3.5 The order problem, finiteness problem and groups and semigroups generated by
3.5.3 The order problem in H n
In this section we shall start to introduce the tools needed to understand the order problem in the groupHn. However, as the techniques are applicable toPenand will be relevant in later sections, we shall work with the monoidPenfor majority of this section. It is standard in the literature to tackle the order problem by investigating the structure of the dual transducer, see for instance [38, 1], and this is what we do below. The exposition and results in this paper are from the paper [44]. The dual at levelk
LetA=hX,Q,π,λibe a synchronous transducer and letk∈N. We form the levelkdual,
A∨k =hX∨k,Qk∨,π∨k,λ∨ki
ofAas follows. The state setQ∨k ofA∨k is the set of all words of lengthkin the input alphabetX. This dual transducer has its input alphabet equal to its output alphabet and they are both equal to
X∨k :=Qthe set of states ofA. The transition functionπ∨k is defined as follows: for statesq,q0∈Q, andΓ,Γ0∈Q∨k we have:
1. π∨k(q,Γ) =Γ0if and only ifλ(Γ,q) =Γ0, and 2. λ∨k(q,Γ) =q0if and only ifπ(Γ,q) =q0.
We observe thatA∨k+1=A∨k ∗A1∨. For suppose thatΓ ia word of lengthk+1 is a state inA∨k+1, andqis any state symbol ofAsuch that after readingqfromΓ iinA∨k+1we are in state∆jand the output isp. Then inAwe haveπ(Γ i,q) =pandλ(Γ i,q) =∆j. We can break up this transition into two steps. Supposeπ(Γ,q) =p0, then we haveλ(Γ,q) =∆,λ(i,p0) =jandπ(i,p0) =p. Hence in
A∨k we readqfromΓ and transition to∆andp0is the outputp0. Moreover inA∨1 we readp0from
iand transition tojwith outputp. Therefore the state(Γ,i)ofA∨k ∗A∨1 , is such that we readq
from this state and transition to the state(∆,j)and the output produced isp.
Notation 3.5.25. As we shall sometimes work simultaneously with a transducerAand its dual, giveni∈N1we shall, occasionally write a state(p1, . . . ,pi)ofAias a wordp1. . .pi.
The following definition gives a tool which connects the synchronizing level of powers of an element ofPento a property of the dual transducer.
Definition 3.5.26(Splits). LetAbe an element ofPen, with synchronizing levelk. Then we say that
A∨r (r>k)splitsif we have the following picture inA∨r :
Γ Γ1 Λ1 Λ2 Γ2 Λl−1 Γl−1 Λl Γl q1|s1 p1|s1 q2|s2 p2|s2 pl|sl ql|sl Figure 3.9: A split
whereΓl∈Wt1andΛl∈Wt2 for distinct statest1andt2ofA, and for all other pairs(Γi,Λi), 16i6l−1,ΓiandΛiare in the sameWui. We say that thel-tuples(p1, . . . ,pl)and(q1, . . . ,ql)
splitA∨r. We shall call {p1,q1}the top of the split, {t1,t2}the bottom of the split, and the triple
Definition 3.5.27. LetAbe an element of Pen, with synchronizing levelk. Letr > kand let
((q1. . . ,ql),(p1, . . . ,pl),Γ)be a split ofA∨r forΓ ∈ Xrn and(q1. . . ,ql),(p1, . . . ,pl) ∈ QlA. Let
{t1,t2}be the bottom of this split. Then we say thatthe bottom of the split((q1. . . ,ql),(p1, . . . ,pl),Γ) depends only on the topif for any other tuplesU1,U2 ∈ Ql−1A we have that((q1,U1),(p1,U2),Γ)
is also a split with bottom{t1,t2}and we have for any u,u0 ∈ Q, πAl+1(Γ,(p1, . . . ,pl,u)) =
πAl+1(Γ,(p1,U2,u0)) and πAl+1(Γ,(q1, . . . ,ql,u)) = πAl+1(Γ,(q1,q2,u0)). The last condition means that ifλAl(Γ,(q1, . . . ,ql))∈Wt1 then so also isλAl(Γ,(q1,U1))and likewise for(p1, . . . ,pl) and(P1,U2).
Definition 3.5.28. For a transducerA, we define ther splitting length ofA(forrgreater than or equal to the minimal synchronizing length) to be minimallsuch that there is a pair ofl-tuples of states which splitA∨r. If there is no such pair the we set the r splitting length ofAto be∞. Remark 3.5.29. LetAbe a transducer with minimalrsplitting lengthl <∞. By minimality oflit follows that for a given pair inQl×Qlwhich splitsA∨r, then the bottom of the split depends only on the top. Therefore the top and bottom of the split have cardinality two. In particular for any split whose bottom depends only on its top, the top and bottom of the split both have cardinality two.
