What is left when the middle is taken out

The group drawn as a map turns multiplication into walking: one dot per element, one step per generator. How fast the ball fills then counts the dots near the start, and the edge that is as big as the ball weighs the outermost layer of that ball against the rest. All three look at the ball. This essay looks at everything else — the infinite remainder of the map once a finite middle has been cut out of it — and asks the simplest question available about that remainder: into how many pieces does it fall?

The question sounds like a matter of taste in drawing, and the answer turns out to be one of four numbers, with a theorem behind each. A group whose picture is a line leaves two pieces running away from the cut, a group whose picture is a plane leaves one, a finite group leaves none at all, and a group whose picture branches like a tree leaves more and more with every larger cut. Nothing leaves three. Exactly three, or five, or seventeen, is impossible for every group there is, and the reason is a short argument about moving the cut.

Counting the pieces that run away #

Take the ball of radius $r$ around the identity — every element reachable in at most $r$ steps — and delete it from the picture, together with every step that touches it. What remains is still infinite whenever the group is, and it may be connected or not. Some of its pieces may be finite, a few stranded dots with no way out; those are ignored. The count that matters is the number of infinite pieces, the ones that run off to infinity.

For the integers the answer is two at every radius. Removing $\{-r, \dots, r\}$ leaves the numbers above and the numbers below, and there is no way from one to the other that avoids the middle. For the integers squared the answer is one: removing a diamond leaves a ring round it, and a ring can be walked all the way round, so everything outside is a single piece. The doubled line — the integers crossed with a two-element group, pictured as two copies of the line with each point joined to its twin — also leaves two, and it is worth noticing that it needed a cut to get there. Remove the identity alone and the doubled line stays connected, because a path can climb to the other rail and walk past the gap; remove a ball of radius one and both rails are cut at once. So the count can rise as the cut grows, and the number that belongs to the group is where the count settles.

A computer cannot hold an infinite remainder, so every figure here counts inside a larger ball: delete the ball of radius $r$, keep everything out to some radius $R$, and count the pieces that reach the outer shell. A piece that reaches the shell has run $R - r$ steps away from the cut. The count is taken again with the outer shell one step further out, and in every figure on this page it does not change. That is evidence and not proof — a piece could in principle close up just beyond any radius checked — and for these particular groups the arguments below settle it.

The settled number is the group’s number of ends, a word Hans Freudenthal introduced in 1931 for spaces and that Heinz Hopf carried over to groups in 1944. An end is a way of going to infinity, and two routes out are the same end when no finite cut can separate them.

A tree that splits at every cut #

The free group on two generators has a picture that is a tree: every element has four neighbours, and there are no loops, because a reduced word in $a$, $b$ and their inverses can only be undone by retracing it letter by letter.

Removing the identity leaves four pieces, one for each first letter. Removing the ball of radius one leaves twelve, since each of the four neighbours has three further branches and each of those is now cut off from the others. Radius two leaves thirty-six, radius three a hundred and eight. The count triples with every step and never settles, and the free group is said to have infinitely many ends.

There is a clean description of what those ends are. A route that runs off to infinity in a tree without doubling back spells an infinite reduced word — $abA^{-1}\dots$, a letter at a time — and two different words give two different ends, because their routes part at the first letter where the words disagree, and removing the element where they part separates them for good. So the ends of the free group are its infinite reduced words, and there are uncountably many of them. Arranged by their first letters, then their second, then their third, they form a set with the structure of the middle-thirds Cantor set: every end is a limit of other ends, and any two can be separated by a finite choice of letters.

Eight groups and only four answers #

The table below runs the count on eight groups, removing balls of radius nought to four.

The finite group, the six turns of a hexagon, gives nought: its ball of radius three is the whole group, and removing it leaves nothing at all. The integers, the doubled line and the group generated by two reflections all settle at two. The integers squared and the integers cubed settle at one. And two rows never settle. The free group triples. The last row, the group made from a reflection $s$ and a third-turn $t$ with no relation between them except $s^2 = 1$ and $t^3 = 1$, runs $2, 3, 4, 6, 8$ — slower than the free group, roughly doubling every two steps, and still climbing without limit.

