Profinite completions of free-by-free groups

 Free-by-cyclic groups are an intensively studied class of groups in geometric group theory, mainly because they behave well enough that one can prove things about them. Free-by-free groups, in contrast, are completely wild. Today we'll see that they can have arbitrarily weird profinite completions, which is something I still sort of refuse to accept. Specifically, there is this theorem of Bridson:

Theorem 1: Given an arbitrary, finitely generated, recursively presented group \(\Gamma\) that is residually finite, one can construct a finitely generated, residually finite free-by-free group \(M_\Gamma = F_\infty \rtimes F_4\) and an embedding \(M_\Gamma \hookrightarrow (F_4 * \Gamma) \times F_4\) that induces an isomorphism of profinite completions.

The free-by-free groups aren't finitely presented, which perhaps explains why one sometimes restricts to finitely presented groups when dealing with profinite completions. In that case, the world is slightly better behaved: Bridson and co proved that the class of finitely presented residually finite groups is recursively enumerable. 

We will try to encode arbitrarily bad behaviour in finitely presented groups. Here are some strange things which can happen in finitely presented groups:

• Higman showed that there exists a 2-generated, finitely presented group that contains an isomorphic copy of every finitely presented group.
• Every finitely presented group can be embedded in a finitely presented group that has no finite quotients. A variant of this has been discussed on the blog before. 
• Baumslag-Dyer-Heller proved that every finitely presented group can be embedded in a finitely presented acyclic group. 

There is in fact some amount of structure to well-behaved properties of groups. 

Lemma 1: Let $\Pi$ be a property of groups that is inherited by direct limits and suppose that every finitely presented group $G$ can be embedded in a finitely presented group $G_\Pi$ that has property $\Pi$. 
Let $\Pi'$ be a second such property. Then there exists a 3-generated group $U^\dagger = K\rtimes\mathbb{Z}$  such that

1. $U^{\dagger}$ is finitely presented;
2. $U^{\dagger}$ contains an isomorphic copy of every finitely presented group;
3. $K$ has property $\Pi$ and property $\Pi'$.

Proof: Let $U_0$ be a finitely presented group that contains an isomorphic copy of every finitely presented group. By hypothesis, there is a finitely presented group $V$ that contains $U_0$ and has property $\Pi$, and there is a finitely presented group $W$ that contains $V$ and has property $\Pi'$. Consider the following chain of embeddings, where the existence of the embedding into $U_1 \cong U_0$ comes from the universal property of $U_0$,
\begin{equation} \label{eq:e1}
U_0 < V < W < U_1. \tag{1}
\end{equation}
We fix an isomorphism $\phi: U_1 \to U_0$ and define $U^{\dagger}$ to be the ascending HNN extension  $(U_1, t \mid t^{-1}ut = \phi(u) \, \forall u\in U_1)$. Let $K$ be the normal closure of $U_1$ in $U^{\dagger}$ and note that this is the kernel of the natural retraction $U^{\dagger}\to \langle t\rangle$. Note too that $t^{-i}U_1t^i < U_0$ for all positive integers $i$. It follows that for each positive integer $d$, we can express $K$ as an ascending union 
$$
K = \bigcup_{i\ge d} t^i U_1t^{-i}  = \bigcup_{i\ge d-1} t^i U_0 t^{-i}.
$$
From (\ref{eq:e1}) we deduce that $K$ is the direct limit of each of the ascending unions  $\bigcup_i t^iVt^{-i}$ and $\bigcup_i t^iWt^{-i}$. The first union has property $\Pi$, while the second has property $\Pi'$. $\blacksquare$

We will be interested in applying this lemma to the properties of being acyclic and having no finite quotients. It is clear that having no non-trivial finite quotients is preserved under passage to direct limits, and acyclicity is preserved because homology commutes with direct limits. 

Thus Lemma 1 provides us with a finitely presented group $U^{\dagger} = K\rtimes \mathbb{Z}$ such that $K$ is acyclic and has no non-trivial finite quotients.

Theorem 2: There is a finitely presented acyclic group $U$ such that

1. $U$ has no proper subgroups of finite index;
2. every finitely presented group can be embedded in $U$.

Proof: Let $B$ be a finitely presented acyclic group that has no non-trivial finite quotients and let $\tau\in B$ be an element of infinite order. For instance, one can take Higman's group or the variants from the post on the Rips construction.

Let $U = U^{\dagger}\ast_C B$ be the amalgamated free product in which $\langle \tau\rangle$ is identified with   $C:=1\times \mathbb{Z} < U^{\dagger}$. A routine spectral sequence argument shows that if $1\to N\to G\to Q\to 1$ is exact and $N$ is acyclic, then $G\to Q$ induces an isomorphism $H_n(G,\mathbb{Z})\to H_n(Q,\mathbb{Z})$ for every $n$. 

