On the Asphericity of a Family of Positive Relative Group Presentations

Abstract

Excluding four exceptional cases, the asphericity of the relative presentation P= ⟨G, x|x m gxh⟩ for m ≥ 2 is determined. If H = ⟨g, h⟩ ≤ G, then the exceptional cases occur when H is isomorphic to C 5 or C 6 .2010 Mathematical subject classification: 20F05, 57M05

Aspherical Relative Presentations all Over Again

“…Asphericity of one-relator relative presentations has been considered in the articles [12], [6], [27], [43], [23], [3], [4], [2], [1], [65], [28] (1) and in [59], [51], [50]. Those listed at (1) contain classifications of aspherical onerelator relative presentations when the relator has free product length four; that is for relative presentations…”

“…Here we have (|g|, |h|, |gh −1 |) = (2,3,6), and so µ = 1. Using GAP [31] we have that G(Q) is a finite, solvable, group of order 27216 = 2 4 · 3 5 · 7 and derived length 6.…”

Cyclically presented groups with length four positive relators

“…The As in other asphericity studies [12,17], these exceptional cases were unresolved for asphericity status in [2]. In recent work, Aldwaik and Edjvet [1] have shown that Q is aspherical in the previously unresolved cases E4 and E5. Also, Williams [28] has informed us of calculations in GAP [15] showing that the group E = a, u | a 6 , u 3 a 3 ua is finite of order 24 530 688 = 2 8 · 3 4 · 7 · 13 2 with second derived subgroup E ′′ isomorphic to the simple group PSL(3, 3) of order 5616.…”

“…Neither 3 nor 4 is a divisor of j + l, so j + l ≡ 2 or 10. The only arrangements meeting these requirements are (j, l) ≡ (7, 3), (11,3), (1,9), or (5,9). In each of these four cases we have l ≡ 9j so w = x 0 x j x 2j x 9j where j ∈ Z * 12 as in (I12).…”

“…Theorem 4],[1,28]) Consider a relative presentationQ = H, u | u 3 αuβ ,where 1 = α, β ∈ H and suppose that if α, β ∼ = Z 6 , then α = β ±2 and β = α ±2 . Then Q is aspherical if and only if none of the following conditions holds: 1. α = β ±1 has finite order; 2. α = β 2 or β = α 2 and α, β ∼ = Z 4 or Z 5 ; 3. α = β 3 or β = α 3 and α, β ∼ = Z 6 ; 4.…”