Weak Pronormality of a finite group

Question

Definition: A subgroup $H$ of $G$ is said to be pronormal in $G$ if for all $g\in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x =H^g$.

Definition: A subgroup $H$ of $G$ is said to be weakly pronormal in $G$ if for $g \in G$, there exists $x \in H^{\langle g \rangle} := \langle g^nHg^{-n} \, |\, n\in \mathbb{Z} \rangle$ such that $H^x =H^g$.

By the inclusion $\langle H, H^g \rangle \leq H^{\langle g \rangle}$, we have that pronormality implies weak pronormality. It is also known that they coincide for finite solvable groups. However it does not hold in general according to the paper

I need to construct an example to show that weak pronormality does not imply pronormality for finite groups in general. This would have to be a non-solvable finite group to start with.

Answer 1

Here is a suggestion on a possible way to look for an example, though I have not pursued it myself: suppose that $H$ is a non-maximal subgroup of maximal order of a finite simple group $G$ and that $H$ is self-normalizing( given the first assumption, the second one is equivalent to requiring that $H$ is not a (proper non-trivial) normal subgroup of maximal order of any maximal subgroup of $G$). Suppose further that there is some $x \in G$ such that $x \not \in \langle H,H^{x} \rangle.$ Then $H$ is weakly pronormal in $G$, but not pronormal.

If $H$ were pronormal, then for $x$ as above we would have $H^{x} = H^{y}$ for some $y \in \langle H,H^{x} \rangle$, and then $xy^{-1} \in N_{G}(H) =H.$ But then $x \in \langle H,H^{x} \rangle,$ contrary to hypothesis. Hence $H$ is not pronormal.

To prove that $H$ is weakly pronormal, we note that $H^{\langle y \rangle }$ is normalized by $y$ for any $y \in G.$

If there were some $y \in G$ with $H$ not conjugate to $H^{y}$ in $H^{\langle y \rangle},$ then $H < H^{\langle y \rangle} < N_{G}(H^{\langle y \rangle})$ is a strictly increasing chain of proper subgoups of $G$. But this is a contradiction, since the choice of $H$ forces $H^{\langle y \rangle}$ to be maximal in $G$, and then $G = N_{G}(H^{\langle y \rangle}),$ contrary to the simplicity of $G$.

Answer 2

Added later: The example described below is not the smallest such example, which is $G={\rm PSU}(3,3)$ with $H \cong S_4$ and $\langle H,H^t \rangle \cong {\rm PSL}(2,7)$. But I will leave the original example.

I have found an example based on Geoff's suggestion. In fact the subgroup $H$ does not have maximal order among non-maximal subgroups, but it is self-normalizing, and all subgroups that properly contain it are maximal, which I believe is all that we need.

For $G$ we take the Mathieu group $M_{12}$ and for $H$ a transitive subgroup isomorphic to $A_5$. (There are other intransitive such subgroups.) Then the minimal overgroups of $H$ in $G$ are two maximal subgroups, both isomorphic to $L_2(11)$.

I used Magma to find the example, but here is a GAP calculation to check it - the $\mathtt{IntermediateSubgroups}$ command in GAP is useful here! I also checked the weak pronormality of $H$ directly from the definition, but that is time-consuming.

gap> G:=Group((1, 4)(3, 10)(5, 11)(6, 12),     
>          (1, 8, 9)(2, 3, 4)(5, 12, 11)(6, 10, 7) );;
gap> H := Subgroup(G, [(1, 5)(2, 11)(3, 4)(6, 7)(8, 12)(9, 10),
>                      (1, 12, 8)(2, 6, 9)(3, 4, 7)(5, 10, 11)]);;
gap> Size(G); Size(H);
95040
60
gap> Normalizer(G,H) = H;
true
gap> IS:=IntermediateSubgroups(G,H);;
gap> List(IS.subgroups,x->Size(x));
[ 660, 660 ]
gap> t := (1, 4, 10, 3, 9, 11, 2, 12)(5, 6, 8, 7);;
#Check that H and H^t are not conjugate in <H, H^t>, so H is not pronormal.
gap> K := Subgroup(G,
>            Concatenation(GeneratorsOfGroup(H),GeneratorsOfGroup(H^t)));;
gap> IsConjugate(K, H, H^t);
false