Remark 3.5.30. LetAbe a transducer such that the minimalrsplitting length ofAis infinite for somer, then the minimalr+1 splitting length ofAis also infinite.
The following lemma demonstrates that forA∈Penandr >2, thersplitting length ofAis bigger than ther−1 splitting length ofA.
Lemma 3.5.31. LetA∈Penbe synchronizing at levelk, and suppose that themksplitting length ofAis finite form∈N,m >0, then the(m+1)ksplitting length ofAis strictly greater than themksplitting length ofA.
Proof. Suppose thatAhasmksplitting lengthl. It suffices to show that for any wordΓ ∈X(m+1)kn , and anyl+1-tuplePinQl+1A , the output ofPthroughΓ depends only onΓ.
First we set up some notation. LetAj:=hXn,QjA,λj,πjiand letA∨j :=hQA,X j
n,λ∨j ,π∨j ifor
j∈N. For a wordγ∈Xnkletqγdenote the state ofAforced byΓ.
Now since Ahasmk splitting lengthl, it follows that for anyP := (p1, . . . ,pl) andT :=
(t1, . . . ,tl)inQlAandΓ ∈ Xmkn we have thatλ∨mk(P,Γ) =λ∨mk(T,Γ). By definition of the dual,
λ∨mk(P,Γ) =πl(Γ,P).
Now let γ ∈ Xkn be arbitrary and letp ∈ QA and P ∈ QlA also be arbitrary. Consider
λ∨(m+1)k(Pp,Γ γ), we have:
λ∨(m+1)k(Pp,Γ γ) =πl+1(Γ γ,Pp) =πl(γ,πl(Γ,P))π1(λl(γ,πl(Γ,P)),π1(λl(Γ,P),p))
However observe that sinceAis synchronizing at levelkthat the suffix
π1(λl(γ,πl(Γ,P)),π1(λl(Γ,P),p))
depends only onλl(γ,πl(Γ,P)). However sinceA∨mkhas minimal splitting lengthlwe have that
πl(Γ,P)depends only onΓ. Therefore we have thatλ∨(m+1)k(Pp,Γ γ)depends only onΓ γ. Remark 3.5.32. It follows from the lemma above that ifA∈Penis synchronizing at levelk, then themksplitting length ofA, if it is finite, is at leastmform∈N,m >0.
The following lemma shows that the minimal splitting length is connected with the synchronizing level of powers of a transducer.
Lemma 3.5.33. LetAbe a transducer with synchronizing level less than or equal tok. IfAhasksplitting lengthl, thenmin(Core(Al+1))has minimal synchronizing levelm>k+1.
Proof. Letlbe as in the statement of the lemma. Now consider Core(Al). The states of Core(Al)
will consist of all lengthloutputs ofA∨k. Moreover by choice ofl, Core(Al)is also synchronizing at levelk.
LetΓ be the word which achieves the minimall, and suppose the picture is exactly as given in Figure 3.9, whereΓl ∈Wt1andΛl∈Wt2 for distinct statest1andt2.
We now considerB:=Core(A∗Core(Al)). It is easy to see that there are states inBof the form
(p1,P),(q1,Q)for appropriateP,Q∈Core(Al). Therefore inBwe have that when we have read
Γ through(p1,P), we are in state(s1, . . . ,sl,t1), and when we have readΓ through state(q1,Q)
we go to state(s1, . . . ,sl,t2). Sincet16=t2these states are notωequivalent. This concludes the proof.
Lemma 3.5.34. An elementA∈Hneither has finite order or for allk∈Nthere is anN∈Nsuch that for allm∈Nwe have thatANmis bi-synchronizing at level greater thank, moreoverNis depends only onAandk.
Proof. Suppose thatAdoes not have finite order. SinceAhas infinite order thenA∨j splits for every
j∈N, therefore there is anm0∈NandN1∈Nsuch that min Core(AN1)is bi-synchronizing at
levelm0. In order to simplify the notation we shall identifyAN1 with the minimal, core transducer
min Core(AN1).
Now consider the permutationANm1
0, we shall assume that it is written as a product of disjoint
cycles. Letdbe the order of this permutation. LetΓ ∈Xm0
n , thenΓ belongs to a cycle(Γ1Γ2. . .ΓdΓ)
where Γ1 := Γ anddΓ|d. LeteΓ = d/dΓ. To this cycle there is associated a tuple of states
(qΓ1,qΓ2, . . . ,qΓdΓ)whereqΓi is the state ofA
N1 forced byΓ
i for 16i6dΓ. Now observe that
(qΓ1,qΓ2, . . . ,qΓdΓ)is a state ofAN1dΓ, moreover sinceλN1(Γi,qΓi) = Γi+1for 16 i < dΓ, and
λN1(ΓdΓ,qΓdΓ) =Γ1, then we have that
πN1d
Γ(Γ1,(qΓ1,qΓ2, . . . ,qΓdΓ)) = (qΓ1,qΓ2, . . . ,qΓdΓ).
