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Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 12 (2017) 145-154
Automorphism group of the balanced hypercube*
Jin-Xin Zhou Jin Ho Kwak Yan-Quan Feng , Zhen-Lin Wu
Mathematics, Beijing Jiaotong University, Beijing 100044, P.R. China Received 11 March 2015, accepted 10 March 2016, published online 12 November 2016
Abstract
Huang and Wu in [IEEE Transactions on Computers 46 (1997), pp. 484-490] introduced the balanced hypercube BHn as an interconnection network topology for computing systems. In this paper, we completely determine the full automorphism group of the balanced hypercube. Applying this, we first show that the n-dimensional balanced hypercube BHn is arc-transitive but not 2-arc-transitive whenever n > 2. Then, we show that BHn is a lexicographic product of an n-valent graph Xn and the null graph with two vertices, where Xn is a Zn-1 -regular cover of the n-dimensional hypercube Qn.
Keywords: Automorphism group, balanced hypercube, Cayley graph, arc-transitive. Math. Subj. Class.: 05C25, 20B25
1 Introduction
The hypercube is widely known as one of the most popular interconnection networks for parallel computing systems. As a variant of the hypercube, the balanced hypercube was proposed by Huang and Wu [8] to enhance some properties of the hypercube. An n-dimensional balanced hypercube, denoted by BHn, is defined as follows.
Definition 1.1. For n > 1, BHn has 4n vertices, and each vertex has a unique n-component vector on {0,1,2,3} for an address, also called an n-bit string. A vertex (a0, a\,..., an_i) is connected to the following 2n vertices:
( ((ao + 1)(mod 4), ai,..., aj_i, a*, ai+i,..., an_i), \ ((ao - 1)(mod 4), ai,..., a_i,ai, ai+i,..., an_i),
*This work was supported by the National Natural Science Foundation of China (11271012, 11571035, 11231008) and the Fundamental Research Funds for the Central Universities (2015JBM110). t Corresponding author.
E-mail addresses: jxzhou@bjtu.edu.cn (Jin-Xin Zhou), jinkwak@postech.ac.kr (Jin Ho Kwak), yqfeng@bjtu.edu.cn (Yan-Quan Feng), 10271079@bjtu.edu.cn (Zhen-Lin Wu)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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f ((ao + 1)(mod 4), ai,..., a4_i, (ai + (-1)°° )(mod 4), ai+i,..., an_i), \ ((ao - 1)(mod 4),ai,.. .,ai_i, (ai + (-1)°° )(mod 4),am,.. .,a„_i),
for 1 < i < n — 1.
(0, 0) (3, 0) (0, 3) (3, 3)
(1, 1)
(2, 1) (1, 2) (2, 2)
Figure 1: Two balanced hypercubes: BH1 and BH2
By now, various properties of the balanced hypercube, such as, Hamiltonian laceabil-ity, bipanconnectivity, super connectivity etc. have been extensively investigated in the literature [7, 8, 9, 14, 16, 17, 18, 19]. In many situations, it is highly desired to use interconnection networks which are highly symmetric. This often simplifies the computational and routing algorithms. It has been shown that the balanced hypercube is vertex-transitive and arc-transitive (see [14, 22]). When dealing with the symmetry of graphs, the goal is to gain as much information as possible about the structure of the full automorphism groups. Recently, several publications have been put into investigation of automorphism groups of Cayley graphs having connection with interconnection networks (see, for example, [5, 10, 20, 21]).
In [22], it was proved that BHn is an arc-transitive Cayley graph.
Definition 1.2. For n > 1, let Hn be an abelian group defined as follows:
Hn
(y) x (zi) x (z2) x ... x (z„_i) = Z2 x Z4 x Z4 x ... x Z4.
The generalized dihedral group of Hn, denoted by Dih(Hn), is the semi-direct product of Hn by a group (x) of order 2 with the involution x inverting every element in Hn. Let
Gn = Dih(Hn) = Hn x (x) and let S = {x, xy, xzi, xyzi | i = 1, 2,..., n — 1}. Let rn be the following Cayley graph:
r„ = Cay(G„,S).
