ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.79635

J. Algebra Comb. Discrete Appl.
3(2) • 61–80

Received: 5 September 2015
Accepted: 19 January 2016

Journal of Algebra Combinatorics Discrete Structures and Applications

Commutative Schur rings over symmetric groups II:
The case n = 6

Research Article

Amanda E. Francis, Stephen P. Humphries

Abstract: We determine the commutative Schur rings over S6 that contain the sum of all the transpositions in
S6. There are eight such types (up to conjugacy), of which four have the set of all the transpositions
as a principal set of the Schur ring.

2010 MSC: 20C05, 20F55

Keywords: Schur ring, Conjugacy class, Symmetric group, Group algebra

1. Introduction

Given a finite group G it is sometimes possible to characterize the Schur rings over G in some
way. This has been accomplished for the family of cyclic groups [3, 6–9]. In [4] we characterized some
commutative Schur rings over symmetric groups. In this paper we use those results to characterize certain
commutative Schur rings over S6.

For a finite group G and X = {x1, . . . ,xk} ⊆ G, |X| = k, we let X = x1 + · · · + xk ∈ CG. We also
let X−1 = {x−1 : x ∈ X}.

Let C2 denote the class of transpositions in the symmetric group Sn. As a consequence of the main
result of [4] we have:

Corollary 1.1. If S is a commutative Schur ring over S6 containing C2, then S determines (up to
conjugacy) one of the following partitions of C2:

(i) C2 = C2;
(ii) C2 = C1 ∪C2 ∪C3 where

C1 = {(1, 2)}, C2 = {(i,j) : 3 ≤ i < j ≤ 6} and C3 = C2 \ (C1 ∪C2);

Amanda E. Francis (Corresponding Author), Stephen P. Humphries; Department of Mathematics, Brigham
Young University, Provo,UT 84602, U.S.A. (email: amanda@mathematics.byu.edu, steve@mathematics.byu.edu).

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(iii) C2 = C1 ∪C2 where

C1 = {(1, 2), (3, 4), (5, 6)} and C2 = C2 \C1;

(iv) C2 = C1 ∪C2 where

C1 = {(1, 2), (1, 3), (2, 3), (4, 5), (4, 6), (5, 6)} and C2 = C2 \C1;

(v) C2 = C1 ∪C2 where

C1 = {(i,j) : 1 ≤ i < j ≤ 5} and C2 = C2 \C1.

If H ≤ G, then the set of orbits of elements of G under the action of conjugation by elements of H
gives a Schur-ring that we denote S(G,H).

In [4] we found all commutative Schur rings over Sn,n ≤ 5, that contain C2. In this paper we do the
same for S6, this being our main result:

Theorem 1.2. The only commutative Schur rings over S6 containing C2 are (up to conjugacy):
(1) Z(CS6);
(2) S(S6,H120), where H120 = 〈(1, 4)(3, 5), (1, 4, 6, 2, 5, 3)〉∼= S5 is a 3-transitive subgroup;
(3) S(S6,H), where H = A6;

(4) S36 (to be constructed in §4).
(5) S(S6,H), where H = S2 ×S4 ≤ S6;
(6) S(S6,H), where H = S3 oS2 ≤ S6;
(7) S(S6,H), where H = S2 oS3 ≤ S6;
(8) S(S6,H), where H = S5.

Of these, (1)-(4) are those where C2 is a principal set of the Schur ring.

These Schur rings have dimensions 11, 19, 12, 12, 34, 26, 34, 19, respectively

We have shown in [5] that the dimension of a commutative Schur-ring over G is bounded by sG :=∑r
i=1 di, where the di, i ≤ r, are the irreducible character degrees of G. In [5] we showed that some

(specific) groups realize this bound, while others do not. For example, we showed that S3,S4,S5 each
have a commutative Schur-ring of this maximal dimension. We note that sS6 = 76. Then Theorem 1.2
allows us to prove

Corollary 1.3. There is no commutative Schur-ring over S6 of dimension sS6 = 76.

Another consequence of our main result is that there are non-Schurian Schur rings over S6 (see (4)
above).

There is of necessity a certain amount of computation involved in the proof of Theorem 1.2, however
we have tried to restrict such computations to: the Schur-ring algorithm that is described in §2; compu-
tations of Gröbner bases for certain ideals of polynomial algebras; a small number of small enumerations
of possibilities.

All computer computations involved in the preparation of this paper were accomplished using Magma
[1].

2. Schur-rings

We now define the concept of a Schur ring [2, 9–11]:

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A Schur-ring (or S-ring) over a finite group G is a subring S of CG that is constructed from a
partition {Γ1, Γ2, . . . , Γm} of the elements of G: G = Γ1 ∪ Γ2 ∪·· ·∪ Γm, with Γ1 = {id}, satisfying:
(1) if 1 ≤ i ≤ m, then there is some j ≥ 1 such that Γ−1i = Γj;

(2) if 1 ≤ i,j ≤ m, then ΓiΓj =
∑m
k=1 λijkΓk, where λijk ∈ Z

≥0 for all i,j,k.

The Γi are called the principal sets of the S-ring.

An S-ring naturally gives rise to a subalgebra of CG by extending coefficients. We will usually think
of S-rings as C-algebras in this way.

We refer to the survey [9] for an account of recent developments and applications of the theory of
Schur rings.

We recall the following fact (called the Schur-Wielandt principle, see Proposition 22.3 of [11]):

Lemma 2.1. Let S be an S-ring over a group G. If C ⊆ G satisfies C ∈ S and
∑
g∈G λgg ∈ S, then for

all λ ∈ R the element
∑
g∈C δλg,λg is in S; here δ is the Kronecker delta function i.e. δx,y = 0 if x 6= y

and is 1 otherwise.

S-ring Algorithm: Suppose that we have a subalgebra H of CG (|G| < ∞) and we wish to find the
smallest S-ring that contains H. Suppose that we start with a (ring) generating set c1, . . . ,cr for the
subalgebra H ⊂ CG. For each ci partition the elements of G according to their coefficients in ci. For
each such subset C of this partition add C and C−1 to your set of generators; do this for each i and
consider this new set of generators, that we will denote by d1, . . . ,dt. We simplify the set d1, . . . ,dt by
eliminating any C-linear dependences. Now the di determine and are determined by subsets of G, so that
t ≤ |G|. Now consider the products didj, 1 ≤ i,j ≤ t. Again we partition the terms of didj according to
their coefficients, and add in C,C−1, for all sets C in the partition. Simplify this generating set using
any linear dependences. This describes the basic step. If, after this basic step, one has a C-basis for an
S-ring, then we are done; otherwise we repeat the basic step. The process is guaranteed to terminate
since |G| < ∞. It is easy to write a program to implement this algorithm (in, say, Magma [1]).

3. Covers of complete graphs

Let Kn be the complete graph on n vertices. Then for λ ∈ N and a graph P, a cover of λKn by P’s
is a set T of subgraphs P1, . . . ,Pm of Kn, each of which is isomorphic to P , and such that every edge of
Kn occurs λ times in P1, . . . ,Pm. We will need the following result in §4.

Lemma 3.1. Any cover T of λK6 by r distinct triangles must have 5|r. Further, the cases r = 5, 15 do
not happen, and the case r = 10 is unique up to a permutation by an element of S6:

{{1, 2, 3},{1, 2, 4},{2, 4, 5},{1, 3, 5},{1, 5, 6},{3, 4, 5},{3, 4, 6},{2, 5, 6},{2, 3, 6},{1, 4, 6}}.

The situation r = 20 is where we have all the triangles in K6.

Proof. Let T be a set of r triangles giving the cover of λK6. Then counting edges we see that 3r = 15λ,
so that 5|r. Further, the number of triangles in K6 is 20. If r = 15, then taking the complement of T in
the set of all triangles gives the case r = 5. Thus we consider the cases r ∈{5, 10}, since if r = 20, then
we have all the triangles.

Case A: r = 5. Here λ = 1. Then (by permuting if necessary) we can assume that {1, 2, 3},{1, 4, 5}∈ T.
Considering the edge {2, 4} one is forced (since λ = 1) to have {2, 4, 6} ∈ T. But now one checks that
the edge {2, 5} cannot be in any triangle. Thus this case does not arise.
Case B: r = 10. Here λ = 2. We consider two cases:

Case B1: there is some K4 ⊂ K6, all of whose triangles are in T. Here we can assume that the
vertices of the K4 are 1, 2, 3, 4. Then one sees that the only possibilities for extra triangles in T are

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

{i, 5, 6}, i = 1, 2, 3, 4 (otherwise we have edges with multiplicities greater than 2). But this makes it
impossible to cover all edges and have the multiplicity of the edge {5, 6} be 2.

