$\Gamma$ with base ${\mathbb{Z}}$ (Figure 18). We allow exceptional surfaces
and non-retractable splittings. These groups were studied in [GLS19] (where
they were called _socket groups with identified sockets_). They are torsion-
free hyperbolic groups, and are one-ended by Lemma 5.24. We shall show
(Corollary 10.24) that they are either prototypes or elementarily free groups.

If the surface $\Sigma$ associated to $\Gamma$ has genus $g$ and $b$ boundary
components, $G$ is presented as

$\left\langle\begin{matrix}a_{1},\dots,a_{b}\\\ z,t_{1},\dots,t_{b}\\\
u_{1},v_{1},\dots,u_{g},v_{g}\end{matrix}\ \middle|\ \begin{matrix}a_{1}\cdots
a_{b}=[u_{1},v_{1}]\cdots[u_{g},v_{g}]\\\ t_{i}z^{d_{i}}t_{i}^{-1}=a_{i}\\\
t_{1}=1\end{matrix}\right\rangle$

or

$\left\langle\begin{matrix}a_{1},\dots,a_{b}\\\ z,t_{1},\dots,t_{b}\\\
u_{1},\dots,u_{g}\end{matrix}\ \middle|\ \begin{matrix}a_{1}\cdots
a_{b}=u_{1}^{2}\cdots u_{g}^{2}\\\ t_{i}z^{d_{i}}t_{i}^{-1}=a_{i}\\\
t_{1}=1\end{matrix}\right\rangle,$

depending on whether $\Sigma$ is orientable or not, with indices $d_{i}\neq 0$
(possibly negative).

Minimality of $\Gamma$ implies $\sum|d_{i}|\geq 2$. The case when
$\sum|d_{i}|=2$ is special, since $G$ is then a closed surface group (the
surface is the union of $\Sigma$ and an annulus or a Möbius band). We have
seen (Example 4.35) that such groups are elementarily free, except that
$\pi_{1}(N_{3})$ is a prototype.

######  Lemma 10.20.

If $\sum|d_{i}|\geq 3$, then $\Gamma$ is the cyclic JSJ decomposition of $G$.
In particular, if $\sum|d_{i}|\geq 3$ and $\Sigma$ is exceptional, then $G$ is
a prototype.

###### Proof.

The proof is based on Bowditch’s construction of the cyclic JSJ tree
$T_{\mathrm{can}}$ from the topology of the Gromov boundary $\partial G$ of
$G$ [Bow98].

Let $T$ be the Bass-Serre tree of $\Gamma$. Let $E$ be the open star of the
vertex with stabilizer $\langle z\rangle$. It consists of exactly
$N=\sum|d_{i}|$ edges, all fixed by a power of $z$.

Let $\Lambda_{z}\subset\partial G$ be the pair of fixed points of $\langle
z\rangle$. By [Bow98], $\partial G\setminus\Lambda_{z}$ can be partitioned
into $N$ disjoint open sets corresponding to the connected components of
$T\setminus E$. It follows that $\partial G\setminus\Lambda_{z}$ has at least
$N$ components.

Since $N\geq 3$, [Bow98, Proposition 5.13] and [Gui00, Lemma 4.3] show that
$\langle z\rangle$ is universally elliptic (elliptic in every cyclic splitting
of $G$). In particular, the edge stabilizers of $T$ are universally elliptic,
so $T$ is dominated by the cyclic JSJ tree $T_{\mathrm{can}}$. On the other
hand, QH vertex stabilizers are elliptic in $T_{\mathrm{can}}$ by [GL17, Prop.
5], so $T$ dominates $T_{\mathrm{can}}$. It follows that $T$ is a JSJ tree.
Since it is its own tree of cylinders, it coincides with $T_{\mathrm{can}}$
(see [GL17]).

If $\sum|d_{i}|\geq 3$ and $\Sigma$ is exceptional, then $G$ has no
retractable centered splitting by Lemma 5.29, so is a prototype (recall that,
by Definition 4.23, a retractable centered splitting is non-exceptional). ∎

######  Remark 10.21.

The argument applies in the following more general setting: $G$ has a centered
splitting $\Gamma$ with all bottom groups cyclic, and for each bottom group
the sum of the indices of the incident edge groups is at least 3.

We proved in [GLS19] that, depending on the values of $g,b$, and the
$d_{i}$’s, either $\Gamma$ is retractable and $G$ is elementarily free, or
$\Gamma$ is not retractable. When retractable, the splitting $\Gamma$
expresses $G$ as an extended étage over ${\mathbb{Z}}$, but not as a simple
étage (see Remark 4.17). As in Proposition 5.1, we get a simple étage if
$\Sigma$ is complicated enough.

######  Proposition 10.22.

Assume that $\Gamma$ is retractable, and the surface $\Sigma$ either has Euler
characteristic satisfying $|\chi(\Sigma)|\geq 3$ or is a twice-punctured
torus. Then $G$ is a simple étage over ${\mathbb{F}}_{2}$.

The conclusion of the proposition implies that $G$ has a proper elementarily
embedded subgroup (isomorphic to ${\mathbb{F}}_{2}$). It applies whenever the
surface $\Sigma$ is not a 4-punctured sphere $S_{0,4}$, a thrice-punctured
projective plane $N_{1,3}$, a twice-punctured Klein bottle $N_{2,2}$, or
$N_{3,1}$ (note that $\Sigma$ cannot be a once-punctured torus or an
exceptional surface since $\Gamma$ is retractable). In these special cases we
reach the opposite conclusion:

######  Proposition 10.23.

Assume that $\Sigma$ has Euler characteristic satisfying $|\chi(\Sigma)|\leq
2$, and $\Sigma$ is not a twice-punctured torus. Then $G$ is not a simple
étage (over a non-abelian group) and $G$ has no proper elementarily embedded
subgroup.

Combining Lemma 5.29, Corollary 7.10, and the propositions, we get:

######  Corollary 10.24.

Let $G$ be a parachute.

  * •

If $\Gamma$ is not retractable (in particular if $\Sigma$ is exceptional),
then $G$ is a prototype.

  * •

If $\Gamma$ is retractable, $G$ is elementarily free; it has a proper
elementarily embedded subgroup if and only if $\Sigma$ satisfies
$|\chi(\Sigma)|\geq 3$ or is a twice-punctured torus. ∎

We shall now prove the propositions, starting with the easier one. They are
true if $\sum|d_{i}|=2$ (i.e. $G$ is a surface group), as explained in
Subsection 10.3. We therefore assume $\sum|d_{i}|\geq 3$. Lemma 10.20 then
implies that $\Gamma$ is the cyclic JSJ decomposition of $G$.

###### Proof of Proposition 10.23.

If $\Sigma$ is exceptional, then $G$ is a prototype by Lemma 10.20 and the
proposition holds. We are left with the surfaces $N_{3,1}$, $N_{2,2}$,
$N_{1,3}$, $S_{0,4}$. By Theorem 8.1, it suffices to show that $G$ is not an
extended étage over a non-abelian group relative to a non-abelian group.
Suppose it is. The associated surface $\Sigma^{\prime}$ may be viewed as a
proper incompressible subsurface of $\Sigma$ (see Lemma 5.26); moreover, the
complement of $\Sigma^{\prime}$ in $\Sigma$ cannot consist only of annuli and
Möbius bands because the étage is relative to a non-abelian group, so
$\chi(\Sigma^{\prime})=-1$. Since $\Sigma^{\prime}$ is non-exceptional and
$\Sigma$ is not a twice-punctured torus, the only possibility is that
$\Sigma^{\prime}$ is a once-punctured torus, $\Sigma$ is $N_{3,1}$, and
$\Sigma\setminus\Sigma^{\prime}$ is a twice-punctured projective plane.

We then write

$G=\langle u,v,u^{\prime},z\mid[u,v]u^{\prime 2}=z^{d}\rangle=\langle
u,v,u^{\prime},z\mid[u,v]=z^{d}u^{\prime-2}\rangle$

where $\pi_{1}(\Sigma^{\prime})=\langle u,v\rangle$ and the bottom group of
the splitting defining the extended étage is $\langle
z,u^{\prime}\rangle\simeq{\mathbb{F}}_{2}$. But this splitting cannot be
retractable since $z^{d}u^{\prime-2}$ is not a commutator in $\langle
z,u^{\prime}\rangle$. This concludes the proof. ∎

######  Remark 10.25.

Some of the groups in Proposition 10.23 are (non-relative) extended étages
over ${\mathbb{F}}_{2}$, but most of them are not. For instance, as in Example
10.11, consider parachutes with $\Sigma=N_{2,2}$. If the indices $d_{1}$ and
$d_{2}$ are even, $G$ is elementarily free. We claim that it is not an
extended tower over ${\mathbb{F}}_{2}$ if $|d_{1}|\neq|d_{2}|$.