Proposition 1.3. [22, Theorem 3.7] For each n > 1, BHn = rn is arc-transitive. Definition 1.4. Let Ln be a subgroup of Hn defined by
(1.1)
Ln
(zi) x (z2) x ... x (zn_i) = Z4 x Z4 x ... x Z4.
i
1
3
2
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Let Tn = Dih(Ln) = Ln x (x). Clearly, Tn is a subgroup of Gn of index 2. Set Q = = 1, 2,..., n — 1}, and define Xn as the following Cayley graph:
Xn = Cay(Tn, Q).	(1.2)
For convenience, in what follows we shall always let rn = BHn. In [3], the authors proved the following result.
Proposition 1.5. [3, Theorem 3.4] For each n > 1, BHn = Xn [2Ki], where Xn is defined as following:
By Proposition 2.1, it is easy to see that Aut(BHn) = Z21 Aut(Xn) *. So, to determine the automorphism group of BHn, the key is to determine the automorphism group of Xn. In this paper, we prove that Xn is a 2-arc-transitive normal Cayley graph, and Aut(Xn) = R(Tn) X Aut(Tn, Q) = Tn X Sn.
As the automorphism group of BHn is known, one may ask: Does BHn have a stronger symmetry property? In this paper, we show that BHn is arc-transitive but not 2-arc-transitive.
As another application, we prove that Xn is a Zn-1-regular cover of the hypercube Qn. This, together with the fact BHn = Xn[2K1], gives a theoretical explanation of the relationship between BHn and Qn.
2 Preliminaries
In this section, we shall introduce some notations, terminology and preliminary results. Throughout this paper only undirected simple connected graphs without loops and multiple edges are considered. Unless stated otherwise, we follow Bondy and Murty [2] for terminology and definitions.
Let n be a positive integer. Denote by Zn the cyclic group of order n, by Sn the symmetric group of degree n and by Kn n the complete bipartite graph of order 2n and valency n, respectively. We also use nKi, Kn and Cn to denote the null graph, the complete graph and the cycle with n vertices, respectively.
In a parallel computing system, processors are connected based on a specific interconnection network. An interconnection network is usually represented by a graph in which vertices represent processors and edges represent links between processors. Let G be a simple undirected connected graph. We denote by Aut(G) the full automorphism group of G, and by V(G) and E(G) the sets of vertices and edges of G, respectively. For u, v e V(G), denote by {u, v} the edge incident to u and v in G. For a vertex v in a graph G, use NG (v) to denote the neighborhood of v, that is, the set of vertices adjacent to v.
*One can also obtain this by using [4, Theorem 5.7]. We thank a referee for pointing out this.
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An s-arc in a graph G is an ordered (s + 1)-tuple (v0, vi,..., vs_i, vs) of vertices of G such that vi_1 is adjacent to vi for 1 < i < s and vi_1 = vi+1 for 1 < i < s - 1. A graph G is said to be s-arc-transitive if Aut(G) is transitive on the set of s-arcs in G. In particular, 0-arc-transitive means vertex-transitive, and 1-arc-transitive means arc-transitive or symmetric. A graph G is edge-transitive if Aut(G) acts transitively on E(G). Clearly, every arc-transitive graph is both edge-transitive and vertex-transitive.
2.1	Wreath products of groups
For a set V and a group G with identity element 1, an action of G on V is a mapping V x G ^ V, (v, g) ^ vg, such that v1 = v and (vg)h = vgh for v e V and g, h e G. The kernel of G acting on V is the subgroup of G fixing V pointwise. For two groups K, H, if H acts on K (as a set) such that (xy)h = xhyh for any x, y e K and h e H, then H is said to act on K as a group. In this case, we use K x H to denote the semi-direct product of K by H with respect to the action.
Let H be a permutation group on a finite set A. For convenience, let A = {1,2, • • • , n}. Let G be a permutation group on a finite set and let
N = G x G x ••• x G.
v-v-'
n times
We define the action of H on N as following:
Vh e H, (g1, g2 • • • , gn)h = (g^-1, g2h-i, • • • , gnh-1), gi e G, i = 1, 2, • • • , n.
