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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) 111-126
Coherent configurations over copies of association schemes of prime order
Reza Sharafdini *
Department of Mathematics, Faculty of Science, Persian Gulf University, Bushehr 75169-13817, Iran.
Mitsugu Hirasaka
Department of Mathematics, College of Natural Sciences, Pusan National University,
Busan 609-735, South Korea.
Received 21 November 2014, accepted 17 May 2016, published online 2 November 2016 Abstract
Let G be a group acting faithfully and transitively on Qi for i = 1,2. A famous theorem by Burnside implies the following fact: If |Qi| = |Q2| is a prime and the rank of one of the actions is greater than two, then the actions are equivalent, or equivalently |(a,3)G| = |Qi| = |Q21 for some (a, ¡3) e Qi x Q2.
In this paper we consider a combinatorial analogue to this fact through the theory of coherent configurations, and give some arithmetic sufficient conditions for a coherent configuration with two homogeneous components of prime order to be uniquely determined by one of the homogeneous components.
Keywords: Coherent configurations, association schemes, prime order, symmetric designs. Math. Subj. Class.: 05C15, 05C10
1 Introduction
A famous theorem by Burnside states that each transitive permutation group of prime degree with rank greater than two is Frobenius or regular. Since any Frobenius group of prime degree is a subgroup of one-dimensional affine group, it follows that such a permutation group is uniquely determined by its rank and degree up to equivalence of group actions. Especially, if a group acts faithfully, transitively but not 2-transitively on each of two sets of the same prime size, then the two actions are equivalent. Let us formulate this fact in the following two paragraphs.
* Corresponding author.
E-mail addresses: sharafdini@pgu.ac.ir (Reza Sharafdini), hirasaka@pusan.ac.kr (Mitsugu Hirasaka)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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Let G be a group acting transitively on Q, for i = 1, 2. Then G acts on Q, x Qj by
(a,|)g = (ag ,lg) for (a,|) G Q, x Qj and g G G,
for all i,j = 1, 2. It is well-known that (e.g., see [6, Lemma 1.6B]) the following are equivalent:
(a)	The action of G on Q1 is equivalent to that on Q2;
(b)	There exists (a, |) G Q1 x Q2 such that Ga = Gp;
(c)	There exists (a, l ) G Qi x Q2 such that |(a,|)G| = |Qi| = |Q21.
Note that the rank of the action of G on Q, is equal to the number of orbits of G on Q, x Q,, and if G acts faithfully on Q,, then G can be identified with a permutation group of Q,.
Suppose that G acts faithfully on Q, with i = 1,2 and |Q11 = |Q21 is a prime. Then, as mentioned in the first paragraph, these actions are equivalent if the rank of one of the actions is greater than two, and so there exists an orbit R of G on Q1 x Q2 such that |R| = | Q11 = |Q21.
In this paper we consider a combinatorial analogy to this fact through the theory of coherent configurations. The concept of coherent configurations was first introduced by Higman who published a series of papers (e.g., [11], [12], [13]) to associate a lot of important criterions with group actions.
Here we define a coherent configuration, its intersection numbers and its fibers according to the notations as in [7].
Definition 1.1. Let V be a finite set and R a partition of V x V. We say that the pair
C = (V, R) is a coherent configuration if it satisfies the following:
1.	The diagonal relation AV is a union of elements of R where we denote {(u,u) | u g U} by Au for a set U.
2.	For each R gR its transpose Rl = {(u, v) | (v, u) G R} is an element of R.
3.	For all R,S,T gR there exists a constant cRS such that
cRS = |R(u) n S4(v)| for all (u, v) G T, where we denote by T (w) the set {z g V | (w, z) G T} for w G V and T gR.
The constants cRS are called the intersection numbers. A subset X of V is called a fiber of C if Ax G R. We denote the set of all fibers of C by Fib(C). By Definition 1.1(i), V is partitioned into the fibers of C, and by Definition 1.1(i),(iii), R is partitioned into
\Rx,y | X,Y g Fib(C) j where Rx,y = [R G R | R C X x
Let U be a union of fibers of C. Then the pair
(U, {R G R | R C U x U^,
is also a coherent configuration, which is denoted by CU.
