*Additive combinatorics* is largely focused on the additive properties of finite subsets A of an additive group . This group can be finite or infinite, but there is a very convenient trick, the *Ruzsa projection trick*, which allows one to reduce the latter case to the former. For instance, consider the set inside the integers . The integers of course form an infinite group, but if we are only interested in sums of at most two elements of A at a time, we can embed A ininside the finite cyclic group without losing any combinatorial information. More precisely, there is a *Freiman isomorphism of order 2* between the set in and the set in . One can view the latter version of as a *model* for the former version of . More generally, it turns out that any finite set A in an additive group can be modeled in the above set by an equivalent set in a finite group, and in fact one can ensure that this ambient modeling group is not much larger than A itself if A has some additive structure; see this paper of Ruzsa (or Lemma 5.26 of my book with Van Vu) for a precise statement. This projection trick has a number of important uses in additive combinatorics, most notably in Ruzsa’s simplified proof of Freiman’s theorem.

Given the interest in non-commutative analogues of Freiman’s theorem, it is natural to ask whether one can similarly model finite sets A in multiplicative (and non-commutative) groups using finite models. Unfortunately (as I learned recently from Akshay Venkatesh, via Ben Green), this turns out to be impossible in general, due to an old example of Higman. More precisely, Higman shows:

Theorem 1.There exists an infinite group G generated by four distinct elements a,b,c,d that obey the relations; (1)

in fact, a and c generate the free group in G. On the other hand, if G’ is a finite group containing four elements a,b,c,d obeying (1), then a,b,c,d are all trivial.

As a consequence, the finite set in G has no model (in the sense of Freiman isomorphisms) in a finite group.

Theorem 1 is proven by a small amount of elementary group theory and number theory, and it was neat enough that I thought I would reproduce it here.

– No non-trivial finite models –

Let’s first show the second part of Theorem 1. The key point is that in a finite group G’, all elements have finite order, thanks to Lagrange’s theorem. From (1) we have

and hence by induction

(2)

for any positive n. One consequence of (2) is that if , then , and thus . Applying this with n equal to the order of b, we conclude that

.

As a consequence, if is divisible by some prime p, then is divisible by p, which forces p to be odd and to be divisible by the multiplicative order of 2 modulo p. This is at most p-1 (by Fermat’s little theorem), and so is divisible by a prime strictly smaller than the prime dividing . But we can cyclically permute this argument and conclude that is divisible by an even smaller prime than the prime dividing , and so forth, creating an infinite descent, which is absurd. Thus none of can be divisible by any prime, and so a,b,c,d are trivial as claimed.

**Remark 1.** There is nothing special here about using four generators; the above arguments work with any number of generators (adapting (1) appropriately). But we will need four generators in order to establish the infinite model below.

**Remark 2. T**he above argument also shows that the group G has no non-trivial finite-dimensional linear representation. Indeed, let a, b, c, d be matrices obeying (1), then b is conjugate to , which by the spectral theorem forces the eigenvalues of b to be roots of unity, which implies in particular that grows at most polynomially in n; similarly for . Applying (2) we see that grows at most polynomially in n, which by the Jordan normal form for a implies that a is diagonalisable; since its eigenvalues are roots of unity, it thus has finite order. Similarly for b,c,d. Now apply the previous argument to conclude that a,b,c,d are trivial.

– Existence of an infinite model –

To build the infinite group G that obeys the relations (1), we need the notion of an *amalgamated free product* of groups.

Recall that the free product of two groups and can be defined (up to group isomorphism) in one of three equivalent ways:

- (Relations-based definition) is the group generated by the disjoint union of and , with no further relations between these elements beyond those already present in and separately.
- (Category-theoretic definition) is a group with homomorphisms from and into , which is universal in the sense that any other group G’ with homomorphisms from will have these homomorphisms factor uniquely through .
- (Word-based definition) is the collection of all words , where each lies in either or , with no two adjacent lying in the same (let’s label here to be disjoint to avoid notational confusion), with the obvious group operations.

It is not hard to see that all three definitions are equivalent, and that the free product exists and is unique up to group isomorphism.

**Example 1.** The free product of the free cyclic group with one generator a, and the free cyclic group with one generator b, is the free (non-abelian) group on two generators.

We will need a “relative” generalisation of the free product concept, in which the groups are not totally disjoint, but instead share a common subgroup H (or if one wants to proceed more category-theoretically, with a group H that embeds into both and ). In this situation, we define the amalgamated free product by one of the following two equivalent definitions:

- (Relations-based definition) is the group generated by the relative disjoint union of and (which is the same as the disjoint union but with the common subgroup H identified), with no further relations between these elements beyond those already present in and separately.
- (Category-theoretic definition) is a group with homomorphisms from and into that agree on H, which is universal in the sense that any other group G’ with homomorphisms from that agree on H will have these homomorphisms factor uniquely through .

**Example 2.** Let be the group generated by two elements with one relation . It is not hard to see that all elements of can be expressed uniquely as for some integers n, m, and in particular that is a free cyclic group. Let be the group generated by two elements with one relation , then again is a free cyclic group, and isomorphic to the previous copy of H. The amalgamated free product is then generated by three elements with two relations .

It is not hard to see that the above two definitions are equivalent, and that exists and is unique up to group isomorphism. But note that I did not give the word-based definition of the amalgamated free product yet. We will need to do so now; I will use the arguments from this 1954 paper of Neumann (who I briefly overlapped with, actually, at ANU), though the basic result I need here (namely, Corollary 1) dates all the way back to the work of Schreier in 1927.

