Grothendieck pairs and the strong profinite genus

 Algebraic geometers are interested in profinite completions of discrete groups because the algebraic fundamental group of a projective variety over $\mathbb{C}$ is precisely the profinite completion of its fundamental group. Motivated by whether the algebraic fundamental group determines the variety, Grothendieck asked whether there exist finitely presented groups $H, G$ and an injection $\iota: H \to G$ such that $\hat{\iota}: \hat H \to \hat{G}$ is an isomorphism. A pair $(G, H)$ with this property became known as a Grothendieck pair.

Platonov and Tavgen made important progress by coming up with pairs of finitely generated groups which induce an isomorphism (because of this, some people call such pairs of only finitely generated groups Grothendieck pairs, so there is a bit of ambiguity in the term). They make use of an important and versatile construction in mathematics: the fibre product.

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 reader can find an introduction to applications of the fibre product to the algorithmic side of things here. We can now state

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.

Actually, this is a mild generalisation of the original criterion due to Bridson-Grunewald. To prove this, we need a couple of auxiliary lemmas.

Lemma 1: If \( G_1 \) and \( G_2 \) are finitely generated and \( Q \) is finitely presented, then \( P \) is finitely generated.

Lemma 2: Let \(H\) be a finitely generated group, and let \(L \subset N\) be normal subgroups of \(H\). Assume \(N/L\) is finite, \(Q = H/N\) has no finite quotients and \(H_2(Q,\mathbb{Z}) = 0\). Then there exists a subgroup \(S_1 \subset H\) of finite index such that \(S_1 \cap N = L\).

The main point is that $H_2(G,A)$ classifies central extensions by abelian groups $A$, and the hypothesis implies every central extension splits.

Proof of Platonov-Tavgen: Let \(\Gamma = G_1 \times G_2\). The surjectivity of \(\hat{u}\) is equivalent to the statement that there is no proper subgroup of finite index \(G \subset \Gamma\) that contains \(P\). If there were such a subgroup, then we would have normal subgroups $N_i$ of $G_i$ such that \(N_1 \times N_2 \subset P \subset G\), and \(G/(N_1 \times N_2)\) would be a proper subgroup of finite index in \((G_1/N_1) \times (G_2/N_2)\), of which there are none by hypothesis.

In order to show that \(\hat{u}\) is injective, it is enough to prove that given any normal subgroup of finite index \(R \subset P\), there exists a subgroup of finite index \(S \subset \Gamma\) such that \(S \cap P \subseteq R\). Note that \(L_1 := R \cap (N_1 \times \{1\})\), which is normal in \(P\) and of finite index in \(N_1' = (N_1 \times \{1\})\), is also normal in \(G_1' = G_1 \times \{1\}\), because the action of \((g_1,1) \in G_1'\) by conjugation on \(L_1\) is the same as the action of \((g_1,g_2) \in P\) for any $g_2 \in G_2$. Similar considerations apply to \(N_2' = (\{1\} \times N_2)\) and \(L_2 = R \cap N_2'\).

By the previous lemma we now have subgroups of finite index \(S_1 \subset G_1 \times \{1\}\) and \(S_2 \subset \{1\} \times G_2\) such that \(S_i \cap N_i' = L_i\) for \(i = 1,2\). Thus \(S := S_1 S_2\) intersects \(N_1' N_2'\) in \(L_1 L_2 \subseteq R \cap N_1' N_2'\).

Consider \(p \in P \smallsetminus R\). Since \(P\) and \(R \cap S\) have the same image in \(Q \times Q = \Gamma / N_1' N_2'\) (namely the diagonal) there exists \(r \in R \cap S\) such that \(pr \in N_1'N_2' \smallsetminus R\). Since \(N_1' N_2' \cap S \subseteq N_1' N_2' \cap R\), we conclude \(p \notin S\). Hence \(P \cap S \subseteq R\). $\blacksquare$

The hypothesis on $Q$ imply that it has separable cohomology in dimension 2, i.e. when $M$ is finite profinite completion induces an isomorphism on $H^2(-, M)$. This implies that for finitely generated $H$, any extension \[1\ \to H \to E \to  \Gamma \to 1\] induces an isomorphism on profinite completions. Hence if the kernel of $p_i$ is finitely generaed, $\ker{p_i} \to G_i $ is a Grothendieck pair (in the broader sense).

We have previously mentioned Higman's group, which satisfies the conditions of the criterion. This gives rise to a Grothendieck pair. In fact, Bridson-Grunewald showed that Higman's group and other similar groups give rise to genuine Grothendieck pairs. They also used these to answer a related question of Grothendieck about representation categories. 