The argument is the same as in the previous proof, but we now have to consider
the case when $\chi(\Sigma^{\prime})=-2$. It may be a three-punctured
projective plane or a 4-punctured sphere, obtained from $\Sigma$ by removing
one Möbius band, two Möbius bands, or a non-separating annulus. Using Lemma
4.9, one checks that the corresponding splittings cannot be retractable.

###### Proof of Proposition 10.22.

Given a non-degenerate boundary-preserving map $p:\pi_{1}(\Sigma)\to G$, let
${\mathcal{C}}$ be a maximal family of pinched curves on $\Sigma$. It is non-
empty by Lemma 4.9.

_Case 0: there exist two boundary components $C_{1}$, $C_{2}$ of $\Sigma$ that
are separated by ${\mathcal{C}}$._ In this case, we may argue as in the proof
of Proposition 5.1, using an arc $\gamma$ from $C_{1}$ to $C_{2}$ meeting a
curve $\delta\in{\mathcal{C}}$ exactly once. The surface $\hat{\Sigma}$
obtained by removing a pair of pants obtained as a regular neighborhood of
$C_{1}\cup\gamma\cup C_{2}$ is non-exceptional because
$\chi(\hat{\Sigma})=\chi(\Sigma)+1\leq-2$ (when $\Sigma$ is a twice-punctured
torus, $\hat{\Sigma}$ is a once-punctured torus, hence non-exceptional). As in
the proof of Proposition 5.1, this defines a retractable simple centered
splitting with base ${\mathbb{F}}_{2}$ and surface $\hat{\Sigma}$, so the
proposition holds in this case.

Figure 19: Case 1: the new étage when $\Sigma$ is orientable

_The remaining cases: ${\mathcal{C}}$ does not separate boundary components._
We shall distinguish several cases. In each case we define a new étage
structure by writing the surface $\Sigma$ as the union of two subsurfaces
$\Sigma=\Sigma^{\prime}\cup\hat{\Sigma}$ intersecting in a simple closed
curve. The surface $\Sigma^{\prime}$ satisfies $\chi(\Sigma^{\prime})=-1$ and
contains at least one component of $\partial\Sigma$, and $\hat{\Sigma}$ is
associated to a simple centered splitting with base $H\simeq{\mathbb{F}}_{2}$.

The retractability of $\Gamma$ implies $d_{1}+\dots+d_{b}=0$ (and in
particular $b\geq 2$) when $\Sigma$ is orientable, and $d_{1}+\dots+d_{b}$
even when $\Sigma$ is non-orientable ([GLS19], Propositions 6.7 and 6.8).
Using this, we construct a retraction $\rho:G\to H$ such that the central
group $\pi_{1}(\hat{\Sigma})$ has non-abelian image.

We use the presentation of $G$ given at the beginning of this subsection, with
$g$ the genus of $\Sigma$ and $b$ the number of boundary components. It
suffices to define $\rho$ on the generators $z,t_{i},u_{j},v_{j}$ since
$a_{i}=t_{i}z^{d_{i}}t_{i}^{-1}$.

_Case 1: $\Sigma$ is orientable_ (see Figure 19). If $g=0$, then any curve in
${\mathcal{C}}$ separates two boundary components of $\Sigma$ and we are done.
Assume therefore that $g\geq 1$. Recall that $b\geq 2$. Decomposing $\Sigma$
into a pair of pants $\Sigma^{\prime}$ with fundamental group $\langle
a_{1},a_{2}\rangle$ and its complement $\hat{\Sigma}$, one gets a simple
centered splitting with bottom group $H=\langle z,t_{2}\rangle$ and central
vertex group $Q=\langle a_{3},\dots,a_{b},u_{1},v_{1},\dots
u_{g},v_{g}\rangle$. As in case 0, the surface $\hat{\Sigma}$ is non-
exceptional.

One then defines a retraction $\rho:G\to H$ by

$\displaystyle z,t_{2}$ $\displaystyle\mapsto z,t_{2}$ $\displaystyle
t_{3},\dots,t_{b},u_{2},\dots,u_{g},v_{2},\dots,v_{g}$ $\displaystyle\mapsto
1$

and there remains to define the images $\tilde{u}_{1}$, $\tilde{v}_{1}\in H$
of $u_{1},v_{1}$ satisfying the equation

$z^{d_{1}}t_{2}z^{d_{2}}t_{2}^{-1}z^{d_{3}+\dots+d_{b}}=[\tilde{u}_{1},\tilde{v}_{1}].$

Since $d_{1}+\dots+d_{b}=0$, we may take $\tilde{u}_{1}=z^{d_{1}}t_{2}$ and
$\tilde{v}_{1}=z^{d_{2}}$. Finally, $\rho(Q)$ contains
$\langle\tilde{u}_{1},\tilde{v}_{1}\rangle$, which is non-abelian since
$d_{2}\neq 0$.

Figure 20: Cases 2 and 3: $\Sigma$ is non-orientable of genus $g\geq 3$

_Case 2: $\Sigma$ is non-orientable with a single boundary component_ (see
Figure 20). Note that $|\chi(\Sigma)|\geq 3$ implies $g\geq 4$, and recall
that $d_{1}$ is even. In this case, we take for $\Sigma^{\prime}$ a twice-
punctured projective plane containing $\partial\Sigma$ (see Figure 20). Its
complement $\hat{\Sigma}$ in $\Sigma$ is homeomorphic to $N_{g-1,1}$, hence
non-exceptional because $g\geq 4$. The group $G$ has a simple centered
splitting with bottom group $H=\langle z,u_{1}\rangle$ and central vertex
group $Q=\pi_{1}(\hat{\Sigma})=\langle u_{2},\dots,u_{g}\rangle$.

One defines $\rho:G\to H$ by

$\displaystyle z,u_{1}$ $\displaystyle\mapsto z,u_{1}$ $\displaystyle
u_{4},\dots,u_{g}$ $\displaystyle\mapsto 1,$

and there remains to define the images $\tilde{u}_{2}$, $\tilde{u}_{3}$ of
$u_{2},u_{3}$ satisfying the equation
$z^{d_{1}}=u_{1}^{2}\tilde{u}_{2}^{2}\tilde{u}_{3}^{2}$. Since $d_{1}$ is even
and non-zero, one may take $\tilde{u}_{2}=u_{1}^{-1}$ and
$\tilde{u}_{3}=z^{d_{1}/2}$, and
$\rho(Q)=\langle\tilde{u}_{2},\tilde{u}_{3}\rangle$ is non-abelian.

_Case 3: $\Sigma$ is non-orientable of genus $g\geq 3$ with $b\geq 2$ boundary
components_ (see Figure 20). As in case 1, we take for $\Sigma^{\prime}$ a
pair of pants, and we get a simple centered splitting of $G$ with bottom group
$H=\langle z,t_{2}\rangle$ and central vertex group $Q=\langle
a_{3},\dots,a_{b},u_{1},\dots,u_{g}\rangle$.

One defines $\rho$ by

$\displaystyle z,t_{2}$ $\displaystyle\mapsto z,t_{2}$ $\displaystyle
t_{3},\dots,t_{b},u_{4},\dots,u_{g}$ $\displaystyle\mapsto 1$

and there remains to define the images
$\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{3}$ of $u_{1},u_{2},u_{3}$ in $H$
satisfying the equation

$\tilde{u}_{1}^{2}\tilde{u}_{2}^{2}\tilde{u}_{3}^{2}=z^{d_{1}}t_{2}z^{d_{2}}t_{2}^{-1}z^{d_{3}+\dots+d_{b}}=[z^{d_{1}}t_{2},z^{d_{2}}]z^{2k}$

with $d_{1}+\dots+d_{b}=2k$. This is possible since $[a,b]c^{2}$ is a product
of three squares in any group (e.g.
$[a,b]c^{2}=(ac^{-1}b)^{2}(b^{-1}c)^{2}(c^{-1}ba^{-1}b^{-1}c^{2})^{2}$, which
can be viewed as a consequence of the homeomorphism $T^{2}\\#P^{2}\simeq
P^{2}\\#P^{2}\\#P^{2}$). One may thus take
$\tilde{u}_{1}=z^{d_{1}}t_{2}z^{d_{2}-k}$, $\tilde{u}_{2}=z^{k-d_{2}}$, and
$\tilde{u}_{3}=z^{-k+d_{2}}t_{2}^{-1}z^{2k-d_{1}-d_{2}}$. Note that
$\langle\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{3}\rangle$ is non-abelian
unless $k=d_{2}$ and $2k-d_{2}=0$, i.e. $k=d_{2}=0$, which is impossible.

