# a long product of elements in hyperbolic group is not a proper power

Let $$G$$ be a hyperbolic group, i.e., there exist $$\delta>0$$ and a finite generating set $$S$$ of $$G$$ such that the Cayley graph $$X$$ of $$G$$ relative to $$S$$ is a $$\delta$$-hyperbolic space. Assume also that $$G$$ is non-elementary (i.e., does not contain a finite-index cyclic subgroup).

In general, given an element $$g\in G$$, we say that $$g$$ is a proper power (in $$G$$) if there exists $$h\in G$$ and an integer $$n>1$$ such that $$h^n = g$$. For example, every torsion element is a proper power. It is a basic fact that every hyperbolic element of $$G$$ is a power of an (hyperbolic) element which is not a proper power.

My question is as follows. Does there exists a word $$w(X,Y)$$ in $$F(X,Y)$$ (the free group on the set of two elements $$\{X,Y\}$$) such that for every non-commuting elements $$r_1,r_2\in G$$, the substitution $$w(r_1,r_2)$$ is not a proper-power in $$G$$? What if we allow to $$w(r_1,r_2)$$ to be torsion (i.e. asking for a word $$w$$ such that [if $$w(r_1,r_2)$$ is hyperbolic then not proper power])?

If somebody has an answer with further restrictions (e.g. torsion-free), then, feel free to add them.

Thank You!

• Is $G$ fixed or can it vary? For example, every word $W\in F(X)$ defines a torsion element of some hyperbolic group. – user1729 Mar 16 '20 at 20:23
• (Specifically, the group $\langle X\mid W^n\rangle$ with $n>1$ is hyperbolic and the element $W$ has order precisely $n$. Neither of these facts are obvious; the element order can be found in the one-relator groups section of the book Combinatorial group theory by Lyndon and Schupp (see also Magnus, Karrass and Solitar's book of the same name), while hyperbolicity follows from the B.B.Newman spelling theorem, which is also in Lyndon and Schupp.) – user1729 Mar 16 '20 at 20:23
• Yes! According to your comment, the required word has to be dependent of the group $G$. Thanks. – Al-Hasan Ibn Al-Hasan Mar 16 '20 at 20:38
• Okay, great. Then the equation $x^py^q=z^r$, $p,q,r$ fixed integers, famously has no non-commuting solutions in free groups (this is due to Lyndon, and I think Schützenberger). Therefore, if $G$ is a free group then $x^2y^2$ is never a proper power, unless $x$ and $y$ commute. – user1729 Mar 16 '20 at 20:53

I'll make my comments into an answer. In short, if we fix the word and vary the group then the answer is "no", while if we fix the group to be an arbitrary torsion-free hyperbolic group then the answer is "yes".

Firstly, if we fix the word and vary the group then the answer is "no". (For references I've given a paper of McCool and Schupp, which has pretty accessible proofs. Lyndon and Schupp's book "Combinatorial group theory" also contains the same results, but I think without proofs and also I don't have it in front of me so can't reference it properly.)

Lemma 1. For all words $$W\in F(\mathbf{x})$$ there exists a hyperbolic group $$G_W$$ such that the word $$W$$ is a proper power in $$G_W$$.

Proof. Fix $$W$$ and take $$G_W:=\langle \mathbf{x}\mid W^n\rangle$$ for some $$n>1$$. Then the word $$W$$ defines an element of order precisely $$n$$ in $$G_W$$ [1, Theorem 2], and is a proper power. Moreover, this presentation is a Dehn presentation [1, Theorem 4], and hence $$G_W$$ is hyperbolic. QED

Secondly, if we fix the group and vary the word then the answer is dependent on the group. The answer is "no" for finite groups (which are all hyperbolic).

Lemma 2. If $$G$$ is a finite group then there exists no word $$W\in F(a, b)$$ such that for every non-commuting elements $$r_1, r_2\in G$$ the substitution $$W(r_1, r_2)$$ is not a proper power in $$G_W$$.

Proof. As $$G$$ is finite, every element is a proper power. QED

Next, the answer is "yes" for torsion-free hyperbolic groups.

