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).
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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 $BS(2,3)$. $\endgroup$– markvsCommented Oct 5, 2021 at 18:32
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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.)
