# Order-conjugate and Hall not implies order-dominating

## Statement

There can exist a finite group $G$ with a Hall subgroup $H$ such that the order and index of $H$ in $G$ are relatively prime, and such that $H$ is conjugate to any subgroup of $G$ of the same order as $H$, but such that there exists a subgroup math>K[/itex] of $G$ such that the order of $K$ divides the order of $H$, but $K$ is not contained in $H$.

## Proof

### Example of the alternating group of degree five

Further information: alternating group:A5, subgroup structure of alternating group:A5

Suppose $G$ is the alternating group on the set $S = \{ 1,2,3,4,5 \}$. Suppose $H$ is the subgroup of $G$ that is the alternating group on $\{ 1,2,3,4 \}$ (it is isomorphic to alternating group:A4). In other words, $H$ is the stabilizer of the point $\{ 5 \}$. Then, $H$ is a Hall subgroup of $G$. Further, $H$ is order-conjugate in $G$: the subgroups of the same order as $H$ are precisely the stabilizers of points in $S$, and these are conjugate to $H$ by suitable $5$-cycles.

On the other hand, consider the subgroup $K$:

$K := \{ (), (1,2,3), (1,3,2), (1,2)(4,5), (2,3)(4,5), (1,3)(4,5) \}$.

$K$ is a group of order six, isomorphic to the symmetric group of degree three. However, $K$ is not contained in any conjugate of $H$, because any conjugate of $H$ stabilizes some element, and $K$ does not stabilize any element.