Figure 21: Case 4: $\Sigma$ is non-orientable with genus $g\leq 2$, and $b\geq
3$

_Case 4: $\Sigma$ is non-orientable of genus $g=2$_ (see Figure 21). Since
$|\chi(\Sigma)|\geq 3$, we have $b\geq 3$. Since $d_{1}+\dots+d_{b}$ is even,
one may assume (after renumbering) that $d_{2}$ or $d_{2}+d_{3}$ is even. As
above, one may take for $\Sigma^{\prime}$ a pair of pants, so that $G$ has a
simple centered splitting with bottom group $H=\langle z,t_{2}\rangle$. One
defines $\rho$ by

$\displaystyle z,t_{2}$ $\displaystyle\mapsto z,t_{2}$ $\displaystyle t_{3}$
$\displaystyle\mapsto 1\text{ or }t_{2}$ $\displaystyle t_{4},\dots,t_{b}$
$\displaystyle\mapsto 1,$

where we map $t_{3}$ to $1$ if $d_{2}$ is even and to $t_{2}$ otherwise. We
treat the case where $d_{2}$ is even, the other case being similar. To
complete the definition of $\rho$, we need to find images
$\tilde{u}_{1},\tilde{u}_{2}$ of $u_{1},u_{2}$ in $H$ which satisfy the
equation

$z^{d_{1}}t_{2}z^{d_{2}}t_{2}^{-1}z^{d_{3}+\dots+d_{b}}=\tilde{u}_{1}^{2}\tilde{u}_{2}^{2}.$

This is possible because $d_{2}$ and $d_{3}+\dots+d_{b}+d_{1}$ are even. Since
$\rho(a_{1}a_{2})=z^{d_{1}}t_{2}z^{d_{2}}t_{2}^{-1}$ does not commute with
$\rho(a_{3})$, which is conjugate to $z^{d_{3}}$, the image $\rho(Q)$ is non-
abelian.

_Case 5: $\Sigma$ is non-orientable of genus 1._ We show that this cannot
happen under the assumption that ${\mathcal{C}}$ does not separate boundary
components. Being two-sided, all curves in ${\mathcal{C}}$ are separating. The
components of $\Sigma\setminus{\mathcal{C}}$ which do not meet
$\partial\Sigma$ are punctured projective planes or spheres bounded by pinched
curves, and their fundamental group has trivial image under
$p:\pi_{1}(\Sigma)\to G$ because $G$ is torsion-free. The connected component
containing $\partial\Sigma$ has cyclic image by Lemma 4.9 and this forces $p$
to have an abelian image, a contradiction. ∎

##### Multi-parachutes.

More generally, suppose that $G$ has a (possibly non-simple) centered
splitting $\Gamma$ with bottom groups all isomorphic to ${\mathbb{Z}}$. We
call such a group a multi-parachute. For instance, one may connect all
boundary components of $\Sigma$ to distinct cyclic vertex groups; groups
obtained in this way were studied in [GLS19] under the name of socket groups,
and elementarily free socket groups are described in [GLS19, Proposition 6.2].

We now study general multi-parachutes. In particular, we show that they are
either prototypes or elementarily free groups, and we determine whether they
have proper elementarily embedded subgroups.

Like parachutes, multi-parachutes are one-ended hyperbolic groups. After
gluing annuli and Möbius bands to the surface if needed, we may assume that
$\sum d_{i}(v)\geq 3$ for each bottom vertex $v$ (we denote by $d_{i}(v)$ the
indices of incident edge groups in the vertex group), so $\Gamma$ is the
cyclic JSJ decomposition of $G$ by Lemma 10.20 and Remark 10.21. The case of
surface groups and parachutes has already been studied, so assume that
$\Gamma$ has at least two (cyclic) bottom groups.

If $\Gamma$ is not retractable (in particular if $\Sigma$ is exceptional), $G$
is a prototype by Lemma 5.29, so we now assume that $\Gamma$ is retractable.
In this case $G$ is an extended étage over a free group, hence elementarily
free by Corollary 7.10. We distinguish three cases, depending on $\Sigma$.

$\bullet$ If $|\chi(\Sigma)|\geq 3$, Proposition 5.1 shows that $G$ is a
simple tower over a non-abelian free group, thus providing an elementarily
embedded proper free subgroup by “Tarski”.

$\bullet$ If $\Sigma$ is a twice-punctured torus, then $\Gamma$ has to be
simple by Proposition 6.2 of [GLS19] (see Remark 4.25). Thus $G$ is a
parachute, and it has an elementarily embedded proper free subgroup by
Proposition 10.22.

$\bullet$ If $|\chi(\Sigma)|\leq 2$ with $\Sigma$ non-exceptional, not a
twice-punctured torus, then $G$ is not an extended étage over a non-abelian
group relative to a non-abelian group (so has no proper elementarily embedded
subgroup by Theorem 8.1). This is proved in the same way as Proposition 10.23.

We have proved the following generalisation of Corollary 10.24.

######  Corollary 10.26.

Let $\Gamma$ be a multi-parachute with $\sum d_{i}(v)\geq 3$ at each bottom
vertex.

  * •

If $\Gamma$ is not retractable (in particular if $\Sigma$ is exceptional),
then $G$ is a prototype.

  * •

If $\Gamma$ is retractable, then $G$ is elementarily free; in this case, it
has a proper elementarily embedded subgroup if and only if $|\chi(\Sigma)|\geq
3$ or $\Sigma$ is a twice-punctured torus. ∎

One may decide under what conditions $\Gamma$ is retractable. This is done in
Propositions 6.2, 6.7, and 6.8 of [GLS19] for parachutes and socket groups.
The other cases may also be worked out, we leave details to the reader. The
complete list of elementarily free multi-parachutes having no proper
elementarily embedded subgroup (called minimal in the next section) is shown
on Figure 22. We will see that, together with ${\mathbb{F}}_{2}$ and
$\pi_{1}(N_{4})$, this describes all minimal finitely generated elementarily
free groups (Theorem 11.4).

Figure 22: The complete list of finitely generated, minimal, elementarily free
groups other than ${\mathbb{F}}_{2}$ and $\pi_{1}(N_{4})$; the sum of the
indices at each cyclic group should be $\geq 3$.

##  11 Minimal and maximal models

Although the definitions and facts that follow make sense in the broader
context of any first-order structure in a countable language, we restrict
ourselves to groups (in the first-order language of groups). We refer to
[Mar02, Hod93] for basic facts.

A group is called minimal if it has no proper elementarily embedded subgroup.
For example, the group of integers is minimal.

A group $H$ is called prime if it elementarily embeds into any other group
elementarily equivalent to it; equivalently, $H$ is countable and, for any
tuple $\underline{h}\in H^{k}$, there is a formula $\sigma$ (without
constants) such that, given another tuple $\underline{h}^{\prime}\in H^{k}$,
one has $H\models\sigma(\underline{h}^{\prime})$ if and only if
$\underline{h}$ and $\underline{h}^{\prime}$ are in the same
${\mathrm{Aut}}(H)$-orbit; in other words, $H$ is countable and, for all
$k\geq 1$ and all tuples $\underline{h}\in H^{k}$, the
${\mathrm{Aut}}(H)$-orbit of $\underline{h}$ is $\emptyset$-definable (i.e.
definable without constants).

We shall characterize prime torsion-free hyperbolic groups and minimal
elementarily free groups.

A countable group $H$ is _homogeneous_ if, for any $k\geq 1$, any two tuples
$\underline{h},\underline{h}^{\prime}\in H^{k}$ satisfying the same first-
order formulas (without constants) are in the same ${\mathrm{Aut}}(H)$-orbit.
Equivalently, the ${\mathrm{Aut}}(H)$-orbit of every finite tuple is
$\emptyset$-type definable, i.e. definable by infinitely many first-order
formulas (without constants). Prime groups are homogeneous; conversely,
finitely generated free groups are homogeneous [PS12] but not prime.

A sequence of elementary embeddings
$H_{1}\preceq_{e}H_{2}\preceq_{e}\dots\preceq_{e}H_{n}\preceq_{e}\cdots$ is
called an _elementary chain_ of groups.

Using a Fraïssé construction, we prove that, given a non-abelian torsion-free
hyperbolic group $G$, there exists a homogeneous countable (infinitely
generated) group ${\mathcal{M}}_{G}\equiv G$ which is the union of an
elementary chain of finitely generated subgroups and in which all finitely
generated groups elementarily equivalent to $G$ embed elementarily; moreover,
${\mathcal{M}}_{G}$ is unique up to isomorphism. This extends Theorem 5.9 of
[KMS18].

###  11.1 Primality

Primality is characterized as follows.

######  Theorem 11.1.

A non-abelian torsion-free hyperbolic group $G$ is prime if and only if $G$ is
a one-ended prototype.

###### Proof.

We first prove that prime groups are prototypes. If $G$ is prime and
elementarily free, it embeds elementarily into ${\mathbb{F}}_{2}$, hence is
isomorphic to ${\mathbb{F}}_{2}$ by Corollary 8.3. But ${\mathbb{F}}_{2}$ is
not prime because it does not embed elementarily into $\pi_{1}(N_{4})$ by
Lemma 10.12 (though $\pi_{1}(N_{4})\equiv{\mathbb{F}}_{2}$). This proves that
no elementarily free group is prime.

