# Zassenhaus isomorphism theorem

From Groupprops

This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic

View other semi-basic facts in group theoryVIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|

VIEW: Survey articles about this

This article is about an isomorphism theorem in group theory.

View a complete list of isomorphism theorems| Read a survey article about the isomorphism theorems

## Contents

## Statement

Suppose are subgroups of a group such that is a normal subgroup of and is a normal subgroup of . Then, is normal in and is normal in , and we have an isomorphism:

.

## Facts used

- Normality satisfies transfer condition: If is normal in and is a subgroup of , is normal in .
- Modular property of groups: If , then and .
- Join lemma for normal subgroup of subgroup with normal subgroup of whole group: If and is normal in , then the subgroup is normal in the subgroup .
- Second isomorphism theorem: If are subgroups of a group such that is normal in , then is isomorphic to .

## Proof

### Hands-on proof

**Given**: are subgroups of a group such that is a normal subgroup of and is a normal subgroup of .

**To prove**: is normal in and , and we have an isomorphism:

.

**Proof**:

- (
**Given data used**: ;**Fact used**: fact (1)): is normal in : This follows from fact (1), setting . - (
**Given data used**: ): and normalize : Since is normal in , the normalizer of contains , hence it also contains . - and are subgroups: This follows directly from step (2).
- is normal in : By step (2), is normal in , and by step (1), is normal in . Thus, by fact (3), we obtain that is normal in . (Here, we set ).
- (
**Facts used**: fact (4)): Setting and , we observe that by step (4), is normal in , so applying fact (4) yields:- .

- (
**Facts used**: fact (2)): Applying fact (2) with yields . (Note that both are equal because of the fact that , being normal in , permutes with . We thus have:- .

- is normal in , both are subgroups, and we have (by analogous reasoning to steps (1) - (6)):
- .

- The right sides for steps (6) and (7) are equal, hence the left sides are isomorphic, completing the proof.

### Proof by application of the correspondence theorem

**PLACEHOLDER FOR INFORMATION TO BE FILLED IN**: [SHOW MORE]