Require: M0[G;I,Λ;P] a finite Rees 0-matrix semigroup Require: Ga group,P a regular matrix
1: procedureLinkedTripleFromPairs(R) 2: N :={1G} 3: S:= ∆I 4: T := ∆Λ 5: for(x, y)∈R do 6: if x=ythen 7: continue 8: else if x= 0ory= 0then
9: returnUniversal Congruence (no linked triple) 10: Letx= (i, a, λ)
11: Lety= (j, b, µ)
12: if (i, j)∈/ εI or(λ, µ)∈/εΛ then
13: returnUniversal Congruence (no linked triple) 14: . Combine row and column classes
15: S ←(S ∪(i, j))e 16: T ←(T ∪(λ, µ))e
17: . Add generators for normal subgroup
18: Chooseν ∈Λ such thatpνi6= 0
19: Choosek∈I such thatpλk 6= 0
20: N ← hhN,(pνiapλk)(pνjbpµk)−1ii
21: forξ∈Λ\ {ν} such thatpξi6= 0do
22: N ← hhN, qνξijii
23: . N is a normal subgroup of Gcontaining everyqνξij considered so far
24: forz∈I\ {k}such thatpλz6= 0do
25: N ← hhN, qλµkzii
26: . N is a normal subgroup of Gcontaining everyqλµkz considered so far
27: . (N,S,T)is linked and its congruence contains every pair(x, y)considered so far
28: return(N,S,T)
We can see the algorithm working in the following example.
Example 3.22. Let S = M0[D
4;{1,2,3,4},{1,2};P] be a Rees 0-matrix semigroup, where
D4 is the permutation grouph(1 2 3 4),(2 4)i, isomorphic to the dihedral group on 4 points,
andP is the 2×4 matrix
0 (1 2)(3 4) 0 (1 4 3 2) (2 4) (1 4)(2 3) (2 4) 0
! .
Letρ be the congruence generated by the single pair (1,(),1),(3,(1 2 3 4),1). We can use
LinkedTripleFromPairsto find the linked triple corresponding toρ, or to determine thatρ
is universal.
We set our triple to {()},∆4,∆2 to start with, where ∆n is the diagonal relation on
(3,(1 2 3 4),1). We do not have x = y, x = 0 or y = 0, so we go on to consider the two elements componentwise. The pair (i, j) = (1,3) lies inεI since columns 1 and 3 contain zeroes
in the same positions, and (λ, µ) = (1,1) certainly lies in εΛ since it is a reflexive pair; hence
we do not have to return the universal congruence. We modify S by joining the classes of 1 and 3 together; we do not have to modify T, since (1,1) is already inT. Finally we have to add generators toN: we can set bothν andkto 2, and then we add toN the element
(pνiapλk)(pνjbpµk)−1= p21()p12 p23(1 2 3 4)p12 −1 = (2 4)()(1 2)(3 4) (2 4)(1 2 3 4)(1 2)(3 4)−1 = (1 2 3 4),
and take the normal closure. Finally we have to add any appropriate q values. There is no value ofξwhich meets the stated requirements, but there is one appropriate value forx: x= 4. Hence we have to add the element qλµkx = q1124 = p12p12−1p14p−141 = (). Since the identity
already lies inN, we do not need to make any changes. There are no more pairs to process, so we return the linked triple (N,S,T) = (C4,(1,3)e,∆2), where
• C4=h(1 2 3 4)iis the subgroup ofD4consisting of the four rotations, isomorphic to the
cyclic group of order 4;
• (1,3)eis the least equivalence on{1,2,3,4}containing the pair (1,3) (its classes are{1,3},
{2}, and{4});
• ∆2 is the diagonal relation on{1,2}(its classes are {1} and{2}).
3.2.4
Generating pairs from a linked triple
Let S be a completely simple or completely 0-simple semigroup, and letρbe a non-universal congruence onS. In Section 3.2.3 we presented an algorithm to find the linked triple ofρ, given only a set of generating pairs for ρ. In this section, we will present the reverse: a method to find a set of generating pairs for ρgiven only its linked triple (N,S,T).
Firstly we require a lemma describing the inclusion of congruences in each other, and how it mirrors an inclusion of linked triples.
Lemma 3.23 ([How95, Lemma 3.5.5], [Tor14b, Lemma 3.9]). Let ρ and σ be non-universal congruences on S with linked triples(Nρ,Sρ,Tρ)and(Nσ,Sσ,Tσ)respectively. We have ρ⊆σ
if and only if Nρ≤Nσ,Sρ⊆ Sσ, andTρ⊆ Tσ.
Now we can state the main theorem which will inform this algorithm. It also relies on ideas from Theorem 3.20.
Theorem 3.24. Let S = M0[G;I,Λ;P] be a finite 0-simple semigroup, and let ρ be a non-
universal congruence with linked triple(Nρ,Sρ,Tρ). LetNρ0 ⊆Nρ,Sρ0 ⊆ Sρ andTρ0 ⊆ Tρ be any
subsets with the following properties:
• Nρ is the normal closure ofNρ0 inG,
• Tρ= (Tρ0) e.
