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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.

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  • $\begingroup$ In your title you write "of a finite group" and in the sequel you don't ask about finite groups. Is your question for finite groups? $\endgroup$
    – YCor
    May 20, 2017 at 14:01
  • $\begingroup$ I think your definition of $H^{\langle g\rangle}$ is mistaken (it cannot make sense since $g$ is given and then is a new variable). I guess you just mean $H^K=\langle kHk^{-1}|k\in K\rangle$. $\endgroup$
    – YCor
    May 20, 2017 at 14:04
  • $\begingroup$ @YCor. Thank you for pointing that out. I have edited my question to reflect what I intended. And yes, I would wnat to find such a counterexample for a finite group, as weak pronormality and pronormality coincide for finite solvable groups $\endgroup$
    – R Maharaj
    May 20, 2017 at 14:15
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    $\begingroup$ You have not clarified about whether you want a finite group. And "this would be a non-solvable group" is not correct. It would not be a finite solvable group. $\endgroup$
    – YCor
    May 20, 2017 at 14:40
  • $\begingroup$ Could you provide a little more context? How do you know that weak pronormality does not imply pronormality in general finite groups, and why do you need an example? Is this an exercise somewhere, or just for interest's sake? $\endgroup$
    – Derek Holt
    May 20, 2017 at 15:00

2 Answers 2

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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 $G$.

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    $\begingroup$ I am not sure that this answer should have been accepted. While it suggests an approach, it gives no example where this actually happens. $\endgroup$ May 22, 2017 at 8:33
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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
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  • $\begingroup$ That's great Derek: yes, that is all we need: the maximal order among non-maximal subgroups condition was "overkill", but the argument works equally well if all (proper) overgroups (strictly) containing $H$ are maximal.Anyway, who needs theory when you have MAGMA/GAP? $\endgroup$ May 23, 2017 at 12:07
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    $\begingroup$ Well I wouldn't have found the example without your suggestion of where to look. There do not seem to be any smaller examples (although I haven't conclusively proved that) and I had given up on more naive searches. Also, once you understand the example, you do not really need a computer to complete the checks. The key fact is that $\langle H,H^t \rangle \cong L_2(11)$ has two classes of $A_5$ subgroups, which are fused in $G=M_{12}$. I am sure there are people around who would be familiar with things like that. $\endgroup$
    – Derek Holt
    May 23, 2017 at 12:25

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