Parafree Groups

Today’s post is about groups which have the same nilpotent quotients as free groups. There are many natural questions about how different these groups can be from the free group. This provides a kind of testing ground for how well equivalence of nilpotent quotients might be expected to preserve group properties.

This post is based on the following references:

Baumslag, Parafree Groups, in Infinite groups: geometric, combinatorial and dynamical aspects, 1-14, Progr. Math., 248, Birkhauser, Basel, 2005

Baumslag, Musings on Magnus, The mathematical legacy of Wilhelm Magnus: groups, geometry and special functions (Brooklyn, NY, 1992), 99–106, Contemp. Math., 169, Amer. Math. Soc., Providence, RI, 1994

Definition 0. The lower central series for a group G is defined recursively as $\gamma_1(G) = G$, $\gamma_i(G) = [\gamma_{i-1}(G), G]$.

Recall that nilpotent groups are those for which $\gamma_i(G) = 1$ for some finite i.

To any group we can associate a lower central sequence, which is the (ordered) set of quotients $G/ \gamma_i(G)$. The kth element of the lower central sequence is nilpotent of class k. Furthermore, every map from $G \to Q$ where Q is nilpotent of class k will factor through the group $G/G^{(k+1)}$, as this factor group only imposes the relations (and their consequences) necessary to make G nilpotent of class k.

We might ask how well the lower central sequence captures properties of a group. If G and H are nilpotent groups and have the same lower central sequence, then, in particular, they are isomorphic groups: because nilpotent groups appear as elements of their lower central sequences, the properties of nilpotent groups are completely described by the properties of elements of their lower central sequence.

We can’t expect to get full information about our group if there are elements which no nilpotent quotients can see. (That is, if there are non-trivial elements in $\cap_{i=1}^{\infty}\gamma_i(G)$.)  Consider, for example, when $G$ a perfect group. Then $\gamma_2(G) = [G,G] = G$ and so $G/\gamma_i(G) = \{1\}$ for all i. Therefore a perfect group has the same set of quotients as the trivial group. Taking nilpotent quotients  has thrown out all of our group elements.

In what follows, we will restrict our attention to groups that are residually nilpotent.

Definition 1. A group G is residually nilpotent if $\cap_{i \in \mathbb{N}} \gamma_i(N) = \{1\}$.

Definition 2. A group G is residually nilpotent if for every $g \neq 1$ there is some nilpotent quotient of G which does not trivialize $g$.

By the facts about the lower central sequence above, these two definitions are equivalent. These are precisely the groups in which all non-trivial elements eventually appear in the lower central sequence.

Proposition 1. The free group $F_2 = \langle a, b \rangle$ is residually nilpotent.

Proof: Let R be the the ring of formal powers series on the non-commuting indeterminants x and y with rational coefficients. Every element of R can be written as $r= r_0 + r_1 + \dots$ where $r_i$ is the sum of all degree i terms. There is a map $F_2 \to R$ which sends $a \to 1+ x$ and $b \to 1 + y$. These elements are units in R with inverses given by $(1 -x + x^2 -x^3+...)$ and $(1 -y + y^2 -y^3+...)$. The elements $(1+x)$ and $(1+y)$ freely  generate a subgroup of rank 2 amoung the units of R. Conveniently, we can put a metric on the units of R which is given by $d(1+ r_n + r_{n+1}+ \dots,1) = 1/n$ where n is the degree of the first non-zero term $r_n$. It is an easy induction argument to show that if $f \in \gamma_k(F)$ that the image is $f = 1 + \phi$ and $\phi_1 = \phi_2= \dots =\phi_{k-1} = 0$. This implies that the intersection of $\gamma_i(G)$ over all $i$ is just the identity. $\Box$

Now that we have a little more comfort with residually nilpotent groups, we can return to the question of how much we can say about groups which have the same lower central sequences. Free groups are a natural beginning, and the following question of Hannah Neumann is the starting point for this study:

Question 1. Are the free groups characterised by their lower central sequence? That is, if $latexG$ is a finitely generated group with the same lower central sequence as the free group $F$, then is G a free group?

Definition 3. A group G is parafree if it is residually nilpotent and if there is some free group $F$ such that $G / \gamma_i(G) = F/\gamma_i(F)$ for every $i$.

The parafree groups are those for which nilpotent quotients are enough to approximate the group, but for which nilpotent quotients aren’t enough to distinguish the group from a free group. Clearly free groups are an example, but there are others too. The first non-free example was the following, discovered by Baumslag.

Proposition 2. The group $G = \langle a, b ,c | a = [c, a][c, b] \rangle$ is a non-free parafree group.

We’ll show that this group has the same lower central sequence as $\langle b,c \rangle$, but first we’ll need to recall a definition and a lemma:

Definition 4. The element $g \in G$ is called non-generating if it is contained in the intersection of all maximal subgroups of $G$. Equivalently, if $g$ ever appears in a presentation as a generator for $G$, it can be removed from the generating set, without changing the group. The set of all non-generating elements for $G$ is actually a subgroup of $G$, called the Frattini subgroup.

Lemma 1. In the nilpotent group $G$, the Frattini subgroup contains the derived subgroup $\gamma_2(G) = G^{(1)}$.

