Normal implies permutable: Difference between revisions
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===Normal subgroup=== | ===Normal subgroup=== | ||
{{further|[[Normal subgroup]]}} | |||
A subgroup <math>H</math> of a group <math>G</math> is a [[normal subgroup]] if for any <math>g \in G</math>, <math>gH = Hg</math> (viz, the [[left coset]]s are the same as the [[right coset]]s). | A subgroup <math>H</math> of a group <math>G</math> is a [[normal subgroup]] if for any <math>g \in G</math>, <math>gH = Hg</math> (viz, the [[left coset]]s are the same as the [[right coset]]s). | ||
===Permutable subgroup=== | ===Permutable subgroup=== | ||
{{further|[[Permutable subgroup]]}} | |||
A subgroup <math>H</math> of a group <math>G</math> is a [[permutable subgroup]] if for any subgroup <math>K \le G</math>, <math>HK = KH</math>. | A subgroup <math>H</math> of a group <math>G</math> is a [[permutable subgroup]] if for any subgroup <math>K \le G</math>, <math>HK = KH</math>. | ||
Revision as of 21:43, 22 February 2008
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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 must also satisfy the second subgroup property
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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 .
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 A subgroup of a group is a permutable subgroup if for any subgroup , .
Proof
Let be a normal subgroup of . We need to show that is permutable in .
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.