Now assume that Case B1 does not happen.

Case B2: there is some K4 ⊂ K6, all but one of whose triangles are in T. So assume that
{1, 2, 3},{1, 2, 4},{1, 3, 4} ∈ T. Then, considering the edge {2, 4} we see that, up to permuting 5, 6, this
forces {2, 4, 5}∈ T. Considering the edge {2, 5} forces either (a) {2, 5, 6}∈ T or (b) {2, 3, 5}∈ T.

Assume that we have (a) {2, 5, 6}∈ T. Considering the edge {1, 6} forces {2, 3, 6}∈ T. Considering
the edge {2, 6} forces {1, 5, 6}∈ T, and that this is the only triangle that can be in T that contains {1, 6}.
Thus this case cannot happen.

If we have (b) {2, 3, 5}∈ T, then considering the edge {2, 6} gives a contradiction.
Now assume that Case B1 and Case B2 do not happen. Then we can assume that {1, 2, 3},

{1, 2, 4} ∈ T, {1, 3, 4},{2, 3, 4} /∈ T. Considering the edge {2, 4} forces {2, 4, 5} ∈ T (up to permut-
ing 5, 6). Considering the edge {1, 5} (and using the fact that Case B2 does not occur) we must have
{1, 3, 5},{1, 5, 6} ∈ T. Considering the edge {3, 4} (and using the fact that Case B2 does not occur)
we must have {3, 4, 5},{3, 4, 6} ∈ T. Similarly, considering the edge {5, 6} we must have {2, 5, 6} ∈ T.
Considering the edge {2, 6} we must have {2, 3, 6}∈ T. This leaves {1, 4, 6} as the remaining triangle.

Let C3 denote the class of 3-cycles in Sn. In general for µ ` n we let Cµ denote the class of elements
of Sn of cycle type µ.

To each element (i,j,k) ∈ C3 there is associated the triangle {i,j,k} in K6. We let the following
subset of C3 represent the set of triangles in the above Lemma:

C3 = {(1, 2, 3), (1, 2, 4), (2, 4, 5), (1, 3, 5), (1, 5, 6), (3, 4, 5), (3, 4, 6), (2, 5, 6), (2, 3, 6), (1, 4, 6)}.

The set of triangles corresponding to the elements of C3 can be represented as the set of triangles
of the hemi-icosahedron H (a polyhedral decomposition of the projective plane). See Figure 1, where we
have drawn the ten triangles (2-simplices) of H, and the outside edges of these 2-simplices are identified
in pairs as usual.

1

2

6

5

4

3

4

5

6

Figure 1. The hemi-icosahedron corresponding to C3

The automorphism group of this 2-complex is A6 ≤ S6, and it acts transitively on the ten triangles
corresponding to the elements of C3.

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4. The case where C2 is a principal set

In this section we assume that C2 is a principal set of a commutative S-ring S. This is the most
difficult case. Since C2 ∈ S it follows from Lemma 4.1 of [5] that Z(CG) is a subring of S.

It follows from this and the Schur-Wielandt principle that if C is a principal set of S, then C is
contained in some conjugacy class, namely the class of one of its elements.

The class C3.
We first consider the principal sets C ⊆C3 of S, with the goal of showing C = C3.
For µ ` n and α =

∑
g∈G λgg we let

αµ =
∑
g∈Cµ

λgg.

Lemma 4.1. If C ⊆ C3 ⊂ Sn is a principal set of S and
(
C · C2

)
(2,1n−2)

= λC2, then C determines a
cover of λKn by triangles. Moreover, we have

3 · |C| = λ · |C2|.

Proof. We have

(i,j,k) · ((i,j) + (j,k) + (i,k)) = (i,j) + (j,k) + (i,k).

Thus each 3-cycle (i,j,k) ∈ C contributes (i,j) + (j,k) + (i,k) to the product C · C2.
Further, for each α = (i,j,k) ∈ C3, there are only three β ∈ C2 (namely β ∈ {(i,j), (j,k), (k,i)})

with αβ ∈ C2. Since each (i,j) ∈ C2 occurs λ times in C · C2 we have a cover of λKn by triangles, and
the first part of the result follows. By counting edges we see that 3 · |C| = λ · |C2|.

If n = 6, then Lemma 3.1 together with the fact that either C = C−1 or C ∩C−1 = ∅, shows that
(up to conjugacy) we have one of:

(I) C = C3;
(II) |C| = 20 where C =

∑
α∈C3 α + α

−1;

(III) |C| = 10 where C =
∑
α∈C3 α

ε(α) with ε(α) ∈{1,−1};

(IV) C =
∑

1≤i<j<k≤6(i,j,k)
ε(i,j,k), for some ε(i,j,k) ∈{1,−1}.

We want to show that (I) is the only possibility. If we have (II), then applying the S-ring algorithm
to the algebra generated by Z(CS6) and C gives an S-ring that is not commutative. Thus (II) does not
happen.

Suppose that we have (III), so that C =
∑
α∈C3 α

ε(α) with ε(α) ∈{1,−1}.
We will use Figure 1 to determine a standard orientation (counter clockwise) for each of the triangles

of C3, so that, for example the order 1, 2, 3 determines a positive orientation for the triangle {1, 2, 3} in
C3.

Since we have (III), every (i,j,k) ∈ C has the form αε(α), for some α ∈ C3. Each such αε(α) =
(i,j,k) ∈ C determines a triangle t(αε(α)) = {i,j,k} of H. In fact αε(α) = (i,j,k) ∈ C determines
an orientation for t(αε(α)), this being i 7→ j 7→ k 7→ i. We denote the set of oriented triangles of H
determined by C as H(C). Further, each oriented triangle determines an orientation on each of its edges.

An edge of H will be said to be oppositely-oriented in H(C) if that edge has different orientations in
the two triangles of H(C) in which it is contained. Let o(C) denote the number of oppositely-oriented
edges in H(C). We note that o(C) < 15, since H has 15 edges.

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Lemma 4.2. (a) If (i,j,k), (r,s,t) ∈C3, then (i,j,k)(r,s,t) ∈C3 if and only if either
(i) (i,j,k) = (r,s,t); or

(ii) {i,j,k}∩{r,s,t} = {u,v} has size 2 and the edge u,v in (i,j,k) is oriented differently from that in
(r,s,t).

(b) Two (oriented) triangles of H(C) that share an edge will contribute two elements of C3 to C
2
if

and only if their common edge is oriented differently in the two triangles, if and only if the two triangles
are oriented the same. In any other case they contribute no elements of C3.

(c)
(
C

2
−C−1

)
(3,13)

is a sum of 2o(C) elements of C3.

Proof. (a) We may assume that i,j,k,r,s,t ≤ 6, and one checks a small number of cases.

(b) Now applying Lemma 4.2 (a) to C
2
|(3,13) we see that there is the contribution that comes from∑

α∈C α
2. We note that none of these elements α2 of C3 are in C, since C∩C−1 = ∅. Any other product

comes from a pair α1,α2 ∈ C that satisfy Lemma 4.2 (a) (ii). Thus α1,α2 determine an orientation of
two of the triangles of H that have opposite orientations on the common edge of these two triangles.

So considering H(C) we see that two triangles of H(C) that share an edge will contribute two elements
of C3 to C

2
if and only if their common edge is oriented differently in the two triangles, if and only if the

two triangles are oriented the same.

(c) This follows from (b) and its proof.

Let Ω(C) be the subgraph of the 1-skeleton H(C)(1) of H(C) consisting of edges that are oppositely-
oriented. Let f+,f− be the number of triangles of C3 that are oriented positively or negatively (respec-
tively) in H(C). Then f+ + f− = 10 and we may assume that f+ ≥ f− (else replace C by C−1).

Lemma 4.3. (a) The graph Ω(C) contains all the vertices of H(C)(1).

(b) o(C) ∈{5, 10}.
(c) f+ ≡ f− ≡ o(C) mod 2. In particular,

if f− = 0, then o(C) = 12;

if f− = 1, then o(C) ∈{9, 11};
if f− = 2, then o(C) ∈{6, 8, 10};
if f− = 3, then o(C) ∈{5, 7, 9};
if f− = 4, then o(C) ∈{6, 8, 10};
if f− = 5, then o(C) ∈{3, 7};
if f− = 6, then o(C) = 6.

