By Jonathan A. Hillman

To assault yes difficulties in four-dimensional knot conception the writer attracts on numerous ideas, concentrating on knots in S^T4, whose basic teams include abelian general subgroups. Their classification includes the main geometrically attractive and top understood examples. furthermore, it truly is attainable to use fresh paintings in algebraic the way to those difficulties. New paintings in 4-dimensional topology is utilized in later chapters to the matter of classifying 2-knots.

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Is a A- torsion module, and the subgroup zA is a Then A Z - torsion finite A-submodule. Proof If we tensor A as A finitely generated A-module on which with the field of fractions QCt) of A we ge t 0, and is finitely generated it must be a torsion module. It is clear that is a submodule, and that t -1 acts invertibly on zA. Moreover since A zA is noetherian zA is also finitely generated, and so has finite exponent m as an abelian group. Suppose first that m is prime. D. Fm [t ,t -1], on which t -1 acts invertibly, and so must be finite.

Represent A = Z(HIH 1 acts on H Hand b be a generator for HIH. e. g = bgb- 1 for g in is finitely generated as a A-module. Since H via is Z-torsion free AIIII(H) is principal (H: page 351, and since H we must have H if necessary) HIH ::; Z H has rank 1 ::; N(mt -II) for some m, II has a presentation in Z. Thus (after relabeling . Since we must have m-II = ±1. Now I has a subgroup of finite index which maps onto Z abelian kernel A. Therefore if I with is finitely presentable, this subgroup is a constructible solvable group by (BB 19761 and (BS 19781, and so I ally torsion free.

In particular, " has cohomological dimension at most 2. Moreover, " has a Wirtinger presenta tion of deficiency