P-constrained not implies p-solvable

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This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., p-constrained group) need not satisfy the second group property (i.e., p-solvable group)
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Definition

It is possible to have a finite group G and a prime number p such that G is a p-constrained group but not a p-solvable group.

Related facts

Converse

Opposite facts

Proof

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 is not p-solvable: For this, note that alternating group:A5, the simple non-abelian composition factor of G, is neither a p-group nor a p'-group.