Hall not implies pronormal

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

Facts used

 * 1) uses::Hall not implies order-conjugate
 * 2) uses::Hall satisfies transitivity

Related facts

 * Hall not implies procharacteristic

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, $$B$$ 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 $$G$$.