---
tag: 64K1A69C80
kind: statement
role: theorem
name: Schur–Zassenhaus theorem
msc: [20D20, 20D40, "20E22"]
uses: [G5S1M9, 0CB0DX5BVV, 0DTXBQPA2S, 0VZ48N083F, 0H3WZXCZNF, 0H11ZPX4ND, MF4N3VGGFV, KPWW9X14V2, 0FXJ7YFMMX]
about: [MF4N3VGGFV, KPWW9X14V2]
sources:
  - url: https://en.wikipedia.org/wiki/Schur–Zassenhaus_theorem
same_as:
  wikidata: Q2226800
  wikipedia: Schur–Zassenhaus theorem
---

## Statement

Let $G$ be a finite [[group|G5S1M9]] and let $N$ be a [[normal subgroup|0CB0DX5BVV]] of $G$ such that $\abs{N}$ and the [[index|0DTXBQPA2S]] $[G : N]$ are [[coprime|0VZ48N083F]].

1. $N$ has a complement in $G$: a subgroup $H$ with $NH = G$ and $N \cap H = \set{e}$. Hence $G$ is a [[semidirect product|0H3WZXCZNF]] $N \rtimes H$.
2. Any two complements of $N$ in $G$ are [[conjugate|0H11ZPX4ND]] in $G$.

## Remarks

In other words, every [[extension|MF4N3VGGFV]] 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|0FXJ7YFMMX]]; 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.

## Explanations

### L1 human

Suppose a finite group has a normal subgroup $N$, and the size of $N$ shares no prime factor with the number of its cosets.
Then the group splits cleanly: there is a second subgroup that meets $N$ only in the identity and fills up the rest.
So the group is built from $N$ and the quotient in the simplest possible way, as a semidirect product.

### L2 human

A normal subgroup $N$ with $\gcd(\abs{N}, [G : N]) = 1$ is a normal [[Hall subgroup|KPWW9X14V2]], 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.