# Existence of n-axial elements in groups with at least 2 ends

Let $$G$$ be a finitely generated group. Fix some symmetric finite generating set $$S$$ for $$G$$, and write $$\Gamma$$ for the Cayley graph of $$G$$ with respect to $$S$$.

Given finite subsets $$X,S,Y$$ of $$G$$, we say that $$S$$ separates $$X$$ from $$Y$$, denoted $$(X:S:Y)$$, if $$X$$ and $$Y$$ lie in distinct connected components of $$\Gamma\setminus S$$. If $$B_n$$ denotes the $$n$$-ball in $$G$$, we say that $$g\in G$$ is $$n$$-axial if

$$(g^a B_n: g^b B_n : g^c B_n)$$

whenever $$a are integers. Hopefully, it is easy to imagine what such elements look like in $$\mathbb{Z}$$ or $$\mathbb{Z}\ast\mathbb{Z}$$, where some power of any nontrivial element will be $$n$$-axial. On the other hand, there are no $$n$$-axial elements in a one ended group like $$\mathbb{Z}^2$$.

My PhD thesis claims that, when $$G$$ has at least two ends, $$n$$-axial elements exist for $$n$$ sufficiently large. Sadly, Ville Salo and Ilkka Törmä recently pointed out to me that my proof is wrong.

Before returning my degree, it seems prudent to ask:

If $$G$$ has at least two ends, must $$G$$ contain an $$n$$-axial element for some $$n$$?

I also wonder about the Bass-Serre theory interpretation of this problem. That is, suppose that $$G$$ acts nontrivially (without global fixed point) with finite edge stabilizers on a tree $$T$$. If $$g\in G$$ acts hyperbolically on $$T$$, must some power of $$g$$ be $$n$$-axial?

• I guess it can be proved without Stallings' theorem. Freudenthal proved in the 40s that there exists some $g$ acting on the end boundary with a north-south dynamics. (without the language, he essentially proved the action on the boundary is convergence). I believe one could deduce this property (possibly for some power of $g$). I haven't checked details. – YCor Sep 19 at 7:35

Let $$G$$ be a multi-ended finitely generated group.

• We know from Stallings' theorem that $$G$$ splits over a finite subgroup. So $$G$$ can be thought of as the fundamental group of a (compact) graph of spaces $$X$$ whose edge-spaces have the form $$Y \times [0,1]$$ where $$Y$$ is a compact $$2$$-dimensional CW-complex such that $$\pi_1(Y)$$ is finite.

• The universal cover $$\widetilde{X}$$ (on which $$G$$ acts geometrically) is now a tree of spaces whose edge-spaces have the form $$\widetilde{Y} \times [0,1]$$, where $$\widetilde{Y}$$ is the universal cover of $$Y$$; notice that $$\widetilde{Y}$$ must be a compact $$2$$-dimensional CW-complex.

• In fact, by collapsing these edge-spaces to $$[0,1]$$ and the vertex-groups to points, you get a tree $$T$$ which coincides with the Bass-Serre tree associted to the initial splitting.

• Consequently, if $$g \in G$$ acts on the Bass-Serre tree as a loxodromic isometry, then up to replacing $$g$$ with a sufficiently large power, we have $$(g^p \widetilde{Y} : g^q \widetilde{Y} : g^r \widetilde{Y})$$ for every $$p. (Here we are just saying that an edge in a tree always separates.) As $$\widetilde{Y}$$ is bounded in $$\widetilde{X}$$ the desired conclusion follows.

Details about graphs of spaces/groups can be found in Scott and Wall's paper Topological methods in group theory.

Working directly in the Cayley graph of $$G$$ should also be possible. An argument could be extracted from this answer.

• If I understand correctly, in your other answer, you find $a \in G$ such that a branch $B$ outside some separating set $A$, which we can take to be an $n$-ball, properly contains the set $aB$. Then $a$ indeed has infinite order, and we also obtain that $a$ is $n$-axial: any path from $1$ to $a^2$ in particular has to step into $aB$, since $a^2B$ is inside $aB$. Translate and apply the lemma that $(B_i : B_{i+1} : B_{i+2})$ implies $(B_i : B_j : B_k)$ for $i < j < k$. – Ville Salo Sep 19 at 8:37

For $$>2$$-ended groups. By geodesics I mean paths in the Cayley graph whose length is equal to the word metric between their initial and terminal vertex.

