2013
DOI: 10.1090/s0002-9947-2013-05908-9
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GGS-groups: Order of congruence quotients and Hausdorff dimension

Abstract: If G is a GGS-group defined over a p-adic tree, where p is an odd prime, we calculate the order of the congruence quotients Gn = G/ StabG(n) for every n. If G is defined by the vector e = (e1, . . . , ep−1) ∈ F p−1 p , the determination of the order of Gn is split into three cases, according as e is non-symmetric, non-constant symmetric, or constant. The formulas that we obtain only depend on p, n, and the rank of the circulant matrix whose first row is e. As a consequence of these formulas, we also obtain the… Show more

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Cited by 41 publications
(67 citation statements)
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“…The next result mimics [4,Lemma 3.2], which applies to GGS-groups. We remark that there are no new exceptions, in addition to the GGS-group…”
Section: 2supporting
confidence: 53%
See 1 more Smart Citation
“…The next result mimics [4,Lemma 3.2], which applies to GGS-groups. We remark that there are no new exceptions, in addition to the GGS-group…”
Section: 2supporting
confidence: 53%
“…We do not know a proof that the GGS-group G in (3.7) is not branch. From properties that were established in [4] we derive the following result. …”
Section: Theorem 38 ([6 Theorem 4])mentioning
confidence: 99%
“…This property will be important for the proofs of Theorem 1 and Theorem 5 but it is also obviously useful for the study of finite quotients of G. In this direction, we point out that in [3] the authors give an explicit formula for the indices |Γ : St Γ (n)| not just for the Gupta-Sidki p-group, but a more general class of groups Γ which act on p-regular rooted trees (GGS groups). They also prove, using different methods, some of the properties stated in the previous lemmas for arbitrary GGS groups.…”
Section: Proposition 24 An Element G ∈ G Is In B If and Only If Thementioning
confidence: 99%
“…Thus in the remainder we assume that e is not constant. By Lemmas 3.2 and 3.4 of [8], we know that G is regular branch over K, where K = γ 3 (G) if e is symmetric and K = G otherwise. Since rst G (n) contains a copy of K × p n · · · × K for every n ∈ N, if we prove that K is not Engel then Theorem C applies to conclude that R(G) = 1.…”
Section: Right Engel Elements In Weakly Branch Groupsmentioning
confidence: 99%