## Infinite groups

Before looking for surface subgroups, it is worth thinking about how to find (or rule out the existence of) simpler classes of subgroups. This is a very general question, and I do not intend to give a complete survey; however, it is instructive to build up to the question of surface subgroups incrementally and to catalog some of the interesting examples and counterexamples along the way.

Question: When is a group infinite?

Already this question is more than hard enough. But first we must examine some unstated assumptions behind the question. We have some group $G$ in mind, and want to know whether it is infinite or not. But in what sense do we “have” the group $G$? There are several things we might mean by this, including:

1. An explicit group $G$ given by generators and relations; i.e. $G = \langle S \; | \; R \rangle$.
2. A group $G$ given together with an action on a set $\Sigma$.
3. A group $G$ not uniquely defined, but described implicitly in terms of its properties (e.g. $G$ is amenable, or left-orderable, or has property $(T)$, or is linear, or is residually $p$, or is a $3$-manifold group, or is finitely presented, or satisfies a law, etc.).

In general, it is hard to learn much about a group from a presentation. However, sometimes one can have some success:

Example: If $G$ is given by a finite presentation $G = \langle S \; | \; R \rangle$, the deficiency of the presentation is the difference between the number of generators and the number of relators; i.e. $|S| - |R|$. The deficiency of $G$ is the maximum of the deficiency of all finite presentations. In practice, it is very difficult to determine the deficiency of a group, but trivial to determine the deficiency of a given presentation. The rank of the abelianization of $G$ (i.e. the rank of $H_1(G;\mathbb{Z})$) is at least as big as the deficiency; hence if the deficiency is positive, $G$ is infinite, and in fact contains a copy of $\mathbb{Z}$.

Example: Daniel Allcock observed that one can do better when some of the relators $R$ as above are proper powers. Geometrically, a relator of order $p$ counts as only “$1/p$ of a relator” for the purposes of computing the rank of $H_1$. Explicitly, Allcock shows that if $G$ is a group with a presentation of the form $G = \langle a_1, \cdots, a_n \; | \; w_1^{r_1} = \cdots = w_m^{r_m} = 1 \rangle$ then if $H$ is a normal subgroup of $G$ of index $N < \infty$ and for each index $j$, one has $w_j^k \notin H$ for $1 \le k \le r_j-1$ then the rank of the abelianization of $H$ is at least $1+ N(n-1 - \sum_i \frac {1} {r_i})$. If this rank is positive, then $H$ is infinite, and therefore so is $G$.

Example: A much more subtle example is the famous Golod-Shafarevich inequality. Let $G$ be a finite $p$-group (i.e. a group in which every element is torsion, with order a power of $p$). Let $n(G)$ be the minimum number of generators of $G$, and $r(G)$ the number of relations between these generators in the corresponding free pro-$p$-group (if $R(G)$ denotes the minimum number of relations defining $G$ as a discrete group then $R(G) \ge r(G)$). The G-S inequality is the inequality $r(G) > n(G)^2/4$. In particular, if $G$ is a nontrivial pro-$p$-group for which $n(G)^2/4 \ge r(G)$ (or $n(G)^2/4 \ge R(G)$ which implies it) then $G$ is infinite. This inequality enabled Golod to give a negative answer to the generalized Burnside’s problem, by showing that for each prime $p$ there is an infinite group $G$ generated by three elements, in which every element has order a power of $p$.

Example: Marc Lackenby has made very nice use of the Golod-Shafarevich inequality in his work on Kleinian groups with finite non-cyclic subgroups. A Kleinian group is a finitely generated discrete subgroup of the group of isometries of hyperbolic $3$-space; such a group is the fundamental group of a hyperbolic $3$-orbifold. Marc shows that if a Kleinian group $G$ contains a finite non-cyclic subgroup, then $G$ is finite, or virtually free, or contains a closed surface subgroup. The argument is very interesting and delicate, and I hope to return to it in a later post. But for the moment I just want to remark that the form of the G-S inequality Marc uses is as follows. Let $G$ be a group with a finite presentation $G = \langle S \; | \; R \rangle$. Let $d_p$ denote the dimension of $H_1(G;\mathbb{F}_p)$ where $p$ is a prime. If $d_p^2/4 > d_p - |S| + |R|$ then $G$ is infinite.

