On ranks of core quandles

In this blogpost, we provide infinitely many counterexamples to a conjecture of Bardakov and Fedoseev [BF24] about minimal generating subsets of core quandles of groups. Specifically, we show that \(\mathrm{rank}(\mathrm{Core}(D_n))=4\) for all \(n\geq 2\). The proof is elementary and eschews the more powerful results of Bergman [Be21].

1. Introduction

In 1982, Joyce [Jo82] and Matveev [Ma82] independently introduced nonassociative algebraic structures called quandles to develop complete invariants of knots. Quandles are idempotent magmas such that every right multiplication map \(y\mapsto y\ast x\) is an automophism. Joyce introduced a particularly important class of involutory quandles called core quandles, which are groups equipped with the operation \[g\ast h:= hg^{-1}h.\] Core quandles were studied extensively in [Be21, SB25].

Given a quandle \(Q\) and a subset \(X\subseteq Q\), define the subquandle generated by \(X\) to be the intersection of all subquandles of \(Q\) that contain \(X\). In analogy with groups, define the rank of \(Q\) to be the minimal size of a subset that generates all of \(Q\). In 2024, Bardakov and Fedoseev [BF24, Question 4.3] posed the question of whether \[\mathrm{rank}(\mathrm{Core}(G))=\mathrm{rank}(G)+1\] for all groups \(G\). A theorem of Bergman [Be21, Thm. 8.1] from 2021 provides a lower bound on \(\mathrm{rank}(\mathrm{Core}(G))\) that disproves this conjecture.

The purpose of this blogpost is to show by elementary means that the conjecture of Bardakov and Fedoseev has infinitely many counterexamples. Specifically, denote the dihedral group of order \(2n\) by \[D_n=\langle r,s\mid r^n=1=s^2,\, rs=sr^{-1}\rangle,\] so in particular \(\mathrm{rank}(D_n)=2\). We show the following.

Theorem 1.1. For all integers \(n\geq 2\), we have \(\mathrm{rank}(\mathrm{Core}(D_n))=4\).

Remark 1.2. Most of Theorem 1.1 can be deduced from the results of Bergman. Specifically, in the statement of [Be21, Thm. 8.2], we have \(n=2\) and \(m=1\), so \(\mathrm{rank}(\mathrm{Core}(D_k))\leq 4\) for all \(k\geq 3\). On the other hand, [Be21, Thm. 8.1] yields the bounds \(\mathrm{rank}(\mathrm{Core}(D_k))\geq 4\) for all even \(k\geq 4\) and \(\mathrm{rank}(\mathrm{Core}(D_k))\geq 3\) for all odd \(k\geq 3\).

2. Upper bound

First, we show that \(\mathrm{rank}(\mathrm{Core}(D_n))\leq 4\) for all \(n\geq 2\). Let \(X\subset D_n\) be the subset \[X:=\{e,r,s,sr\},\] and let \(Q\) be the subquandle of \(\mathrm{Core}(D_n)\) generated by \(X\). We show that \(Q=D_n\) as sets.

Recall that elements of the form \(r^k\in D_n\) are called rotations, while all other elements \(sr^k\in D_n\) are called reflections. For all integers \(k\), note that \[r^k\ast r^{k+1}=r^{k+2}.\] Since \(X\) contains \(e=r^0\) and \(r^1\), induction on \(k\) shows that \(Q\) contains every rotation \(r^k\in D_n\). Moreover, \[(sr^k)\ast(sr^{k+1})=sr^{k+2},\] so a similar argument shows that \(Q\) contains every reflection \(sr^k\in D_n\). Hence, \(Q=D_n\).

3. Lower bound

Conversely, let \(X\subset D_n\) be a subset such that the subquandle \(Q\) generated by \(X\) is \(\mathrm{Core}(D_n)\). To show that \(\# X\geq 4\), we show the following.

Proposition 3.1. \(X\) contains at least two reflections and at least two rotations.

To prove Proposition 3.1, we begin with the following straightforward observation.

Lemma 3.2. For all elements \(g\in D_n\) and integers \(k\), the element \(g\ast (sr^k)\) is a reflection, while \(g\ast r^k\in D_n\) is a rotation.

In particular, Lemma 3.2 shows that \(X\) must contain at least one reflection and one rotation.

Suppose for the sake of contradiction that \(X\) contains only one reflection, say \(sr^k\). We claim that \(sr^{k+1}\notin Q\), which will contradict the assumption that \(Q=D_n\). Indeed, for every rotation \(r^m\in D_n\), we have \[r^m\ast (sr^k)=sr^k r^{-m} sr^k = sr^{k-m}r^{-k}s=sr^{-m}s=r^m,\] so \(sr^k\) acts trivially on rotations. Similarly, \[(sr^k)\ast r^m = r^m sr^k r^m=sr^k,\] so rotations act trivially on \(sr^k\). Of course, \((sr^k)\ast(sr^k)=sr^k\). It follows that \(sr^k\) acts trivially on \(X\) and vice versa, so \(sr^{k+1}\) either lies in the subquandle generated by rotations in \(X\) or does not lie in \(Q\). By Lemma 3.2, the latter must be true.

Now, suppose for the sake of contradiction that \(X\) contains only one rotation, say \(r^k\). Since \(r^k\ast r^k=r^k\), the above calculations show that \(r^k\) acts trivially on \(X\) and vice versa. Similarly to the conclusion of the previous paragraph, Lemma 3.2 implies that \(r^{k+1}\notin Q\), which is a contradiction. Hence, \(X\) contains at least two rotations and at least two reflections, which is what we needed to show. QED.

References

[BF24] V. G. Bardakov and D. A. Fedoseev. Products of quandles, Algebra Logic 63 (2024), no. 2, 75–97. MR4918447

[Be21] G. M. Bergman. On core quandles of groups, Comm. Algebra 49 (2021), no. 6, 2516–2537. MR4255023

[Jo82] D. Joyce, A classifying invariant of knots, the knot quandle, J. Pure Appl. Algebra 23 (1982), no. 1, 37–65. MR638121

[Ma82] S. V. Matveev, Distributive groupoids in knot theory, Mat. Sb. (N.S.) 119(161) (1982), no. 1, 78–88, 160. MR672410

[SB25] F. Spaggiari and M. Bonatto. On core quandles, Algebra Universalis 86 (2025), no. 4, Paper No. 28, 18 pp. MR4951428