(d) If o(C) = 5, then in Figure 2 we show the two possibilities for C (up to a permutation of 1, . . . , 6)
where we have shaded the positively (say) oriented triangles, and where the oppositely oriented edges are
drawn dashed.

Proof. (a) This follows, since every edge of H(C)(1) has degree 5, so that for every vertex v of H(C),
at least two adjacent triangles that share the vertex v must have the same orientation in H(C).

(b) We know that every principal set contained in C3 has size a multiple of 5. Lemma 4.2 (c) shows
that 10|o(C). But o(C) < 15 then gives the result.

(c) Changing the orientation of any triangle of H changes the orientation of the three edges of that
triangle. It follows that f+ ≡ f− ≡ o(C) mod 2. The rest follows by looking at each particular case.

(d) We use (c) to check this.

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1

2

6

5

4

3

4

5

6

1

2

6

5

4

3

4

5

6

Figure 2. Orientations of H(C) giving o(C) = 5.

Thus Case (III) has been reduced to showing that the two S-rings determined by Lemma 4.3 (d)
do not give commutative S-rings. One checks this latter fact using the S-ring algorithm. This concludes
consideration of (III).

Now assume that we have (IV), so that any subset of C3 that is a principal set has size 20. Thus
C−1 6= C is also a principal set. Thus we have

C =
∑

1≤i<j<k≤n

(i,j,k)ε(i,j,k) where ε(i,j,k) ∈{1,−1}.

Lemma 4.4. Let a,b,c ∈C3. Then at most one of the eight products a±1b±1,b±1a±1 can be equal to c.

Proof. (a) We may assume that c = (1, 2, 3), and then one only has to check the cases where a,b ∈
C3 ∩S4.

For c = (i,j,k) ∈ C we now wish to determine the coefficient of c in C
2
as a polynomial function of

the ε(r,s,t).

First assume that ε(i,j,k) = 1. Suppose a = (r,s,t)δ1, b = (u,v,w)δ2, are distinct where δ1 =
δ1,i,j,k,r,s,t,u,v,w,δ2 = δ2,i,j,k,r,s,t,u,v,w ∈ {1,−1}, r < s < t,u < v < w, and c is one of ab,ba. Then by
Lemma 4.4 (a) the coefficient of c in (a + b)2 is 1. Thus if a,b ∈ C, then they will contribute 1 to the
coefficient of c in C

2
. So when a = (r,s,t)ε(r,s,t),b = (u,v,w)ε(u,v,w) ∈ C are distinct, then they will

contribute 1 to the coefficient of c in C
2
if and only if ε(r,s,t) = δ1,ε(u,v,w) = δ2.

Thus we have proved (i) of the following result:

Lemma 4.5. (i) When c = (i,j,k) ∈ C, the coefficient of c = (i,j,k) in C
2
coming from the pair

(r,s,t)ε(r,s,t), (u,v,w)ε(u,v,w) is

(1 + δ1ε(r,s,t))

2
·

(1 + δ2ε(u,v,w))

2
.

(ii) When c = (i,j,k) /∈ C, the coefficient of c in C
2

coming from the pair
(r,s,t)ε(r,s,t), (u,v,w)ε(u,v,w is

(1 −δ1ε(r,s,t))
2

·
(1 − δ2ε(u,v,w))

2
.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

(iii) For any c = (i,j,k)ε(i,j,k), 1 ≤ i < j < k ≤ 6, the coefficient of c in C
2
coming from the pair

a = (r,s,t)ε(r,s,t),b = (u,v,w)ε(u,v,w is

(1 + δ1ε(r,s,t)ε(i,j,k))

2
·

(1 + δ2ε(u,v,w)ε(i,j,k))

2
.

Proof. The proof of (ii) is similar to what we did above for (i), and (iii) is a restatement of (i) and
(ii).

Now let P be a polynomial ring over Q with generators X(i,j,k) where 1 ≤ i < j < k ≤ 6 and
define, for c = (i,j,k)ε(i,j,k),a = (r,s,t)ε(r,s,t),b = (u,v,w)ε(u,v,w),a 6= b, where c is one of ab,ba. We
define the polynomial

S(a,b,c) =
(1 + δ1X(r,s,t)X(i,j,k))

2
·

(1 + δ2X(u,v,w)X(i,j,k))

2
.

Here δ1 = δ1,i,j,k,r,s,t,u,v,w,δ2 = δ2,i,j,k,r,s,t,u,v,w ∈ {1,−1}, r < s < t,u < v < w, as in the above. Thus
the X(i,j,k) correspond to the ε(i,j,k). We let Eε : P → Q be the evaluation map X(i,j,k) 7→ εi,j,k.

For each c = (i,j,k)ε(i,j,k), i < j < k, we let S(c) be the sum of all S(a,b,c) where a =
(r,s,t)ε(r,s,t),b = (u,v,w)ε(u,v,w) ∈ C are distinct and one of ab,ba is c. Thus S(c) is a sum of nine
of the S(a,b,c)s. It is clear that if c = (i,j,k)ε(i,j,k), i < j < k, and c′ = (i′,j′,k′)ε(i

′,j′,k′), i′ < j′ < k′,
then Eε(S(c)) = Eε(S(c′)), since C is a principal set.

We now define the ideal I whose generators are all S(c)−S(c′), for c,c′ as above, and all X(i,j,k)2−1
for 1 ≤ i < j < k ≤ 6. Since we can always conjugate C so that ε(1, 2, 3) = 1 = ε(4, 5, 6) we also let
X(1, 2, 3) − 1,X(4, 5, 6) − 1 be in I.

Now if there is such a subset C, then the ideal I will not be equal to P . However a Gröbner basis
calculation ([1]) shows that I = P , and so (IV) is not possible. Thus we now have:

Proposition 4.6. Any commutative S-ring over S6 that contains C2 as a principal set, also contains C3
as a principal set.

The class C(22,12).
We let C2,2 denote C(22,12). Here we have:

Lemma 4.7. If C ⊂C2,2 satisfies C ∈ S, where S is a commutative S-ring over Sn containing C2 as a
principal set, then C determines a cover of λKn by pairs of disjoint edges.

Proof. This follows from

[(i,j)(k,m) · C2](2,1n−2) = [(i,j)(k,m) · [(i,j) + (k,m) + · · · ]](2,1n−2)
= (i,j) + (k,m),

together with the fact that C2 is a principal set, so that [C · C2](2,1n−2) = λC2.

In the above situation we have: 2|C| = λ|C2|.
For each of n = 4,λ = 1; n = 5,λ = 1, and n = 5,λ = 2, one can check that there is a unique

such cover. For n = 6 we have 2|C| = 15λ, so that λ must be even, and 15 divides |C|. We note that
|C(22,12)| = 45, so that we can assume |C| < 45. If |C| = 30, then we replace C by C(22,12) \C, so that we
can assume |C| = 15. We now show

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Lemma 4.8. For C ⊂C(22,12), |C| = 15, as above, the only possibilities for C are those conjugate to the
following element (or its complement in C(22,12))

(1, 2)(3, 4) + (1, 2)(5, 6) + (3, 4)(5, 6)

+ (1, 3)(2, 5) + (1, 3)(4, 6) + (2, 5)(4, 6)

+ (1, 5)(2, 4) + (1, 5)(3, 6) + (2, 4)(3, 6)

+ (1, 6)(2, 3) + (1, 6)(4, 5) + (2, 3)(4, 5)

+ (1, 4)(2, 6) + (1, 4)(3, 5) + (2, 6)(3, 5).

The stabilizer of this element (under conjugation by elements of S6) is

H120 = 〈(1, 4)(3, 5), (1, 4, 6, 2, 5, 3)〉,

so that [S6 : H120] = 6, and there are 6 conjugates of this element.

Further, the S-ring generated by Z(CG) and the above element is S(S6,H120).

Proof. The last two paragraphs are simple computations once we have the first part.

For the first part, let Ken denote the graph with vertices the edges {i,j} of Kn, and where we have an
edge of Ken whenever the corresponding vertices (edges of Kn) are disjoint. Thus each edge {i,j}−{k,m}
of Ken corresponds to an element (i,j)(k,m) ∈ C(22,1n−4), so that we can refer to an edge of Ken by that
element.