Example: Another way to show a group $G$ is infinite is if the relators are very long. This is the method of small cancellation theory, and can be implemented in many different ways. From the modern perspective, a group presentation satisfies a small cancellation condition if one can build a $2$-complex from the presentation which is manifestly non-positively curved in some explicit sense. For example, if $G = \langle S \; | \; R \rangle$ is a symmetrized presentation (i.e. one in which elements of $R$ are cyclically reduced, and $R$ is closed under taking cyclic permutations and inverses), a piece is a word $b$ in the generators if there are distinct relations $ba_1, ba_2$ in $R$. If no relation is a product of fewer than $p$ pieces, one says that $G$ satisfies the small cancellation condition $C(p)$. So, for example, if $G$ is $C(6)$, one can build a $2$-complex presenting $G$ built from polygons, each of which has at least $6$ sides, and is non-positively curved (and therefore $G$ is infinite).

Example: Instead of showing that a particular group is infinite, one can show that certain groups whose presentations are obtained by a statistical process, are infinite with overwhelming probability. Yann Ollivier wrote an introduction to Gromov’s theory of Random Groups, in which it is made precise what one means by a “random group”, and many important properties of such groups are delineated. There is a parameter in the theory which governs the density of relations added to a generating set to determine the random group. The most striking aspect of the theory (in my opinion) is the existence of a phase transition. Gromov showed that if $G$ is a random group at density $d$ then if $d < 1/2$, with overwhelming probability, $G$ is infinite, hyperbolic, torsion-free and of geometric dimension $2$ (i.e. it is not free, but admits a $2$-dimensional $K(G,1)$). However, if $d \ge 1/2$, with overwhelming probability, $G$ is either trivial or $\mathbb{Z}/2\mathbb{Z}$.

Example: A group $G$ which admits a finite dimensional $K(G,1)$ is torsion-free, and therefore either trivial or infinite. This follows from the fact that the $K(\mathbb{Z}/p\mathbb{Z},1)$‘s are the infinite dimensional Lens spaces, which have nontrivial homology in infinitely many dimensions, together with elementary covering space theory. This example begs the question: how do you tell if a group has a finite dimensional $K(G,1)$? Well, one way is to exhibit a free, properly discontinuous action of $G$ on a finite dimensional contractible space; of course, given such an action, it is probably easier to directly find elements in $G$ of infinite order.

Example: A function $f:G \to \mathbb{R}^+$ is said to be a length function if it satisfies $f(\text{id})=0$, if it is symmetric (i.e. $f(g) = f(g^{-1})$ for all $g$) and if it is subadditive: $f(gh) \le f(g) + f(h)$. A group $G$ is said to be strongly bounded if every length function on $G$ is bounded. The strongly bounded property was introduced by George Bergman in this paper. A countable group is strongly bounded if and only if it is finite (the fact that finite groups are strongly bounded is obvious). Moreover, a group which admits an unbounded length function is evidently infinite. However, it turns out that there are many interesting uncountable but strongly bounded groups! Bergman showed that the group of permutations of any set is strongly bounded. Yves de Cornulier, in an appendix to a paper of mine with Mike Freedman, showed that the same is true for $Homeo(S^n)$, the group of homeomorphisms of an $n$-sphere.

Example: One of the most spectacular proofs of the finiteness of a (certain class of) group(s) is Margulis’ proof of the normal subgroup theorem, which says that if $G$ is a lattice in a higher rank Lie group, then every normal subgroup $H$ in $G$ is either finite, or of finite index. The proof has three steps: first, one shows that if $H$ is infinite, then $G/H$ is amenable. Second, since $G$ has property $(T)$, the same is true for $G/H$. Third, an amenable group with property $(T)$ is finite. The second and third steps are not very complicated: a group has property $(T)$ if the trivial representation is isolated in the space of all irreducible unitary representations, in a certain topology. A quotient of a group by a closed normal subgroup certainly has no more unitary representations than the original group itself, so the second step is not hard to show. An amenable group $G/H$ has almost invariant vectors in $L^2(G/H)$; since it has property $(T)$, it has an invariant vector in $L^2(G/H)$; but this implies that $G/H$ is finite. So the hard part is to show that $G/H$ is amenable. This is done using what is now known as boundary theory, and is described in Chapter VI of Margulis’ book.

I would be curious to hear other people’s favorite tricks/techniques to show that a group is or is not infinite.

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