To upgrade the fibre product to be finitely presented, we apply the following: 

Effective 1-2-3 theorem: There exists an algorithm that, given the following data describing group homomorphisms \(f_i : \Gamma_i \to Q\) (\(i = 1, 2\)), will output a finite presentation of the fibre product \(P\) of these maps, together with a map \(P \to \Gamma_1 \times \Gamma_2\) defined on the generators, provided that both the \(f_i\) are surjective and at least one of the kernels \(\ker f_i\) is finitely generated. (If either of these conditions fails, the procedure will not halt.)
Input:

• A finite presentation \(Q \equiv \langle X \mid R \rangle\) for \(Q\).
•  A finite presentation \(\langle a^{(i)} \mid r^{(i)} \rangle\) for \(\Gamma_i\) (\(i = 1, 2\)).
• For each \(a \in a^{(i)}\), a word \(\tilde{a}\) in the free group on \(X\) such that \(\tilde{a} = f_i(a)\) in \(Q\).
• A finite set of identity sequences that generates \(\pi_2 Q\) as a \(\mathbb{Z}Q\)-module.

In fact, Bridson later upgraded his result with Grunewald by proving:

Theorem 3: There exists a finitely presented, residually finite group $\Gamma$ and a recursive sequence of finitely presented subgroups $u_n: P_n \to \Gamma$ such that $(\Gamma, P_n)$ is a Grothendieck pair but $P_m \cong P_n$ iff $m=n$.

Recall that $S:=BS(2,3)$, defined by the presentation $\langle a, t  |ta^2t^{-1}a^{-3}\rangle$ is a Baumslag-Solitar group that is one of the first examples of a non-Hopfian group. This is witnessed by the fact that $\psi: a \mapsto a^2, t \mapsto t$ defines an epimorphism with non-trivial kernel: $[a,tat^{-1}]$ is a non-trivial element by Britton's lemma. It is well-known that one-relator presentations are aspherical, as are HNN extensions of free groups. Now the point is that in fact $\psi^n$ gives different kernels for different $n$, and we will attempt to push some of this up to a hyperbolic special group via the fibre product and Rips constructions.

(For completeness we remark that $BS(m,n)$ is Hopfian iff $m,n$ have the same prime factors, and residually finite iff $|m|=1$, $|n|=1$, or $|m|=|n|$. This is a result due to Baumslag-Solitar and Meskin. The latter fixed a mistake in the original paper. We could just as well take any non-Hopfian (Baumslag-Solitar) or (torsion-free one relator) group.)

Let \(B\) be an infinite acyclic group with no non-trivial finite quotients that has an aspherical presentation. For instance, the Higman group or one of its variants introduced in the post on the Rips construction. Fix an element of infinite order \(b \in B\) and define
\[
Q = S *_{\mathbb{Z}} B
\]
where the amalgamation identifies \(t \in S\) with \(b \in B\). This is the group that we will feed into the Platonov-Tavgen criterion.

Proposition 4: \(Q\) is a non-Hopfian group that has a balanced, aspherical presentation and no non-trivial finite quotients. In particular, \(Q\) is torsion-free and \(H_1(Q, \mathbb{Z}) = H_2(Q, \mathbb{Z}) = 0\).

Proof: Let \(B = \langle X \mid R \rangle\) be a balanced aspherical presentation and let \(\beta\) be a word in the generators that equals \(b^{-1} \in B\). Then
\[
Q = \langle a, t, X \mid ta^2t^{-1}a^{-3}, t\beta, R \rangle
\]
is an aspherical balanced presentation for \(Q\). In any finite quotient of \(Q\), the image of \(B\) is trivial, hence the image of \(t = b\) is trivial. And since \(S\) is in the normal closure of \(t\), it too has trivial image.

To see that \(Q\) is non-Hopfian we consider the homomorphism \(\Psi : Q \to Q\) whose restriction to \(S\) is the epimorphism \(\psi\) and whose restriction to \(B\) is the identity: \(\Psi\) is well-defined because \(\psi(t) = t\) and \(S \cap B = \langle t \rangle\); it is onto because \(S\) and \(B\) lie in the image; and it has non-trivial kernel because \(\psi\) does. $\blacksquare$

We will need a couple more auxiliary results for the proof of theorem 3. A subgroup \(H < G_1 \times G_2\) of a direct product is termed a subdirect product if the coordinate projections map it onto \(G_1\) and \(G_2\), and it is said to be full if both of the intersections \(H \cap G_i\) are non-trivial. The fibre product of any pair of epimorphisms \(G_1 \to Q\) and \(G_2 \to Q\) is a subdirect product, and it is full provided both maps have non-trivial kernel. (In fact, all subdirect products of \(G_1 \times G_2\) arise in this way.)