Then the semi-direct product of N by H with respect to this action is called the wreath product of G and H, denoted by G ^ H. Clearly,
G 2 H = {(g1, g2, • •• ,gn; h) | gi e G, h e H}.
Moreover, G 2 H can be viewed as a permutation group on $ x A as following:
(x, i) (g1 'Bn,h) = (xgi ,ih).
Let G and H be two graphs. The lexicographic product G[H] is defined as the graph with vertex set V(G) x V(H) and for any two vertices (u1 ,v1), (u2,v2) e V(G) x V(H), they are adjacent in G[H] if and only if either u1 = u2 and v1 is adjacent to v2 in H, or u1 is adjacent to u2 in G. In view of [13, Theorem.], we have the following.
Proposition 2.1. [13, Theorem.] Let X and Y betwographs. Then Aut(X [Y]) = Aut(Y)2 Aut(X) if and only if
(1)	if there are two distinct vertices u, v e V (X) such that NX (u) = NX (v), then Y is connected;
(2)	if there are two distinct vertices u, v e V (X) suchthat NX (u)U{u} = NX (v)U{v}, then the complement Y of Y is connected.
2.2	Cayley graphs
Let G be a permutation group on a set Q and a e 0. Denote by Ga the stabilizer of a in G, that is, the subgroup of G fixing the point a. We say that G is semiregular on Q if
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Ga = 1 for every a e Q and regular if G is transitive and semiregular. Given a finite group G and an inverse closed subset S C G \ {1}, the Cayley graph Cay(G, S) on G with respect to S is defined to have vertex set G and edge set {{g, sg} | g e G, s e S}. A Cayley graph Cay(G, S) is connected if and only if S generates G. Given a g e G, define the permutation R(g) on G by x ^ xg, x e G. Then R(G) = {R(g) | g e G}, called the right regular representation of G, is a permutation group isomorphic to G. It is well-known that R(G) < Aut(Cay(G, S)). So, Cay(G, S) is vertex-transitive. In general, a vertex-transitive graph X is isomorphic to a Cayley graph on a group G if and only if its automorphism group has a subgroup isomorphic to G, acting regularly on the vertex set of X (see [1, Lemma 16.3]).
For two inverse closed subsets S and T of a group G not containing the identity 1, if there is an a e Aut(G) such that Sa = T then S and T are said to be equivalent, denoted by S = T. The following proposition is easy to obtain.
Proposition 2.2. If S and T are equivalent then Cay(G, S) = Cay(G, T).
A Cayley graph Cay(G, S) is said to be normal if R(G) is normal in Aut(Cay(G, S)) (see [15]). Let Cay(G, S) be a Cayley graph on a group G with respect to a subset S of G. Set A = Aut(Cay(G, S)) and Aut(G, S) = {a e Aut(G) | Sa = S}.
Proposition 2.3. [15, Proposition 1.5] The Cayley graph Cay(G, S) is normal if and only if Ai = Aut(G, S), where Ai is the stabilizer of the identity 1 of G in A.
2.3 Covers of graphs
An important tool in studying symmetry properties of graphs is the covering technique. An epimorphism p : X ^ X of graphs is called a regular covering projection if there is a semiregular subgroup CT(p) of the automorphism group Aut(X) of X whose orbits in V (X) coincide with the vertex fibers p-i(v), v e V (X), and the arc and edge orbits of CT(p) coincide with the arc fibers p-i((u, v)), u ~ v, and the edge fibers p-i({u, v}), u ~ v, respectively. In particular, we call the graph XX a regular cover of the graph X. The semiregular group CT(p) is the covering transformation group. If CT(p) is isomorphic to an abstract group N then we speak of XX as a regular N-cover of X. For more results on the covering of graphs, we refer the reader to [6, 12].
