Boundedly Generated Group - Free Groups Are Not Boundedly Generated - Gromov Boundary

Gromov Boundary

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 Fn, 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 Fn 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 wn tends to wg ±∞, then gn·wn tends to g+∞ as n tends to ∞.

If Fn were boundedly generated, it could be written as a product of cyclic groups Ci generated by elements hi. Let X0 be the countable subset given by the finitely many Fn-orbits of the fixed points hi ±∞, the fixed points of the hi 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 (gm) of (gn)

gm = h1n(m,1) ··· hkn(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·wn, the limit of the right hand side applied to x is necessarily a fixed point of one of the conjugates of the hi's. On the other hand, this limit also must be g+∞, which is not one of these points, a contradiction.

Read more about this topic:  Boundedly Generated Group, Free Groups Are Not Boundedly Generated

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