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By Ollivier Y.

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So the isomorphism problem for generic one-relator groups reduces to the problem of knowing when an element in the free group is the image of another under some automorphism of the free group. 4): especially, if two elements lie in the same orbit and are of minimal length within this orbit, they can be transformed into each other by means of non-length-increasing Whitehead moves. But for generic elements it can be easily shown that the action of the Whitehead moves increases length except for some trivial cases, so that generic elements do not lie in the same orbit.

Density 1/2 is the case when the cardinal of the set of relators R is more than the square root of the cardinal of the set S of words of length . In particular, this means that we very probably picked twice the same relator in R. e. r1 = wa±1 with |w| = − 1. But i , r2 = waj in the group G = Fm / R , we have by definition r1 =G r2 =G e, so that a±1 =G a±1 i j . Since there are only a finite number of generators, this will eventually occur for every value of i and j and every sign of the exponent, so that in G any generator will be equal to any other and to its inverse, implying that the group has only one or two elements.

We refer to [HV89, BHV, Val02a] for reviews and basic properties. t. some generating set. The neatest ˙ ˙ ´ statement is to be found in [Zuk03], see also [Zuk96, BS97, Pan98, Wan98, Val02a]. Gromov (part 3 of [Gro03]) put this result in a more general context, which allowed Ghys to write a very simple proof [Ghy03, Oll-d]. It happens that in the density model, after suitable manipulations of the presentation, this criterion is satisfied as soon as d > 1/3. ). Theorem 27 – Let d > 1/3 and let G be a random group at density d and at lengths , + 1 and + 2.

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