Now let C ∈ C(22,12) determine a cover of 2K6 by subgraphs isomorphic to K2 ∪ K2. Then every
vertex of Ke6 is in exactly two edges. Thus C determines a set of disjoint cycles Γ1, . . . , Γr in K

e
6. Each

cycle Γi, i ≤ r, has length at least three. We will think of the Γi as subsets of C. We have:

Lemma 4.9. Let n = 6. (i) If {i,j}−{k,m}−{r,s} represent three consecutive vertices in a cycle Γu,
then (up to a permutation) we have one of

(a) {1, 2}−{3, 4}−{5, 6};
(b) {1, 2}−{3, 4}−{1, 5}.

(ii) If {i,j}−{k,m}−{r,s}−{u,v} represent four consecutive vertices in a cycle Γu, then (up to
a permutation) we have one of

(a) {1, 2}−{3, 4}−{1, 5}−{2, 6};
(b) {1, 2}−{3, 4}−{1, 5}−{2, 4};
(c) {1, 2}−{3, 4}−{1, 5}−{3, 4};
(d) {1, 2}−{3, 4}−{5, 6}−{3, 6}.

Further, in a path Γu of length at least 4 with distinct non-consecutive edges {i,j}− {k,m} and
{i′,j′}−{k′,m′}, the product (i,j)(k,m) · (i′,j′)(k′,m′) is not in C(22,12).

(iii) If Γi, Γj are distinct cycles, then no term of Γi · Γj is in C(22,12).
(iv) If Γi has length greater than 4, then the number of elements of the form αβ, (α,β ∈ Γi), that

are in C(22,12) is exactly |Γi|, and none of these elements are in C. If Γi has length 4, then the number
of elements of the form αβ, (α,β ∈ Γi), that are in C(22,12) is 2, and none of these elements are in C.

Proof. (i) In such a path {i,j}−{k,m}−{r,s} we have |{r,s}∩{i,j}| = 0, 1, which give the two
cases.

(ii) The first part is easy to check using (i). If all the vertices of {i,j}−{k,m},{i′,j′}−{k′,m′}
are distinct, then one sees that (i,j)(k,m) · (i′,j′)(k′,m′) is not in C(22,12).

(iii) If Γi, Γj are distinct cycles with α = α1α2 ∈ Γ1,β = β1β2 ∈ Γj,αk,βk ∈ C2,k = 1, 2, then
{α1,α2}∩{β1,β2} = ∅, since Γi, Γj do not share a vertex. Thus no element of Γi · Γj is in C(22,12).

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

(iv) This now follows from (ii), which shows that only products of the form (i,j)(r,s) · (r,s)(u,v)
will produce elements of C2,2.

If we have a cycle Γi of size greater than three, then Lemma 4.9 (iv) shows that all cycles Γj have
size greater than 3. The same result shows that no cycle has length four (since then we would have a
principal set in C(22,12) of size less than 15). Thus (C

2
)(22,12) = 0C + D, where |D| = 15, C ∩D = ∅. We

then similarly have (D
2
)(22,12) = 0D + C. Thus C(22,12) \ (C ∪D) has size 15, and all of its cycles must

have size 3. Thus we may assume that all cycles Γi of C have size 3.

In this case it is easy to see that the components Γi of C are as shown in Figure 3 (up to conjugacy):

(1,2)

(5,6) (3,4)

(1,3)

(4,6) (2,5)

(1,4)

(2,6) (3,5)

(1,5)

(3,6) (2,4)

(1,6)

(4,5) (2,3)

Figure 3.

One now finds that the S-ring determined by C and Z(CS6) is S(S6,H120), and that the complement
of C in C(22,12) is also a principal set (of size 30). Thus we cannot have any principal sets C in S with
|C| = 15 and with cycles Γi of length greater than three.

This completes the proof of Lemma 4.8 and our discussion of the case λ = 2.

For n = 6 the element C(22,12) gives a λ = 6 cover, thus if we have a cover C with λ = 4 then
C(22,12) −C has λ = 2, and the S-ring generated by Z(CS6) and C is the same as the S-ring generated
by Z(CS6) and C(22,12) −C. Thus we have:
Proposition 4.10. Any commutative S-ring S over S6 that contains C2 as a principal set, also contains
C3 as a principal set, and either C2,2 is a principal set or S contains a conjugate of S(S6,H120).

Lastly, any principal set of S that is properly contained in C2,2 is an orbit of one of its elements
under the action of a conjugate of H120.

The C(23) case
Lemma 4.11. The only non-empty and proper subsets C of C(23) ⊂ S6 that satisfy(

C · C2,2
)
(2,14)

= λC2

for some λ ≥ 1 are conjugates of

(i) C1 = {(1, 2)(3, 4)(5, 6), (1, 6)(2, 3)(4, 5), (1, 5)(2, 4)(3, 6), (1, 4)(2, 6)(3, 5),
(1, 3)(2, 5)(4, 6)}; or

(ii) C2 = C(23) \C1.

Further, any conjugate of (i) or (ii) generates (together with Z(CS6)) an S-ring conjugate to S(S6,H120).

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

Proof Since |C(23)| = 15 we may assume that |C| ≤ 7 and that (1, 2)(3, 4)(5, 6) ∈ C. For α ∈ C(23) one
sees that [α · C(22,12)](2,14) is a sum of three elements. Thus 3|C| = 15λ i.e. |C| = 5λ. Since |C| ≤ 7
we see that λ = 1 and |C| = 5. It is now easy to see that up to a permutation we can assume that
(1, 6)(2, 3)(4, 5) ∈ C. Then there are two cases: (a) (1, 4)(2, 6)(3, 5) ∈ C; and (b) (1, 4)(2, 5)(3, 6) ∈ C.
Either case quickly gives a unique C of size 5, these Cs being conjugate. One checks that C1 and C2 are
principal sets of S(S6,H120). This gives the result.

This gives:

Proposition 4.12. Let S be a commutative S-ring over S6 that contains C2 as a principal set. Then
either

(i) C3,C(23) and C2,2 are principal sets of S, or
(ii) S contains a conjugate of S(S6,H120).

Lastly, any principal set of S that is properly contained in C(23) is an orbit of one of its elements
under the action of a conjugate of H120.

The C(3,2,1) case
Let S be a commutative S-ring over Sn,n ≥ 5, that contains C2,C3 as principal sets. For a principal

set C ⊆C(3,2,1n−5) ⊂ Sn we define sets Di,j ⊂C3 ⊂ Sn by

C =
∑

1≤i<j≤n

(i,j)Di,j.

Since [(i,j)(r,s,t) ·C3](2,1n−2) = (i,j), and C2 is a principal set of S we see that there is some λ ∈ N
such that (

C · C3
)
(2,1n−2)

= λC2.

Further, from the definition of the Di,j we see that λ = |Di,j| for all 1 ≤ i < j ≤ n.

Since [(i,j)(r,s,t) · C2](3,1n−3) = (r,s,t) and C3 is a principal set of S we also have µ ∈ N such that(
C · C2

)
(3,1n−3)

= µC3.

But we also have (
C · C2

)
(3,1n−3)

=
∑

1≤i<j≤n

Di,j.

Thus from µC3 =
∑

1≤i<j≤n Di,j we see that for all 1 ≤ i < j ≤ n we have(
n

2

)
|Di,j| =

(
n

2

)
λ = µ|C3| = µ

n(n− 1)(n− 2)
3

.

From this we obtain

Lemma 4.13. λ = |Di,j| and

3 · |Di,j| = 2µ(n− 2).

Since any (r,s,t) ∈ Di,j must satisfy i,j /∈{r,s,t} we see that

|Di,j| ≤
(n− 2)(n− 3)(n− 4)

3
.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

When n = 6 this gives |Di,j| ≤ 8. But if n = 6, then Lemma 4.13 shows that 3 divides µ so that

|Di,j| = 2 ·
µ

3
· 4 ≥ 8.

It follows from these two equations that |Di,j| = 8 for all 1 ≤ i < j ≤ n = 6, so that Di,j is maximal,
i.e. Di,j = {(r,s,t)±1 : 1 ≤ r < s < t ≤ 6, i,j /∈{r,s,t}}, and C = C(3,2,1). Thus we now have shown

Proposition 4.14. Let S be a commutative S-ring over S6 that contains C2 as a principal set. Then
C(3,2,1) is a principal set of S, and either
(i) C3,C(23),C(3,2,1) and C2,2 are principal sets of S, or
(ii) S contains a conjugate of S(S6,H120).