In order to analyse these groups, we will need to study how they act on various spaces. If G is a group, we define a *G-space* to be a set X together with an action of G on X (or equivalently, a homomorphism from G to the permutation group of X). Thus for instance G is itself a G-space. A G-space X is *transitive* if for every , there exists such that . A *morphism* from one G-space X to another G-space Y is a map such that for all and . If a morphism has an inverse that is also a morphism, we say that it is an *isomorphism*.

The first observation is that a G-space with certain properties will necessarily be isomorphic to G itself.

Lemma 1.(Criterion for isomorphism with G) Let G be a group, and let X be a non-empty transitive G-space, and suppose there is a morphism from the G-space X to the G-space G. Then is in fact an isomorphism of G-spaces.

**Proof.** It suffices to show that is both injective and surjective. To show surjectivity, observe that the image is G-invariant and non-empty. But the action of G on G is transitive, and so as desired. To show injectivity, observe from transitivity that if are distinct elements of X then for some non-identity , thus , thus , establishing injectivity.

Now we can give the word formulation of the amalgamated free product.

Lemma 2.(Word-based description of amalgamated free product) Let be two groups with common subgroup H, and let be the amalgamated free product. Let , be some partitions of into right-cosets of H. Let X be the space of all formal words of the form , where , each lies in either or , and no two adjacent lie in the same . let be the obvious evaluation map. Then there is an action of G on X for which becomes an isomorphism of G-spaces.

**Proof.** It is easy to verify that and act separately on X in a manner consistent (via ) with their action on G, and these actions agree on H. Hence the amalgamated free product G also acts on this space and turns into a morphism of G-spaces. From construction of X we see that the G-action is transitive, and the claim now follows from Lemma 1.

Corollary 1.Let be two groups with common subgroup H, and let be the amalgamated free product. Let and be such that the cyclic groups are infinite and have no intersection with H. Then generate a free subgroup in G.

**Proof. ** By hypothesis (and the axiom of choice), we can find a partition where contains the infinite cyclic group , and similarly we can find a partition . Let X be the space in Lemma 2. Each reduced word formed by then generates a distinct element of X, and thus (by Lemma 2) a distinct element of G. The claim follows.

**Remark 3.** The above corollary can also be established by the ping-pong lemma (which is not surprising, since the proof of that lemma uses many of the same ideas, and in particular exploiting an action of G on a space X in order to distinguish various words in G from each other). Indeed, observe that map those words in X with into words with , and similarly for , which is the type of hypothesis needed to apply the ping-pong lemma. [Thanks to Ben Green for this observation.]

Now we can finish the proof of Theorem 1. As discussed in Example 2, the group is the amalgamated free product of and relative to . By Corollary 1, a and c generate the free group here, thus contains as a subgroup. Similarly, the group also contains as a subgroup. We may then form the amalgamated free product

and another application of Theorem 1 shows that b,d generate the free group (and are in particular distinct); similarly a, c are distinct. Finally, the group embeds into , which embeds into G, and so a, b, are also distinct; cyclically permuting this we conclude that all of the a, b, c, d are distinct as claimed.

## 8 comments

Comments feed for this article

7 October, 2008 at 3:47 am

ben greenTerry,

Thanks for explaining this! The example is potentially disturbing in that if one aspires towards a Freiman-type theory for arbitrary groups (a theory which would only involve finite sets) then it may be necessary to understand infinite groups and even infinite-dimensional representations.

I say only *potentially* disturbing because there is some hope that a weaker statement holds. Namely, if is a constant and if is a set for which then there is a subset with which has a Freiman-isomorphic copy inside some finite group . I can’t see it being easy to produce a counterexample to this!

One cannot hope for to have size comparable to (thus , say) as is the case in the abelian setting. I worked out the slightly painful details of an example a few months ago and have posted it here

http://www.dpmms.cam.ac.uk/~bjg23/papers/nonabelianmodels.pdf

since I don’t have any intention of publishing it.

Best wishes

Ben

7 October, 2008 at 9:54 am

Anonymousprofessor tao,

thank you for your well written posts. i am learning a lot of math mainly through your posts.

if it is not too much trouble can you plaese include some references, ideally to books that can

with easy availability. i don’t have access to jstor or a good math library with easy access to

journals.

again thanks for writing up beautiful math.

10 October, 2008 at 9:40 am

AnonymousIn the beginning of the proof of theorem 1: “The key point is that in a finite group G’, all elements have finite order, thanks to Cauchy’s theorem”.

Apart from this fact being obvious, could you explain how it follows from Cauchys theorem?

10 October, 2008 at 11:00 am

Terence TaoOops! I meant Lagrange’s theorem rather than Cauchy’s theorem.

11 October, 2008 at 5:01 pm

David FisherHi Terry,

That is a cool example.

A remark: all linear groups are residually finite i.e. have finite quotients

that separate any finite set.

So Remark 2 actually follows from the the fact that G has no finite quotients.

Your proof that G is non-linear is easier than the proof that linear groups

are residually finite.

Best,

David

23 October, 2008 at 8:02 am

Gabor ElekGordon and Vershik have a paper that characterizes those groups G such that

all finite subset of G can be modelled in finite groups.

They call these groups LEF (locally finitely embeddable). http://www.esi.ac.at/preprints/esi334.pdf

28 October, 2012 at 11:37 am

van Kampen’s theorem via covering spaces « What’s new[…] in and using the combinatorial description of the amalgamated free product (which was discussed in this previous blog post). But I recently learned (thanks to the responses to this recent MathOverflow question of mine) […]

10 February, 2014 at 10:02 am

AnonJoining a little late, but regarding remark 2: every linear group is residually finite, and so the remark follows immediately from the theorem.