For R G Rx,y we denote c^t by dR. Then, by two-way counting we have
|R| = d^X| = dRt ^|.	(1.1)
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For X G Fib(C), CX is nothing but an association scheme, i.e., a coherent configuration with only one fiber (see [2] or [20] for its background). For short we shall write RX,X as RX and CX is called a homogeneous component of C.
A general question here is formulated as follows: what can be said about the coherent configuration if its homogeneous components are known. For example, it is a well-known fact that the coherent configuration corresponds to a system of linked block designs if |RX| = 2 for all X G Fib(C). After the seminal Hanaki-Uno theorem on association schemes of prime order (see [10] or Theorem 3.1), it seems quite natural to ask on a possible structure of a coherent configuration each homogeneous component of which is of prime order. The following is our first main result answering to this question:
Theorem 1.2. Let X, Y G Fib(C) such that |X| = |Y| is a prime. Then |RX,Y| G {1, |RX |}. In particular, if |RX,Y | > 1, then
|Rx,y | = |Rx | = |Ry |.
In order to state our second main theorem we need to recall the following observation. Let G be a group acting on a finite set Q. Then G acts on Oxfi componentwise, and an orbit of G on Q x Q is called an orbital (or 2-orbit) of G. We denote the set of orbitals of G by OG. Then it is well-known that CG = (Q, OG) is a coherent configuration, and Fib(CG) is the set of orbits of G on Q. In this sense, a coherent configuration is a combinatorial object to generalize the orbitals of a group action.
Now we assume that C = ( V, R) is a coherent configuration with exactly two fibers X, Y. Then (1.1) proves the equivalence of the first two statements of the following (see [16] for the remaining):
(d)	There exists R G RX,Y such that |R| = |X| = |Y|.
(e)	1 G {dR | R G Rx,y} n {dR | R G Ry,x}.
(f)	C is isomorphic to CX <g> T2 where 7n = ({1,2,... ,n}, {{(i, j )} | 1 < i, j < n} (see Section 2 for the definition of isomorphism and 0).
We notice the following:
(d) is a combinatorial analogy to (c), and such R is a matching between X and Y ; (e) is a simple arithmetic condition on intersection numbers; (f) implies that CX and CY are isomorphic, and C is uniquely determined by CX.
In this paper we aim to obtain the analogous conclusion (d)-(f) to (a)-(c). The following is our second main result to generalize the fact as in the first paragraph under certain arithmetic conditions on intersection numbers:
Theorem 1.3. Suppose that C = ( V, R) is a coherent configuration with exactly two fibers X, Y satisfying
|X| = |Y| isaprime, |Rx,x | > 2 and |Rx,y | > 1.	(1.2)
Then there exists R G RX,Y such that |R| = |X | = |Y | if one of the following conditions X1
holds with k = —--—- :
|Rx,x | - 1
(i) |RX,X | > k2 (k + e — 2) where e is the number of prime divisors of k;
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(ii)	k G {q, 2q, 3q} for some prime power q;
(iii)	k = 4q for some prime power q with 3 { q +1.
Let us show the reason why we exclude the case of |RX,X | = 2. Each symmetric design induces the coherent configuration with exactly two fibers and eight relations (see [14] or [16, Example 1.3]), and if the design is a non-trivial one on a prime number of points, like the Fano plane, then the induced coherent configuration does not satisfy (d)-(f).
Of course, if |Rx,y| — 1, then none of (d)-(f) hold, while C is the direct sum of Cx and CY (see [16] for the definition of direct sum).
Remark 1.4. Applying Theorem 1.3 for CXUY with |X| < 100 we obtain the same conclusion as Theorem 1.3 except for the case (|X|,k) — (71, 35) (see Section 5 for the details).
Suppose that
(|X|, k) — (71,35) and 1 G {dr | R G Rx,y}.	(1.3)
Then by Theorem 1.2, |RX,Y | — 3. The three elements of RX,Y must form three symmetric designs whose parameters (v, k, A) are (71, 35,17), (71,21,6) and (71,15,3), respectively. Though each of such symmetric designs exists (see [1], [3], [5], [9] and [17] or [4, II.6.24,VI.16.30]), it does not guarantee the existence of a coherent configuration satisfying (1.3).