Now suppose that $G$ is prime and not elementarily free (so that
$\mathrm{\widetilde{c}}(G)=\mathrm{c}(G)\neq\\{1\\}$). Since $G$ is
elementarily equivalent to $\mathrm{\widetilde{c}}(G)$ by Corollary 7.7, it
embeds elementarily into $\mathrm{\widetilde{c}}(G)$ by primeness. But
prototypes have no proper elementarily embedded subgroup by Corollary 8.9, so
this embedding is an isomorphism and $G$ must be a prototype. This proves that
only prototypes may be prime.

Theorem 10.13 implies that infinitely-ended prototypes are not prime, and we
complete the proof by showing that any one-ended prototype $P$ is prime. By
Corollary 8.8, $P$ elementarily embeds into any _finitely generated_ (hence
hyperbolic by [Sel09]) group $G$ elementarily equivalent to $P$. This is not
enough because it does not apply if $G$ is an infinitely generated group, so
we quote [PS16, Th. 5.3] to deduce that $P$ is prime. ∎

Though infinitely-ended prototypes are not prime, they satisfy a weak “prime
model” property.

######  Proposition 11.2 (Weak prime models).

A non-abelian torsion-free hyperbolic group $G$ is the free product of a
prototype and a free group if and only if, for any finitely generated group
$K$ elementarily equivalent to $G$, the group $G$ elementarily embeds into
$K*{\mathbb{F}}_{r}$ for some $r\geq 0$.

######  Example 11.3.

This applies to the free group $G={\mathbb{F}}_{n}$ for $n\geq 2$ (see
Corollary 7.10).

###### Proof.

Suppose $G=P*F^{1}$ with $P$ a prototype and $F^{1}$ free. If $K\equiv G$,
then $P$ is the core of $K$, so $K\mathrel{\geqslant_{t}}P$. By Corollary 5.4,
there exist free groups $F^{2},F^{3}$ such that $K*F^{2}$ is a simple tower
over a subgroup isomorphic to $P*F^{3}$, so $K*F^{2}*F^{1}$ is a simple tower
over a subgroup isomorphic to $P*F^{3}*F^{1}$ and the result follows.

To prove the converse, we let $K$ be the core $c(G)$ and we assume that $G$
elementarily embeds into $c(G)*{\mathbb{F}}_{r}$. By Proposition 8.10, $G$ is
a co-free free factor of $c(G)*{\mathbb{F}}_{r}$, hence is isomorphic to the
free product of $c(G)$ with a free group. ∎

###  11.2 Minimality

We saw that prototypes are minimal (Corollary 8.9). Conversely:

######  Theorem 11.4.

Let $G$ be a non-abelian torsion-free hyperbolic group.

  1. 1.

If $\mathrm{c}(G)$ is one-ended, then $G$ is minimal if and only if
$G=\mathrm{c}(G)$ (i.e. $G$ is a prototype).

  2. 2.

If $\mathrm{c}(G)$ is infinitely-ended, there are infinitely many minimal
hyperbolic groups elementarily equivalent to $G$.

  3. 3.

If $G$ is elementarily free, then $G$ is minimal if and only if $G$ is
isomorphic to ${\mathbb{F}}_{2}$, $\pi_{1}(N_{4})$, or one of the multi-
parachutes shown on Figure 22.

In particular, there are infinitely many minimal groups elementarily
equivalent to $G$ (up to isomorphism) if and only if $\mathrm{c}(G)$ is
trivial or infinitely ended.

###### Proof.

Assertion 1 follows from Corollaries 8.8 and 8.9. If $\mathrm{c}(G)$ is
infinitely-ended, we write its Grushko decomposition $\mathrm{c}(G)\simeq
A_{1}*\cdots*A_{p}$ with $p\geq 2$ and we consider
$G(a_{1},a_{2})=K(A_{1},A_{2},a_{1},a_{2})*A_{3}*\cdots*A_{p}$ as in Theorem
10.13. We have seen that $G(a_{1},a_{2})$ is minimal and elementarily
equivalent to $G$ if $a_{i}$ is a square and $A_{i}$ has no cyclic splitting
relative to $a_{i}$.

In order to get infinitely many different groups, we fix $a_{1},a_{2}$ and for
$n\geq 1$ we consider
$G_{n}=G(a_{1}^{n},a_{2}^{n})=K(A_{1},A_{2},a_{1}^{n},a_{2}^{n})*A_{3}*\cdots*A_{p}$.
These groups are non-isomorphic because by Remark 10.6 their canonical JSJ
decompositions are different: they are distinguished by indices of edge groups
in vertex groups.

To prove Assertion 3, suppose that $G$ is elementarily free, minimal, not
${\mathbb{F}}_{2}$. It cannot have a cyclic free factor, so it is an extended
étage of surface type. Let $\Gamma$ be the associated splitting. First suppose
that the associated surface satisfies $|\chi(\Sigma)|\geq 3$. By Propositions
5.1 (if $\Gamma$ is not simple) and 10.22 (if $G$ is a parachute), $G$ is a
simple étage over a non-abelian group so is not minimal. Thus
$|\chi(\Sigma)|\leq 2$.

Since $G$ is elementarily free, all bottom groups have trivial core. If some
bottom group $H$ is non-abelian, then $G$ is a tower over $H$ relative to $H$
so is not minimal by Theorem 8.1. It follows that $G$ is an étage with all
bottom groups isomorphic to ${\mathbb{Z}}$, i.e. a multi-parachute.

Let $\Sigma^{\prime}$ be the surface obtained by gluing to $\Sigma$ all annuli
or Möbius bands corresponding to bottom vertices $v$ with $\sum d_{i}(v)=2$.
It satisfies $|\chi(\Sigma^{\prime})|\leq 2$.

If it is closed, it must be $N_{4}$ because $\pi_{1}(N_{3})$ is not
elementarily free and $\pi_{1}(S_{2})$ contains an elementarily embedded
subgroup isomorphic to ${\mathbb{F}}_{2}$. Thus $G\simeq\pi_{1}(N_{4})$.
Otherwise, we get a new multi-parachute with $\sum d_{i}(v)\geq 3$ at each
vertex. As explained at the end of Section 10, $G$ is on the list of Figure
22.

This proves one direction of Assertion 3. The converse follows from Corollary
8.3 (if $G$ is free), Lemma 10.12 (if $G\simeq\pi_{1}(N_{4})$), and Corollary
10.26 (if $G$ is a multi-parachute). ∎

###  11.3 Universality

This subsection is devoted to the following result.

######  Theorem 11.5.

Given any non-abelian torsion-free hyperbolic group $G$, there exists a unique
countable (infinitely generated) group ${\mathcal{M}}_{G}$ elementarily
equivalent to $G$ with the following properties:

  1. 1.

${\mathcal{M}}_{G}$ is homogeneous;

  2. 2.

there exists an elementary chain of finitely generated subgroups
$G_{1}\preceq_{e}G_{2}\preceq_{e}\dots\preceq_{e}G_{n}\preceq_{e}\cdots$ such
that ${\mathcal{M}}_{G}=\bigcup_{i\geq 1}G_{i}$;

  3. 3.

all torsion-free hyperbolic groups elementarily equivalent to $G$ elementarily
embed into ${\mathcal{M}}_{G}$.

The proof relies on a Fraïssé construction. The following definition and
Theorem 11.9 extend Fraïssé’s original result. The theorem can be proved
following the same line of thought and is considered folklore; nevertheless, a
more detailed explanation is given in Section 2.3 of [KMS18].

######  Definition 11.6 (strong elementary Fraïssé).

Let $\mathcal{K}$ be a non-empty class of finitely generated groups which is
countable (up to isomorphism) and has the following properties:

  * •

the class $\mathcal{K}$ is closed under isomorphism _(isomorphism property,
IP)_ ;

  * •

the class $\mathcal{K}$ is closed under passing to finitely generated
elementarily embedded subgroups, i.e. if $K\in\mathcal{K}$ and
$K^{\prime}\preceq_{e}K$ is a finitely generated elementarily embedded
subgroup of $K$, then $K^{\prime}\in\mathcal{K}$ _(elementary hereditary
property, e-HP)_ ;

  * •

if $K_{1}$, $K_{2}$ are in $\mathcal{K}$, then there is $K$ in $\mathcal{K}$
and elementary embeddings $f_{i}:K_{i}\rightarrow K$ for $i\leq 2$
_(elementary joint embedding property, e-JEP)_ ;

  * •

if $K_{0}$, $K_{1}$, $K_{2}$ are in $\mathcal{K}$ and $f_{i}:A\rightarrow
K_{i}$, for $i=1,2$, are partial elementary maps of some finitely generated
subgroup $A\subset K_{0}$, then there is a group $K$ in $\mathcal{K}$ and
elementary embeddings $g_{i}:K_{i}\rightarrow K$ for $i=1,2$ with $g_{1}\circ
f_{1}=g_{2}\circ f_{2}$ _(strong elementary amalgamation property, strong
e-AP)_.

Then $\mathcal{K}$ is a _strong elementary Fraïssé class_ (strong $e$-Fraïssé
class for short).

See Definition 9.1 for partial elementary maps.

######  Remark 11.7.

Note that, by e-JEP, all groups in ${\mathcal{K}}$ are elementarily
equivalent. When $A=\\{1\\}$, strong e-AP reduces to e-JEP.