We note that in (ii) the only non-trivial conjugacy classes that do not split are C2,C3 and C(3,2,1).

The C(4,12) case
Let C ⊆C(4,12) be a principal set. Then the fact that(

(i,j,k,m) · C3
)
(2,14)

= (i,j) + (j,k) + (k,m) + (m,i),

shows that C determines a cover of λKn by 4-cycles. Thus, if [C · C3](2,14) = λC2, then

4 · |C| = 15λ.

Thus 15 divides |C|. Now there are twelve α ∈ C(4,12) such that (1, 2, 3, 4)α is a 3-cycle. Thus
considering

(
C · C4

)
(3,13)

we see that (since |C3| = 40) there is some µ ≥ 1 such that:

12 · |C| = 40µ,

(since C3 is a principal set) which shows that 10 divides |C|.
Thus |C| is divisible by 30. Note that |C(4,12)| = 90. If |C| = 60, then we replace C by C(4,12) \C, so

as to have |C| = 30.
Now if C−1 6= C, then C(4,12) \ (C ∪ C−1) is a principal set (since it has size 30) that is its own

inverse. Thus we may assume that the principal set C satisfies |C| = 30 and C = C−1.

Write C = α1 + α
−1
1 + · · · + α15 + α

−1
15 , so that (C

2
)(22,12) = 2α

2
1 + 2α

2
2 + · · · + 2α215, which shows

that {α21,α22, . . . ,α215}⊂C(22,12) is a principal set of S of size 15. Thus Lemma 4.8 shows that such a set
is conjugate to the set of 15 elements given in that lemma. Call this set of 15 elements W15.

We note that the map

ι2 : C(4,12) →C(22,12), α 7→ α2,

is a 2 to 1 surjection, where (ι2)−1((i,k)(j,m)) = {(i,j,k,m), (i,m,k,j)}, is an inverse pair. Since
C−1 = C we see that C is completely determined by ι2(C), which is conjugate to W15. Thus C must be
a conjugate of

{(1, 2, 5, 4), (1, 4, 5, 2), (1, 4, 3, 6), (2, 6, 5, 4), (1, 3, 4, 5), (1, 4, 6, 5), (2, 5, 3, 4), (4.1)
(2, 3, 4, 6), (3, 5, 4, 6), (1, 4, 2, 3), (1, 2, 4, 6), (1, 5, 4, 3), (3, 6, 4, 5), (2, 3, 6, 5),

(1, 6, 2, 5), (2, 4, 5, 6), (1, 3, 5, 6), (2, 4, 3, 5), (1, 3, 6, 2), (1, 2, 3, 5), (1, 6, 3, 4),

(2, 5, 6, 3), (1, 5, 3, 2), (1, 5, 6, 4), (1, 2, 6, 3), (1, 3, 2, 4), (2, 6, 4, 3), (1, 6, 4, 2),

(1, 5, 2, 6), (1, 6, 5, 3)}.

Now one finds that the S-ring generated by Z(CS6) and C is S(S6,H120). Using the fact that any
conjugate of W15 that is not equal to W15 does not commute with W15, this now easily gives:

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

Proposition 4.15. Let S be a commutative S-ring over S6 that contains C2 as a principal set. Then
either

(i) C3,C(23),C(3,2,1),C(4,12) and C2,2 are principal sets of S, or
(ii) S contains a conjugate of S(S6,H120).

Lastly, any principal set of S that is properly contained in C(42,12) is an orbit of one of its elements
under the action of a conjugate of H120, and is either a conjugate of (4.1) or of its complement in
C(4,12).

The C(4,2) case
There is a natural bijection C4,2 →C4, (i,j,k,m)(r,s) 7→ (i,j,k,m). So that the image of the set of

elements (4.1) of C(4,12) is

{(1, 2, 5, 4)(3, 6), (1, 4, 5, 2)(3, 6), (1, 4, 3, 6)(2, 5), (2, 6, 5, 4)(1, 3), (1, 3, 4, 5)(2, 6), (4.2)
(1, 4, 6, 5)(2, 3), (2, 5, 3, 4)(1, 6), (2, 3, 4, 6)(1, 5), (3, 5, 4, 6)(1, 2), (1, 4, 2, 3)(5, 6),

(1, 2, 4, 6)(3, 5), (1, 5, 4, 3)(2, 6), (3, 6, 4, 5)(1, 2), (2, 3, 6, 5)(1, 4), (1, 6, 2, 5)(3, 4),

(2, 4, 5, 6)(1, 3), (1, 3, 5, 6)(2, 4), (2, 4, 3, 5)(1, 6), (1, 3, 6, 2)(4, 5), (1, 2, 3, 5)(4, 6),

(1, 6, 3, 4)(2, 5), (2, 5, 6, 3)(1, 4), (1, 5, 3, 2)(4, 6), (1, 5, 6, 4)(2, 3), (1, 2, 6, 3)(4, 5),

(1, 3, 2, 4)(5, 6), (2, 6, 4, 3)(1, 5), (1, 6, 4, 2)(3, 5), (1, 5, 2, 6)(3, 4), (1, 6, 5, 3)(2, 4)}.

This case then follows by following the proof of the C4 case just completed. This then gives:

Proposition 4.16. Let S be a commutative S-ring over S6 that contains C2 as a principal set. Then
either

(i) C3,C(23),C(3,2,1),C(4,12),C(4,2) and C2,2 are principal sets of S, or
(ii) S contains a conjugate of S(S6,H120).

Lastly, any principal set of S that is properly contained in C(42,2) is an orbit of one of its elements
under the action of a conjugate of H120, and is either a conjugate of (4.2) or of its complement in
C(4,2).

The C(32) case
Consider a principal set C ⊆ C(32). Note that if α = ab ∈ C(32) where a,b ∈ C3, and if c ∈ C3 with

α · c ∈C3, then c ∈{a−1,b−1}.
Let C =

∑r
i=1 αi,αi = aibi ∈C(32),ai,bi ∈C3. Then

(
C · C3

)
(3,13)

=

r∑
i=1

(ai + bi).

Since C3 is a principal set we see that
r∑
i=1

(ai + bi) = λC3.

Since |C3| = 40 this gives 20λ = r. Now r ≤ |C(32)| = 40, so that we have λ ≤ 2. If λ = 2, then r = 40
and C = C(32).

So assume that λ = 1,r = |C| = 20. We may also assume (by conjugating) that (1, 2, 3)(4, 5, 6) ∈ C.
We have [C · C3](3,13) = 1C3, thus for each α ∈ C3 there is a unique α∗ ∈ C3 such that αα∗ ∈ C. Since
(1, 2, 3)(4, 5, 6) ∈ C we cannot have (1, 3, 2)(4, 5, 6) ∈ C, and so we must have (1, 3, 2)(4, 6, 5) ∈ C. Thus
C = C−1.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

For α = ab ∈ C(32),a,b ∈ C3, we let S(α) = {ab,a−1b,ab−1,a−1b−1}. Then the S(α),α ∈ C3,
partition C(32) into ten subsets S1, . . . ,S10 of size 4. Then any set C ⊂ C(32) as above is a union of an
inverse pair {αi,α−1i }⊂ Si from each Si. There are thus 2

10 possible such C’s, but only 29 if one assumes
that (say) (1, 2, 3)(4, 5, 6) ∈ C

Of these 29 choices for C one finds that only twelve satisfy
(
C

2
)
(3,3)

= λC + µC(32) \C for some
λ,µ ≥ 0. These twelve elements are in three conjugacy classes. Of the representatives from these three
conjugacy classes one checks that the subalgebra of CS6 generated by such a C and Z(CS6) is only an
S-ring if (up to conjugacy) we have one of the following two cases.

