Notational confusion about HNN-extensions: $G=K \ast_{H,t}$.

Question

Let $G=K \ast_{H,t}$ denote an HNN-extension, i.e., $$H \le K \le G, H^t \le K.$$

Is it true that $\{K, t\}$ is a generating system for $G$? In particular, $G/K$ is cyclic?

Answer 1

An HNN-extension is not a triple $K_{H, t}\sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', \phi, t)$ where $H, H'\leq K$ and $\phi$ is an isomorphism $\phi: H\rightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=\langle K, t\mid t^{-1}ht=\phi(h)\:\forall h\in H\rangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $\phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=\phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.

So yes, $\{K, t\}$ is a generating system for $G$. This is basically by definition.

On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:

Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $\phi\in\operatorname{Aut}(K)$ and $G$ is the mapping torus of $\phi$, so $G=K\rtimes_{\phi}\mathbb{Z}$.

The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $K\lhd G$ then $K=H=H'$, but if $K\lhd G$ then for all $k\in K$ we have $t^{-\epsilon}kt^{\epsilon}\in K$ for $\epsilon=\pm1$, so for all $k\in K$ we have $t^{-\epsilon}kt^{\epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".

On the other hand, the following lemma holds.

Lemma 2. Let $G$ be an HNN-extension as above, and let $\langle\langle K\rangle\rangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/\langle\langle K\rangle\rangle$ is infinite cyclic.