That is the whole range, and it is a theorem rather than the luck of the rows chosen. Freudenthal and Hopf showed that a finitely generated group has 0, 1, 2 or infinitely many ends. Nought for the finite groups, and then one, two or infinitely many for the infinite ones. There is no group with three.

It also does not matter which generators are used to draw the picture. Changing the generating set changes distances by at most a constant factor, which is the observation behind the growth rate’s independence of the generators, and a constant factor cannot turn one piece into two: a cut that separates two routes in one picture becomes a somewhat larger cut that separates them in the other. The number of ends is a property of the group.

Why a third end drags in infinitely many #

The proof that three is impossible uses nothing but the one thing every Cayley graph has and most graphs do not: symmetry. Multiplying every element on the left by a fixed $g$ moves the whole picture onto itself, steps and all, so whatever is true of the picture near the identity is true of it near $g$.

Suppose some finite cut $K$ leaves exactly $k$ infinite pieces, where $k$ is at least three, and suppose no finite cut leaves more. Pick one of the pieces and walk a long way into it, to an element $g$ far enough out that the moved cut $gK$ lies entirely inside that piece. Now remove both $K$ and $gK$.

The moved cut $gK$ also leaves $k$ infinite pieces around it, since the picture near $g$ is the picture near the identity. One of those pieces contains $K$ and everything on the far side of it. The other $k - 1$ lie beyond $gK$, inside the piece that was chosen, and they are separated from each other by $gK$. So removing both cuts leaves the $k - 1$ pieces of $K$ that were not chosen, untouched, together with at least $k - 1$ new pieces beyond $gK$ — at least $2k - 2$ in all. When $k \ge 3$ that is more than $k$, which contradicts the choice of $k$ as the most that any finite cut leaves. So either no finite cut ever leaves three or more pieces, or cuts leave more and more and the count is unbounded.

The argument fails at exactly one value, and the figure shows where. When $k = 2$, the count $2k - 2$ is two again. The line’s second cut produces a new piece beyond it, but the stretch between the two cuts, which was part of an infinite piece before, is now finite and drops out of the count. A line has room for two ends and a second cut simply trades one for another. In the tree the stretch between the cuts is not all there is between them: branches leave it sideways, each running to infinity on its own, and the count grows.

Two ends means a line in disguise #

Every group in the table with two ends looks like the integers once the fine detail is blurred. The doubled line is a line with a second rail. The group generated by two reflections of a line, $s$ and $t$, with $s^2 = t^2 = 1$ and nothing else, has a picture that is literally a line: its elements are the alternating words $\dots stst\dots$, laid out left and right of the identity by first letter.

That is not a coincidence of the examples. A finitely generated group has two ends exactly when it contains the integers as a subgroup with only finitely many cosets — when a finite number of translated copies of $\mathbb{Z}$ cover the whole group, as the two rails cover the doubled line and the words $st$, $stst$, … together with their images under $s$ cover the reflection group. Two ends is the signature of a group that is a line up to a bounded amount of fuzz.

The same drawing holds the other extreme. The reflection $s$ and third-turn $t$ generate a group whose picture is a tree of triangles: every coset of the third-turn is a triangle, $\{g, gt, gt^2\}$, and the reflection joins a corner of each triangle to a corner of another. Cut out the middle triangle and three branches run away; cut further and each branch splits again. This group has infinitely many ends. It is also the modular group — the group of transformations $z \mapsto (az + b)/(cz + d)$ with whole-number entries and $ad - bc = 1$, taken up to sign — and the tree of triangles is a picture that already exists elsewhere under another name. The modular group permutes the curved triangles that semicircles over neighbouring fractions cut the upper half-plane into, and a route out through the tree crosses those triangles one after another, turning left or right at each — which is exactly the Stern–Brocot descent towards a number. A route that never settles into turning the same way for ever homes in on an irrational number, so the ends of the modular group and the irrational numbers are, all but a countable few, two descriptions of the same set.

Infinitely many ends means a finite seam #

What the free group and the modular group share is not that they are trees but that each is assembled from two smaller groups glued along almost nothing. The free group is the free product $\mathbb{Z} * \mathbb{Z}$: words alternate between powers of $a$ and powers of $b$, with no relation linking them. The modular group is $\mathbb{Z}/2 * \mathbb{Z}/3$: words alternate between the reflection and powers of the third-turn. In each case the two factors meet only in the identity, and cutting the picture at a single element separates the words by how they begin.

