Normal implies permutable
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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) must also satisfy the second subgroup property (i.e., permutable subgroup)
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Statement
Verbal statement
Any Normal subgroup (?) of a group is a Permutable subgroup (?).
Symbolic statement
Let be a group and a normal subgroup of . Then is a permutable, or quasinormal, subgroup of . In other words, for any subgroup of (or equivalently, the Product of subgroups (?) is a subgroup for any subgroup ).
Property-theoretic statement
The subgroup property of being normal is stronger than the subgroup property of being permutable.
Definitions used
Normal subgroup
Further information: Normal subgroup A subgroup of a group is a normal subgroup if for any , (viz, the left cosets are the same as the right cosets).
Permutable subgroup
Further information: Permutable subgroup, permuting subgroups A subgroup of a group is a permutable subgroup if for any subgroup , (or equivalently, is a group). In other words, and are permuting subgroups for every . Here denotes the product of subgroups:
Proof
Hands-on proof
Given: Let be a normal subgroup of .
To prove: is permutable in .
Proof: Let be any subgroup of . For every , . Now we have:
and
Since , we conclude that .
Notice that the above proof does not anywhere use the fact that is a subgroup.
Converse
Permutable subgroups need not be normal
Further information: Permutable not implies normal
A permutable subgroup need not be normal. There are many counterexamples for finite groups.
References
Textbook references
- Topics in Algebra by I. N. Herstein, ^{More info}, Page 53, Problem 3 (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)
- Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, ^{More info}, Page 75, Exercise 5 of Section 5 (Restriction of a homomorphism to a subgroup) (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)