In [14], Higman gave a result to eliminate the case of (|X|, k) — (71,35) as in the previous paragraph. But, the proof given in [14, (3.2)] contains a serious gap, so the result may not be recognized to be true, while we have not found any counterexample. We would be able to disprove [14, (3.2)] if there exists a coherent configuration satisfying (1.3).
In Section 2 we prepare several basic results on intersection numbers and introduce the concepts of complex products and equitable partitions. In Section 3 we give a proof of Theorem 1.2. In Section 4 we give a proof of Theorem 1.3. We add Section 5 for the elimination of coherent configurations on at most 200 points satisfying (1.2).
2 Preliminaries
Throughout this section we assume that C — (V, R) is a coherent configuration. Let Ci — (Vi, Rj) be a coherent configurations, i — 1, 2.
An isomorphism from Ci to C2 is defined to be a bijection ^ : Vl URi —> V2 U R2 such that for all u,v G Vl and R G Rl,
(u,v) G R ^ (V>(u), V(v)) G V(R).
We say that Cl is isomorphic to C2 and denote it by Cl ~ C2 if there exists an isomorphism from Cl to C2. We set
Rl <g>R2 — {Rl < R2 | Rl g Rl, R2 G R2j,
where
Ri < R2 — \ ((ui, U2), (vi,v2^ | (ui, vi) G Ri, (u2,v2) G R2 k
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Then (V1 x V2, R1 ® R2) is a coherent configuration called the tensor product of C1 and C2 and denoted by C1 (g) C2.
Following [20] we define the complex product on the power set of R. For all subsets S and T of R we define the complex product ST of S and T to be the subset
{r gR| 3(S,T ) GSxT ; cRT > o}.
The complex product is an associative binary operation on the power set of R where the proof is parallel to that for association schemes (see [20]). For convenience we shall write S{T}, {S}T and {S}{T} as ST, ST and ST, respectively.
In this paper we need intersection numbers cRS for R e Rx,y, S e RY,Z and T e RX,Z under the assumption |X| = |Y| = |Z|. The following is a collection of simplified equations on such intersection numbers (see [19] or [16, Lemma 2.2] for general formed equations1). For U C R we shall write du instead of ^veu dv.
Lemma 2.1. For all X, Y, Z e Fib(C) with |X| = |Y| = |Z| and all R e Rx,y, S e Ry,z and T e RX,Z we have the following:
1.	dRds =	cRsdr;
T eUx,z
2.	cRSdT = cRSt dR = cRtTds and lcm(dR, ds) | cRSdT;
3.	|{U e R | cRs > 0}| < gcd(dR, ds), i.e., |RS| < gcd(dfl, ds);
4.	|X| = dRx,X = dRx.y.
The following lemmata were proved in [18, Lemma 2.3, Lemma 2.2]2:
Lemma 2.2. For all S,T e RX,Y with |X| = |Y|, we have
SSt n tt 1 C {Ax } if and only if c§tT < 1 for each R e R.
Lemma 2.3. Let Z e Fib(C) such that |Z | is a prime. Then for each R e RZ \ {AZ} we have:
Z1
1. dR = k where k =	•
|Rz |- 1'
2. seRZ css4 = k — 1.
According to [8] or [15] we define an equitable partition of a homogeneous component. Definition 2.4. Let X e Fib(C) and n = {C1, C2,..., Cm} be a partition of X, i.e.,
m
X = J Ci, Ci n Cj = 0 if i = j, and C, = 0 for each i = 1, 2,..., m.
i= i
An element of n is called a cell. We say that n is an equitable partition of CX if, for all
i, j = 1, 2,... ,m and each R e RX, |R(x) n Cj | is constant whenever x e C,.
1 We missed to assume that all fibers of c have the same size at Lemma 2.2 in [16] where the lemma is used only for such coherent configurations in [16].
2Though it is a statement for association schemes, a parallel way to the proof can be applied for balanced
coherent configurations.
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For example, {X} and {{x} | x G X} are equitable partitions of CX. For each Y G Fib(C) and each y G Y we define
ny := {T(y) | T G Ry,x }.	(2.1)
Then ny is an equitable partition of CX, since
|R(x) n S(y)| = cRSt whenever x G T(y).
