Finite solvable not implies p-normal

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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about finite solvable group|Get more facts about p-normal group

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property finite solvable group but not p-normal group|View examples of subgroups satisfying property finite solvable group and p-normal group

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 $P$.