Lemma 5: Let \(\Gamma_1\) and \(\Gamma_2\) be torsion-free, non-elementary hyperbolic groups, let \(P, P' < \Gamma_1 \times \Gamma_2\) be full subdirect products, let \(N_i = P \cap \Gamma_i\) and let \(N_i' = P' \cap \Gamma_i\). Then every isomorphism \(\phi : P \to P'\) maps \(N_1 \times N_2\) isomorphically onto \(N_1' \times N_2'\) (sending the direct summands to direct summands) and extends uniquely to an isomorphism \(\Phi : \Gamma_1 \times \Gamma_2 \to \Gamma_1 \times \Gamma_2\).

Lemma 6: Let \(G\) be a group, let \(K \triangleleft G\) be a normal subgroup and let \(\phi:G \to G\) be an automorphism. If \(K \subsetneq \phi(K)\), then \(\phi\) has infinite order in \(\operatorname{Out}(G)\).

The proof of this is left as an exercise to the reader.

Lemma 7: Let \(G\) be a torsion-free hyperbolic group, let \(N \triangleleft G\) be a non-trivial finitely generated normal subgroup. If \(\operatorname{Out}(G)\) is infinite then $G/N$ is virtually cyclic.

This follows from the theory of actions on $\mathbb{R}$-trees and Paulin's theorem. These have been mentioned on this blog before.

Proof of theorem 3: Let \(\mathcal{Q}\) be the aspherical presentation described in Proposition 4 and let
\[
1 \to N \to \Gamma \xrightarrow{\pi_0} \mathcal{Q} \to 1
\]
be the short exact sequence obtained by applying the Rips construction to it. Let \(\Psi:\mathcal{Q} \to \mathcal{Q}\) be the epimorphism described in the proof of Proposition 4 and define \(\pi_n = \pi_0 \circ \Psi^{n}\). Let \(P_n < \Gamma \times \Gamma\) be the fibre product of the maps \(\pi_0:\Gamma \times \{1\} \to \mathcal{Q}\) and \(\pi_n:\{1\} \times \Gamma \to \mathcal{Q}\). The kernel of \(\pi_0\) is finitely generated, so we have all of the data required to apply the 1-2-3 Theorem. Thus we obtain, in a recursive manner, finite presentations for the fibre products \(P_n\). The Platonov-Tavgen criterion assures us that the inclusion \(P_n \hookrightarrow \Gamma \times \Gamma\) induces an isomorphism of profinite completions. Thus the following claim completes the proof.

Claim: \(P_n \cong P_m\) if and only if \(m = n\).

The intersection of \(P_n\) with \(\Gamma \times \{1\}\) is \(\ker \pi_0\) while its intersection with \(\{1\} \times \Gamma\) is \(\ker \pi_n\). Thus \(P_n\) contains the subgroup \(K_n := \ker \pi_0 \times \ker \pi_n\), which is normal in \(\Gamma \times \Gamma\). Note that \(K_n \subsetneq K_m\) if \(m > n\).

Lemma 5 tells us that any isomorphism \(\phi:P_n \to P_m\) is the restriction to \(P_n\) of an automorphism \(\Phi\) of \(\Gamma \times \Gamma\). The automorphism group of \(\Gamma \times \Gamma\) contains \(\operatorname{Aut}(\Gamma) \times \operatorname{Aut}(\Gamma)\) as a subgroup of index \(2\), and Lemma 7 tells us that the group of inner automorphisms has finite index in this. In particular, \(\Phi\) has finite order in the outer automorphism group of \(\Gamma \times \Gamma\). But then \(\Phi(K_n) = K_m\), by Lemma 5, which contradicts Lemma 6 unless \(m = n\). $\blacksquare$


Remark: The pro-nilpotent completion of a group \(G\) is the inverse limit of its system of nilpotent quotients; equivalently, it is the inverse limit of the sequence \(G/G_c \to G/G_{c-1}\) where \(G_c\) is the \(c\)-th term of the lower central series of \(G\). If a homomorphism of finitely generated groups induces an isomorphism of profinite completions, then it induces an isomorphism of pro-nilpotent completions. 

Now here is a trick. We can arrange for the group \(\Gamma\) of the previous section to be special on the nose, rather than virtually special, and hence residually torsion-free-nilpotent. The point is that the image of the finite index special subgroup is a finite index subgroup of the quotient, so it must be the whole group. Hence we may assume that the $u_n$ induces an isomorphism of pro-nilpotent completions.

The strong profinite genus of a group $G$ refers to the set of subgroups inducing an isomorphism on profinite completions. Bridson thus exhibits many groups with infinite strong profinite genus. The paper, dedicated to the memory of Fritz Grunewald who passed away rather suddenly and too soon, also contains a second proof of theorem 3 that I leave to the interested reader to explore.