3 Proof of Theorem 1.2
In [10] Hanaki and Uno proved the following brilliant theorem:
Theorem 3.1. All non-principal irreducible characters of an association scheme of prime order are algebraic conjugate and of degree one.
The following proposition is obtained as a consequence of the previous theorem:
Proposition 3.2. Let C = (V, R) be an association scheme of prime order and n be an equitable partition of C. Then |n| = 1 mod |R| — 1.
Proof. Let A denote the adjacency algebra of C over C. Then the subspace W spanned by the characteristic vectors of the cells in n is a left A-module with respect to the ordinary matrix product. Since A is semi-simple, W is a direct sum of irreducible submodules.
Note that the subspace spanned by the all-one vector is an A-submodule of W affording the principal character, and its multiplicity is one.
Since the character afforded by W is integral valued, it is left invariant from any algebraic conjugate action. It follows from Theorem 3.1 that all non-principal irreducible submodules of W have the same multiplicity, say m. Since
dime(W) = |n| and dimc(A) = |R|,
it follows that
|n| = 1 + m(|R| — 1).
□
Proof of Theorem 1.2. Let C = (V, R) be a coherent configuration with X, Y G Fib(C) such that |X | = | Y | is a prime. Recall that ny is an equitable partition of CX where y G Y. By (2.1), |ny | = |Rx,y |. Then it follows from Proposition 3.2 that
|Rx,y | = 1 mod |Rx | — 1.
Since |Rx,y| < |Rx| (see [13, p.223] or [16, Proposition 2.7]), |Rx,y| G {1, |Rx|}.
Applying the first statement for CY with |RX,Y | < |RY |, we obtain the second statement.
□
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4 Proof of Theorem 1.3
For the remainder of this paper we assume that C = (V, R) is a coherent configuration with
X, Y e Fib(C) such that
m = |X| = |Y| is a prime, r = |RX| > 2 and |Rx,y| > 1. By Theorem 1.2, we have
r = |Rx | = |Rx,y | = |Ry |. For the remainder of this paper we set
m — 1
k = -.
r — 1
By Lemma 2.3(i) the multi-set (dfl | R e Rz ) with Z e {X, Y} coincides with (1, k,..., k) by a suitable ordering. In this section we aim to show that 1 e {dR | R e RX,Y}, which implies that the multi-set (dR | R e RX,Y) coincides with (1, k,..., k) by a suitable ordering, since the complex product SR is a singleton with dSR = ds whenever S e RX and dR = 1 by Lemma 2.1(iii).
Lemma 4.1. For all S, T e RX,Y with S = T we have the following:
(i)	ds ds = ds mod k;
(ii)	ds dT = 0 mod k.
Proof. (i) Applying Lemma 2.1(i) for S and S4 with ds = dst and c^Jft = ds, we obtain that
ds ds = ds + k ^ cSSt.
T eRx,x T= Ax
(ii) Applying Lemma 2.1(i) for S and T4 with dT = dTt and AX e ST4, we obtain
that
ds dT = k	cST t.
T eRv
□
We set
Si := {T e Rx,y | k t dT}, S2 := {T e Rx,y | dT = k} and S3 := {T e Rx,y | k | dT, k < dT}.
Lemma 4.2. Let k = p^1 • • • Pae where pj are the distinct prime divisors of k and are positive integers. Then we have the following:
1.	For each i = 1,..., e there exists a unique S e RX,Y such that pi t dS;
2.	|S11 < e;
3.	kjSsj + ds1 < 1 + k(e — 1).
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Proof. (i) By Lemma 2.1(iv) and Lemma 2.3(i),
m = 1 + (r — 1)k = 1 mod p».
Since m = dRx,Y, there exists an S G RX,Y such that pj \ ds. The uniqueness of such S is a direct consequence of Lemma 4.1(ii).
(ii)	The correspondence given in (i) gives a function from jpi,p2,... ,pe} to Si. It remains to show that this function is onto.
Let S G S1. By the definition of S1, there exists p» such that p^ does not divide ds. By Lemma 4.1(i),
dsds = ds mod k.
Therefore ds(ds — 1) is divided by k. Since ds and ds — 1 are relatively prime, p"i \ ds implies that p» \ ds .It follows from (i) that ds lies in the range of the function.