######  Remark 11.8.

Strong e-AP can be reformulated as follows with no reference to $K_{0}$: given
$K_{1},K_{2}\in{\mathcal{K}}$, a finitely generated subgroup $A_{1}\subset
K_{1}$ and a partial elementary map $f:A_{1}\rightarrow K_{2}$, there exists a
group $K\in{\mathcal{K}}$ and elementary embeddings $g_{i}:K_{i}\rightarrow K$
for $i=1,2$ with $g_{1}=g_{2}\circ f$.

######  Theorem 11.9.

Let $\mathcal{K}$ be a strong $e$-Fraïssé class. There exists a countable
group $\mathcal{M}$ such that:

  * •

the class of finitely generated groups that elementarily embed in
${\mathcal{M}}$ is exactly $\mathcal{K}$;

  * •

the group $\mathcal{M}$ is homogeneous;

  * •

the group $\mathcal{M}$ is the union of an elementary chain of groups in
$\mathcal{K}$.

Moreover, any other countable group with the above properties is isomorphic to
$\mathcal{M}$. ∎

Using this theorem, we prove Theorem 11.5 by showing that, given a torsion-
free hyperbolic group $G$, the class $\mathcal{K}_{G}$ of all finitely
generated groups elementarily equivalent to $G$ is an elementary Fraïssé
class. We just have to check the four properties of Definition 11.6. Recall
that ${\mathcal{K}}_{G}$ consists of torsion-free hyperbolic groups [Sel09].

The isomorphism property (IP) and the elementary hereditary property (e-HP)
are obvious. The elementary joint embedding property (e-JEP) follows from the
first assertion of Proposition 8.11, which was proved as follows: if
$G_{1}\equiv G_{2}$, then their cores $H_{1},H_{2}$ are isomorphic by Theorem
7.2, and elementarily embedded in $G^{\prime}_{1}=G_{1}*{\mathbb{F}}_{r_{1}}$,
$G^{\prime}_{2}=G_{2}*{\mathbb{F}}_{r_{2}}$ for some free groups
${\mathbb{F}}_{r_{1}},{\mathbb{F}}_{r_{2}}$. We then defined
$G^{\prime}=G^{\prime}_{1}*_{H_{1}=H_{2}}G^{\prime}_{2}$.

The strong elementary amalgamation property (strong e-AP) may be viewed as a
relative version of (e-JEP). It is proved in a similar way, using Remark 11.8
and replacing Theorem 7.2 by Theorem 9.7.

##  Appendix A Appendix: Towers are limit groups

The goal of this section is to prove Corollary A.2 saying that towers over a
torsion-free hyperbolic group $H$ are $H$-limit groups. This basic fact is
stated in [Sel09], but some details of the proof are missing. The same
arguments show that, if a finitely generated group has a strict resolution
over a hyperbolic group $H$, then it is an $H$-limit group.

######  Theorem A.1.

If $G$ is a simple étage of surface type over a torsion-free hyperbolic group
$H$, there exists a discriminating sequence of retractions
$\rho_{n}:G\rightarrow H$.

Discriminating means that, given any finite subset $F\subset G$, the map
$\rho_{n}$ is injective on $F$ for all $n$ large enough. Recall that $H$ must
be non-abelian (Remark 4.17).

######  Corollary A.2.

If $G$ is a tower (simple or extended) over a non-abelian torsion-free
hyperbolic group $H$, then it is an $H$-limit group.

We first deduce the corollary from the theorem. The remainder of this appendix
will be devoted to the proof of the theorem.

###### Proof.

First suppose that $G>G_{1}>\dots>G_{n}=H$ is a simple tower. All groups
$G_{i}$ are hyperbolic, and there exists a discriminating sequence of
retractions from $G_{i+1}$ to $G_{i}$ by Theorem A.1 (it is well known that
this also holds if $G_{i+1}=G_{i}*{\mathbb{Z}}$ is an étage of free product
type with $G_{i}$ non-abelian, by sending the generator of the cyclic factor
to a sequence of small cancellation elements, see for instance [And19, Th.
1.9]). Composing these retractions gives a discriminating sequence of
retractions from $G$ to $H$, so $G$ is an $H$-limit group.

If the tower is not simple, some $G*{\mathbb{F}}_{r}$ is an $H$-limit group by
Corollary 5.4, and $G$ is an $H$-limit group because it is a finitely
generated subgroup of an $H$-limit group. ∎

###  A.1 Applying Baumslag’s lemma

To prove Theorem A.1, we first need the following generalization of Baumslag’s
lemma [Bau67]. For $g\neq 1$ in $H$, we denote by
$g^{\pm\infty}=\lim_{n\to\pm\infty}g^{n}$ the limit of $g^{n}$ in the Gromov
boundary $\partial H$.

######  Lemma A.3 ([And18a, Prop. 2.19]).

Let $a_{0},a_{1},\dots,a_{k},c_{1},\dots,c_{k}$ be elements of a torsion-free
hyperbolic group $H$, with $c_{i}\neq 1$. Assume that
$a_{i}\,c_{i+1}^{+\infty}\neq c_{i}^{-\infty}$ for $i=1,\dots,k-1$. Then there
exists $C\geq 0$ such that the element

$a_{0}c_{1}^{p_{1}}a_{1}\cdots c_{k}^{p_{k}}a_{k}$

is non-trivial for all $p_{1},\dots,p_{k}\geq C$.∎

######  Corollary A.4.

Suppose that a torsion-free hyperbolic group $G$ is the fundamental group of a
graph of groups $\Gamma$ such that all edge groups are infinite cyclic, and
maximal cyclic in adjacent vertex groups (hence in $G$). Let
$\hat{\tau}\in{\mathrm{Out}}(G)$ be the product of the twists around all edges
of $\Gamma$, and $\tau$ a representative of $\hat{\tau}$ in
${\mathrm{Aut}}(G)$.

If $f:G\rightarrow H$ is a homomorphism which is injective on each vertex
group of $\Gamma$, with $H$ a torsion-free hyperbolic group, then the sequence
of homomorphisms $f\circ\tau^{n}:G\rightarrow H$ is discriminating.

Edge groups $G_{e}$ being infinite cyclic, each edge $e$ carries a twist
$\hat{\tau}_{e}$; the pair $\\{\hat{\tau}_{e},\hat{\tau}_{e}^{-1}\\}$ is well-
defined in ${\mathrm{Out}}(G)$, and it does not matter for the theorem whether
we use $\hat{\tau}_{e}$ or $\hat{\tau}_{e}^{-1}$ to define $\hat{\tau}$, or
which representative $\tau$ of $\hat{\tau}$ in ${\mathrm{Aut}}(G)$ we use.

The proof of Corollary A.4 is a standard application of Baumslag’s lemma (see
for instance [CG05, Prop. 4.11]). We prove it in the case of an amalgam
$G=A*_{\langle c\rangle}B$ and leave the general case to the reader.

###### Proof in the case of an amalgam.

We choose the representative $\tau$ of $\hat{\tau}$ equal to the identity on
$A$ and to conjugation by $c$ on $B$. It suffices to prove that, given any
$g\in G\setminus\\{1\\}$, we have $(f\circ\tau^{n})(g)\neq 1$ for all $n$
large enough. This is obvious if $g$ is contained in $A$ or $B$ (up to
conjugacy). Otherwise, up to conjugation, $g$ may be written as
$g=a_{1}b_{1}\dots a_{k}b_{k}$ with $a_{i}\in A_{i}\setminus\langle c\rangle$
and $b_{i}\in B_{i}\setminus\langle c\rangle$, and $k\geq 1$.

Now

$\tau^{n}(g)=a_{1}c^{n}b_{1}c^{-n}a_{2}c^{n}\dots c^{-n}a_{k}c^{n}b_{k}c^{-n}$

and

$(f\circ\tau^{n})(g)=f(a_{1})f(c)^{n}f(b_{1})f(c)^{-n}\cdots
f(a_{k})f(c)^{n}f(b_{k})f(c)^{-n}.$

By assumption, $\langle c\rangle$ is maximal abelian in $A$ and in $B$, so
$a_{i}$ and $b_{i}$ do not commute with $c$. Since $f$ is injective on $A$ and
$B$, the elements $f(a_{i})$ and $f(b_{i})$ do not commute with $f(c)$, so
none of the elements $f(a_{i}).f(c)^{\pm\infty},f(b_{i}).f(c)^{\pm\infty}$
belongs to $\\{f(c)^{\pm\infty}\\}$. Thus Lemma A.3 applies and concludes the
proof. ∎

The idea to prove Theorem A.1 is to apply Corollary A.4 to the centered
splitting $\Gamma$ associated to the étage, with $f$ a retraction $\rho:G\to
H$, but there are two problems. First, the edge groups of $\Gamma$, though
maximal cyclic subgroups of the central group, do not have to be maximal
cyclic in bottom groups (as in Example 10.2). This is taken care of by
introducing surfaces with sockets. The second problem is that the retraction
$\rho:G\to H$ to the bottom group of $\Gamma$ does not have to be injective on
the central group. To fix this we will have to refine $\Gamma$, using a
multicurve on the surface associated to $\Gamma$, so that $\rho$ is injective
on the new vertex groups.