C1 = (1, 3, 5)(2, 4, 6) + (1, 5, 3)(2, 6, 4) + (1, 6, 4)(2, 5, 3) + (1, 4, 5)(2, 3, 6) (4.3)
+ (1, 2, 6)(3, 5, 4) + (1, 5, 2)(3, 4, 6) + (1, 3, 4)(2, 5, 6) + (1, 6, 5)(2, 4, 3)

+ (1, 6, 3)(2, 4, 5) + (1, 3, 6)(2, 5, 4) + (1, 3, 2)(4, 6, 5) + (1, 5, 6)(2, 3, 4)

+ (1, 4, 6)(2, 3, 5) + (1, 5, 4)(2, 6, 3) + (1, 2, 5)(3, 6, 4) + (1, 4, 2)(3, 6, 5)

+ (1, 2, 3)(4, 5, 6) + (1, 2, 4)(3, 5, 6) + (1, 4, 3)(2, 6, 5) + (1, 6, 2)(3, 4, 5); or

C2 = (1, 4, 5)(2, 3, 6) + (1, 2, 6)(3, 5, 4) + (1, 5, 2)(3, 4, 6) + (1, 6, 4)(2, 3, 5)

+ (1, 6, 5)(2, 3, 4) + (1, 6, 3)(2, 4, 5) + (1, 4, 3)(2, 5, 6) + (1, 3, 6)(2, 5, 4)

+ (1, 5, 6)(2, 4, 3) + (1, 3, 2)(4, 6, 5) + (1, 5, 4)(2, 6, 3) + (1, 5, 3)(2, 4, 6)

+ (1, 4, 6)(2, 5, 3) + (1, 2, 5)(3, 6, 4) + (1, 3, 4)(2, 6, 5) + (1, 4, 2)(3, 6, 5)

+ (1, 2, 3)(4, 5, 6) + (1, 2, 4)(3, 5, 6) + (1, 3, 5)(2, 6, 4) + (1, 6, 2)(3, 4, 5).

However in each of these two cases the S-ring that one obtains is just a conjugate of S(S6,H120) again.

Thus we now have:

Proposition 4.17. Let S be a commutative S-ring over S6 that contains C2 as a principal set. Then
either

(i) C3,C(23),C(3,2,1),C(4,12),C(4,2), C2,2 and C(32) are principal sets of S, or
(ii) S contains a conjugate of S(S6,H120).

Lastly, any principal set of S that is properly contained in C(32) is an orbit of one of its elements
under the action of a conjugate of H120, and is either a conjugate of one of the elements shown in (4.3)
or of its complement in C(32).

The C(5,1) case
Consider a principal set C ⊆C5 = C(5,1). Now there are five elements α ∈C2 such that (1, 2, 3, 4, 5)α ∈

C(3,2,1); since C(3,2,1) is a principal set we see that 5 · |C| = λ · |C(3,2,1)| = 120λ. This gives |C| = 24λ.
Also, there are five α ∈ C3 such that (1, 2, 3, 4, 5)α ∈ C(3,13). Since C(3,13) is a principal set we see

that 5 · |C| = µ · |C(3,13)| = 40µ.

Case 1: |C| = 24. Here λ = 1,µ = 3, and so we have (i) [C · C2](3,2,1) = C(3,2,1) and (ii) [C · C3](3,13) =
3C(3,13).

Let α1, . . . ,α72 ∈C(5,1) represent the inverse pairs in C(5,1), so that C(5,1) = ∪72i=1{αi,α
−1
i }.

Assume first that C−1 = C. Then we can write C =
∑72
i=1 xi(αi + α

−1
i ). Here xi = 0, 1 satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 72;

(b)
∑72
i=1 xi = 12;

(c)
(∑72

i=1 xi(αi + α
−1
i ) · C2

)
(3,2,1)

= C(3,2,1);

(d)
(∑72

i=1 xi(αi + α
−1
i ) · C3

)
(3,13)

= 3C3.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

If we now think of the xi as indeterminates in the polynomial ring R = Q[x1, . . . ,x72], then each of
(a)–(d) gives relations satisfied by the xi. Let I be the ideal of R generated by these relations. Finding a
Gröbner basis for I ([1]) we see that all but five of the xi are determined (in fact by a conjugacy we can
assume that one of these five is equal to zero). This leaves a small number of possibilities for C, and one
checks that they are all equal to a conjugate of the H120 orbit of some element of C(5,1). Further each such
element, together with the class sums, generates an S-ring that is just some conjugate of S(S6,H120).
This does the case |C| = 24,C = C−1.

So now assume |C| = 24,C−1 6= C. Then we write C =
∑72
i=1 x2i−1αi + x2iα

−1
i . Here xi = 0, 1

satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 144, and x2i−1x2i = 0 for 1 ≤ i ≤ 144;

(b)
∑144
i=1 xi = 24;

(c)
(∑72

i=1(x2i−1αi + x2iα
−1
i ) · C2

)
(3,2,1)

= C(3,2,1);

(d)
(∑72

i=1(x2i−1αi + x2iα
−1
i ) · C3

)
(3,13)

= 3C3.

Finding the ideal of R = Q[x1, . . . ,x144] determined by these equations gives no solutions. The above
shows that if there is a C ⊂ C(5,1), |C| = 24, then the S-ring contains a conjugate of S(S6,H120), and a
principal set of size 24 in C(5,1) is an orbit under the same conjugate of H120.
Case 2: |C| = 48. Here λ = 2,µ = 6, and we aim to show that no such set exists. First note that if
C 6= C−1, then D = C(5,1)\(C∪C−1) satisfies D = D−1, |D| = 48, and D is in the S-ring. If D is a union
of disjoint sets of size 24, then these sets are conjugates of the sets of size 24 determined in Case 1. One
checks, however, that no two of these sets (there are six of them) commute, and so we see that D must
also be a principal set. Thus we can assume that C = C−1. Then we can write C =

∑72
i=1 xi(αi + α

−1
i ).

Here, as in Case 1, the xi = 0, 1 satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 72;

(b)
∑72
i=1 xi = 24;

(c)
(∑72

i=1 xi(αi + α
−1
i ) · C2

)
(3,2,1)

= 2C(3,2,1);

(d)
(∑72

i=1 xi(αi + α
−1
i ) · C3

)
(3,13)

= 6C3.

Let I be the ideal of R = Q[x1, . . . ,x72] generated by the polynomials determined by these relations.
Finding a Gröbner basis for I ([1]) we see that all but twelve of the xi are determined (again by a
conjugacy we can assume that one of these twelve is equal to zero). Choosing values in {0, 1} for these
eleven variables, one checks the 211 cases, and one finds that there are only 30 (non-trivial) elements C
(any other case gives I = R), and for each of these 30 cases one finds those such that

(
C

2
)
(3,13)

has the

form mC3,m ≥ 0. There are 10 of these, and one finds that they are all conjugate. Thus, if C is one of
these, then we check that the S-ring generated by Z(CS6) and C is commutative and has dimension 34,
but does not contain C2 as a principal set. (Note: each of these ten conjugate Cs does give a commutative
S-ring containing C2, however C2 partitions as {(1, 2), (3, 4), (5, 6)} and the complement; see Corollary 1.1
(iii).) Thus there are no principal sets of size 48 in C(5,1) if C2 is a principal set.
Case 3: |C| = 72. Here λ = 3,µ = 9.

Assume first that C = C−1. Then we can write C =
∑72
i=1 xi(αi + α

−1
i ). Here, as in the above,

xi = 0, 1, satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 72;

(b)
∑72
i=1 xi = 36;

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

(c)
(∑72

i=1 xi(αi + α
−1
i ) · C2

)
(3,2,1)

= 3C(3,2,1);

(d)
(∑72

i=1 xi(αi + α
−1
i ) · C3

)
(3,13)

= 9C3;

(e)
(∑72

i=1 xi(αi + α
−1
i ) · C4

)
(2,14)

= 24C2;

(f) Let
(
C

2
)
(3,13)

=
∑
α∈C3 xαα, where xα ∈ R = Q[x1, . . . ,x72]. Then we have xα = xβ for all α,β ∈C3,

since C3 is a principal set.
Here (e) is obtained in the same way as (c) and (d). Also, (f) is a consequence of the fact that C3

is a principal set of S. Constructing the ideal I (including all xα − xβ as in (f)) again one finds that
there are 8 variables that determine the rest (7 if one uses conjugacy to set one of them to be zero).
Choosing values in {0, 1} for these variables gives 27 cases to consider, of which all but eleven give I = R.
These eleven Cs are in two conjugacy classes of sizes 1 and 10. The element C1, representing the single
conjugacy class, is the orbit of some element of C(5,1) under the action of A6. Let C2 be one of the ten
conjugate elements. Then there is a subgroup H36 of size 36 such that the orbits of H36 on C(5,1) are
O1,O2,O3,O4 where |Oi| = 36, i = 1, 2, 3, 4. Further, these sets Oi can be numbered so that

C1 = O1 ∪O2, C2 = O1 ∪O3.