Let X be a connected k-valent graph and let G < Aut(X) act transitively on the 2-arcs of X. Let N be a normal subgroup of G. The quotient graph XN of X relative to N is defined as the graph with vertices the orbits of N in V(X) and with two orbits adjacent if there is an edge in X between those two orbits. In view of [11, Theorem 9], we have the following.
Proposition 2.4. If N has more than two orbits in V(X), then N is semiregular on V(X), XN is a k-valent graph with G/N as a 2-arc-transitive group of automorphisms, and X is a regular N-cover of XN.
3 Automorphism group of the balanced hypercube
In this section, we shall determine the full automorphism group of the balanced hypercube. From Proposition 1.5 we know that rn = Xn[2Ki], and by Proposition 2.1, Aut(T„ ) = Z21 Aut(Xn). So, the key step is to determine the automorphism group of Xn.
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Lemma 3.1. For n > 1, Xn is a 2-arc-transitive normal Cayley graph, and furthermore, Aut(Xn) = R(Tn) x Aut(Tn, 0), where R(Tn) = Tn = Dih(Zn-1) and Aut(Tn, 0) =
Sn.
Proof. Clearly, X1 = K2 and X2 = C8. It is easy to see that the statement is true for these two cases. In what follows, assume that n > 3. We first prove the following two claims.
Claim 1 Aut(Tn, 0) = Sn.
Since 0 generates Tn, Aut(Tn, 0) acts faithfully on 0, and hence Aut(Tn, 0) < Sn. It is easy to verify that xz1, z-1zi(2 < i < n - 1), z-1 generate Tn and they satisfy the same relations as x, zj(1 < i < n - 2), zn-1. This implies that the map
a : x ^ xz1, zi ^ z-1zi+1 (1 < i < n — 2), zn-1 ^ z-1,
induces an automorphism of Tn. Clearly, for each 1 < i < n — 2, (xzj)a = xzi+1, and x ^ xz1 and (xzn-1)a = x. This implies that a cyclicly permutates the elements in 0, and so a G Aut(Tn, 0).
Similarly, for each 2 < i < n — 1, we define a map as the following:
Pi : x ^ x, z1 ^ zj, zj ^ z1, zj ^ zj(1 < i,j < n — 1 and i = j).
Then pi induces an automorphism of Tn, and furthermore, Pi interchanges xz1 and xzi and fixes all other elements in 0. Hence, G Aut(Tn, 0) and by elementary group theory, we obtain that the subgroup generated by Pi(2 < i < n — 1) is isomorphic to Sn-1. Since Sn-1 is maximal in Sn, one has (a, p | 2 < i < n — 1} = Sn. It follows that Aut(Tn, 0) = (a, | 2 < i < n — 1} = Sn.
Claim 2 For any xzi, there are (n — 2) 6-cycles in Xn passing through the 2-arc (x, 1, xzi), namely, Cj'j = (1, x, z—1, xziz—1, z—1zi, xzi, 1) with j = i and 1 < j < n — 1.
By Claim 1, Aut(Tn, 0) acts 2-transitively on 0. It is well-known that a vertex-transitive graph is 2-arc-transitive if and only if the vertex-stabilizer Aut(Xn)v is 2-transitive on the set of vertices adjacent to v. So, Xn is 2-arc-transitive. To prove the claim, it suffices to show that the statement is true for the case when i = 1.
First, for any 2 < j < n — 1, one may easily check that C1j = (1, x, z—1, xz1z—1, z1z—1, xz1,1) is a 6-cycle passing through the 2-arc (x, 1,xz1). Let C' be an arbitrary 6-cycle passing through (x, 1,xz1). Then there exist s1,s2,t1,t2 G 0 such that C' = (1, x, s1x, s2«1x = ¿2^1xz1, t1xz1, xz1,1), where S1 = x, s2 = s1, ¿1 = xz1 and ¿1 = ¿2. Clearly, s1 = xzj for some 1 < j < n — 1. In the rest of the proof of Claim 2 the following well-known fact will be frequently used.
Fact Every element in (z1} x (z2} x ... x (zn-1} can be uniquely written in the following form
z?1 z?2 . ..zn--1 G Z4(1 < i < n — 1).
