Let $G=\langle H, t; A^t=B\rangle$ by an $HNN$-extension of $H$, $A$ and $B$ isomorpic subgroups of $H$ where conjugation by $t$ induces the isomorphism.

Assuming $H$ is a finite group it is a well-known consequence of Britton's Lemma that if $\hat{H}$ is a subgroup of $G$ with $\hat{H}\cong H$ then $\hat{H}$ is conjugate to $H$ in $G$.

I was wondering under what other conditions on $H$ this result would hold. Specifically,

What conditions can we place on $H$ such that if $\hat{H}$ is a subgroup of $G$ with $\hat{H}\cong H$ then $\hat{H}$ is conjugate to a subgroup of $H$.


If H is a finitely generated torsion group, or more generally has Serre's property FA (fixed point property on trees) then what you want is true. Indeed H will have a fixed point in the Bass-Serre tree of the HNN extension. Since all vertex stabilizers are conjugate on this case, you are done.

  • $\begingroup$ So...groups with property FA are the only groups which have the property in my question? (I.e. having property FA is equivalent to the property I am wondering about?) $\endgroup$ – ADL Nov 24 '11 at 11:41
  • $\begingroup$ Property FA is equivalent to having your property for both HNN extensions and amalgamations I believe. I am not sure whether having this for HNN alone is enough though I expect it is by contracting an appropriate forest in the Bass-Serre tree. $\endgroup$ – Benjamin Steinberg Nov 24 '11 at 11:58

Property FA is equivalent to your property for groups $H$ that do not decompose as a free product. Indeed, suppose that your property holds but $H$ acts non-trivially on a simplicial tree. Then $H$ decomposes as a non-trivial amalgamated product $A*_CB$ (it is either that or an HNN extension which your property rules out). That is $H$ is not conjugate to a subgroup of either $A$ or $B$. Consider the free product $F=A*B$ and an HNN extension $E$ of $F$ conjugating two copies of $C$ there (one in $A$ and one in $B$) with free letter $t$. Then $A*_CB$ is isomorphic to the subgroup of $E$ generated by $tAt^{-1}$ and $B$ (it is proved in Lyndon and Schupp). Hence $H$ is a subgroup of $E$. By your assumption, $H$ is a subgroup of $F$. Hence by Kurosh's theorem $H$ is a non-trivial free product.

For groups that are free products, the properties are most probably also equivalent. Here is an idea. Consider $H=A*B$. Take a factor-group $H'$ of $H$ containing $A, B$, so that $H'$ does not contain a copy of $H$. One can assume also that all subgroups that are not conjugate to subgroups from, $A$ or $B$ are cyclic. That can be done by a result from Olshanskii's book "Geometry of defining relations in groups". Now consider the HNN extension $U$ of $H'$ where the free letter $t$ centralizes $B$: $tbt^{-1}=b, b\in B$. Then $H=A*B$ should be isomorphic to the subgroup of $U$ generated by $tAt^{-1}$ and $B$. This needs to be checked of course.

Update. If both $A,B$ are of order 2, Olshanskii's method does not work. But in that case, one can take the HNN extension of the symmetric group $S_3$ instead. In general, Olshanskii's method works if $A$ and $B$ do not contain involutions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.