Normality satisfies transfer condition: Difference between revisions

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A [[subgroup property]] <math>p</math> is said to satisfy transfer condition if whenever <math>H, K</math> are subgroups of <math>G</math> and <math>H</math> has property <math>p</math> in <math>G</math>, <matH>H \cap K</math> has property <math>p</math> in <math>K</math>.
A [[subgroup property]] <math>p</math> is said to satisfy transfer condition if whenever <math>H, K</math> are subgroups of <math>G</math> and <math>H</math> has property <math>p</math> in <math>G</math>, <matH>H \cap K</math> has property <math>p</math> in <math>K</math>.
==Related facts==
* [[Second isomorphism theorem]]: This result equates the quotient of the non-normal subgroup, by the intersection, with the quotient of the product of subgroups, by the normal subgroup.


==Generalizations==
==Generalizations==

Revision as of 21:18, 13 June 2008

This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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This article gives the statement, and possibly proof, of a basic fact in group theory.
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Statement

Verbal statement

If a subgroup is normal in the group, its intersection with any other subgroup is normal in that subgroup.

Symbolic statement

Let HG be a normal subgroup and let K be any subgroup of G. Then, HKK.

Property-theoretic statement

The subgroup property of being normal satisfies the transfer condition.

Definitions used

Normal subgroup

A subgroup H of a group G is said to be normal if for any gG and hH, ghg1H.

Transfer condition

A subgroup property p is said to satisfy transfer condition if whenever H,K are subgroups of G and H has property p in G, HK has property p in K.

Related facts

  • Second isomorphism theorem: This result equates the quotient of the non-normal subgroup, by the intersection, with the quotient of the product of subgroups, by the normal subgroup.

Generalizations

Stronger metaproperties satisfied by normality

Proof

Hands-on proof

Given: A group G, a normal subgroup HG and a subgroup KG

To prove: HKK. In other words, we need to prove that given any gK and hHK, ghg1HK.

Proof: Since hHK, we in particular have hH. Since HG (viz H is normal in G), ghg1H.

But we also have that gK and hK. Since K is a subgroup, ghg1K.

Combining these two facts, ghg1HK.

References

Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 88, Exercise 24
  • Topics in Algebra by I. N. Herstein, More info, Page 53, Problem 5