If s2 = x, then xxzjx = t2t1xz1. It follows that zjx = t2t1xz1 and hence zjz1 = t2t1. If t2 = x, then t1 = xzk for some 1 < k < n — 1, and so zjz1 = zk. By Fact, this is impossible. If t2 = xz« for some 1 < i < n — 1, then we have either t1 = x or t1 = xzp for some 1 < p < n — 1. For the former, we have zj z1 = z—1, and for the latter, we have t2t1 = xz«xzp = z-1zp = zjz1. From the above Fact, both of these cannot happen.
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If s2 = xzj for some 1 < i < n — 1, then xz^xzjx = t^xz^ It follows that z—1 zjx = t2t1xz1 and hence z— 1zj-z1 = t211. If t1 = xzk and t2 = xzp for some 1 < k,p < n — 1, then t2t1 = z—1zk = z— 1zj-z1. This is also impossible. If t1 = x and t2 = xzp for some 1 < p < n—1,then t2t1 = z—1 = z— 1zj-z1. This is also impossible. So, we must have t1 = xzk and t2 = x for some 1 < k < n — 1. Then t2t1 = zk = z— 1zj-z1. Clearly, s1 = s2. Then zk = zj and zj = z1. That is s2 = xz1, t2 = x, t1 = s1 = xzj. It follows that C' = C 1'j = (1,x, z— 1,xz1z— 1,z— 1z1,xz1,1).
1
Figure 3: 6-cycles passing through (x, 1, xzj)
Now we are ready to complete the proof. Let A = Aut(Xn) and let A1 be the stabilizer of the identity 1 in A. Let A^ be the kernel of A1 acting on Q. Then A^ fixes every element in Q. For any xzj (1 < i < n — 1), by Claim 2, there are exactly (n — 2) 6-cycles in Xn passing through the 2-arc (x, 1, xzj), namely, Cj,j = (1, x, z— 1, xzjz— 1, z— 1zi, xzj, 1) with j = i and 1 < j < n — 1 (see Fig. (3)). Note that the neighborhood of x is {1, z—1 11 < i < n — 1} and the neighborhood of xzj is {1, zj, z— 1zj | 1 < i, j < n — 1, j = i}. Since there are no 6-cycles passing through z—1, x, 1, xzj and zj, it follows that A1 fixes z—1 and zj (1 < i < n — 1) .
By [3, Lemma 4.2], Xn has girth 6, and so Cj j is the unique 6-cycle passing through z— 1, x, 1, xzj, z— 1zj. As A| fixes z— 1, x, 1 andxzj, A| mustfix z— 1zj. By the arbitrariness of i, j, we obtain that A| fixes every vertex of the set {z— 1,zj,z— 1zj | 1 < i,j < n—1, j = i} which is just the set of vertices at distance 2 from the identity 1. By the vertex-transitivity and connectivity of Xn, A| fixes all vertices of Xn. It follows that A1 = 1, and so A1 acts faithfully on Q. Therefore, A1 < Sn. By Claim 1, Aut(Tn, Q) = Sn, and since Aut(Tn, Q) < A1, one has Aut(Tn, Q) = A1. By Proposition 2.3, Xn is normal, and so A = R(Tn) x Aut(Tn, Q). □
Now we are ready to determine the automorphism group of BHn.
Theorem 3.2. For n > 1, Aut(BHn) = Z21 (Tn x Sn).
Proof. By Proposition 1.5, BHn ^ Xn[2Kij. By Proposition 2.1, Aut(BHn) = Z2 \ Aut(Xn). From Theorem 3.1 we obtain that Aut(Xn) = R(Tn) x Aut(Tn, = Tn x Sn. It follows that Aut(BHn) = Z2 I (Tn x Sn). □
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4 Related results
As the automorphism group of BHn is known, we can obtain more information about the symmetry properties of BHn. By Proposition 1.3, BHn is arc-transitive, and by Theorem 3.1, Xn is 2-arc-transitive. It is natural to ask: whether BHn has much stronger symmetry property? We answer this in negative by showing that BHn is not 2-arc-transitive.
