Difference between revisions of "Contranormality is upper join-closed"
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Latest revision as of 12:58, 25 September 2008
This article gives the statement, and possibly proof, of a subgroup property (i.e., contranormal subgroup) satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about contranormal subgroup |Get facts that use property satisfaction of contranormal subgroup | Get facts that use property satisfaction of contranormal subgroup|Get more facts about upper join-closed subgroup property
Statement with symbols
Suppose is a subgroup, and , is an indexed family of subgroups with for each . Then, if is contranormal in each , is also contranormal in the [join of subgroups|join]] of the s.
Further information: contranormal subgroup
is a contranormal subgroup if for any containing such that is normal in , .
- Normality satisfies transfer condition: If is a normal subgroup, and , then is normal in .
Given: , family of subgroups with , and contranormal in each .
To prove: is normal in the join of all the s.
Proof: Suppose is a normal subgroup of the join of the s, containing . Then, for each , is a subgroup of containing , and by fact (1), it is normal in . Since is contranormal in , for each , so for each . Thus, must equal the join of the s.