Schur–Zassenhaus theorem

Remarks

In other words, every extension of a group of order $m$ by a group of order $n$ with $\gcd(m, n) = 1$ splits. When $N$ is abelian, the proof is cohomological. The obstruction to a complement lies in $H^2(G/N, N)$, and the obstruction to conjugacy lies in $H^1(G/N, N)$. Both cohomology groups are zero, because they are killed by both $\abs{N}$ and $[G : N]$. The general case of part 1 reduces to the abelian case by induction, using a Sylow subgroup of $N$ and the Frattini argument. For part 2 in general one needs $N$ or $G/N$ to be solvable; one of them has odd order, so this follows from the Feit–Thompson theorem.

Example: $G = S_3$ and $N = A_3$, with $\abs{N} = 3$ and $[G : N] = 2$. Each subgroup of order $2$, such as $\set{e, (1\,2)}$, is a complement, and the three of them are conjugate.

The coprimality is necessary. In $G = \Z/4\Z$ with $N = \set{0, 2}$, both $\abs{N}$ and $[G : N]$ are $2$, and $N$ has no complement, because the only subgroup of order $2$ is $N$ itself.

Going deeper

For a student

A normal subgroup $N$ with $\gcd(\abs{N}, [G : N]) = 1$ is a normal Hall subgroup, so the theorem says that normal Hall subgroups have complements, unique up to conjugacy. It is a key step in the proof of Hall's theorem on solvable groups and in many arguments with coprime actions. Without coprimality, extensions may fail to split, and then one needs cohomology to describe them.