J Algebr Comb (2015) 42:907–915 DOI 10.1007/s10801-015-0610-5 Finite soluble groups satisfying the swap conjecture Andrea Lucchini1 Received: 8 February 2015 / Accepted: 26 May 2015 / Published online: 9 June 2015 © Springer Science+Business Media New York 2015 Abstract For a d-generated finite group G, we consider the graph d (G) (swap graph) in which the vertices are the ordered generating d-tuples and in which two vertices (x1,...,xd ) and (y1,...,yd ) are adjacent if and only if they differ only by one entry. It was conjectured by Tennant and Turner that d (G) is a connected graph. We prove that this conjecture is true if G is a soluble group satisfying some extra conditions, for example if the derived subgroup of G has odd order or is nilpotent. Keywords Generating graph · Swap conjecture · Soluble groups Mathematics Subject Classification 20D10 · 20F05 · 05C25 1 Introduction Let G be a finite group, and let d(G) be the minimal number of generators of G. For any d integer d ≥ d(G), let Vd (G) ={(g1,...,gd ) ∈ G |g1,...,gd =G} be the set of all generating d-tuples of G. The generating d-tuples γ1 and γ2 aresaidtobeNielsen equivalent if there is a sequence of elementary Nielsen transformations leading from γ1 and γ2. By an elementary Nielsen transformation, here we mean a replacement in γ = ( ,..., ) ∈ ( ) , ( = ) −1 g1 gd Vd G of a component gi by gi g j g j gi i j or by gi .Inthe free group Fn of rank n ≤ d, every γ = (g1,...,gd ) ∈ Vd (Fn) is Nielsen equivalent to ( f1,..., fn, 1,...,1), where ( f1,..., fn) is a fixed free basis for Fn [7]. Thus, the graph whose vertices are the elements of Vd (Fn) and in which two vertices are adjacent if and only if they are Nielsen equivalent is connected. In [9], Tennant and B Andrea Lucchini [email protected] 1 Dipartimento di Matematica, Università degli Studi di Padova, Via Trieste 63, 35121 Padova, Italy 123 908 J Algebr Comb (2015) 42:907–915 Turner introduced the notion of “swap equivalence”, which is wider than the notion of Nielsen equivalence. The d-uples γ1 and γ2 ∈ Vd (G) are said to be swap equivalent if there is a sequence of elementary swaps leading from γ1 to γ2. An elementary swap is thought of as a transformation changing one element of the sequence to an arbitrary element of G. The property of this equivalence relation can be encoded in the “swap graph” d (G): two vertices (x1,...,xd ), (y1,...,yd ) ∈ Vd (G) are adjacent in the swap graph if and only they differ only by one entry. Tennant and Turner proposed the conjecture that d (G) is connected (“swap conjecture”). In [8], it is proved that the free metabelian group of rank 3 does not satisfy this conjecture, but no counterexample is known in the class of finite groups. In [3], it was proved that the conjecture is true if d ≥ d(G) + 1. The case when d = d(G) is much more difficult. In [4], it is proved that the 2-generated finite soluble groups satisfy the swap conjecture and the methods used in the proof of this result are adapted in [3] to prove a weaker version of the swap conjecture in the soluble case, considering the graph d (G) whose vertices are the generating d-tuples and in which two vertices (x1,...,xd ), (y1,...,yd ) are adjacent if and only if there exists i ∈{1,...,d} such that xi = yi . In this paper, we prove that the swap conjecture is true if G is a finite group whose derived subgroup has odd order (so in particular it is true if G itself has odd order). Theorem 1 Let G be a finite group whose derived subgroup has odd order. If d ≥ d(G), then the swap graph d (G) is connected. Indeed, this is a corollary of a more general result. Theorem 2 Suppose that a finite soluble group G has the following property: if A is a nontrivial irreducible G-module G-isomorphic to a complemented chief factor of G, then | EndG (A)| > 2.Ifd≥ d(G), then the swap graph d (G) is connected. Another consequence of Theorem 2 is the following: Corollary 3 Let G be a finite group whose derived subgroup is nilpotent. If d ≥ d(G), then the swap graph d (G) is connected. In particular, by the previous corollary, the swap conjecture is true if G is a finite metabelian or supersoluble group. The