HNN-extension as amalgamated free product

Question

Let $G_{\ast_P}$ be an HNN-extension over a subgroup $P$ which is isomorphic to a subgroup of the another subgroup $P''$ of $G$. Then can we write HNN-extension in the following way:?

$G_{\ast_P} = ({G \ast_{P} P''})_{\ast_{P''}}$

Thanks.

Answer 1

Here's a way to do this, using Bass-Serre theory and graphs of groups.

The group you have denoted $G_{*_P}$ needs some added details in its description, namely an isomorphism $f : P \to P''$. Once this is provided, then one can use graph of groups notation to describe the HNN extension.

Start with a graph $\Gamma$ having one vertex $V$ and edge $E$, so as a topological space $\Gamma$ is homeomorphic to a circle. Next label $V$ with the group $\Gamma_V = G$, and label $E$ with the group $\Gamma_E = P$. Then associate to each orientation of $E$ a homomorphism from $\Gamma_E$ to $\Gamma_V$: choose one orientation of $E$ called the positive orientation, and associate to it the inclusion homomorphism $i_P : \Gamma_E = P \hookrightarrow G = \Gamma_V$; and associated to the other orientation of $E$, called the negative orientation, is is the composed homomorphism $$j : \Gamma_E = P \xrightarrow{f} P'' \xrightarrow{i_{P''}} G = \Gamma_V $$ One can now use the machinery of graphs of groups to define $\pi_1(\Gamma)$, which does not mean the ordinary fundamental group, but instead means the ``graph of groups'' fundamental group. In this case, there is utterly no difference between $\pi_1(\Gamma)$ and the amalgamated free product, so what we've done so far is just a formality.

But now we alter the picture. Change the graph of groups structure on $\Gamma$ by inserting a new vertex $W$ into the middle of $E$, subdividing the edge $E$ into a pair of edges $E = E_1 E_2$, producing a new graph $\Gamma'$. The groups that label the edges and vertices of $\Gamma'$ are defined to be $$\Gamma_V = G, \quad \Gamma_W = P'', \quad \Gamma_{E_1} = P, \quad \Gamma_{E_2} = P'' $$ and the edge-group to vertex-group homomorphisms are $$\Gamma_{E_1} = P \hookrightarrow G = \Gamma_V $$ $$\Gamma_{E_1} = P \xrightarrow{f} P'' = \Gamma_W $$ $$\Gamma_{E_2} = P'' \xrightarrow{Id} P'' = \Gamma_W $$ $$\Gamma_{E_2} = P'' \hookrightarrow G = \Gamma_V $$ Again one can use Bass-Serre theory to define its graph-of-groups fundamental group $\pi_1(\Gamma')$. This is no longer just a formality: graphs-of-groups with more than one edge provide a robust theory of amalgamations that simultaneously generalizes amalgamated free products and HNN amalgamations.

The relationship between $\Gamma'$ and $\Gamma$ is that there is a quotient map $\Gamma' \mapsto \Gamma$ which maps $E_2$ to $V$ and $E_1$ to $E$. Since the homomorphism from the $E_2$ label group (which is $P''$) to the $W$ label group (which is also $P''$) is an isomorphism (namely the identity), one obtains an isomorphism from the fundamental group of the subgraph-of-groups $\pi_1(E_2)$ to the group $\Gamma_V = G$; in other words, $G *_{P''} P'' \approx G$. One can use this to prove that there is an induced isomrophism $\pi_1(\Gamma') \to \pi_1(\Gamma)$.

Similarly, since the homomorphism from the $E_1$ label group (which is $P$) to the $W$ label group (which is $P''$) is an isomorphism (namely $f$), one can similarly define a quotient map $\Gamma' \mapsto \Gamma''$ by defining $\Gamma''$ to be a graph with one vertex $V''$ and one edge $E''$, mapping $E_1$ to $V''$, and mapping $E_2$ to $E''$. The vertex group $V''$ gets labelled with the fundamental group of the sub-graph-of-groups $\pi_1(E_1)$ which is $G *_P P''$. The $E''$ edge is labelled with the same label as the $E_2$ edge, namely the group $P''$. One orientation on $E''$ takes $P''$ to the second factor of $G *_P P''$, and the other orientation on $E''$ takes $P''$ to the $P''$ subgroup of the first factor of $G *_P P''$.

So you're right: assuming one takes great care to properly define all the amalgamations needed for your question to make sense, which includes all of the edge-group to vertex-group homomorphisms needed for all the graphs-of-groups to be defined, one can indeed write $G_{*_P}$ as $(G *_P P'')_{*_{P''}}$.