Finite solvable not implies p-normal

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

It is possible to have a finite solvable group G and a prime p such that G is not a p-normal group. In other words, there exists a p-Sylow subgroup P of G whose center Z(P) is not weakly closed in it.

Proof

Example of the symmetric group of degree four

Further information: symmetric group:S4

Let G be the symmetric group on the set \{ 1,2,3,4 \} and let p = 2. Then:

  • G is solvable.
  • G is not 2-normal: The center of a 2-Sylow subgroup is not weakly closed in it. For instance, consider the 2-Sylow subgroup:

P = \{ (), (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,3), (2,4), (1,2)(3,4), (1,4)(2,3) \}.

The center of this is:

Z(P) = \{ (), (1,3)(2,4) \}.

This is not weakly closed in P. For instance, it is conjugate in G to the subgroup generated by (1,2)(3,4), which is another subgroup inside P.