(iii)	Note that r = |Si | + S | + |Ss| and
3
m = ^ ds = ^dSi > d5l + k|S2| + 2k|S31.
seRx.Y	¿=1
Since k|S21 + k|S3| = k(r — |Si |) and m = 1 + k(r — 1), it follows that
1 + k(|Si| — 1) > dsi + k|Ss|.
By (ii), we have
1 + k(e — 1) > ds1 + k|Ss|. This completes the proof of (iii).	□
Lemma 4.3. We have max{ds | S G RX,Y} < k • min{ds | S G RX,Y}. Proof: Let S, T G RX,Y such that
ds = min{ds | S G RX,Y} and dT := max{ds | S G RX,Y}.
Then T G RS for some R G RX since T G RX S. Applying Lemma 2.1(i) we have
dT < kds. □
For S g RX,Y we define
Us := {R G RX | R4Rn SS1 = {AX}}. Lemma 4.4. For each S G RX,Y we have the following:
1.	r — |US| < (ds — 1)(k — 1).
2.	If R G Us — {AX }, then k divides dT for each T G RS.
3.	If Us S n S2 = 0, then r < ds (k + e — 2).
Proof. (i) Note that
rx —Us = (J {R G rx | Ri G R4R}.
Riess'-{Ax }
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By Lemma 2.1(iii) with c^X > 0,
ISS4 -{Ax }| < ds - 1. It follows from Lemma 2.3(ii) that
|{R e Rx I Ri € R4R}| < £ cR1r = k - 1.
ReR
This implies that
r -|Us| = |Rx -Us|< (ds - 1)(k - 1).
(ii)	It is an immediate consequence of Lemma 2.1(ii) and Lemma 2.2.
(iii)	Suppose that
Us S ns2 = 0.
Then we have
UsS C Rx,y - S2.
It follows from (ii) that
(Us -{Ax })S C S3. By Lemma 4.2(iii) and Lemma 4.3,
dsa < dsk|S3| < ds[1 + k(e - 1) - d5l].	(4.1)
On the other hand, applying Lemma 2.2 and Lemma 2.1(iv) for the first inequality and (i) for the second one,
duss > 1 + (|Us| - 1)k - ds > 1 + [r - (ds - 1)(k - 1) - 1]k. (4.2) Since (Us - {Ax})S C S3,
dUss - ds < d(Us-{Ax})s < dS3 . It follows from (4.1) and (4.2) that
1 + [r - (ds - 1)(k - 1) - 1]k - ds < ds[1 + k(e - 1) - d^],
and hence,
Thus,
r < dS[2 + k(e - 1) - dSl] - k + (ds - 1)(k - 1) + 1.
r < ds[k2 + e - 1 - d|1 + k - 1] - k + 2 - k < ds(k + e - 2).
This completes the proof of (iii).	□
Proposition 4.5. If r > k2 (k + e — 2) where e is the number of prime divisors of k, then 1 e {ds | s g }.
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Proof. We claim that
min{ds | S e RX,Y} < k.
If not, then
1 + k(r — 1) = m ^^ ds > kr,
seUx,v
a contradiction. By Lemma 4.3,
max{ds | S e RX,Y} < k2.
Applying the contraposition of Lemma 4.4(iii) we have Us S n S2 = 0 for each S e Rx,y , and hence, T e RS for some R e Us and T e S2. Since dT = k and cRs = 1 by Lemma 2.2, ds divides k for each S e RX,Y. This implies that |S3| = 0.
We claim S | = 1. Suppose not. Since 1 + (r — 1)k = m = dSl + k(r — |Si |),
1 + k|S1| < k + ^ ds < k + k/2 + k/2+ (|S1| — 2)k,
sesi
a contradiction. By the claim we have S1 = {S} for some S e RX,Y. Since
1 + k(r — 1) = m = k|S2| + ds = k(r — 1) + ds, we have ds = 1. This completes the proof.	□
Lemma 4.6. If S, T e Rx,y with ST4 = {R}, then
cRr > dT for each R1 e SS4 and cRr > ds for each R2 e TT
Proof. Let y e Y, x1,x2 e S4(y) and z e T4(y). Note that (xj,z) e R for i = 1,2 since ST4 = {R}. Since z e T4(y) is arbitrarily taken, we have T4(y) Ç R(x1) n R(x2), which proves the first statement. By the symmetric argument the second statement can be proved.	□
Proposition 4.7. There exist no S, T e RX,Y such that
ST4 = {R}, ds + dT > k + 1 and 1 < ds < dT.	(4.3)
Proof. Suppose that S, T e RX,Y satisfies (4.3).
