# Difference between revisions of "Zassenhaus isomorphism theorem"

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'''Given''': <math>A, B, C, D</math> are subgroups of a group <math>G</math> such that <math>A</math> is a [[normal subgroup]] of <math>B</math> and <math>C</math> is a [[normal subgroup]] of <math>D</math>. | '''Given''': <math>A, B, C, D</math> are subgroups of a group <math>G</math> such that <math>A</math> is a [[normal subgroup]] of <math>B</math> and <math>C</math> is a [[normal subgroup]] of <math>D</math>. | ||

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# ('''Facts used''': fact (2)): Applying fact (2) with <math>L = C, M = D, N = A</math> yields <math>C(D \cap A) = D \cap CA = D \cap C</math>. (Note that both are equal because of the fact that <math>C</math>, being normal in <math>D</math>, permutes with <math>D \cap A</math>. We thus have: | # ('''Facts used''': fact (2)): Applying fact (2) with <math>L = C, M = D, N = A</math> yields <math>C(D \cap A) = D \cap CA = D \cap C</math>. (Note that both are equal because of the fact that <math>C</math>, being normal in <math>D</math>, permutes with <math>D \cap A</math>. We thus have: | ||

#* <math>\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap CA} = \frac{D \cap B}{D \cap B \cap AC}</math>. | #* <math>\frac{C(D \cap B)}{C(D \cap A)} \cong \frac{D \cap B}{D \cap B \cap CA} = \frac{D \cap B}{D \cap B \cap AC}</math>. | ||

− | # <math>A ( | + | # <math>A (B \cap C)</math> is normal in <math>A(B \cap D)</math>, both are subgroups, and we have (by analogous reasoning to steps (1) - (6)): |

− | #* <math>\frac{A(B \cap D)}{A( | + | #* <math>\frac{A(B \cap D)}{A(B \cap C)} \cong \frac{B \cap D}{B \cap D \cap AC} = \frac{B \cap D}{B \cap D \cap CA}</math>. |

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

## Latest revision as of 20:11, 30 May 2016

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

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This article is about an isomorphism theorem in group theory.

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## 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

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