Sylow does not satisfy transfer condition: Difference between revisions
(New page: {{subgroup metaproperty dissatisfaction in| group property = finite group| property = Sylow subgroup| metaproperty = transfer condition}} ==Statement== It is possible to have a [[finite ...) |
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It is possible to have a [[finite group]] <math>G</math>, a Sylow subgroup <math>H</math>, and a subgroup <math>K</math> of <math>G</math> such that <math>H \cap K</math> is not a Sylow subgroup of <math>K</math>. | It is possible to have a [[finite group]] <math>G</math>, a Sylow subgroup <math>H</math>, and a subgroup <math>K</math> of <math>G</math> such that <math>H \cap K</math> is not a Sylow subgroup of <math>K</math>. | ||
==Related facts== | |||
* [[Hall does not satisfy transfer condition]] | |||
==Proof== | ==Proof== | ||
Revision as of 21:43, 19 September 2008
This article gives the statement, and possibly proof, of a subgroup property (i.e., Sylow subgroup) not satisfying a subgroup metaproperty (i.e., transfer condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about Sylow subgroup|Get more facts about transfer condition|
Statement
It is possible to have a finite group , a Sylow subgroup , and a subgroup of such that is not a Sylow subgroup of .
Related facts
Proof
Hands-on proof
Property-theoretic proof
We know that the property of being a Sylow subgroup is transitive (a Sylow subgroup of a Sylow subgroup is Sylow). Thus, if the property of being Sylow satisfies the transfer condition, we have that the property of being a Sylow subgroup is intersection-closed, by the general fact Transitive and transfer condition implies intersection-closed.
On the other hand, an intersection of Sylow subgroups need not be Sylow.