We explore transversals of finite index subgroups of finitely generated groups. We show that when H is a subgroup of a rank-n group G and H has index at least n in G, we can construct a left transversal for H which contains a generating set of size n for G; this construction is algorithmic when G is finitely presented. We also show that, in the case where G has rank n ≤ 3, there is a simultaneous left-right transversal for H which contains a generating set of size n for G. We finish by showing that if H is a subgroup of a rank-n group G with index less than 3 · 2 n−1 , and H contains no primitive elements of G, then H is normal in G and G/H C n 2 .2010 Mathematics subject classification: primary 20F05; secondary 20E99.
We give a uniform construction that, on input of a recursive presentation $P$ of a group, outputs a recursive presentation of a torsion-free group, isomorphic to $P$ whenever $P$ is itself torsion-free. We use this to re-obtain a known result, the existence of a universal finitely presented torsion-free group; one into which all finitely presented torsion-free groups embed. We apply our techniques to show that recognising embeddability of finitely presented groups is $\Pi^{0}_{2}$-hard, $\Sigma^{0}_{2}$-hard, and lies in $\Sigma^{0}_{3}$. We also show that the sets of orders of torsion elements of finitely presented groups are precisely the $\Sigma^{0}_{2}$ sets which are closed under taking factors.Comment: 11 pages. This is the version submitted for publicatio
It is well known that the triviality problem for finitely presented groups is unsolvable; we ask the question of whether there exists a general procedure to produce a non-trivial element from a finite presentation of a non-trivial group. If not, then this would resolve an open problem by J. Wiegold: `Is every finitely generated perfect group the normal closure of one element?' We prove a weakened version of our question: there is no general procedure to pick a non-trivial generator from a finite presentation of a non-trivial group. We also show there is neither a general procedure to decompose a finite presentation of a non-trivial free product into two non-trivial finitely presented factors, nor one to construct an embedding from one finitely presented group into another in which it embeds. We apply our results to show that a construction by Stallings on splitting groups with more than one end can never be made algorithmic, nor can the process of splitting connect sums of non-simply connected closed 4-manifolds.Comment: 16 pages, revised version, minor corrections made, additional results on algorithms to construct embeddings and enumerate subgroups. v3: minor changes to typesetting and theorem numberings, minor corrections mad
We construct a 2-generator recursively presented group with infinite torsion length. We also explore the construction in the context of solvable and word-hyperbolic groups.
Abstract. Let H, K be subgroups of G. We investigate the intersection properties of left and right cosets of these subgroups.If H and K are subgroups of G, then G can be partitioned as the disjoint union of all left cosets of H, as well as the disjoint union of all right cosets of K. But how do these two partitions of G intersect each other? Definition 1. Let G be a group, and H a subgroup of G. A left transversal for H in G is a set {t α } α∈I ⊆ G such that for each left coset gH there is precisely one α ∈ I satisfying t α H = gH. A right transversal for H in G in defined in an analogous fashion. A left-right transversal for H is a set S which is simultaneously a left transversal, and a right transversal, for H in G.A useful tool for studying the way left and right cosets interact, and obtaining transversals, is the coset intersection graph which we introduce here.Definition 2. Let G be a group and H, K subgroups of G. We define the coset intersection graph Γ G H,K to be a graph with vertex set consisting of all left cosets of H ({l i H} i∈I ) together with all right cosets of K ({Kr j } j∈J ), where I, J are index sets. If a left coset of H and right coset of K correspond, they are still included twice. Edges (undirected) are included whenever any two of these cosets intersect, and the edge aH − Kb corresponds to the set aH ∩ Kb.Observing that left (respectively, right) cosets do not intersect, we see that Γ G H,K is a bipartite graph, split between {l i H} i∈I and {Kr j } j∈J .For H a finite index subgroup of G, the existence of a left-right transversal is well known, sometimes presented as an application of Hall's marriage theorem [3]. When G is finite H will have size n, so any set of k left cosets of H intersects at least k right cosets of H (or their union would have size < kn).
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