Theorem 4.1. For n > 2, BHn is arc-transitive but not 2-arc-transitive.
Proof. Suppose, by way of contradiction, that BHn is 2-arc-transitive. Recall that BHn = Cay(Gn, S). Then the vertex-stabilizer Aut(BH„)i of the identity 1 of Gn in Aut(BHn) is 2-transitive on S. That is, for any two distinct ordered pairs from S x S, say (u1, v1) and (u2,v2), there exists a e Aut(BHn)1 such that (u1,v1)a = (u2,v2). In particular, there exists a e Aut(BHn)1 such that (x, xy)a = (x, xz1). This implies that x and xz1 have the same neighborhood because x and xy have the same neighborhood. However, from [22, Lemma 3.8], we see that xy is the unique vertex which has the same neighborhood as x, a contradiction. □
By Proposition 1.5, BHn = Xn[2K1]. As a consequence of Theorem 3.1, we can also prove that Xn is a Z^-1-regular cover of the hypercube Qn. This reveals the relationship between the balanced hypercube BHn and the hypercube
Lemma 4.2. For n > 1, let N = Zf. Let G = Cay(N, S) be a connected n-valent Cayley graph. Then G is isomorphic to the n-dimensional hypercube Qn.
Proof. It is well-known that Qn is a Cayley graph on N with respect to the subset
T = {(1,0,0, ••• , 0), (0,1,0, ••• , 0), ••• , (0,0,0, ••• , 1)}.
Viewing N as an n-dimensional vector space on the field Z2, one may see that T is a basis of N. Since G is an n-valent Cayley graph, one has |S | = n, and since G is connected, one has N = (S). This means that S is also a basis of N. So, there is an automorphism of N which maps S to T. By Proposition 2.2, G = Qn, as desired. □
Theorem 4.3. For n > 3, Xn is a Zf-1 -regular cover of Qn.
Proof. By Theorem 3.1, R(Tn) is normal in Aut(Xn). Remember that Tn = Dih(Ln) = Ln x (x), where
Ln = (z1) x ... x (zn-1) = Z4 x ... x Z4,
and x is an involution inverting every element in Ln. Set Z = (R(z2)} x ... x (R(zn_1)).
t,—1 times
is an involution inveiung every elemeutiu Ln. set z — (R^^ ^ /i?^2
Then
Z = Z2 x ... x Z2, v-^-'
n— 1 times
and Z is just the center of R(Tn). It follows that Z is characteristic in R(Tn). Since R(Tn) < Aut(Xn), one has Z < Aut(Xn). Consider the quotient graph Yn of Xn relative to Z. Clearly, Z is semiregular on the vertex-set of Xn, and so it has more than 2 orbits on V(X). Since Xn is 2-arc-transitive, by Proposition 2.4, Yn is also an n-valent graph with
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Aut(Xn)/Z as a 2-arc-transitive automorphism group, and Xn is a Z-regular cover of Yn. To complete the proof, it suffices to prove that Yn = Qn.
Noting that Z < R(Tn) and R(Tn) is regular on V(Xn), R(Tn)/Z is regular on V(Yn). It follows that Yn is a Cayley graph on R(Tn)/Z. As R(Tn) = Dih(Ln), one has R(Tn)/Z = Zn. Since Yn has valency n, by Lemma 4.2, one has Yn = Qn. □
Conclusion
In [14], the authors introduced the balanced hypercube to enhance some properties of the hypercube. Graph symmetry is an important factor in the design of an interconnection network. In 1997, it has been shown that the balanced hypercube is vertex-transitive. Recently, it was shown that the balanced hypercube is also arc-transitive. However, the full automorphism group of the balanced hypercube remained unknown. In this paper, we solve this problem. As applications, we first analyze the symmetry properties of the balanced hypercube and show that the balanced hypercube is not 2-arc-transitive. Then, we give a theoretical explanation of the relationship between the balanced hypercube and the hypercube.
Acknowledgements: The authors are indebted to the anonymous referees for many valuable comments and constructive suggestions.
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