We claim that SS4 = {AX,R1} for some R1 e RX — {AX}. Suppose not, i.e., SS4 — {AX } has at least two elements R1, R2. By Lemma 2.1(i),
k2 = dfldflt > k + cRRtdfll + cRRdfl2 = k + k + k.
It follows from Lemma 4.6 and ds + dT > k +1 that
k2 > k(k + 2),
a contradiction.
We claim that SS4 n TT4 = {Ax, R1}. Suppose not, i.e., SS4 n TT4 = {Ax}. Then, by Lemma 2.2, cRTt = 1. It follows from Lemma 2.1(i) that k = dR = dsdT, which contradicts ds + dT > k +1 and 1 < ds < dT.
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We claim that R = R4. Suppose not, i.e., R = R4. Then, by Lemma 2.3(ii),
cR1 > cR1 I cR1 > da
cr2r2 - cRflt + cRtfl - dS
k - != E cR1r2 - CRRt + ^R - dS + dT - k + 1,
a contradiction.
We claim that TT4 = {AX,RJ. If R2 G TT4 - {AX,RJ, then cRR - ds by Lemma 4.6 with R = R4. By Lemma 2.1(i),
k2 = dfldR - k + cRRk + cRRk,
which implies that k - 1 + dT + dS, a contradiction to dS + dT - k +1.
We claim that cR^ - dT - 2. By the previous claim, for all zi, z2 G Twith
zi = z2 we have (z1, z2) G R1. Thus,
cRU* = |Ri(zi) n Ri(z2)| - |T4(y) - {zi, Z2}| - dT - 2.
Since Cr1 Rt + cRRt - dT - 2 + dT - k by Lemma 4.6, it follows from Lemma 2.3(ii) that R = Ri. Thus, cRRt = k - 1 since 1 < dS and
S4(y) U T4(y) \ {xi,x2} Ç R(xi) n R(x2) for xi,x2 G S4(y).
Since {AX, R} is closed under the complex product, 1 + k divides |X|. Since |X| is a prime, it follows that {AX, R} = , and hence |RX | = 2, a contradiction.	□
Lemma 4.8. Suppose that k = 4q for some prime power q and 1 G {ds | S G }. Then S1 =0, |Si| = 2, and {ds | S G Si} = {3q, q + 1}.
Proof. By Lemma 4.2(iii) and the assumption, |S3| = 0. By Lemma 4.2(ii), |Si | < 2. Let S G Si. Then, by Lemma 4.1, dS = 1 mod q. By the assumption, 1 < dS < 4q. Since ds < dS1 < 1 + 4q Lemma 4.2(iii), it follows from Lemma 4.1 that
ds G {q + 1, 3q + 1}.
Let T G	with S = T. Since dSdT = 0 mod 4q by Lemma 4.1, q | dT. Since
m = 1 + k(r - 1) = dS1 + dS2 = dS + dT + k(r - 2), we have dS + dT = k + 1. Therefore, we conclude from Proposition 4.7 that {dS | S G Si} = {3q, q + 1}.	□
Proof of Theorem 1.3. (i) is a direct consequence of Proposition 4.5. (ii) Suppose on the contrary that
1 G {ds | S gRx,y}.
Note that e < 2 if k G {q, 2q, 3q} for some prime power q. By Lemma 4.2(iii), |S3| = 0, and dS1 < k + 1. Since
1 + k(r - 1) = ds1 + ds2 < k + 1 + ds2,
we have dS2 - k(r - 2), and, hence, |S2| - r - 2.
Suppose k = q. Then the statement follows from Lemma 4.2(iii) since e =1.
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Suppose k = 2q. Then |Si| < 2 and {ds | S G Si} = {q, q +1} by Lemma 4.2(ii),(iii) and Lemma 4.1. Without loss of generality we assume that
Si = {S, T}, ds = q + 1 and dT = q.
