Boundedly generated group

Since bounded generation is preserved under taking homomorphic images, if a single finitely generated group with at least two generators is known to be not boundedly generated, this will be true for the free group on the same number of generators, and hence for all free groups. To show that no (non-cyclic) free group has bounded generation, it is therefore enough to produce one example of a finitely generated group which is not boundedly generated, and any finitely generated infinite torsion group will work. The existence of such groups constitutes Golod and Shafarevich's negative solution of the generalized Burnside problem in 1964; later, other explicit examples of infinite finitely generated torsion groups were constructed by Aleshin, Olshanskii, and Grigorchuk, using automata. Consequently, free groups of rank at least two are not boundedly generated.

The symmetric group S_n can be generated by two elements, a 2-cycle and an n-cycle, so that it is a quotient group of F₂. On the other hand, it is easy to show that the maximal order M(n) of an element in S_n satisfies

log M(n) ≤ n/e

where e is Euler's number (Edmund Landau proved the more precise asymptotic estimate log M(n) ~ (n log n)^1/2). In fact if the cycles in a cycle decomposition of a permutation have length N₁, ..., N_k with N₁ + ··· + N_k = n, then the order of the permutation divides the product N₁ ··· N_k, which in turn is bounded by (n/k)^k, using the inequality of arithmetic and geometric means. On the other hand, (n/x)^x is maximized when x = e. If F₂ could be written as a product of m cyclic subgroups, then necessarily n! would have to be less than or equal to M(n)^m for all n, contradicting Stirling's asymptotic formula.

There is also a simple geometric proof that G = SL₂(Z) is not boundedly generated. It acts by Möbius transformations on the upper half-plane H, with the Poincaré metric. Any compactly supported 1-form α on a fundamental domain of G extends uniquely to a G-invariant 1-form on H. If z is in H and γ is the geodesic from z to g(z), the function defined by

${\displaystyle F(g)\equiv F_{\alpha ,z}(g)=\int _{\gamma }\,\alpha }$

satisfies the first condition for a pseudocharacter since by the Stokes theorem

${\displaystyle F(gh)-F(g)-F(h)=\int _{\Delta }\,d\alpha ,}$

where Δ is the geodesic triangle with vertices z, g(z) and h⁻¹(z), and geodesics triangles have area bounded by π. The homogenized function

${\displaystyle f_{\alpha }(g)=\lim _{n\rightarrow \infty }F_{\alpha ,z}(g^{n})/n}$

defines a pseudocharacter, depending only on α. As is well known from the theory of dynamical systems, any orbit (g^k(z)) of a hyperbolic element g has limit set consisting of two fixed points on the extended real axis; it follows that the geodesic segment from z to g(z) cuts through only finitely many translates of the fundamental domain. It is therefore easy to choose α so that f_α equals one on a given hyperbolic element and vanishes on a finite set of other hyperbolic elements with distinct fixed points. Since G therefore has an infinite-dimensional space of pseudocharacters, it cannot be boundedly generated.

Dynamical properties of hyperbolic elements can similarly be used to prove that any non-elementary Gromov-hyperbolic group is not boundedly generated.

Robert Brooks gave a combinatorial scheme to produce pseudocharacters of any free group F_n; this scheme was later shown to yield an infinite-dimensional family of pseudocharacters (see Grigorchuk 1994). Epstein and Fujiwara later extended these results to all non-elementary Gromov-hyperbolic groups.

This simple folklore proof uses dynamical properties of the action of hyperbolic elements on the Gromov boundary of a Gromov-hyperbolic group. For the special case of the free group F_n, the boundary (or space of ends) can be identified with the space X of semi-infinite reduced words

g₁ g₂ ···

in the generators and their inverses. It gives a natural compactification of the tree, given by the Cayley graph with respect to the generators. A sequence of semi-infinite words converges to another such word provided that the initial segments agree after a certain stage, so that X is compact (and metrizable). The free group acts by left multiplication on the semi-infinite words. Moreover, any element g in F_n has exactly two fixed points g^±∞, namely the reduced infinite words given by the limits of gⁿ as n tends to ±∞. Furthermore, gⁿ·w tends to g^±∞ as n tends to ±∞ for any semi-infinite word w; and more generally if w_n tends to w ≠ g^±∞, then gⁿ·w_n tends to g^+∞ as n tends to ∞.

If F_n were boundedly generated, it could be written as a product of cyclic groups C_i generated by elements h_i. Let X₀ be the countable subset given by the finitely many F_n-orbits of the fixed points h_i ^±∞, the fixed points of the h_i and all their conjugates. Since X is uncountable, there is an element of g with fixed points outside X₀ and a point w outside X₀ different from these fixed points. Then for some subsequence (g_m) of (gⁿ)

g_m = h₁^n(m,1) ··· h_k^n(m,k), with each n(m,i ) constant or strictly monotone.

On the one hand, by successive use of the rules for computing limits of the form hⁿ·w_n, the limit of the right hand side applied to x is necessarily a fixed point of one of the conjugates of the h_i's. On the other hand, this limit also must be g^+∞, which is not one of these points, a contradiction.