of the commutator subgroup of a knot group
of the commutator subgroup of a knot group
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结群的换向子群的
DOI:
10.1090/s0002-9939-1971-0275416-9
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发表时间:
1971
期刊:
影响因子:
--
通讯作者:
D. Sumners
中科院分区:
文献类型:
--
作者:
D. Sumners
A short topological proof is given for the well-known theorem that if G is a knot group and G' its commutator subgroup, then H2(G'; Z) =0. The purpose of this note is to give a short topological proof of the following well-known theorem [1], [2 ], [6 ], [7 ]: THEOREM. If G is a knot group and G' is its commutator subgroup, then H2(G'; Z) = 0. PROOF. Let S denote the bounded complement of a tamely embedded S' in S3. S is a compact 3-manifold-with-boundary, and is homotopy equivalent to a finite 2-dimensional simplicial complex K. Let G=w7r(K). As is well known [5], K is aspherical (ri(K) =0, i> 2), hence K is the Eilenberg-MacLane space K(G, 1). Let K denote the infinite cyclic covering space of K; that is, 7r(K) =G' (the commutator subgroup of G), and HI(K; Z) = J(t) (the infinite cyclic multiplicative group generated by t) acts on K as the group of simplicial covering translations. K is also aspherical, and is the EilenbergMacLane space for G'. Let r denote the rational group ring of J(t). Following [3], [4] we have for all q that the simplicial chain groups Cq(k; Q) are finitely generated free r-modules, with generators in 1-1 correspondence with the q-simplexes of K. Since r is a principal ideal domain, then Hq(k; Q) is a f.g. r-module for all q. Now collapsing out the infinite cyclic group of covering translations on k yields the orbit space K. Following Milnor [4 ], this is expressed algebraically by the short exact sequence of chain complexes (as r-modules) (t 1) wihil C*(K; Q) l ) C*(s ; Q) o ec C*f(K; Q) hOmolo which yields the long exact sequence of homology Received by the editors May 24, 1970 and, in revised form, August 14, 1970. AMS 1970 subject classifications. Primary 55A25, 18H10.