As $K$ is acyclic, $ U^\dagger\to C$ induces an isomorphism $H_*(U^\dagger, \mathbb{Z}) \cong H_*(C,\mathbb{Z})$. In particular, $H_n(U^\dagger, \mathbb{Z}) = 0$ for $n\ge 2$, and in the Mayer-Vietoris sequence for $U = U^{\dagger}\ast_C B$ the only potentially non-zero terms are $$0\to H_2(U,\mathbb{Z}) \to H_1(C,\mathbb{Z}) \to H_1(U^\dagger,\mathbb{Z}) \oplus H_1(B,\mathbb{Z})  \to H_1(U,\mathbb{Z}) \to 0.$$ $H_1(B,\mathbb{Z})=0$ and $H_1(C,\mathbb{Z}) \to H_1(U^\dagger,\mathbb{Z})$ is an isomorphism, so we deduce that $U$ is acyclic.

Each subgroup of finite index $S<U$ will intersect both $U^\dagger$ and $B$ in a subgroup of finite index. Since neither has any proper subgroups of finite index, $S$ must contain both $U^\dagger$ and $H$. Hence $S=U$. $\blacksquare$

Remark: One can take this $U$ to be 4-generated: Ol'shanskii-Sapir construct a 2-generator, 2-relator group with no finite quotients. Using this instead of Higman's group, that the group from the previous lemma is 3-generated, and that the amalgamation identifies an element of infinite order with the stable letter, one obtains a group that is 4-generated.

Although we won't need it, I think the following corollary is quite nice.

Corollary 3: Let $\mathcal{A}=(A_n)_n$ be a sequence of abelian groups, the first of which is finitely generated. If the $A_n$ are given by a recursive sequence of recursive presentations, each of which is untangled, then there is a finitely presented group $Q_{\mathcal{A}}$ with no proper subgroups of finite index and $H_n(Q_{\mathcal{A}}, \mathbb{Z})\cong A_{n-1}$ for all $n\ge 2$.

This relies on a nice theorem of Baumslag-Dyer-Miller. 

Definition: A recursive presentation $(X\,|\, R)_{\rm{Ab}}$ of an abelian group is said to be {\em untangled} if the set $R$ is a basis for the subgroup $\langle R\rangle$ of the free abelian group generated by $X$. The following corollary can be deduced from Theorem using the Baumslag-Dyer-Miller construction.

Theorem 4: If $\mathcal{A}=(A_n)$ is as described in Corollary 3 then there is a finitely generated, recursively presented group $G_{\mathcal{A}}$ with $H_n(G_{\mathcal{A}},\mathbb{Z})\cong A_n$ for all $n\ge 1$.

Proof of Corollary 3: By the Higman Embedding Theorem, $G_{\mathcal{A}}$ can be embedded in the universal finitely presented group $U$ constructed in the preceding proof. We form the amalgamated free product of two copies of $U$ along $G_{\mathcal{A}}$,
 $$ Q_{\mathcal{A}}:=U\ast_{G_{\mathcal{A}}} U.$$
Note that because $G_{\mathcal{A}}$ is finitely generated, $Q_{\mathcal{A}}$
is finitely presented. As in the preceding proof, since the  factors of the amalgam have no proper
subgroups of finite index, neither does $ Q_{\mathcal{A}}$.

The Mayer-Vietoris sequence for this amalgam yields, for all $n\ge 2$, an exact sequence
(where the $\mathbb{Z}$ coefficients have been suppressed):
$$
 H_n U\oplus H_n U\to H_n Q_{\mathcal{A}}  \to H_{n-1} G_{\mathcal{A}}\to H_{n-1}U \oplus H_{n-1}U .
$$
Thus, since $U$ is acyclic,  $H_nQ_{\mathcal{A}} \cong H_{n-1} G_{\mathcal{A}} \cong A_{n-1}$ for all $n\ge 2$. $\blacksquare$

For the proof of theorem 1, recall the following from last time:

Definition: The fibre product of a pair of epimorphisms \( p_i : G_i \to Q \) (\( i = 1, 2 \)) is the subgroup \( P = \{(g_1, g_2) \mid p_1 (g_1) = p_2 (g_2)\} < G_1 \times G_2 \). 

The relevance to the profinite world is 

Platonov-Tavgen criterion: Let $G_1, G_2$ be finitely generated groups, and $Q$ be a finitely presented group with no finite quotients and $H_2(Q,\mathbb{Z})=0$. For epimorphisms $p_i: G_i \to Q$, the inclusion of the fibre product $P \to G_1 \times G_2$ induces an isomorphism of profinite completions.

