0
$\begingroup$

We define a profinite HNN extension as the profinite completion of the abstract HNN extension. In the abstract case, the homomorphim of the base group to the HNN extension is always a monomorphism. However, it is not true for profinite groups (or pro-C) in general. When this homomorphism is injective in the profinite case we say that the HNN extension is proper. Is there any example of a non-proper profinite HNN-extension? (Or pro-C in general, where C is a class of finite groups).

$\endgroup$
1
  • 1
    $\begingroup$ HNN extensions of residually finite groups are very rarely residually finite, so very rarely embed into profinite groups. For example, Baumslag-Solitar group $BS(2,3)$. $\endgroup$
    – markvs
    Oct 5, 2021 at 18:32

1 Answer 1

3
$\begingroup$

Take $G=\mathbf{Z}_2$ (2-adics), and the isomorphism from $G=\mathbf{Z}_2$ to $2\mathbf{Z}_2$, given by multiplication by $2$.

Let $H$ be a profinite group with an element $t$, and a homomorphism $f:G\to H$ such that $f(2x)=tf(x)t^{-1}$ for all $x\in\mathbf{Z}_2$. (The HNN extension comes with such data, being universal for it.) Then the subgroup $\bigcup_n t^{-n}f(\mathbf{Z}_2)t^n$ is abelian divisible. But any abelian divisible subgroup of a profinite group is trivial. Hence the image of $G$ in the HNN extension is trivial.

(Note however that an HNN extension given by a topological isomorphism between two open subgroups of a locally compact group, is a nice locally compact group. In the above case, then HNN extension is $\mathbf{Q}_2\rtimes\mathbf{Z}$, acting by powers of $2$. But such a group doesn't "fit" in any profinite group.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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