Normality satisfies transfer condition
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 be a normal subgroup and let be any subgroup of . Then, .
Propertytheoretic statement
The subgroup property of being normal satisfies the transfer condition.
Definitions used
Normal subgroup
A subgroup of a group is said to be normal if for any and , .
Transfer condition
A subgroup property is said to satisfy transfer condition if whenever are subgroups of and has property in , has property in .
Related facts
Further facts
 Second isomorphism theorem: This result equates the quotient of the nonnormal subgroup, by the intersection, with the quotient of the product of subgroups, by the normal subgroup. Specifically, it states that if is normal in and is any subgroup of , we have .
Related metaproperties satisfied by normality
 Normality satisfies intermediate subgroup condition: The intermediate subgroup condition is weaker. It says that if are subgroups are is normal in , then is normal in .
 Normality satisfies inverse image condition: The inverse image condition is stronger. It says that the inverse image of a normal subgroup under a homomorphism is normal.
Other metaproperties satisfied by normality, that are somewhat related:
 Normality satisfies image condition: The image of a normal subgroup under a surjective homomorphism is normal in the image.
 Normality is upper joinclosed
Transfer condition for other subgroup properties
 Subnormality satisfies transfer condition: This follows directly from the fact that normality satisfies the transfer condition, and the fact that transfer condition is subordinationclosed.
 Permutability satisfies transfer condition
Analogues in other algebraic structure
Proof
Handson proof
Given: A group , a normal subgroup and a subgroup
To prove: . In other words, we need to prove that given any and , .
Proof: Since , we in particular have . Since (viz is normal in ), .
But we also have that and . Since is a subgroup, .
Combining these two facts, .
References
Textbook references
 Abstract Algebra by David S. Dummit and Richard M. Foote, 10digit ISBN 0471433349, 13digit ISBN 9780471433347, ^{More info}, Page 88, Exercise 24
 Topics in Algebra by I. N. Herstein, ^{More info}, Page 53, Problem 5