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In what follows, all groups are assumed to be finite.

Recall that if $K \leq H \leq G$ are groups, $K$ is said to be a weakly closed subgroup of $H$ in $G$ if, for all $g \in G$, $g^{-1}Kg \leq H$ implies that $g^{-1}Kg = K$.
The subgroup property I am interested in is this one: what kind of subgroup of $H$ is $K$ when, for all groups $G$ with $H \leq G$, $K$ is a weakly closed subgroup of $H$ in $G$?
It is trivial to verify that $1$ and $H$ are subgroups of $H$ with this property. (Here, already, the finiteness condition is important: if $H$ is infinite, one could have $g^{-1}Hg < H$ instead of having $g^{-1}Hg = H$. So I won't go there.)

It is clearly necessary that $K$ be a characteristic subgroup of $H$ for this unnamed property to hold:
If $H$ is finite, so is its holomorph $Hol(H) = H \rtimes Aut(H)$. So let $G = Hol(H)$. Then, regarding $Aut(H)$ as a subgroup of $G$, let $\alpha \in Aut(H)$. Then $\alpha^{-1}K\alpha \leq H$ because $\alpha$ is an automorphism of $H$. Since $K$ is, by assumption, a weakly closed subgroup of $H$ in $G$, this means $\alpha^{-1}K\alpha = K$. Since $\alpha \in Aut(H)$ was arbitrary, it follows that $K$ is a characteristic subgroup of $H$.

On the other hand, merely being characteristic is not sufficient:
Let $G = <(1243675), (4657)(23)>$, so that $G \cong GL_{3}(2)$. Let $H = <(4657)(23), (67)(23)>$ and $K = <(45)(67)>$. Then $K = Z(H)$, so $K$ is a characteristic subgroup of $H$. It is also true that $H \leq G$, and that $g = (13)(57) \in G$.
But then $g^{-1}Kg = <(47)(65)> \leq H$, but $<(47)(65)> \neq < (45)(67) >$.
I have heard of fully invariant subgroups, but this also shows that a fully invariant subgroup need not have this property: in the example above, $Z(H)$ is a fully invariant subgroup of $H$ because $Z(H)$ is the subgroup generated by the squares in $H$.

What I can say is that if a subgroup $K$ of $H$ is the only subgroup of $H$ with elements having the orders they do, then $K$ is this kind of subgroup of $H$. (So, in the above example, $<(4657)(23)>$ is the unique cyclic subgroup of order 4 in $H$ and it is this kind of subgroup of $H$, unlike $Z(H)$.)
Does this kind of subgroup have a name? What other sets of conditions on a characteristic subgroup imply this stronger property?

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A tempting conjecture is that there must be no other subgroup of $H$ isomorphic to $K$. Certainly this property implies your property. On the other hand if $K$ is isomorphic to some other subgroup $K'$ of $H$ then one can form an HNN extension $G$ of $H$ wherein $K$ and $K'$ are conjugate. This is not a complete answer because HNN extensions are infinite. –  Sean Eberhard Jul 2 at 22:42
    
Geoff, I am impressed by the nontrivial extent to which you read my mind -- I did not know the Thompson subgroup always did this, but I was trying to get a better handle on some local analysis. –  DavidLHarden Jul 3 at 0:22
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Oops, I just deleted the comment. I don't know what sort of local analysis you were looking at, but the search for characteristic subgroups of a Sylow $p$-subgroup which might be normal in a $p$-constrained group often involves looking for the sort of subgroup you are looking at above, and the Thompson subgroup (and various relatives) was one of the first attempts to do this. –  Geoff Robinson Jul 3 at 10:13

2 Answers 2

up vote 8 down vote accepted

Instead of going via HNN extensions, one can work entirely with finite groups to show that if $X \cong Y$ are subgroups of a finite group $H$, then there exists a finite group $G \supseteq H$ such that $X$ and $Y$ are conjugate in G.

The trick is to take $G$ to be the full symmetric group on he elements of $H$, with $H$ embedded as the permutations induced by right multiplication (as in Cayley's theorem). Let $\theta$ be an isomorphism from $X$ to $Y$. Now choose a set $T$ of representatives for the left cosets of $X$ in $H$, and similarly, a set $S$ of representatives for the left cosets of $Y$. Note that $|T| = |H:X| = |H:Y| = |S|$, and let $\sigma$ be an arbitrary bijection from $T$ to $S$. Extend $\sigma$ to a permutation of $H$ by setting $(tx)^\sigma = t^\sigma x^\theta$.

Now $\sigma$ is an element of $G$, and I argue that conjugation in $G$ by $\sigma$ carries right multiplication by $x$ to right multiplication by $x^\theta$. This will show that the copies of $X$ and $Y$ in $G$ are conjugate in $G$.

Write $y = x^\theta$ and let $r_x$ and $r_y$ be the corresponding right multiplication maps on $H$. I want to show that $\sigma^{-1}r_x\sigma = r_y$, or equivalently, that $r_x\sigma = \sigma r_y$. To check this, we apply each side to an arbitrary element $h$ in $H$. Applying $r_x\sigma$ yields $(hx)^\sigma$, and applying $\sigma r_y$ yields $h^\sigma y$. We thus want $(hx)^\sigma = h^\sigma y$. Now write $h = tu$, with $t \in T$ and $u \in X$. Then $$\ (hx)^\sigma = (tux)^\sigma = t^\sigma(ux)^\theta = t^\sigma u^\theta x^\theta = (tu)^\sigma y = h^\sigma y , $$ as wanted.

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Thanks. I learnt something new and interesting today. –  P Vanchinathan Jul 8 at 3:23

There must be no other subgroup of $H$ isomorphic to $K$. Certainly this property implies your property, as $g^{-1}Kg$ is always isomorphic to $K$. Conversely if $K'\leq H$ is isomorphic to $K$ let $\alpha:K\to K'$ be an isomorphism, and let $G$ be the HNN extension $H\ast_\alpha$. The natural map $H\to G$ is an injection, and $K$ and $K'$ are conjugate in $G$.

If moreover $H$ is finite then $G$ is residually finite. Thus there is a finite quotient $G'$ of $G$ such that the composition $H\to G\to G'$ is injective. Since $K$ and $K'$ were conjugate in $G$ they remain so in $G'$.

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