# Schur-Zassenhaus Theorem

## Theorem

Let $G$ be a finite group and $N$ be a normal subgroup in $G$.

Let $N$ be a Hall subgroup of $G$.

Then there exists $H$, a complement of $N$, such that $G$ is the semidirect product of $N$ and $H$.

## Proof

The proof proceeds by induction.

By definition, $N$ is a Hall subgroup if and only if the index and order of $N$ in $G$ are relatively prime numbers.

Let $G$ be a group whose identity is $e$.

We induct on $\order G$, where $\order G$ is the order of $G$.

We may assume that $N \ne \set e$.

Let $p$ be a prime number dividing $\order N$.

Let $\Syl p N$ be the set of Sylow $p$-subgroups of $N$.

By the First Sylow Theorem:

- $\Syl p N \ne \O$

Let:

- $P \in \Syl p N$
- $G_0$ be the normalizer in $G$ of $P$
- $N_0 = N \cap G_0$.

- $G = G_0 N$

By the Second Isomorphism Theorem for Groups and thence Lagrange's Theorem (Group Theory), it follows that:

- $N_0$ is a Hall subgroup of $G_0$
- $\index {G_0} {N_0} = \index G H$

Suppose $G_0 < G$.

Then by induction applied to $N_0$ in $G_0$, we find that $G_0$ contains a complement $H \in N_0$.

We have that:

- $\order H = \index {G_0} {N_0}$

and so $H$ is also a complement to $N$ in $G$.

So we may assume that $P$ is normal in $G$ (that is: $G_0 < G$).

This article, or a section of it, needs explaining.In particular: What is the significance of $G_0 < G$ here? It has already been stated above. What is its purpose at this point in the argument?You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Let $Z \paren P$ be the center of $P$.

By:

- Center is Characteristic Subgroup
- $P$ is normal in $G$
- Characteristic Subgroup of Normal Subgroup is Normal

$Z \paren P$ is also normal in $G$.

Let $Z \paren P = N$.

Then there exists a long exact sequence of cohomology groups:

- $0 \to H^1 \paren {G / N, P^N} \to H^1 \paren {G, P} \to H^1 \paren {N, P} \to H^2 \paren {G / N, P} \to H^2 \paren {G, P}$

which splits as desired.

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Otherwise:

- $Z \paren P \ne N$

In this case $N / Z \paren P$ is a normal (Hall) subgroup of $G / Z \paren P$.

By induction:

- $N / Z \paren P$ has a complement $H / Z \paren P$ in $E // Z \paren P$.

This article, or a section of it, needs explaining.In particular: The meaning of $E // Z \paren P$ and definition of $E$ in this context.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

This article, or a section of it, needs explaining.In particular: Although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Let $G_1$ be the preimage of $H // Z \paren P$ in $G$ (under the equivalence relation).

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Then:

- $\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N} \times \order {Z \paren P}$

This article, or a section of it, needs explaining.In particular: The definition of $K$ in $\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N}\ \times \order {Z \paren P}$.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Therefore, $Z \paren P$ is normal Hall subgroup of $G_1$.

By induction, $Z \paren P$ has a complement in $G_1$ and is also a complement of $N$ in $G$.

This article, or a section of it, needs explaining.In particular: Again, although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

$\blacksquare$

## Also known as

Some sources refer to this theorem as **Schur's theorem**, but that name is also used for an unrelated result in Ramsey theory.

## Source of Name

This entry was named for Issai Schur and Hans Julius Zassenhaus.