Generators of the free product with amalgamation of two infinite cyclic groups
Generators of the free product with amalgamation of two infinite cyclic groups
复制标题
两个无限循环群合并的自由积生成器
DOI：
10.1007/bf01361856
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发表时间：
1977
影响因子：
1.4
通讯作者：
H. Zieschang
中科院分区：
文献类型：
--
作者：
H. Zieschang
群G的两个有限元集合称为Nielsen等价，如果存在从自由群F到G的同态，使得这些集合是F的两个自由生成元系的映象，这是一个长期悬而未决的问题[6，3.6；16]对于一个群G，两个生成元系是否Nielsen等价。对于有限阶自由阿贝尔群的极小基数生成元集[6，3.5]，这一点早已为人所知。1；T7]对于曲面的基本群[10，12，27，Satz6]和由于这个问题而变得或多或少有趣的其他群[9，13，14，18，20-22]。类似的说法对于有限循环群显然是错误的，但最近出现了有趣的群族，其中一个定义关系也是错误的[1--4，7，8，13，19，20，27，28]。在[1，3]和[13]中，描述了具有无限多个Nielsen等价类的极小基数生成元集的群。推广了[3]，确定了(环结)群(S，TISp=Tq)，p，q>；2，p+q>；4的生成偶的所有Nielsen等价类。在定理5.1中：生成偶的每个Nielsen等价类包含一个且仅有一个系统S“，Tb使得gcd(a，p)=gcd(b，q)=gccl(a，b)=gccl(a，b)=1和0<；2a<；pb，0<；2b<；qa。(等式可能只有当pb或qa等于2时才成立)。
Two finite sets of elements of a group G are called Nielsen equivalent if there is a homomorphism from a free group F to G such that the sets are the images of two systems of free generators of F. It is a question of long standing [6, 3.6; 16] whether for a group G two systems of generators are Nielsen equivalent. This has long been known to be true for sets of generators of minimal cardinality for free abelian groups of finite rank [6, 3.5. 1; t7] for the fundamental groups of surfaces [10, 12, 27, Satz 6] and for other groups which became more-or-less interesting because of this problem [9, 13, 14, 18, 20-22]. The analogous statement is obviously wrong for finite cyclic groups, but recently there have been emerged interesting families of groups with one defining relation for which it is also wrong [1--4, 7, 8, 13, 19, 20, 27, 28].In [1 3] and [13] are described groups with infinitely many Nielsen equivalence classes of sets of generators of minimal cardinality. We extend [3] and determine all Nielsen equivalence classes of generating pairs for the (torus knot) groups (S, TIS p= Tq), p, q> 2, p+ q> 4 in Theorem 5.1: Each Nielsen equivalence class of generating pairs contains one and only one system S", T b such that gcd (a, p)= gcd (b, q)= gccl (a, b)= 1 and 0< 2a< pb, 0< 2b< qa.(Equality may hold only ifpb or qa equals 2.)