Hyperbolicity of T(6) Cyclically Presented Groups

Abstract

We consider groups defined by cyclic presentations where the defining word has length three and the cyclic presentation satisfies the T(6) small cancellation condition. We classify when these groups are hyperbolic. When combined with known results, this completely classifies the hyperbolic T(6) cyclically presented groups.

Hyperbolic groups of Fibonacci type and T(5) cyclically presented groups

“…The Gilbert-Howie groups H(n, n/2 + 2) [10,Lemma 3.3]) and the corresponding cyclic presentation satisfies the small cancellation condition T(6) when n = 8 or n ≥ 12.…”

“…The non-elementary hyperbolic T(6) groups G n (m, k) were classified in [10]. The non-hyperbolic groups are those in Theorem 2.12, with the remainder being non-elementary hyperbolic, and hence SQ-universal.…”

“…(For conciseness we omit the case m = 0 from Table 1, since this is finite cyclic by Theorem 2.17.) We need the following corollary to results from [10], [27] concerning the T (6) group H(8, 4).…”

“…In all but one of the cases where the group contains a non-abelian free subgroup, we show that it is SQ-universal. (A group G is SQ-universal if every countable group can be embedded in a quotient of G.) We set out the appropriate background precisely in Section 2, but essentially the key prior results are a classification of the hyperbolic Fibonacci groups in [21], [8], a result concerning hyperbolicity of the groups H(n, 3) from [27], and the classification of the hyperbolic T(6) cyclically presented groups given in [10], together with computations using KBMAG [25] and MAF [44]. These results reduce the problem to the case k = sn/p where p ∈ {3, 4, 5} and (s, p) = 1.…”

“…Thus we may assume that w is a non-positive word of length 3, and so G n (w) ∼ = G n (m, k) for some 0 ≤ m, k < n. As in [27,Theorem 10], [10,Table 2], consideration of the girth of the star graph of P n (m, k) shows that the presentation satisfies T(5) if and only if A ± B ≡ 0 mod n and tA ≡ 0, tB ≡ 0 mod n for any 1 ≤ t ≤ 4. Corollary B(d),(e) and Table 1 then classify the T(5) hyperbolic groups G n (m, k) and show that the Tits alternative holds (note that there are no unresolved cases here, since the cyclic presentations of H(9, 4) and H(9, 7) do not satisfy T (5)).…”