Since q and q +1 are relatively prime, it follows from Lemma 2.1(iii) that ST4 = {R} for some R gR, which contradicts Proposition 4.7.
Suppose k = 3q. Then we have either
{ds | S G Si} = {q, 2q + 1} or {ds | S G Si} = {2q, q + 1}.
The first case is done by Proposition 4.7.
For the last case we assume that Si = {S, T}, ds = q +1 and dT = 2q. By Lemma 2.1(i),(ii), SS4 = {AX,R} for some R G R with R = R4. This implies that k = dR is even since |X | is an odd prime, so q is a power of two. Thus, ds and dT are relatively prime. Therefore, the statement follows from Lemma 2.1(iii) and Proposition 4.7.
(iii) Suppose k = 4q. Then, by Lemma 4.8, {ds | S G RX,Y} = {q, 3q + 1} or {ds | S G Rx,y} = {3q, q + 1}. The statement follows from the assumption and Proposition 4.7.	□
5 Appendix
In this section we show how Theorem 1.3 is applied to small configurations CXUY with |X| = |Y| < 100.
First, we denote by M the set of primes m less than 100. Second, we take the set K of positive integers k such that
k | m — 1 for some m G M with k < m — 1 and
k G {q, 2q, 3q | q is a prime power} U {4q | q is a prime power with 3 \ q +1}. Then K = {20,30, 35,44}. Lemma 5.1. If k = 20, then 1 G {ds | S G Rx,y}.
Proof. Suppose not. By Lemma 4.8, {ds | S G Si} = {15,6}. Let S G Rx,y with ds = 6. By Lemma 2.1(ii), 6 | cfstk for R G SS1 \ {AX}. Thus, 3 | cRSt, which contradicts Lemma 2.1(ii).	□
Lemma 5.2. Suppose that each element of RY = {AY, R, R'} is symmetric and nx = {C1,	} is the equitable partition of (Y, RY) as in Section 2 for x G X .We define
{ßij}i<i,j<3 and {Yj}i<i,j<3
such that ßij = |R(y) n Cj | with y G Ci and Yj := |R'(y) n Cj | with y G Ci. Then we have the following:
1.	For each i we haveYl3=1 ßj = k;
2.	For all i, j with i = j we have ßij + Yij = |Cj |;
3.	For each i we have ßii + Yii = |Ci| — 1;
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4.	For all i, j we have dfaj = ¡3ji\Cj|;
5.	We have ¡11 + ¡22 + ¡33 = k - 1.
Proof. The first four statements can be proved by checking the definition of equitable partitions and using a double-way counting for (Ci x Cj) n R.
Let A be the adjacency algebra of CY and W the subspace spanned by the characteristic vectors of the cells of nx. Then W is a left A-module corresponding to the algebra homomorphism defined by AR ^ (¡ij-), AR ^ (7^).
We claim that W affords the regular character. Let x be the character afforded by W, i.e., the value of the adjacency matrix of R is equal to J23=i ¡ii. Note that the character afforded by W is integral valued but not a sum of principal character. Since dim( W) = 3, it follows that x is the sum of irreducible characters of A. This implies that x is the regular character of A, and, hence, the trace of the matrix (¡ij) is equal to k - 1 by Lemma 2.3(ii) with Lemma 2.1(ii).	□
Proposition 5.3. If (k, m) = (30,61), then 1 G {ds | S G Rx,y}.
Proof. Suppose not.
By Lemma 4.2(ii),(iii), |S11 < 3 and |S3| < 1. if |S3| = 1, then 2k < dS3 < m = 2k + 1, a contradiction. Thus, |S3| = 0.
Since |S21 < 1, it follows from Lemma 4.1 that the following are only possible cases of {ds | S gRx,y}:
{30, 25, 6}; {30,15,16}; {30,10, 21};{15, 36,10};{15, 6,40}.
The first three cases do not occur by Proposition 4.7 since each of them contains a pair of relatively prime numbers.
Note that {|Ci| | i = 1,2,3} = {dS | S G RX,Y} where nx = {C1, C2, C3} as in Lemma 5.2. Without loss of generality we may assume that
Ci = Si(x) for i = 1,2,3.