Lemma 3. If $$G$$ is a torsion-free hyperbolic group then there exists a word $$W\in F(a, b)$$ such that for every non-commuting elements $$r_1, r_2\in G$$ the substitution $$W(r_1, r_2)$$ is not a proper power in $$G$$. In fact, for all $$m\geq4$$ the word $$W$$ can be chosen to be of length $$\geq m$$.

Proof. Firstly, assume that $$G$$ is free. Lyndon and Schützenberger proved that in a free group, if $$x^i = y^jz^k$$ with $$i, j, k\geq2$$ then the elements pairwise commute [2]. Therefore, taking $$j, k\geq2$$ such that $$j+k=m$$, we have that the word $$W:=y^jz^k$$ is never a proper power, as required.

Next, let $$G$$ be an arbitrary torsion-free hyperbolic group. Then there exists some number $$N\in\mathbb{N}$$ such that for all $$a, b, c$$ pairwise non-commuting elements of $$G$$ and numbers $$p, q, r>N$$ the subgroup $$\langle a^p, b^q, c^r\rangle$$ is free on the given basis (this is standard). Therefore, take both $$j$$ and $$k$$ to be $$\max(m ,N)$$, and suppose $$x^i=y^jz^k$$. If $$i\geq N$$ then the result holds by the above paragraph. Otherwise, take powers to $$N$$ to get $$x^{Ni}=(y^jz^k)^N$$, and note that this identity holds in the subgroup $$\langle x^N, y^j, z^k\rangle$$. Hence, the subgroup $$\langle x^N, y^j, z^k\rangle$$ is not free, and so either $$[x, y]=1$$, $$[y, z]=1$$ or $$[x, z]=1$$. Applying the fact that $$x^i=y^jz^k$$ also holds, and that we are in a torsion-free hyperbolic group so centralisers of elements are cyclic, we have that $$x$$, $$y$$ and $$z$$ pairwise commute, as required. QED

Note that there exists an infinite family of hyperbolic groups for which the equation $$x^iy^jz^k=1$$ has non-commuting solutions [3] (but here $$i, j, k).

[1] McCool, James, and Paul E. Schupp. "On one relator groups and HNN extensions." Journal of the Australian Mathematical Society 16.2 (1973): 249-256.

[2] Lyndon, Roger C., and Marcel-Paul Schützenberger. "The equation $$a^ M= b^ Nc^ P$$ in a free group." The Michigan Mathematical Journal 9.4 (1962): 289-298.

[3] Brady, N., Ciobanu, L., Martino, A. and O Rourke, S., "The equation $$x^{p} y^{q}= z^{r}$$ and groups that act freely on $$\Lambda$$-trees." Transactions of the American Mathematical Society 361.1 (2009): 223-236.

• Given any torsion-free hyperbolic non-elementary group $G$, the quotient $G_\infty=G/G^n$ is infinite for some large odd integer $n=n(G)$. $G^n$ stands for the normal closure of all the $n$-th powers in $G$. Moreover, $G_\infty$ is viewed as the direct limit of a sequence $(G_k)_k$ of hyperbolic groups, where $G_0=G$. We can find two elements $r_1,r_2 \in G$ which do not commute in $G_\infty$. Given a word $W\in F_2$, we have that $W(r_1,r_2)^n$ is trivial in $G_\infty$, thus in some $G_k$. This gives an alternative proof for your Lemma1. – Al-Hasan Ibn Al-Hasan Mar 18 '20 at 12:22
• Note that the second part of the question (allowing $W(r_1,r_2)$ to be torsion, which i added in accordance with your comments) still not answered. Thank You. – Al-Hasan Ibn Al-Hasan Mar 18 '20 at 12:22
• That's a neat proof of the Lemma 1! I missed your edit, but I'll have a think about it now. – user1729 Mar 18 '20 at 12:58
• So if the torsion-free Tarski monster groups (finitely generated non-cyclic groups where every proper subgroup is infinite cyclic) are limits of torsion free hyperbolic groups (which I think they are), then you can alter your proof of lemma 1 to answer this. In particular, every element of a Tarski monster $G_{\infty}$ is a proper power, so the word $W$ corresponds to a proper power in one of the $G_i$. – user1729 Mar 18 '20 at 14:52
• The groups $G_i$ cannot be all torsion-free (otherwise $G_\infty$ is torsion-free). – Al-Hasan Ibn Al-Hasan Mar 18 '20 at 15:00