If $$G$$ is a group with finite generating set $$S$$, $$A$$ finite and $$G \setminus A$$ has all components infinite and has at least three components (in the Cayley graph w.r.t. $$S$$). Suppose $$h \notin A$$, then there exists $$k$$ such that $$A \cap kA = \emptyset$$ and $$kh$$ is connected to $$h$$ in $$G \setminus (A \cup kA)$$.

Proof. Consider the set $$P \subset G$$ consisting of all geodesics in $$G \setminus A$$ that begin from $$\partial A$$ (the boundary of $$A$$, i.e. elements adjacent to $$A$$ but not inside it), in the component of $$h$$. Write $$P_n$$ for the set of nodes first reached by some geodesic in $$P$$ after exactly $$n$$ steps. Since every translate of $$A$$ splits $$G$$ into at least three components, it is easy to show that for some $$\alpha > 1$$, we have $$|P_n| \geq \alpha^n$$ for all large enough $$n$$. For all $$g \in P_n$$, fix a geodesic representation $$w_g \in S^n$$, representing a the path from some element $$e_g \in \partial A$$ to $$g$$ arbitrarily.

Now pick $$h'$$ in $$G \setminus A$$ in $$P_m$$ for large $$m$$. Suppose that for all $$g \in P_n$$, the set $$g(h')^{-1}A$$ intersects the range of the geodesic $$w_g$$ (in the obvious sense of partial products starting from $$e_n$$). Then, intuitively, we can use this fact to compress the information in the paths in $$P_n$$. More precisely, if $$g \in P_n$$ then $$g = g'a^{-1}h'$$ for some $$a \in A$$ where $$g'$$ lies on the range of $$w_g$$. Because $$h' \in P_m$$ has word norm at least $$m - |e_g|$$ and $$w_g$$ is a geodesic, $$g'$$ must be in $$P_{n-m+k}$$ where $$k \in [-3r,3r]$$ where $$r$$ is the maximal word norm among elements of $$A$$. In particular, the distance of $$g$$ from $$\partial A$$ is reduced by at least $$m-3r$$ when we multiply it by a suitable $$h^{-1}a$$. We can now describe any element of $$P_n$$ by listing at most $$\lceil n/(m-3r) \rceil$$ elements $$a$$, encountered when iteratively applying this observation, starting from $$g$$, and listing also the element of $$P_{n'}$$ for $$n' < m+3r$$ obtained in the end. Now, $$C |S|^{m+3r+1} |A|^{\lceil n/(m-r) \rceil}$$ (where $$C |S|^{m+3r}$$ is an upper bound for the number of elements obtained in the end, $$C$$ is the boundary size of $$A$$) has slower asymptotic growth in $$n$$ than $$\alpha^n$$, as soon as $$|A|^{1/(m-3r)} < \alpha$$, a contradiction if $$m$$ was picked large enough.

This means we can necessarily find some $$g \in P_n$$ (for any sufficiently large $$n$$) such that $$g(h')^{-1} A$$ does not intersect the geodesic $$w_g$$. Assuming $$n$$ and $$m$$ are large enough, we have both $$A \cap g(h')^{-1}A = \emptyset$$, and that the connected component of $$g$$ in $$G \setminus (A \cup g(h')^{-1}A)$$ contains both $$h$$ and $$g(h')^{-1}h$$. The first fact is obvious, and for the second, observe that as $$h$$ and $$g(h')^{-1}h$$ are, respectively, in the same connected components as $$h'$$ and $$g(h')^{-1}h' = g$$ in $$G \setminus A$$ and $$G \setminus g(h')^{-1}A$$ respectively, it is suffices that $$n$$ and $$m$$ be larger than the length of a minimal path between $$h$$ and $$h'$$ in $$G \setminus A$$. This means we can pick $$k = g(h')^{-1}$$. End square.

Let $$G$$ have at least $$3$$ ends. Then $$G$$ has an $$n$$-axial element for some $$n$$.

