I know that Schur Zassenhaus theorem is valid for any finite group, but my professor said that if a group is nilpotent group, the proof of Schur Zassenhaus theorem becomes very easy. However, I couldn't find the proof. Can you help me about that?
The question: Let G is finite nilpotent group, $N\vartriangleleft G$, $\gcd(| N|,\left[G:N\right])=1$. Show that there exist a $H\leq G$ such that $G=NH$, $N\cap H=1$.
A finite nilpotent group $G$ is the product of its Sylow $p$-subgroups, $G = P_1\times\cdots \times P_k$, with $P_i$ the Sylow $p_i$-subgroup.
Because $|N|$ is relatively prime to its index, if $p_i$ does not divide $|N|$, then the projection of $N$ onto the $P_i$ factor is trivial (otherwise, you could lift it to an element with an order a multiple of $p_i$, and from there get an element of order $p_i$).
Moreover, for any prime $p$ that divides $|N|$, a Sylow $p$-subgroup of $N$ is a Sylow $p$-subgroup of $G$... and there is one and only one of those. Therefore, $N$ is the product of the Sylow $p$-subgroups for all primes that divide $|N|$. We can then take $H$ to be the product of the remaning Sylow $p$-subgroups of $G$.
(The fact that a finite nilpotent group is the direct product of its Sylow subgroups follows from the fact that they are all normal; and that fact follows from the fact that a proper subgroup of a nilpotent group is always contained in its normalizer. Since $N_G(N_G(P))=N_G(P)$ for any Sylow subgroup $P$, it follows that $N_G(P)=G$)