Proof of Theorem 1: Let \( U \) be a 4-generator group that satisfies Theorem 2. We fix an epimorphism \( \mu : F_4 \to U \). Given a finitely generated, recursively presented group \(\Gamma\), we fix an embedding \( \psi : \Gamma \hookrightarrow U \) and extend this to an epimorphism \( \Psi : F_4 * \Gamma \to U \) that restricts to \( \mu \) on \( F_4 \) and \( \psi \) on \(\Gamma\). Consider the fibre product of \( \Psi \) and \( \mu \),
\[
P < (F_4 * \Gamma) \times F_4.
\]
It is not difficult to see that \( P \) is finitely generated and the Platonov-Tavgen criterion tells us that the inclusion \( P \hookrightarrow (F_4 * \Gamma) \times F_4 \) induces an isomorphism of profinite completions.

The restriction of \( \Psi \) to each conjugate of \(\Gamma\) is injective, so by the Kurosh subgroup theorem ker \( \Psi \) is free. The projection from \( P \) to the second factor of \((F_4 * \Gamma) \times F_4\) is onto and has kernel ker \( \Psi \). Thus, \( P \) is free-by-free; more precisely, it is of the form \( F_{\infty} \rtimes F_4 \). Define \( M_\Gamma = P \). $\blacksquare$

The fibre products \( M_\Gamma \) that we are considering are free-by-free (and not free) and hence have cohomological dimension 2. But if \(\Gamma\) has cohomological dimension \(d\) then \(D(\Gamma) := (F_4 * \Gamma) \times F_4\) has cohomological dimension \(d + 1\). Thus, Theorem 1 yields pairs of finitely generated, residually finite groups that have the same profinite completion but have an arbitrary difference in their cohomological dimensions; for example, we can take \(\Gamma \cong \mathbb{Z}^d\). Moreover, \(D(\mathbb{Z}^d)\) is good in the sense of Serre and it retracts onto \(\mathbb{Z}^{d+1}\), so \(\overline{D(\mathbb{Z}^d)} \cong \hat{M}_\Gamma\) also has cohomological dimension \(d + 1\). Thus, Theorem 1 provides us with examples of groups of cohomological dimension 2 whose profinite completions have cohomological dimension \(d + 1\), where \(d\) is arbitrary. One can also arrange for \(\hat{M}_\Gamma\) to have infinite cohomological dimension (even if it is torsion free).

Moreover, despite being torsion-free itself, a free-by-free group can have all manner of torsion in its profinite completion. One can pick $\Gamma$ to be any manner of infinite torsion group such as the Grigorchuk group. 

The theorem also tells us that, with the possible (but unlikely) exception of certain free-by-free groups \(H\), no statement of the following form can be valid for all pairs of finitely generated, residually finite groups \(\Gamma_1\) and \(\Gamma_2\): ``if \(\hat{\Gamma}_1 \cong \hat{\Gamma}_2\) and \(\Gamma_1\) has a subgroup isomorphic to \(H\), then \(\Gamma_2\) has a subgroup isomorphic to \(H\).'' Furthermore, the theorem tells us that if a property \(\mathcal{P}\) is common to the subgroups of free-by-free groups but not to the subgroups of all finitely presented, residually finite groups, then \(\mathcal{P}\) is not a profinite invariant. Such properties include: being torsion-free, being locally indicable (i.e. every finitely generated subgroup maps onto \(\mathbb{Z}\)), being left-orderable (and all its consequences such as satisfying the Kaplansky unit conjecture), all 2-generator subgroups being finitely presented (or coherent), etc.

Remarks on torsion:

Earlier work by Lubotzky showed that any there exist finitely generated, residually finite groups such that any separable profinite group embeds in their profinite completion. In particular, they have arbitrary amounts of torsion. The construction was based on arithmetic lattices of higher rank and used a number of beautiful and powerful results, including that they have CSP, superrigidity, and Dirichlet's theorem on primes in AP. It amounts to looking at a direct product of $SL_n(\mathbb{Z})$ as $n$ increases and saying that one can find finite index torsion-free subgroups of $SL_n(\mathbb{Z})$ whose profinite completion contains the alternating group by CSP and then finding any finite group as a subgroup of $A_n$ for large enough $n$. One also knows from this sort of argument that torsion-free lattices with CSP are never good in the sense of Serre: torsion-free lattices have finite cohomological dimension, but the profinite completion has torsion so has infinite cohomological dimension.