If Ris a subset ofρsuch that
(i) for each pair (i, j)∈ S0
ρ there existλ, µ∈Λ anda, b∈Gsuch that (i, a, λ)R(j, b, µ);
(ii) for each pair (λ, µ)∈ T0
ρ there existi, j∈I anda, b∈Gsuch that(i, a, λ) R(j, b, µ);
(iii) for each element n∈Nρ0 there exist i, j, x ∈I andλ, µ, ξ ∈Λ such that pξi and pλx are
both non-zero and
(i, p−ξi1np−λx1, λ)R (j, p−ξj1p−µx1, µ);
thenR]=ρ.
Proof. AssumeRis as stated. Sinceρis a congruence andR⊆ρ, we know thatR]⊆ρ. Hence we only need to show that ρ⊆R].
Let (NR],SR],TR]) denote the linked triple associated with R]. We will show thatNρ ⊆
NR],Sρ⊆ SR], andTρ⊆ TR], and therefore thatρ⊆Rby Lemma 3.23.
Recall the relationsR|I andR|Λfrom Definition 3.19. By (i) we can see thatSρ0 ⊆R|I and
hence (S0
ρ)e⊆(R|I)e. Meanwhile by Theorem 3.20 we have (R|I)e=SR]. In total this gives usSρ = (Sρ0)e⊆(R|I)e=SR], so Sρ⊆ SR]. Similarly by (ii) we haveTρ⊆ TR].
Now we turn our attention toNρ, and its generating setNρ0 – we wish to show thatNρ ⊆
NR]. Letn∈Nρ0. By (iii), there existi, j, x∈Iandλ, µ, ξ∈Λ such thatpξi andpλx are both
non-zero and (i, a, λ)R(j, b, µ), where
a=p−ξi1np−λx1 and b=p−ξj1p−µx1.
Note that pξj and pµx must also be non-zero since (i, j) ∈ εI and (λ, µ) ∈ εΛ. To see that
n∈NR], observe thatpξiapλx=nandpξjbpµx= 1G. Hencensatisfies the condition that
n= (pξiapλx)(pξjbpµx)−1
for somei, j, x∈I, someλ, µ, ξ∈Λ, and somea, b∈Gsuch that (i, a, λ)R(j, b, µ) andpξiand
pλxare non-zero; this is precisely the requirement in Theorem 3.20 which means thatn∈NR]. Hence Nρ0 ⊆NR]. Since Nρ is the normal closure of Nρ0, and NR] is a normal subgroup, we have Nρ⊆NR].
SinceNρ⊆NR],Sρ⊆ SR] andTρ⊆ TR], Lemma 3.23 gives usρ⊆R], as required. Theorem 3.24 is enough to justify thePairsFromLinkedTriplealgorithm, which is pre- sented in this thesis as Algorithm 3.26. Given a linked triple (N,S,T), we only need to choose applicable subsets N0 ⊆ N, S0 ⊆ S and T0 ⊆ T and we have a good idea of what pairs are
necessary to generate a congruence. In the algorithm, we assume that a generating set N0 is known for N (line 2) – this is certainly likely to be the case in a computational setting, for example inGAP[GAP18] where groups almost always have a known generating set. We should note that this set should act as a set of normal subgroup generators, meaning that it might be even smaller than a standard set of subgroup generators. For S0 we use as few pairs as
possible for each class of S: for each class C = {i1, . . . , in} we include the pair (i1, il) for all
l∈ {2, . . . , n}(lines 3–8). These pairs relate all elements in the class to each other by transitiv- ity (line 7), so we do not need to add any other elements. Hence each class requires a number
of pairs inS0 equal to one less than the size of the class; and so|S0| =|I| −k
S, where kS is
the number of classes inS. We similarly choose as small a set as possible for T0 (lines 9–14),
so |T0|=|Λ| −k
T, wherekT is the number of classes inT.
Once our three generating sets have been calculated, we collate them into the set of pairs
Ras efficiently as possible: each pair we add toRcan satisfy the conditions in Theorem 3.24 for one pair (i, j) ∈ S0, one pair (λ, µ) ∈ T0, and one element n ∈ N0. The while-loop on
lines 19–30 steps through the three lists, on each iteration taking a new elementafrom N0, a new pair of columns (i, j) fromS0, and a new pair of rows (λ, µ) from T0. Then, after fixing
appropriateξandkin lines 26–27 as in Theorem 3.24 condition (iii), we add the pair (i, p−ξi1ap−λk1, λ),(j, p−ξj1p−µk1, µ)
,
which can be seen by inspection to satisfy (i), (ii) and (iii) for the three generators in question. On each iteration of the while-loop, we remove one item from each ofN0, S0 andT0 (lines
20–25), and we add one pair to R. This is repeated until all three sets are exhausted, and so the total number of pairs returned by the algorithm is equal to the size of the largest of the three sets – that is,
max(|N0|,|I| −kS,|Λ| −kT).