Proof that G is free. Clearly

$\langle a \gamma_k,~ b \gamma_k, ~ c \gamma_k \rangle = \langle a^{-1}C\gamma_k, ~ b\gamma_k, ~ c\gamma_k, ~ C\gamma_k ~|~ C\gamma_k=[c,C(a^{-1}C)^{-1}][c,b]\gamma_k\rangle,$

where the equality is via a Tietze transformation and $C$ is shorthand for $[c,a][c,b]$.  By the lemma above, since we are in a nilpotent group, we can remove $[c, a][c,b]\gamma_k$ from the generating set. This also removes the relator. Therefore the set $a^{-1}[c, a][c, b] \gamma_k, b\gamma_k, c\gamma_k$ is a generating set for the free rank 3 nilpotent group of nilpotence class k. There is a map from $F_3 / F_3^{(k)} \to F_2 / F_2^{(k)}$ which just kills the first generator.  This means that trivializing $a^{-1}[c, a][c, b]\gamma_k(G)$ maps us to the free rank 2 nilpotent group of class k. On the other hand, $G/ \gamma_k(G)$ is the freest nilpotent group of class k which satisfies that $a^{-1}[c, a][c, b] =1$. Therefore these two groups are isomorphic.

If G was a free group, it would be the group $F_2$, as its nilpotent quotients have rank 2. However, notice that we can express G as an HNN extension with stable letter $b$.

$\langle a, c, b | c^b = c[a,c]a\rangle$

By the method of Magnus, this implies that there is a short exact sequence

$N \hookrightarrow G \to \mathbb{Z}$

where N is the group $\langle a_i, c_i | c_{i+1} = a_i[c_i,a_i]\rangle$, and $a_i = a^{b^{i}},~~ c_i = c^{b^{i}}$. From the abelianization, we see that if G was free that the element a had to be trivial in G. However from this decomposition, we have that the element $a= a_0$ maps non-trivially into G$\Box$

(The hardest part of proving that a group is parafree seems to almost always be proving residual nilpotence, which we skip here.)

Definition 5. The rank of a parafree group G is defined to be the minimum number of generators of the abelianization $G/G^{(1)}$. In other words, a parafree group has the same set of nilpotent quotients as some free group F. The rank of G is defined to be the rank of F.

For example, the group above has rank 2.

Proposition 3. If $G$ is an infinite, rank-1 parafree group, then it is isomorphic to $\mathbb{Z}$.

Proof. Indeed, this follows from the fact that the lower central sequence stabilized immediately. By the residual nilpotence of $G$, if all non-trivial elements of $G$ had to remain non-trivial in some element of the lower central sequence, then in fact the map $G \to G/G^{(1)}$ couldn’t have any kernel, so it was an isomorphism, and $G/G^{(1)}$ was by assumption isomorphic to $\mathbb{Z}$$\Box$

Magnus proved that a rank n parafree group which can be generated with an $n$ element set is in fact a free group. This result was later generalized by Bridson and Reid, who showed that if a parafree group has rank $r\geq 2$ that there is some $r$ element subset of its generating set that generates $F_r$ as a subgroup of $G$.

The following is an important tool for finding parafree groups:

Theorem 1 (Stallings). If $h: A \to B$ is a homomorphism which induces an isomorphism on the abelianizations (ie $H_1(A)$ and $H_1(B)$ are isomorphic), then if $H_2(A)$ surjects onto $H_2(B)$ then for all $i$, $A/ \gamma_i(A) \cong B/\gamma_i(B)$. In particular, this implies that if there is a homomorphism $F_n \to G$ that induces an isomorphism on their abelianizations, and G has trivial second homology, then if G is residually nilpotent, it is parafree.

The second homology of a group $G$, $H_2(G; \mathbb{Z})$, which is often just abbreviated to $H_2(G)$, is the second homology of an Eilenberg-Maclane space for $G$. Another interpretation of the second homology of a group comes from the Hopf isomorphism, which says that $H_2(F/R) = R \cap [F,F] / [R,F]$ where $F$ is a free group, and is a normal subgroup. We can see clearly from this equation that free groups have trivial second homology. It’s also easy to see that this holds in our example above. Indeed, the index sum of the a must be zero to intersect with $[F,F]$ and this happens exactly when the index sum of the relator r is equal to zero, so we can express it as an element of $[F,R]$. Therefore Stallings’ Theorem provides another proof for our example being parafree (once residual nilpotence is verified).

Next we consider some properties of free groups that also hold amoung parafree groups:

Theorem 2. If $G$ is a (para)free group, then G embeds into the units of a power series ring over $\mathbb{Z}$.

Theorem 3. If $G$ is a (para)free group, then $G$ embeds into a (para)free group of rank 2.

Theorem 4 (Bridson and Reid). If $G$ is a (para)free group, then every finitely generated normal subgroup of $G$ has finite index.

The following is a property of free groups which is conjectured to hold for parafree groups:

Conjecture 1 (Baumslag’s Parafree Conjecture). If $G$ is a finitely generated parafree group, then  $H_2(G; \mathbb{Z}) = 0$, that is, the second homology of an Eilenberg-Maclane space for $G$ will be trivial.

This conjecture is related to a  number of topological questions. It is related to the length of 3-manifold groups, that is, the ordinal at which the lower central series of a 3-manifold grous stabilizes. It is also related to Whitehead’s Asphericity Conjecture, that every connected subcomplex of a 2 dimensional aspherical CW complex is also aspherical (it has contractible universal covering space).

Although this conjecture has not been resolved, Bridson and Reid have recently shown that having the same set of nilpotent quotients does not imply having the same second homology. They have constructed an example of two finitely presented residually torsion free nilpotent groups with the same set of quotients for which one has finitely generated second homology and the other has infinitely generated second homology.

Of course, not all properties of free groups hold in the case of free groups. For example,

Theorem 5 (Baumslag). Parafree groups are not classified by their rank. In particular, there are continuum many parafree groups.

Similarly, there are many properties of free groups for which it is not yet known whether or not they hold amoung all parafree groups.

Open Questions:

• Are parafree groups linear?
• Are parafree groups hyperbolic?
• Suppose that $G$ is a finitely generated, residually finite group with the same finite images as a free group. Is $G$ parafree?