ G / \bigcap_{i=1}^n Z_i \cong N_1 / Z_1 \times \ldots \times N_n / Z_n. $$ So by numbering the set of primes $q_1, \ldots, q_k$ and setting $Z_q := N_q \cap \prod_{q\ne r} N_r$ this yields here $$ N / \bigcap_q Z_q \cong N_{q_1} / Z_{q_1} \times \ldots \times N_{q_1} / Z_{q_k}. $$ Also $\bigcap_q Z_q$ is a normal $\pi$-group in $N$, so by definition $\bigcap_q Z_q \le O_{\pi}(N)$. Applying one of the isomorphism theorems we have $$ \left( N / \bigcap_q Z_q \right) / \left( O_{\pi}(N) / \bigcap_q Z_q \right) \cong G / O_{\pi}(N) $$ But here I do not see that this yields a decomposition as a direct product? (For this maybe $O_{\pi}(N) / \bigcap_q Z_q$ has to be direct, which would be if $O_{\pi}(N)$ is direct, but why should $O_{\pi}(N)$ be a direct product?)
2) My second question regards the application of the Schur-Zassenhaus Theorem.
Is it applied to $O_{\pi}(G)$? So that we have a $\pi$-complement $K$ with $O_{\pi}(G) = HK$? But why should the orders of $H$ and its index in $O_{\pi}(G)$ be coprime? Or is it applied to $H$ and $N$, but same questions, why should the order of $H$ and the index $|N : H|$ be coprime?
Further the conclusions about the conjugates I do not understand? Thanks for any explanations!