A representative for H36 up to conjugacy is

H36 = 〈(1, 6)(2, 3, 4, 5), (1, 2, 5, 4)(3, 6)〉.

We note that if one uses O1 ∪O4, then one obtains a commutative S-ring of dimension 26 that contains
{(1, 3), (3, 5), (1, 5), (2, 6), (4, 6), (2, 4)} and its complement in C2, as principal sets - see Corollary 1.1 (iii).

One finds that each of the S-rings 〈Z(CS6),C1〉,〈Z(CS6),C2〉 has dimension 12 (recall that Z(CS6)
has dimension 11) where Ci,C(5,1) \ Ci, i = 1, 2, are principal sets of these S-rings. This concludes the
case where |C| = 72,C = C−1. We will let C36 denote the element C2 in what follows, and we will denote
the S-ring 〈Z(CS6),C36〉 by S36.

Now if C 6= C−1, then we can write C =
∑72
i=1 x2i−1αi + x2iα

−1
i . Here xi = 0, 1 satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 144 and x2i−1x2i = 0 for 1 ≤ i ≤ 72;

(b)
∑72
i=1 xi = 72;

(c)
(∑72

i=1(x2i−1αi + x2iα
−1
i ) · C2

)
(3,2,1)

= 3C(3,2,1);

(d)
(∑72

i=1(x2i−1αi + x2iα
−1
i ) · C3

)
(3,13)

= 9C3;

(e)
(∑72

i=1(x2i−1αi + x2iα
−1
i ) · C4

)
(2,14)

= 24C2;

(f) Let
(
C

2
)
(3,13)

=
∑
α∈C3 xαα, where xα ∈ R = Q[x1, . . . ,x72]. Then we have xα = xβ for all α,β ∈C3,

since C3 is a principal set.
One then finds that the ideal determined by (a)–(f) is the whole ring, so that there are no solutions

in this situation.

Now if C ( C(5,1) is a principal set of S with |C| > 72, then there are certainly principal sets D
of S with |D| < 72. However in the above considerations of Cases 1,2 we have shown that the S-ring
generated by D and Z(CS6) does not have a principal set of size greater than 72. Thus the situation
|C| > 72 does not occur, and we have now proved:

Proposition 4.18. Let S be a commutative S-ring over S6 containing C2. Then any principal set
C ⊂C(5,1) of S is either a conjugate of C36, or is the orbit of one of its elements under the action of A6
or of a conjugate of H120.

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The C(6) case
Consider a principal set C ( C(6). Now there are six elements α ∈C(4,12) such that (1, 2, 3, 4, 5, 6)α ∈

C(3,13); since C(3,13) is a principal set we see that 6 · |C| = λ1 · |C(3,13)| = 40λ1. This gives (i) 3|C| = 20λ1.
Similarly, by considering

(ii)
(
C · C(5,1)

)
(2,14)

= λ2C2, we see that 2|C| = 5λ2;

(iii)
(
C · C(3,2,1)

)
(3,13)

= λ3C3, we see that 3|C| = 10λ3;

(iv)
(
C · C2,2

)
(3,2,1)

= λ4C(3,2,1), we see that 3|C| = 20λ4.

One sees that |C| is divisible by 20. Also |C(6)| = 120. Let αi, i = 1, . . . , 60 be representatives for the
inverse pair sets in C(6).
Case 1: |C| = 20. Here λ1 = 3,λ2 = 8,λ3 = 6,λ4 = 3.

First assume that C = C−1. Then we can write C =
∑60
i=1 xi(αi + α

−1
i ). Here, as in the above,

xi = 0, 1 satisfy

(a) x2i −xi = 0 for 1 ≤ i ≤ 60;

(b)
∑60
i=1 xi = 10;

(c) relations for each of (i)–(iv) above;

(d) let
(
C

2
)
(3,13)

=
∑
α∈C3 xαα, where xα ∈ R = Q[x1, . . . ,x10]. Then we have xα = xβ for all α,β ∈C3,

since C3 is a principal set.

(e) Let
(
C

2
· C2,2

)
(3,13)

=
∑
α∈C3 xαα, where xα ∈ R = Q[x1, . . . ,x10]. Then we have xα = xβ for all

α,β ∈C3, since C3 is a principal set.
Constructing the ideal I (including all xα−xβ as in (d),(e)) again one finds that there are 5 variables

that determine the rest (4 if one uses conjugacy to set one of them to be zero). Looking at the 24 cases
one finds that there are 5 possibilities for C. These are all in a single conjugacy class, so we need only
consider one of them, C say. One finds that the S-ring generated by Z(CS6) and C is a conjugate of
S(S6,H120). This does the case where C = C−1.

Now if C 6= C−1, then we can write C =
∑60
i=1 x2i−1αi + x2iα

−1
i . Using the same tests as in the

C = C−1 case above one creates an ideal that has nine variables that determine the rest. Enumerating
the various possibilities for C gives ten non-conjugate elements. One checks that each of these generates
CS6 as an S-ring. This shows that this case cannot occur. We further note

Lemma 4.19. If C1,C2 ⊆ C6 are distinct principal sets of a commutative S-ring over S6 that contains
C2 as a principal set, then at most one of C1,C2 can have size 20.

Proof. If |C1| = 20, then the above shows that C2 is a conjugate of C1; but there are only six such
conjugates and it is easy to check that no two such distinct commute.

Case 2: |C| = 40. Here λ1 = 6,λ2 = 16,λ3 = 12,λ4 = 6.

First assume that C = C−1. Then we can write C =
∑60
i=1 xi(αi + α

−1
i ). Applying the same ideal

calculation as in the |C| = 20 case one obtains an ideal having 4 variables that determine the rest.
Enumerating the various possibilities for C gives four elements, all of them conjugate to each other.
Considering one of them, C say, one finds that the S-ring generated by Z(CS6) and C is a conjugate of
S(S6,H120). This does the case where C = C−1.

Repeating the above for the situation where |C| = 40,C 6= C−1, one finds that there are no solutions.
Case 3: |C| = 60. Here λ1 = 9,λ2 = 24,λ3 = 18,λ4 = 9.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

First assume that C = C−1. Then we can write C =
∑60
i=1 xi(αi + α

−1
i ). Applying the same ideal

calculation as in the |C| = 20 case one obtains an ideal having 5 variables that determine the rest.
Enumerating the various possibilities for C gives six possibilities for C, and there are two conjugacy
classes of such elements. If C say, represents either of these classes, one finds that the S-ring generated
by Z(CS6) and C is a conjugate of S(S6,H120). This does the case where C = C−1.

Repeating the above for the situation where |C| = 60,C 6= C−1, one finds that there are no solutions.
The cases |C| > 60 are dealt with as in the situation |C| > 72 of Case 2.
This concludes consideration of all conjugacy classes of S6 where S has C2 as a principal set.
The following is mostly a summary of what we have done:

Proposition 4.20. Let S be a commutative S-ring over S6 containing C2 as a principal set. Then
(i) C(3,13) and C(3,2,1) are principal sets of S.

(ii) If C ( C2,2 is a principal set of S, then |C| = 15 or |C| = 30, and C is either a conjugate of
the element shown in Lemma 4.8, or is C2,2 − C for such a set of size 15. There are six conjugates of
each such set C. No two distinct such conjugates commute. The S-ring generated by Z(CS6) and C is
S(S6,H120).

(iii) If C ( C(23) is a principal set of S, then |C| = 5 or |C| = 10. If |C| = 5, then C is conjugate to
the element shown in Lemma 4.11 (i). There are six conjugates of each such set C. No two distinct such
conjugates commute. The S-ring generated by Z(CS6) and C is a conjugate of S(S6,H120).

(iv) If C ( C(4,12) is a principal set of S, then |C| = 30 or |C| = 60. If |C| = 30, then C is a conjugate
of the element shown in (4.1), otherwise it is the complement in C(4,12) of such an element. There are
six conjugates of each such set C. No two distinct such conjugates commute. The S-ring generated by
Z(CS6) and C is a conjugate of S(S6,H120).

(v) If C ( C(4,2) is a principal set of S, then |C| = 30 or |C| = 60. If |C| = 30, then C is a conjugate
of the element shown in (4.2), otherwise it is the complement in C(4,2) of such an element. There are
six conjugates of each such set C. No two distinct such conjugates commute. The S-ring generated by
Z(CS6) and C is a conjugate of S(S6,H120).