###  A.2 Surfaces with sockets

A _surface group with sockets_ $\hat{Q}$ is the amalgam of the fundamental
group of a compact hyperbolic surface $\Sigma$ with cyclic groups over
boundary subgroups of $\pi_{1}(\Sigma)$. More precisely, it is the fundamental
group of a (possibly non-minimal) tree of groups $\Lambda$ with a central
surface-type vertex $v$ (in the sense of Definition 3.5) carrying the group
$Q=\pi_{1}(\Sigma)$ for some hyperbolic surface $\Sigma$, and terminal
vertices $v_{1},\dots,v_{n}$ carrying ${\mathbb{Z}}$. We say that $\hat{Q}$ is
_non-exceptional_ if the surface $\Sigma$ is non-exceptional.

This is similar to a centered splitting with all bottom groups ${\mathbb{Z}}$,
with two differences. First, there is a single edge $e_{i}$ joining $v$ to a
given $v_{i}$. Second, $\Lambda$ does not have to be minimal: $G_{e_{i}}$ may
be equal to $G_{v_{i}}$; if this does not happen, $\hat{Q}$ is a particular
kind of a multi-parachute in the sense of Subsection 10.6.

The group carried by $e_{i}$ is the fundamental group of a component $C_{i}$
of $\partial\Sigma$, and $C_{i}\neq C_{j}$ for $i\neq j$. It has finite index
$d_{i}$ in the vertex group $G_{v_{i}}$. We allow $d_{i}=1$ (i.e.
$G_{e_{i}}=G_{v_{i}}$) and $d_{i}=2$ (this corresponds to gluing a Möbius band
to $C_{i}$).

In topological terms, $\hat{Q}$ is the fundamental group of a graph of spaces
$X_{\hat{Q}}$ obtained by gluing each boundary component $C_{i}$ to a circle
by a covering of degree $d_{i}$.

The _socket subgroups_ of $\hat{Q}$ are the conjugates of the vertex groups
$G_{v_{i}}$. A homomorphism from $\hat{Q}$ to an arbitrary group is injective
on each boundary subgroup of $\pi_{1}(\Sigma)$ if and only if it is injective
on each socket subgroup of $\hat{Q}$.

An _essential multicurve_ in $\Sigma$ is a collection ${\mathcal{C}}$ of
disjoint non-parallel 2-sided simple closed curves, each of which is not
boundary parallel and does not bound a Möbius band or a disc. The cyclic
splitting of $\pi_{1}(\Sigma)$ dual to ${\mathcal{C}}$ induces a cyclic
splitting of $\hat{Q}$. The vertex groups of this splitting are fundamental
groups of connected components of $X_{\hat{Q}}\setminus{\mathcal{C}}$, they
have a natural structure of a surface with sockets (curves in ${\mathcal{C}}$
give rise to sockets of index 1).

###  A.3 A reduction

Following Sela, the proof of Theorem A.1 is based on the following result.

######  Theorem A.5.

Let $\hat{Q}$ be a non-exceptional surface group with sockets, with underlying
surface $\Sigma$, and let $H$ be a torsion-free hyperbolic group. Let
$f:\hat{Q}\rightarrow H$ be a homomorphism with non-abelian image, which is
injective on each socket subgroup of $\hat{Q}$.

There exists an essential multicurve ${\mathcal{C}}$ in $\Sigma$ such that the
restriction of $f$ to the fundamental group of each connected component of
$X_{\hat{Q}}\setminus{\mathcal{C}}$ is injective.

######  Remark A.6.

Theorem A.5 also applies if $\Sigma$ is the closed non-orientable surface
$N_{3}$ of genus $3$ (with no sockets), see Remark A.10.

###### Proof of Theorem A.1 from Theorem A.5.

Let $\Gamma$ be the simple centered splitting of $G$ associated to the étage,
with bottom group $H$ and central group $Q$. Perform folds to transform
$\Gamma$ into a splitting $\hat{\Gamma}$ over maximal cyclic subgroups; this
replaces $Q$ by a surface group with sockets $\hat{Q}$ without changing $H$
(compare Remark 10.6, where one obtains $\hat{\Gamma}$ from
$\Gamma_{\\!\mathrm{can}}$ by collapsing the edges $vv_{i}$).

Let $\rho:G\rightarrow H$ be a retraction associated to the étage, as in
Proposition 4.14. Its restriction to $\hat{Q}$ satisfies the assumptions of
Theorem A.5, so we can refine the splitting $\hat{\Gamma}$ using the essential
multicurve provided by the theorem. This provides a new splitting
$\hat{\Gamma}_{c}$ over maximal cyclic subgroups such that $\rho$ is injective
in restriction to every vertex group.

We apply Corollary A.4 to this splitting. Since $H$ is a vertex group of
$\hat{\Gamma}_{c}$, the outer automorphism $\hat{\tau}$ has a representative
$\tau$ in ${\mathrm{Aut}}(G)$ which is the identity on $H$ (the assumption
that the étage is simple is used here). Thus the maps
$\rho_{n}:=\rho\circ\tau^{n}:G\rightarrow H$ are retractions. By Corollary
A.4, this sequence of retractions is discriminating. ∎

The following subsections are devoted to the proof of Theorem A.5. We first
prove in subsection A.4 that one can decompose the surface $\Sigma$ into pairs
of pants and twice-punctured projective planes whose fundamental groups all
have non-abelian image. We then prove the theorem in Subsection A.5 by using
the fact that $\Sigma$ is non-exceptional and reducing the general case to two
special cases: the once-punctured torus and the 4-punctured sphere. The proof
in these special cases requires ping-pong lemmas which are proved in
Subsection A.6.

###  A.4 Decomposing into pieces with non-abelian image

The first step in the proof of Theorem A.5 is the following lemma. Recall
(Definition 3.4) that a two-sided simple closed curve is pinched under a
homomorphism if its fundamental group is contained in the kernel.

######  Lemma A.7.

Let $H$ be a torsion-free CSA group, and let $f:\hat{Q}\rightarrow H$ be a
homomorphism with non-abelian image, with $\hat{Q}$ a surface group with
sockets. Assume that no component of $\partial\Sigma$ is pinched.

There exists an essential multicurve ${\mathcal{C}}_{0}$ consisting of non-
pinched curves such that every connected component of
$\Sigma\setminus{\mathcal{C}}_{0}$ is a pair of pants or a twice-punctured
projective plane whose fundamental group has non-abelian image under $f$.

We do not assume that the underlying surface $\Sigma$ is non-exceptional.

######  Remark A.8.

As a preliminary remark, we note that, if $\hat{Q}$ is obtained from a group
$Q=\pi_{1}(\Sigma)$ by adding sockets, and $f:\hat{Q}\rightarrow H$ takes
values in a torsion-free commutative transitive group, then $f(\hat{Q})$ is
non-abelian if and only if $f(Q)$ is non-abelian. We give the argument when
there is only one socket. Write $\hat{Q}=\langle Q,u\rangle$ with $u^{d}\in
Q$, and assume that $f(Q)$ is abelian. If $f(u^{d})$ is trivial, so is $f(u)$
because $H$ is torsion-free, and $f(\hat{Q})=f(Q)$ is abelian. Otherwise, both
$f(Q)$ and $f(u)$ commute with $f(u^{d})$, hence commute by commutative
transitivity, so $f(\hat{Q})$ is abelian.

This remark shows that we may forget about sockets in Lemma A.7: it suffices
to prove the lemma for the fundamental group of a surface with boundary. It is
an immediate consequence of the following.

######  Lemma A.9 ([Sel01, Lemma 5.13], [Sel09, Lemma 1.32]).

Let $\Sigma$ be a compact surface which is not homeomorphic to a pair of pants
or a twice-punctured projective plane. Let $H$ be a torsion-free CSA group,
and $f:\pi_{1}(\Sigma)\rightarrow H$ a homomorphism with non-abelian image.
Assume that no component of $\partial\Sigma$ is pinched.

There exists a non-pinched essential two-sided simple closed curve
$\gamma\subset\Sigma$ such that the fundamental group of every connected
component of $\Sigma\setminus\gamma$ has non-abelian image under $f$.

###### Proof.

Let $Q=\pi_{1}(\Sigma)$. We start with two observations.

Obervation 1: _if there exists a non-separating two-sided non-pinched curve
$\gamma$, then the fundamental group $J$ of $\Sigma\setminus\gamma$
automatically has non-abelian image._ Indeed, one has an HNN decomposition
$Q=J*_{\pi_{1}(\gamma)}$ and, if $f(J)$ is abelian, then the image of the
stable letter has to commute with $f(J)$ by the CSA property.