proof of Theorem 2 depends on the solution of a combinatorial problem in linear algebra. Denote by Mr×s(F) the set of the r × s matrices with coefficients over the field F.Leta be a nonnegative integer, b a positive integer and let n = a + b. Let A ∈ Ma×n(F) with rank A = a. Moreover, let A be the set of matrices B ∈ Mb×n(F) with the property that A det = 0. B We define a graph A whose vertices are the matrices in A and in which two vertices B1 and B2 are adjacent if and only if they differ only by one column. Theorem 4 If F is a finite field and |F| > 2, then the graph A is connected. Unfortunately, the previous theorem is not true if |F|=2. Consider, for example, the matrix A = 11 with coefficients over the field with two elements. In this case, the graph A consists only of two isolated vertices: B1 = 10 and B2 = 01. 123 J Algebr Comb (2015) 42:907–915 909 2 Proof of Theorem 4 In the particular case, when a = 0, our theorem is equivalent to the following lemma (and no assumption on |F| is needed): Lemma 5 Let V be a vector space of dimension n over the field F and consider the graph whose vertices are the ordered sequences (u1,...,un) of n linearly independent vectors of V and in which two sequences (v1,...,vn) and (w1,...,wn) are adjacent if and only if they differ only by one entry. Then, is a connected graph. Proof First, we prove the following claim: Claim 1 If (u1,...,un) is a vertex of and σ ∈ Sym(n), then there is a path in joining (u1,...,un) and (u1σ ,...,unσ ). In order to prove this claim, it suffices to show that if 1 ≤ i < j ≤ n, then there is a path in between (u1,...,ui ,...,u j ,...,un) and (u1,...,u j ,...,ui ,...,un). However, these vertices are joined by the following path (u1,...,ui ,...,u j ,...,un), (u1,...,ui + u j ,...,u j ,...,un), (u1,...,ui + u j ,...,ui ,...,un), (u1,...,u j ,...,ui ,...,un), proving Claim 1. Next, we prove the following claim. Claim 2 Let {v1,...,vn} and {w1,...,wn} be two bases of the vector space V and sup- pose that {v1,...,vi ,w1,...,wn−i } is a basis of V . Then, there exists j ∈{1,...,i} such that {v1,...,vj−1,vj+1,...,vi ,w1,...,wn−i+1} is a basis of V . In order to prove this claim, suppose that {v1,...,vi ,w1,...,wn−i } is a basis of V . There exists α1,...,αi ,β1,...,βn−i ∈ F with wn−i+1 = αr vr + βsws. 1≤r≤i 1≤s≤n−i There exists j such that α j = 0 (otherwise wn−i+1 would be a linear combination of w1,...,wn−i ) hence v j belongs to the subspace of V spanned by the vectors v1,...,vj−1,vj+1,...,vi ,w1,...,wn−i+1: but then v1,...,vj−1,vj+1,...,vi ,w1,...,wn−i+1 is a generating set of V, and consequently a basis. Now, we can conclude the proof. By Claim 1, it suffices to prove the connectivity of the graph whose vertices are the bases of V and in which two bases are adjacent if and only if their intersection has cardinality n − 1. This follows from iterated applications of Claim 2. ˜ Proof of Theorem 4 Clearly, A and A˜ coincide if A and A have the same row space, so we may substitute A with its reduced row echelon form. Moreover, the graph A 123 910 J Algebr Comb (2015) 42:907–915 ˜ and the graph A˜ are isomorphic if A is obtained from A by applying a columns permutation. So it not restrictive to assume ⎛ ⎞ 10··· 0 a1,a+1 ··· a1,n ⎜ ⎟ ⎜01··· 0 a2,a+1 ··· a2,n ⎟ A := ⎜. ⎟ . ⎝. ··· . ··· ⎠ 00··· 1 aa,a+1 ··· aa,n Let B = B1 ,..., Bn ∈ A (where with Bi we denote the i-th column of B). In the following, we will denote by Ir the identity r × r matrix and by 0r×s the zero matrix in the ring Mr×s(F). First, we prove the following claim: Claim 1 If det Ba+1 ··· Bn = 0, then there is a path in the graph A joining B to the matrix 0b×a Ib . The case a = 0 follows from Lemma 5.Sowemayassumea > 0. Let A∗ be the (a − 1) × (n − 1) matrix obtained by removing the first row and the first column from ∗ A and letB be the b × (n − 1) matrix obtained by removing the first column from A∗ B. If det = 0, then by induction the graph ∗ contains a path B∗ A ∗ C0 = B ,...,Cu = 0b×a−1 Ib . Since ⎛ ⎞ 1 a , ··· a , 1 2 1 n A det ⎝0 A∗ ⎠ = det = 0 0 Ci 0 Ci for each i ∈{0,...,u}, we can conclude that ∗ B, 0 B , 0 C1 , ··· , 0 Cu = 0b×a Ib A∗ is a path in the graph A.