If the sets have different sizes, then on some of the later runs through the loop, one or two of the three sets will be empty. This does not present a problem: if a set is empty, it is not popped, and the last value from the loop is simply used again. The set of pairs still satisfies the necessary condition from the theorem. The only remaining case is that one of the sets may be empty to start with. To account for this case, we give each of the three variables a default value in lines 16–18, so that even if a set is never popped, there is a sensible default value that does not invalidate the condition: forawe can use the identity 1Gwhich must always be inN;
for (i, j) or (λ, µ) we can use a reflexive pair, from ∆I or ∆Λ respectively.
It is natural to ask whether a set of generating pairs returned byPairsFromLinkedTriple
is minimal – that is, to ask whether any smaller set of pairs could be found which generates the same congruence.
Theorem 3.25. If|N0| ≤ |I| −kS or|N0| ≤ |Λ| −kT, then PairsFromLinkedTriplereturns
a set of generating pairs which is minimal.
Proof. The number of pairs returned by PairsFromLinkedTriple depends solely on the sizes ofN0,S0 andT0: it is simply the maximum of these three sizes. The generating setN0 is
assumed by the algorithm to have been known in advance, and hence is not guaranteed to be minimal in any way. However,S0andT0 are created in the algorithm, each one consisting of a
set of pairs inI×I or Λ×Λ which makes as few links as possible between elements in a class. In other words, S0 and T0 contain the smallest possible number of pairs such that (S0)e =S
and (T0)e=T. Note that |S0|=|I| −k
S and|T0|=|Λ| −kT.
LetRdenote the output of the algorithm, and letR|I be as in Definition 3.19. We can see
from the definition of our algorithm thatR|I =S0, and Theorem 3.20 tells us thatSR] = (R|I)e;
that is, we require that (S0)e = S for the algorithm for the output R to be valid. Since we
have already seen thatS0has as few pairs as possible such that (S0)e=S, we know that every
bound for the size of a set of generating pairs for the congruence. By similar reasoning,|T0|is
also a lower bound.
Assume|N0| ≤ |I| −kS or|N0| ≤ |Λ| −kT. The number of pairs returned by the algorithm
will be either |I| −kS or |Λ| −kT. Since we know that these are both lower bounds for the
possible size of a generating set, we can conclude that the size ofRequals the minimum possible size, soRis minimal.
In the case thatN0 is larger than both|I| −kS and|Λ| −kT, a claim to minimality cannot
be made so easily. Again referring to Theorem 3.20, we see that whereas SR] and TR] are determined entirely by theI and Λ parts ofRrespectively, the normal subgroupNR] contains elements that may be implied by all three components of pairs in R. Indeed, it may be that some elements inN0are in fact implied to be inN by someqλµij, and so could be removed from
N0 without any loss. Identifying which elements are required in N0 and which are not could be difficult computationally, but could be an interesting area of further research that would guarantee minimality in all cases. Note that, so long as N0, S0 andT0 are all finite, the set of
pairs is guaranteed to be finite, as is the number of steps in the algorithm. Hence we can be certain that the algorithm will terminate, since we can never enter an infinite loop.
We can see how the algorithm performs on the following example.
Example 3.27. Consider the semigroup S =M0[S
4;{1,2,3},{1,2,3,4};P] where S4 is the
symmetric group of degree 4, and P is the 4×3 matrix 0 0 (1 2)(3 4) (1 4) () 0 (1 3 2 4) (2 3 4) 0 0 (1 4 2) 0 .
This semigroup has 8 congruences: the universal congruence ∇S, and 7 congruences defined
by linked triples – this can be calculated slowly by hand, but much more quickly using the Semigroupspackage [M+19]. One such congruence is given by the linked triple (A
4,∆3,(2,3)e),
where A4 is the alternating group of degree 4, ∆3 is the diagonal relation on the column set
{1,2,3}, and (2,3)e is the equivalence on the row set{1,2,3,4} which only unites rows 2 and 3.
If we callPairsFromLinkedTriple(A4,∆3,(2,3)e), the algorithm first produces the three
generating componentsN0,S0 andT0. The alternating groupA
4 can be generated by the set
{(1 2 3),(2 3 4)}, so we may choose this to be our generating setN0; since ∆3 is diagonal we
produceS0 =
∅and since only two rows are united by (2,3)e we produce T0 ={(2,3)}. Now
we collate these three sets to make generating pairs for the congruence.
For the first pair, we have a = (1 2 3), the default column values of (i, j) = (1,1), and (λ, µ) = (2,3). We use the lowest possible values forξ and k: ξ= 2 andk = 1. The pair we add is
(i, pξi−1ap−λk1, λ),(j, pξj−1p−µk1, µ)= (1, p21−1(1 2 3)p−211,2),(1, p21−1p−311,3)
= (1,(1 4)(1 2 3)(1 4),2),(1,(1 4)(1 4 2 3),3) = (1,(2 3 4),2),(1,(1 2 3),3).