Normality satisfies transfer condition
This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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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,
.
Property-theoretic 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 non-normal 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 join-closed
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 subordination-closed.
- Permutability satisfies transfer condition
Analogues in other algebraic structure
Proof
Hands-on 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, 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