Sylow does not satisfy transfer condition

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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).
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It is possible to have a finite group G, a Sylow subgroup H, and a subgroup K of G such that H \cap K is not a Sylow subgroup of K.

Related facts


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 finite-intersection-closed.

On the other hand, an intersection of Sylow subgroups need not be Sylow.