The lemma follows in the case when $\Sigma$ is orientable with at most one
boundary component, because $\pi_{1}(\Sigma)$ is generated by fundamental
groups of non-separating essential simple closed curves.

Observation 2: _if $\Sigma$ contains a two-sided non-separating simple closed
curve $\gamma$ which is pinched, and $\partial\Sigma\neq\emptyset$, one can
find another such curve $\gamma^{\prime}$ which is non-pinched_ as follows:
take an arc $I$ joining $\gamma$ to a boundary component $b$, and consider the
curve $\gamma^{\prime}$ obtained as the connected sum of $b$ and $\gamma$
along $I$. Since $\gamma$ is pinched, the elements of $\pi_{1}(\Sigma)$
corresponding to $b$ and $\gamma^{\prime}$ have conjugate images under $f$, so
$\gamma^{\prime}$ is not pinched. Moreover, if $I$ is chosen disjoint from
some simple closed curve $\delta$ with $i(\delta,\gamma)=1$, then
$i(\delta,\gamma^{\prime})=1$ so $\gamma^{\prime}$ is non-separating.

Using these two observations repeatedly, we see that there remain three cases
to analyze: the $n$-punctured sphere ($n\geq 4$), the $n$-punctured projective
plane ($n\geq 3$), and the closed non-orientable surfaces.

$\bullet$ If $\Sigma$ is an $n$-punctured sphere with $n\geq 4$, then
$Q=\langle a_{1},\dots,a_{n}\mid a_{1}\cdots a_{n}=1\rangle$. Since $H$ is
CSA, there exists $i$ such that $f(a_{i})$ does not commute with $f(a_{i+1})$,
and we may assume $i=1$ without loss of generality. The subgroup $\langle
a_{1},a_{2}\rangle$ is the fundamental group of a pair of pants contained in
$\Sigma$; it has non-abelian image and its third boundary curve, whose
fundamental group is generated by $a_{1}a_{2}$, is not pinched.

The fundamental group of the complement of this pair of pants is generated by
$a_{3},\dots,a_{n}$ and contains $a_{1}a_{2}$. We are done if it has non-
abelian image under $f$, so assume otherwise. Then
$f(a_{3}),\dots,f(a_{n}),f(a_{1}a_{2})$ are non-trivial commuting elements.
Let $A$ be the maximal abelian group containing $f(\langle
a_{3},\dots,a_{n}\rangle)$ (hence also $f(a_{1}a_{2})$).

We have $f(a_{2})\notin A$ since otherwise $f(Q)=A$, contradicting the non-
abelianity of $f(Q)$. By commutative transitivity, $f(a_{2})$ does not commute
with $f(a_{3})$, and similarly $f(a_{n})$ does not commute with $f(a_{1})$. We
then let $\gamma$ be a curve bounding a pair of pants with fundamental group
$\langle a_{2},a_{3}\rangle$.

$\bullet$ If $\Sigma$ is an $n$-punctured projective plane with $n\geq 3$, we
claim that there exists an embedded Möbius band $M\subset\Sigma$ whose
fundamental group is not killed by $f$. If the claim is true, then one can
write $\Sigma$ as an $(n+1)$-punctured sphere $\Sigma^{\prime}$ with $M$ glued
along a boundary component, which can be viewed as $\Sigma^{\prime}$ with a
socket of index 2. Since $n+1\geq 4$ and boundary components of
$\Sigma^{\prime}$ are not pinched, one reduces to the previous case using
Remark A.8.

To prove the claim, consider a one-sided simple closed curve $\gamma$ (whose
regular neighborhood is a Möbius band), and assume that $\pi_{1}(\gamma)$ is
killed. Let $I$ be an arc joining $\gamma$ to a boundary component $b$. Then
the concatenation $\gamma.I.b.\overline{I}$ is homotopic to a simple closed
curve which is one-sided, and whose image under $f$ is conjugate to that of
$b$ hence is non-trivial.

$\bullet$ If $\Sigma$ is a closed non-orientable surface (of genus at least 3
because its fundamental group has non-abelian image in $H$), one can write it
as the union of an orientable surface $\Sigma^{\prime}$ with one boundary
component and a Möbius band or a punctured Klein bottle. If $\Sigma^{\prime}$
contains a non-pinched non-separating curve, we are done. Otherwise, being
generated by such curves, $\pi_{1}(\Sigma^{\prime})$ has trivial image under
$f$, so $f$ factors through the fundamental group of a projective plane or a
Klein bottle. Since $H$ is CSA, this implies that the image of $f$ is abelian,
a contradiction. ∎

###  A.5 Proof of Theorem A.5

Consider the decomposition of the surface $\Sigma$ (assumed to be non-
exceptional) into pairs of pants and twice-punctured projective planes given
by Lemma A.7. We first claim that, up to adding extra sockets, we may assume
that all pieces of this decomposition are pairs of pants.

Indeed, one can view a twice-punctured projective plane $P$ as a pair of pants
$P^{\prime}$ with a Möbius band glued to one of its boundary components; this
corresponds to adding a socket of index 2 to $P^{\prime}$. The fundamental
group of the Möbius band (and of the corresponding boundary component of
$P^{\prime}$) have non-trivial image by $f$ since otherwise $\pi_{1}(P)$ would
have abelian image. By Remark A.8, the image of the fundamental group of
$P^{\prime}$ is also non-abelian. The claim is then proved by replacing each
twice-punctured projective plane by a pair of pants with sockets.

We thus have a decomposition of the space $X_{\hat{Q}}$ into pairs of pants
with sockets $P_{i}$ whose fundamental groups have non-abelian images under
$f$. We are done if $f$ is injective on every $\pi_{1}(P_{i})$, so let
$P=P_{i}$ be a pair of pants with sockets such that $f$ is not injective on
$\pi_{1}(P)$.

If the Euler characteristic of $\Sigma$ is $-1$, then $\Sigma$ is a once-
punctured torus because it is assumed to be non-exceptional. This is a special
case which will be treated below. Otherwise, there exists a pair of pants
$P^{\prime}\neq P$ adjacent to $P$ in the decomposition, and $P\cup
P^{\prime}$ is a 4-punctured sphere (possibly with boundary components
identified). If we prove the theorem when $\Sigma$ is either a once-punctured
torus or a 4-punctured sphere, the general case follows by induction on the
number of pairs of pants $P$ such that $f$ is not injective on $\pi_{1}(P)$.

$\bullet$ The punctured torus.

Let $d\geq 1$ and let $\hat{Q}=\langle a,b,c\mid[a,b]=c^{d}\rangle$ be the
fundamental group of a punctured torus with a socket of index $d$ (with $d=1$
if there is no socket), where the fundamental group $Q$ of the punctured torus
$\Sigma$ is the subgroup $\langle a,b\rangle$, the curve $C$ provided by Lemma
A.7 has fundamental group $\langle a\rangle$, and cutting $\Sigma$ along $C$
produces a pair of pants $P$ whose fundamental group $\langle
a,bab^{-1}\rangle=\langle a,c^{d}\rangle=\langle bab^{-1},c^{d}\rangle$ has
non-abelian image.

Let $\gamma_{n}$ be the curve obtained by twisting the curve representing $b$
$n$ times around $C$. Cutting the graph of spaces $X_{\hat{Q}}$ representing
$\hat{Q}$ along $\gamma_{n}$ creates a pair of pants with socket whose
fundamental group is a free group $\hat{P}_{n}=\langle ba^{n},c\rangle$.

When applied to $(u,v,w)=(f(a),f(b),f(c))$, Corollary A.13, stated and proved
in the next subsection, ensures that $f(\hat{P}_{n})$ is free of rank 2 for
all $n$ large enough, so the restriction of $f$ to $\hat{P}_{n}$ is injective.
This proves Theorem A.5 when $\Sigma$ is a once-punctured torus.

$\bullet$ The 4-punctured sphere.

Using Lemma A.7, we may assume that $G$ has a presentation of the form
$\langle a_{1},\dots,a_{4}\mid
a_{1}^{d_{1}}a_{2}^{d_{2}}a_{3}^{d_{3}}a_{4}^{d_{4}}=1\rangle$, with
$\pi_{1}(\Sigma)=\langle a_{1}^{d_{1}},\dots,a_{4}^{d_{4}}\rangle$, such that
the simple closed curve $\gamma$ represented by
$g:=a_{1}^{d_{1}}a_{2}^{d_{2}}=(a_{3}^{d_{3}}a_{4}^{d_{4}})^{-1}$ is not
pinched by $f$, and the associated pairs of pants with sockets $\langle
a_{1},a_{2}\rangle$ and $\langle a_{3},a_{4}\rangle$ have non-abelian image by
$f$. Note that $f(g)$ does not commute with any $f(a_{i})$: if for instance it
commutes with $f(a_{1})$, then it also commutes with
$f(a_{2}^{d_{2}})=f(a_{1}^{-d_{1}}g)$ and $f(\langle a_{1},a_{2}\rangle)$ is
abelian by commutative transitivity, a contradiction.