Proof. Let $$A$$ finite and $$G \setminus A$$ has all components infinite and has at least three components, you can always find such by flood filling the finitely many finite ones. Let $$h_1 \notin A$$ and $$h_2 \notin A$$ be in distinct components of $$G \setminus A$$. Now, use the previous lemma to find $$k_1$$ such that $$h_1$$ and $$k_1h_1$$ are in the same component of $$G \setminus (A \cup k_1A)$$ and $$A$$ and $$k_1A$$ are disjoint. Then find $$k_2$$ so that $$A$$ and $$k_2A$$ are disjoint and $$h_2$$ is in the same component of $$G \setminus (A \cup k_2A)$$ as $$k_2h_2$$. Then $$k_2^{-1}k_1$$ is $$n$$-axial, if $$n$$ is bigger than the size of $$A$$.

Now consider the sequence $$B_0 = A, B_1 = k_2^{-1}A, B_2 = k_2^{-1} k_1 A, B_3 = k_2^{-1} k_1 k_2^{-1} A, B_4 = k_2^{-1}k_1k_2^{-1}k_1 A...$$ i.e. you multiply the left translation element from the right alternately by $$k_2^{-1}$$ and $$k_1$$.

We have $$(B_0 : B_1 : B_2)$$: The separation $$(k_2A : A : k_1A)$$ is clear, and gives $$(A : k_2^{-1}A : k_2^{-1}k_1A)$$ by translating.

We have $$(B_1 : B_2 : B_3)$$: We prove that $$(A : k_1A : k_1k_2^{-1}A)$$ holds: suppose there is a path from $$A$$ to $$k_1k_2^{-1}A$$ that does not go through $$k_1A$$. Then, there is a path from $$k_1h_1$$ to $$k_1h_2$$ that does not go through $$k_1A$$ (which would be a contradiction). To see this, take a path from $$k_1h_1$$ to $$h_1$$ (using the assumption on $$k_1$$) that does not visit any of the three sets, then move to $$k_1k_2^{-1}A$$ and then $$k_1k_2^{-1}h_2$$ without visiting $$k_1A$$ (using the counterassumption that $$(A : k_1A : k_1k_2^{-1}A)$$ does not hold). Now, observe that since $$h_2$$ and $$k_2h_2$$ are in the same connected component of $$G \setminus (A \cup k_2A)$$, by translating by $$k_1k_2^{-1}$$ we get that $$k_1k_2^{-1}h_2$$ and $$k_1h_2$$ are in the same connected component of $$G \setminus (k_1k_2^{-1} A \cup k_1A)$$, so we have found a path from $$k_1h_1$$ to $$k_1h_2$$ that avoids $$k_1A$$. From $$(A : k_1A : k_1k_2^{-1}A)$$, we get $$(B_1 : B_2 : B_3) = (k_2^{-1}A : k_2^{-1}k_1A : k_2^{-1}k_1k_2^{-1}A)$$ by translating, as desired.

We have shown $$(B_0 : B_1 : B_2)$$ and $$(B_1 : B_2 : B_3)$$, and by translating by $$k_2^{-1}k_1$$ from the left we get $$(B_i : B_{i+1} : B_{i+2}) \implies (B_{i+2} : B_{i+3} : B_{i+4})$$, so by induction $$(B_i : B_{i+1} : B_{i+2})$$ holds for all $$i$$. By a basic connectivity lemma (see e.g. Cohen's thesis) we have $$(B_i : B_j : B_k)$$ for all $$i < j < k$$. The sequence $$B_{2i}$$ proves that $$k_2^{-1}k_1$$ is axial. End square.

Edit: The axial part had a chirality issue. That's what you get for trying to be fast, I should maybe have drawn a picture. Fixed some typos as well.

• I prove this in two parts above because the first lemma was what I needed myself and the second is just to show you can get an axial from it. But I think the first lemma could be strengthened so axial elements drop more directly, by instead for any $A$ finding $k$ so $h_1$ is connected to $kh_2$ in $G \setminus (A \cup kA)$. The proof should be exactly the same, and then $k$ seems clearly axial if $h_1$ and $h_2$ are in distinct components of $G \setminus A$. – Ville Salo Sep 19 at 10:33