From now on we shall use Lemma 5.2 many times without mentioning. Suppose that
(|Ci|, |C2|, |Cs|) = (10,15, 36).
Since |C2|£23 = |C3|&2, we have 12 | ¡23. If ¡23 G {0,36}, then |S2S31 = 1, which contradicts Proposition 4.5. Replacing R G RY by R' if necessary we may assume that ¡23 = 24, and hence, ¡32 = 10. Since
|Ci |¡13 = |C3|^31,
we have 18 | ¡13. If ¡13 G {0,36}, then |S3Sf | = 1, which contradicts Proposition 4.7. Thus, ¡13 = 18, and, hence, ¡31 = 5. By Lemma 5.2(i),
¡33 = 15, ¡21 + ¡22 = 6 and ¡11 + ¡12 = 12.
By Lemma 5.2(v), ¡11 + ¡22 = 23. Thus, ¡12 + ¡21 = 19, which contradicts 10^2 = 15y#21. Therefore, (dS, dT, dy) = (10,15,36) does not occur.
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Ars Math. Contemp. 12 (2017) 111-126
Suppose
(|Ci|, |C2|, |Cs|) = (15,6,40).
Since |C2|^23 = |Cs|^32, we have 20 | £23. If £23 € {0,40}, then ^St | = 1, which contradicts Proposition 4.7. We may assume that £23 = 20, and hence, £32 = 3. Since
|Ci|£i2 = |C2|&1,
we have 5 | £21. By Lemma 5.2(i),
£21 + £22 +20 = 30.
Thus, 5 | £22. Replacing R € RY by R' if necessary we may assume that £22 = 5, and hence, £21 = 5 and £12 = 2. By Lemma 5.2(i),(v), we have
£11 + £13 = 28, £31 + fe = 27 and £11 + £33 = 24.
Thus, £13 + £31 = 31, which contradicts 15£13 = 40£31.
This completes the proof.	□
Proposition 5.4. If (k, m) = (44,89), then 1 € {ds | S € Rx,y}.
Proof. Suppose not. By Lemma 4.8, the following is a unique possible case of {ds | S €
Rx,y }:
{12,33,44}.
Without loss of generality we may assume that
Ci = Si(y) for i = 1,2, 3 and (|C1|, |C2|, |C3|) = (12, 33,44).
Since 12£12 = 33£21, £12 € {0,11, 22, 33}. Proposition 4.7 forces £12 € {11, 22}, and we may assume that £12 = 22 by replacing R € RY by R'. Then £21 = 8.
Note that 11 divides £13 and so does £11 by Lemma 5.2(i). We divide our consideration into the following two cases £11 = 11 or 0.
Suppose £11 = 11. Then £13 = 11 and £31 = 3. By Lemma 5.2(i),(v),
£22 + £23 = 36, £32 + £33 = 41 and £22 + £33 = 32.
Therefore, £23 + £32 = 45, which contradicts 33£23 = 44£32. Suppose £11 = 0. Then
£13 = 22 and £31 = 6.
By Lemma 5.2(i),(v),
£22 + £23 = 36, £32 + £33 = 38 and £22 + £33 = 43.
Therefore, £23 + £32 = 34, which contradicts 33£23 = 44£32.
This completes the proof.	□
Lemma 5.5. If (k, m) = (35, 71) then {ds | S € RX,Y} = {15,21,35}.
Proof. Applying Lemma 4.2(i),(iii) and Lemma 4.1 we conclude that {15, 21, 35} is a unique case of {ds | S € RX,Y}.	□
We notice that the lemmata given in this section justify the elimination given in Introduction.
R. Sharafdini and M. Hirasaka: Coherent configurations over copies of association schemes .
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Acknowledgements
First, the authors would like to express gratitude to Professor Akihiro Munemasa for his valuable suggestions in revising Theorem 1.3.
Second, the authors are grateful to anonymous referees for their kind attention to our work, carefully checking and error corrections.
The introduction of the current article was revised according to comments and suggestions of one of the referees. We greatly appreciate his/her contribution helped clarify and improve our article.
The first author was supported by the Persian Gulf University Research Council under the contract No: PGU/FS/49-1/1390/1468.
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