(vi) If C ( C(32) is a principal set of S, then |C| = 20 and C is one of the two elements C1, C2 shown in
(4.3). Now C1 and C2 commute. Each of C1,C2 has six conjugates; no two distinct conjugates of each
Ci, i = 1, 2, commute, and each C1

g
only commutes with a single conjugate of C2. For i = 1, 2, the S-ring

generated by Z(CS6) and Ci is a conjugate of S(S6,H120).

(vii) If C ( C(5,1) is a principal set of S, then either
(viia) |C| = 24 and C is an orbit of an element of C(5,1) under the action of a conjugate of H120; or
(viib) |C| = 120 and C is an orbit of an element of C(5,1) under the action of a conjugate of H120; or
(viic) |C| = 72 and C is an A6-orbit of an element of C(5,1).
(viid) |C| = 72 and C is a conjugate of C36.
Further, no two distinct conjugates of C, for C of type (viia) commute; no two distinct conjugates

of type (viib) commute; no conjugate of type (viia) commutes with a conjugate of type (viib) unless their
sum is C(5,1). Any conjugate of type (viic) commutes with any conjugate of type (viia) or (viib).

Any element of type (viia) or (viib) generates (with Z(CS6)) an S-ring which is a conjugate of
S(S6,H120). Any element of type (viic) generates (with Z(CS6)) an S-ring which is a conjugate of
S(S6,A6).

(viii) If C ( C(6) is a principal set of S, then |C| ∈ {20, 40, 60}, and in each case the S-ring generated by
Z(CS6) and C is a conjugate of S(S6,H120). Let O20,O40,O60 denote the sets of conjugates (of the sums

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

C) for each case, so that each Oi has size six. Then for i ∈ {20, 40, 60} no two distinct elements of Oi
commute, and for distinct i,j ∈{20, 40, 60} elements of Oi,Oj commute if and only if the corresponding
sets are disjoint.

Further, the only principal elements of S(S6,H120) that C36 commutes with are C2,C3 and C(3,2,1).

Proof. Here (i) follows from Proposition 4.6 and Proposition 4.14. Part (ii) follows from Lemma 4.8,
the fact that [S6 : H120] = 6, and a calculation to show that distinct conjugates don’t commute. (iii)
follows from Lemma 4.11, the fact that [S6 : H120] = 6, and a calculation to show that distinct conjugates
don’t commute.

Part (iv) follows from the proof of Proposition 4.15, and a calculation to show that distinct conjugates
don’t commute. Part (v) follows from the discussion of the C(4,2) case, and a calculation to show that
distinct conjugates don’t commute. Part (vi) follows from the discussion of the C(32) class that resulted in
the two cases shown in (4.3) (up to conjugacy), and a calculation to show that distinct conjugates don’t
commute. Parts (vii) and (viii) follow from the discussion of the C(5,1) and C(6) classes, and a calculation
to show that certain conjugates of these elements don’t commute.

We use this to prove:

Theorem 4.21. The only commutative Schur rings over S6 containing C2 as a principal set are:
(1) S(S6,S6);

(2) S(S6,H120);

(3) S(S6,A6);

(4) S36.

Proof. The last statement of Proposition 4.20 shows that such an S-ring cannot have split S6 classes
that are orbits under a conjugate of H120, and also have C36 (or its complement) in it. One checks that
the S-ring generated by any conjugate of S36 and S(S6,A6) is not commutative. Given the fact that,
according to Proposition 4.20, not very many principal elements of conjugates of S(S6,H120) commute,
it is now easy to prove the theorem.

5. When C2 is not a principal set

We consider each case as enumerated in Corollary 1.1.

Case (ii): C2 = C1∪C2∪C3, where C1 = {(1, 2)}; C2 = {(i,j) : 3 ≤ i < j ≤ 6}, and C2 = C2\(C1∪C2).
One finds that the S-ring generated by Z(CS6), C1,C2 and C3 is S(S6,S2 × S4), where S2 × S4

is the subgroup 〈(1, 2), (3, 4), (3, 4, 5, 6)〉 ≤ S6. There are 34 principal sets O1, . . . ,O34 of sizes
1, 3, 6, 8, 12, 16, 24, 48. The idea is to show that no proper, non-empty subset of each Oi can be a principal
set of a commutative S-ring containing S(S6,S2 ×S4).

One can check this directly (using [1]) for each Oi with |Oi| ≤ 8. For the rest let Oi = {o1, . . . ,om}.
Let R = Q[x1, . . . ,xm] be a polynomial ring, and let e =

∑m
j=1 xjoj. One considers the ideal I of R

generated by all the coefficients of the elements eOk − Oke, these being linear polynomials in the xj.
One finds that a Gröbner basis for I has only elements of the form Ph − Puk, where the Puk are free
variables, and there are at most six of the Puk. One chooses a subset of the Puk, and puts these equal to
1, while one puts the rest equal to 0. This produces an ideal, that determines an element in CS6. One
then sees whether this element (together with Z(CS6)) generates a commutative S-ring. One finds that
S(S6,S2 ×S4) is the only possible case.
Case (iii): C2 = C1 ∪C2 where C1 = {(1, 2), (3, 4), (5, 6)} and C2 = C2 \C1.

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A. Francis, S. Humphries / J. Algebra Comb. Discrete Appl. 3(2) (2016) 61–80

One finds that the S-ring generated by Z(CS6), C1 and C2 is S(S6,S2 o S3), where S2 o S3 is the
(wreath product) subgroup 〈(1, 2), (3, 4), (5, 6), (1, 3)(2, 4), (3, 5)(4, 6)〉 ≤ S6. There are 34 principal sets
O1, . . . ,O34 of sizes 1, 3, 6, 8, 12, 16, 24, 48.

We note that S6 has an outer automorphism α such that α(S(S6,S2 ×S4)) = S(S6,S2 oS3). Thus
this case follows from Case (ii).

Case (iv): C2 = C1 ∪C2 where C1 = {(1, 2), (1, 3), (2, 3), (4, 5), (4, 6), (5, 6)} and C2 = C2 \C1.
One finds that the S-ring generated by Z(CS6), C1 and C2 is S(S6,S3oS2), where S3oS2 is the (wreath

product) subgroup 〈(1, 2), (2, 3), (5, 6), (1, 4)(2, 5)(3, 6)〉 ≤ S6. There are 26 principal sets O1, . . . ,O26 of
sizes 1, 4, 6, 9, 12, 18, 36, 72. The idea is to again show that no proper, non-empty subset of each Oi can
be a principal set of a commutative S-ring containing S(S6,S3 oS2).
Case (v): C2 = C1 ∪C2 where C1 = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6)} and C2 = C2 \C1.

One finds that the S-ring generated by Z(CS6), C1 and C2 is S(S6,S5). There are 19 principal sets
O1, . . . ,O19 of sizes 1, 5, 10, 15, 20, 24, 30, 40, 60, 120. We note that S6 has an outer automorphism, and
that S(S6,H120) and S(S6,S5) are related by this automorphism. Thus this case follows from (iv).

This concludes the proof of Theorem 1.2.

Proof of Corollary 1.3 In [5] it is shown that any commutative S-ring S, of maximal dimension sG, over
a group G contains Z(CG). Thus the principal sets of S give a partition of G that is a refinement of
the partition of G by conjugacy classes. In particular, for G = S6, we must have C2 ∈ S. Thus any such
S-ring must be in the list given in Theorem 1.2. However none of the S-rings listed in Theorem 1.2 has
dimension 76 = sS6. This proves Corollary 1.3.

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http://www.mathnet.ru/links/26300f6e5e5113706596d3dd3372a943/aa940.pdf
http://www.mathnet.ru/links/26300f6e5e5113706596d3dd3372a943/aa940.pdf
http://dx.doi.org/10.1007/s10801-015-0613-2
http://dx.doi.org/10.1007/s10801-015-0613-2
http://dx.doi.org/10.1080/00927872.2014.974258
http://dx.doi.org/10.1080/00927872.2014.974258
http://dx.doi.org/10.1007/BF02773471
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http://dx.doi.org/10.1016/j.ejc.2008.11.006

	Introduction
	Schur-rings
	Covers of complete graphs
	 The case where C2 is a principal set
	When C2 is not a principal set
	References