# Hall not implies pronormal

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., Hall subgroup) need not satisfy the second subgroup property (i.e., pronormal subgroup)
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## Statement

We can have a finite group $G$ and a Hall subgroup $H$ of $G$ such that $H$ is not a pronormal subgroup of $G$.

## Proof

### General proof

By fact (1), construct a finite group $G$ and Hall subgroups $H, K$ of $G$ such that $H$ and $K$ are Hall of the same order but are not conjugate.

Let $p$ be a prime that does not divide the order of $G$. Consider the wreath product of $G$ with the cyclic group of order $p$ acting regularly. This is a group $A$ given by: $A = (G \times G \times G \times \dots \times G) \rtimes \mathbb{Z}/p\mathbb{Z}$.

Now consider the subgroups of $A$ given by: $B = K \times H \times H \times \dots \times H, \qquad C = G \times G \times \dots \times G$.

• $B$ is a Hall subgroup of $A$: $B$ is clearly a Hall subgroup of $C$. $C$ is a Hall subgroup of $A$ because $p is relatively prime to its order. Thus, [itex]B$ is a Hall subgroup of $A$ (fact (2)).
• $B$ is not pronormal in $A$: Consider $B$ and its conjugate $B_1$ by a generator of the $\mathbb{Z}/p\mathbb{Z}$. We have $B_1 = H \times K \times H \times \dots \times H$. These are both subgroups in $C$, hence if they are conjugate in the subgroup they are generate, they are conjugate in $C$. However, if $B$ and $B_1$ are conjugate in $C$, then the conjugating element acts coordinate-wise, so the first coordinate of $B$ (which is $K$) is conjugate to the first coordinate of $B_1$ (which is $H$) in $G$. But this is contradictory to the assumption that $H$ and $K$ are not conjugate in $G$.