John Stallings proved between 1968 and 1971 that this is the only way it happens. A finitely generated group has more than one end exactly when it splits along a finite subgroup — as a free product of two groups glued along a finite subgroup they share, or as a group extended by a new generator that conjugates one finite subgroup to another. So the count is a structural fact rather than a geometric one: a group has infinitely many ends when it can be taken apart along a finite seam into pieces neither of which is almost everything, and it has two when the pieces are one line’s worth of each other.

The theorem is the hard part of the subject, and it has a companion that makes it usable. Stallings’s own method for folding a graph until it decides is how the subgroups of a free group are read off pictures, and the ends of a free group’s subgroups are what the subgroup that is freer than the group turns on. His theorem about ends came first, and it is the reason a question about the shape of a picture at infinity has an answer in algebra.

Growth does not decide it #

It is tempting to think that the count is just growth read differently — that polynomial growth means few ends and exponential growth means many. The integers squared and cubed have polynomial growth and one end; the free group has exponential growth and infinitely many. But the pairing breaks as soon as it is tested.

The group of symmetries of a two-holed surface is the counterexample worth knowing. A closed surface with two handles, of the kind the classification builds from a sphere, has a fundamental group generated by four loops with one relation between them, and that group acts on the hyperbolic plane by the symmetries of a tiling by octagons. Its ball grows exponentially, like the free group’s, because the hyperbolic plane’s circles grow exponentially. Yet it has one end: the hyperbolic plane with a disc removed is still connected, just as the flat plane is, and a group has the same number of ends as any space it acts on with compact quotient and finite stabilisers. So a surface group grows like a tree and is shaped at infinity like a plane.

The two invariants measure different things. Growth counts how much of the group lies near the identity. Ends count how the far-away part of the group hangs together once a finite amount of it is removed — whether a finite fence can divide it, and into how many regions. The first is about volume and the second is about connectedness, and a group can have much of one and little of the other.

What the pictures cannot certify #

Every count on this page is made inside a finite ball, and the claim is about the infinite remainder. A piece that reaches the outer shell of the drawing has run a few steps away from the cut; it has not been followed to infinity, and nothing drawn rules out a finite piece that happens to be larger than the drawing. The figures check each count again with the outer shell one step further out and find it unchanged, which catches a piece that closes up immediately beyond the edge and cannot catch one that closes up further on. For the groups drawn the gap is closed by argument — a tree has no loops to close a piece, and a ring in the plane can always be walked round — and not by the pictures.

The same limit applies more sharply to the theorem. No figure shows that a group cannot have three ends; a figure can only show that the groups drawn do not. The translation argument is what excludes three, and the figure beside it is an instance, not a proof. Stallings’s theorem is quoted here and not drawn at all: it says that every group with more than one end splits along a finite subgroup, which is a statement about every such group, and its proof builds the seam from the cut by an argument about cohomology that no drawing of a ball can carry.

And the space of ends itself is invisible. For the free group it is a Cantor set of infinite words — uncountable, with no isolated points — and every drawing here shows at most a few hundred of its first letters.

Still open: what a one-ended group looks like at infinity #

Ends are the coarsest description of a group at infinity: they say how many directions there are, not what shape the far-away part has. For the hyperbolic groups — the ones whose pictures have thin triangles, like the free group and the surface groups — Mikhail Gromov defined a finer object in 1987, the boundary, which records every direction of escape as a point of a space. The free group’s boundary is the Cantor set of its ends. A surface group’s boundary is a circle, the circle at the edge of the hyperbolic plane.

A one-ended hyperbolic group has a connected boundary, and the question is which connected spaces occur and what each forces. The one with the clearest statement is Cannon’s conjecture, from 1991: a hyperbolic group whose boundary is a two-dimensional sphere acts on three-dimensional hyperbolic space as the symmetries of a tiling, just as a group with a circle for its boundary acts on the hyperbolic plane. The circle case is a theorem; the sphere case is proved for several special classes of group and open in general. It would say that the shape of a group at infinity, when it is a sphere, determines a geometry for the whole group — which is the same kind of statement as the one this essay proves at its crudest level, where the number of ways to infinity decides whether the group is a line, a plane or a tree.