We consider a second decomposition of $X_{\hat{Q}}$ into two pairs of pants
with sockets, corresponding to $P_{0}=\langle a_{2},a_{3}\rangle$ and
$P^{\prime}_{0}=\langle a_{4},a_{1}\rangle$. Twisting it $n$ times around
$\gamma$, we obtain a decomposition into $P_{n}=\langle
a_{2},g^{n}a_{3}g^{-n}\rangle$ and $P^{\prime}_{n}=\langle
g^{n}a_{4}g^{-n},a_{1}\rangle$. Since $f(g)$ commutes with no $f(a_{i})$,
Lemma A.12 applies and ensures that $f$ is injective on $P_{n}$ and
$P^{\prime}_{n}$ for $n$ large.

######  Remark A.10.

To prove the theorem when $\hat{Q}$ is the fundamental group of the closed
exceptional surface $N_{3}$, one views $\hat{Q}$ as a once-punctured torus
with a socket of index $d=2$. One just has to check that $f$ is injective on
the socket group. This is true since otherwise $f$ would factor by
${\mathbb{Z}}^{2}$, the fundamental group of the closed torus, and would
therefore have abelian image.

###  A.6 Ping-pong lemmas

Let $H$ be a hyperbolic group (torsion is allowed). We fix a finite generating
set and we let ${\mathcal{G}}$ be the corresponding Cayley graph. Let
$\bar{\mathcal{G}}={\mathcal{G}}\cup\partial{\mathcal{G}}$ be its
compactification, obtained by adding the Gromov boundary
$\partial{\mathcal{G}}=\partial H$ (see [GdlH90]). The action of $H$ on itself
by left-translations extends to a continuous action on $\bar{\mathcal{G}}$.

For $g$ of infinite order, we denote by
$g^{\pm\infty}=\lim_{g\to\pm\infty}g^{n}$ the attracting and repelling fixed
points of $g$ in $\partial{\mathcal{G}}$. It is a standard fact that $g$ has
North-South dynamics on $\bar{\mathcal{G}}$: given neighborhoods $W^{\pm}$ of
$g^{\pm\infty}$, one has $g^{n}(\bar{\mathcal{G}}\setminus W^{-})\subset
W^{+}$ and $g^{-n}(\bar{\mathcal{G}}\setminus W^{+})\subset W^{-}$ for $n$
large. If $g,h$ have infinite order and $\langle g,h\rangle$ is not virtually
cyclic, the four points $g^{\pm\infty},h^{\pm\infty}$ are distinct (this
applies in particular if $H$ is torsion-free and $g,h$ do not commute).

The following lemmas are proved by standard arguments, but we have not found
them in the literature.

######  Lemma A.11.

Let $u,v,w\in H$, with $u,w$ of infinite order (we allow $v=1$). Let
$\xi^{+}=vu^{+\infty}$ and $\xi^{-}=u^{-\infty}$. Assume that
$\xi^{+},\xi^{-},w^{+\infty},w^{-\infty}$ are distinct, and
$w^{i}\xi^{-}\neq\xi^{+}$ for all integers $i$. For $n$ large enough, $\langle
vu^{n},w\rangle$ is a free subgroup freely generated by $vu^{n}$ and $w$.

######  Lemma A.12.

Let $u,v,w\in H$ have infinite order. Assume that
$\\{w^{\pm\infty}\\}\cap\\{u^{\pm\infty}\\}=\\{w^{\pm\infty}\\}\cap\\{v^{\pm\infty}\\}=\emptyset$
(we allow $u^{\pm\infty}=v^{\pm\infty}$). For $n$ large enough, $\langle
u,w^{n}vw^{-n}\rangle$ is a free subgroup freely generated by $u$ and
$w^{n}vw^{-n}$.

###### Proof of Lemma A.11.

Let $x$ be a basepoint in ${\mathcal{G}}$. Using North-South dynamics of $w$,
we can find neighborhoods $W^{\pm}$ of $\xi^{\pm}$ in $\bar{\mathcal{G}}$ such
that the neighborhoods $w^{i}W^{\pm}$ for $i\in{\mathbb{Z}}$ and the points
$w^{j}x$ for $j\in{\mathbb{Z}}$ are all disjoint. Denoting $v_{n}=vu^{n}$, we
have $v_{n}(\bar{\mathcal{G}}\setminus W^{-})\subset W^{+}$ and
$v_{n}^{-1}(\bar{\mathcal{G}}\setminus W^{+})\subset W^{-}$ for $n$ larger
than some $n_{0}$: this follows from North-South dynamics of $u$, noting that
$v^{-1}W^{+}$ is a neighborhood of $u^{+\infty}$. Note that, for $n>n_{0}$,
one has $v_{n}^{i}(\bar{\mathcal{G}}\setminus W^{-})\subset W^{+}$ and
$v_{n}^{-i}(\bar{\mathcal{G}}\setminus W^{+})\subset W^{-}$ for all $i>0$. In
particular, $v_{n}$ has infinite order.

We can now play ping-pong. For $n>n_{0}$, consider an element

$g=v_{n}^{n_{k}}w^{p_{k-1}}\cdots
v_{n}^{n_{2}}w^{p_{1}}v_{n}^{n_{1}}w^{p_{0}}$

in $\langle v_{n},w\rangle$, with all exponents nonzero and $k\geq 1$. Let
$\varepsilon_{i}\in\\{\pm\\}$ be the sign of $n_{i}$. We show $gx\in
W^{\varepsilon_{k}}$ (this implies $gx\neq x$, hence $g\neq 1$). The point
$w^{p_{0}}x$ is not in $W^{-\varepsilon_{1}}$, so $v_{n}^{n_{1}}w^{p_{0}}x\in
W^{\varepsilon_{1}}$. We then have $w^{p_{1}}v_{n}^{n_{1}}w^{p_{0}}\in
w^{p_{1}}W^{\varepsilon_{1}}$, so $w^{p_{1}}v_{n}^{n_{1}}w^{p_{0}}\notin
W^{-\varepsilon_{2}}$ and $v_{n}^{n_{2}}w^{n_{2}}v_{n}^{n_{1}}w^{p_{0}}\in
W^{\varepsilon_{2}}$. Iterating gives the desired result. ∎

###### Proof of Lemma A.12.

The proof is similar. We choose neighborhoods $W^{\pm}$ of $w^{\pm\infty}$ in
$\bar{\mathcal{G}}$ such that no $u^{i}x$ belongs to $W^{+}$, the
neighborhoods $u^{i}W^{+}$ for $i\in{\mathbb{Z}}$ are disjoint, and the
neighborhoods $v^{i}W^{-}$ are disjoint. For $n$ larger than some $n_{0}$, we
have $w^{n}(\bar{\mathcal{G}}\setminus W_{-})\subset W_{+}$ and
$w^{-n}(\bar{\mathcal{G}}\setminus W_{+})\subset W_{-}$.

We consider an element

$g=w^{n}v^{p_{k}}w^{-n}\cdots u^{n_{2}}w^{n}v^{p_{1}}w^{-n}u^{n_{1}}$

in $\langle u,w^{n}vw^{-n}\rangle$ and we show $gx\in W^{+}$ (as above, $k\geq
1$ and all exponents are nonzero). We have $u^{n_{1}}x\notin W^{+}$, hence
$w^{-n}u^{n_{1}}x\in W^{-}$, so $v^{p_{1}}w^{-n}u^{n_{1}}x\notin W^{-}$ and
$w^{n}v^{p_{1}}w^{-n}u^{n_{1}}x\in W^{+}$. Applying $u^{n_{2}}$ takes the
point outside of $W^{+}$ and we can iterate. ∎

######  Corollary A.13.

Let $H$ be torsion-free. Let $u,v,w\in H$, with $u,w$ non-trivial. Assume that
$w$ commutes with none of $u$ or $vuv^{-1}$. For $n$ large enough, $\langle
vu^{n},w\rangle$ is a free subgroup freely generated by $vu^{n}$ and $w$.

###### Proof.

We check that Lemma A.11 applies. The points $w^{+\infty},w^{-\infty}$ are
distinct from $\xi^{-}=u^{-\infty}$ and $\xi^{+}=(vuv^{-1})^{+\infty}$ because
$w$ does not commute with $u$ and $vuv^{-1}$. Since $H$ is torsion-free, no
element exchanges the points $u^{+\infty}$ and $u^{-\infty}$, so $\xi^{+}$ and
$\xi^{-}$ are not in the same $H$-orbit. ∎

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Vincent Guirardel

Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France

_e-mail:_<EMAIL_ADDRESS>

Gilbert Levitt

Laboratoire de Mathématiques Nicolas Oresme, Université de Caen et CNRS (UMR
6139)

(Pour Shanghai : Normandie Univ, UNICAEN, CNRS, LMNO, 14000 Caen, France)

_e-mail:_<EMAIL_ADDRESS>

Rizos Sklinos

Department of Mathematical Sciences

Stevens Institute of Technology

1 Castle Point Terrace, Hoboken, 07030, NJ, USA

_e-mail:_<EMAIL_ADDRESS>