P-constraint is not subgroup-closed

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This article gives the statement, and possibly proof, of a group property (i.e., p-constrained group) not satisfying a group metaproperty (i.e., subgroup-closed group property).
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It is possible to have a finite group G, a subgroup H, and a prime number p such that G is a p-constrained group but H is not a p-constrained group.

Related facts

Facts used

  1. Constrained for a prime divisor implies not simple non-abelian


Let G be the wreath product of Z2 and A5 defined as the wreath product with base group cyclic group:Z2 and acting group alternating group:A5, where we use the natural permutation action of the acting group on a set of five elements. More explicitly, G is the external semidirect product of elementary abelian group:E32 and alternating group:A5 where the latter acts on the former by coordinate permutations induced by the permutations on a set of five elements.

The group G has order \! 2^5 \cdot 60 = 1920 = 2^7 \cdot 3 \cdot 5

Let p = 2.

We note that:

  1. G is p-constrained: Indeed, O_{p'}(G) is trivial, and O_{p',p}(G) is the base of the semidirect product, i.e., a normal subgroup isomorphic to elementary abelian group:E32. In particular, this is contained in any p-Sylow subgroup P, so P \cap O_{p',p}(G) is also the normal subgroup that forms the base of the semidirect product. The subgroup is a self-centralizing normal subgroup, because it is an abelian normal subgroup and the induced action by the quotient is faithful. Thus, we get the condition C_G(P \cap O_{p',p}(G)) \le O_{p',p}(G).
  2. G has a subgroup isomorphic to alternating group:A5, namely the non-normal part of its realization as a semidirect product. This subgroup is a simple non-abelian group with p dividing its order (i.e., 2 divides 60) and hence, by Fact (1), is not a p-constrained group.