Now lettΓ be the order of the permutation induced by(qΓ1,qΓ2, . . . ,qΓdΓ)eΓ onXmn0.
Lettbe the lowest common multiple of the set{tΓ|γ∈ Xmn0}. In order to keep the notation concise letPΓ represent the state(qΓ1,qΓ2, . . . ,qΓdΓ)eΓt. Notice thatPΓ acts locally as the identity for allΓ ∈Xm0
n . MoreoverPΓ is the state ofAN1dtsuch thatπN1dt(Γ,PΓ) =PΓ.
Now letN=N1dt, and letm>N. Suppose thatAm, (where againAmis identified with the minimal core transducer ofAm) is synchronizing at levelm0. Sincem>Nwe may writem=rN
for somer∈Nand 06s < N. Therefore states ofAmlook likePforPa state ofAN.
Now observe once more that all the statesPΓ are locally the identity for allΓ ∈ Xmn0 and
πm(Γ,PΓ) =PΓ. Now sinceAm is synchronizing at levelN, we must have that the state ofAm forced byΓ is preciselyPΓ. Therefore the states ofAmcan be identified with the statesPΓ. Now as all of these states are locally identity it follows thatAmis the identity. Which is a contradiction of our initial assumption thatAdoes not have finite order. ThereforeAmmust be synchronizing at level greater thanm0.
Lemma 3.5.35. LetA ∈ Pn be a core, minimal transducer such that|A| > n(n+1). LetBbe any transducer synchronizing at level1. Thenmin Core(AB)is synchronizing at level strictly greater than 1. Proof. For eachi ∈ Xn letIi := {πA(i,p)|p ∈ QA}. Notice since Ais synchronizing it is also strongly connected, therefore for allp∈QAthere is a setIifor somei∈Xnsuch thatp∈Ii. It now follows that∪i∈XnIi=QA.
Now if|Ii|< n+1 for allithen we have that:
|A|=|∪i∈XnIi|6
n
X
i=1
|Ii|< n∗(n+1)<|A|
which is a contradiction. Therefore there must be ani∈Xnsuch that|Ii|> n+1. Fix such an
Now since|Ii|> n+1, there must be statesp10,p20,p1,p2∈QAsuch thatp16=p2andp10 6=p20 and such that the following transitions are valid:
p10−→i|j p1 p20 −→i|j p2 for somej∈Xn.
Now observe that there are states(p10,q10)and(p20,q20)in the core ofABwhereq1andq2are states ofB. LetπB(j,q10) =qjandπB(j,q20) =qj(sinceBis synchronizing at level 1).
Therefore the following transitions are valid:
(p10,q10)−→i|l1 (p1,qj) (p20,q20) i|l1
−→(p2,qj)
wherel1 =λB(j,q10)anl2= λB(j,q20)Now if min Core(AB)is synchronizing at level 1, then
(p1,qj)an(p2,qj)would beω-equivalent, since(p10,q10)and(p20,q20)are states in the core ofAB. However(p1,qj)=∼ω(p2,qj)implies thatp1=∼ωp2, but by assumptionp1andp2are distinct and
Ais minimal and sop1=∼ωp2is a contradiction.
Therefore min Core(AB)is not synchronizing at level 1.
Now suppose thatA ∈ Pen and the semigrouphAi+ := {Ai|i ∈ N}is finite. Notice that if
A∈Henand the semigrouphAi+is finite then it coincides with the group generated byA. The next result demonstrates that in the case where the semigrouphAi+is finite, there must be some
j∈ N,j > 0 for which thejsplitting length ofAis infinite. From this result one may deduce Lemma 3.5.22.
Lemma 3.5.36. LetA∈ePnbe synchronizing at levelk. Suppose that the semigrouphAi+is finite, and thatjis the maximum of the minimal synchronizing level of the elements ofhAi+. ThenAhas infinitej
splitting length.
Proof. This is a consequence of Lemma 3.5.33. Since if A has j splitting length l, then by Lemma 3.5.33 min(Core(Al+1))has minimal synchronizing levelj+1, which is a contradiction. Remark 3.5.37. The above means that we can partitionA∨j into componentsD1, . . . ,Disuch that to each component there is a pair of wordsWi,1andWi,2in the states ofAsuch that the only possible outputs from the componentDifor any input have the formu(Wi,2)lvwhereuis any suffix ofWi,1Wi,2, including the empty suffix, andvis a prefix ofWi,2including the empty prefix. Below are some examples of finite order bi-synchronizing, synchronous transducers witnessing Lemma 3.5.36.
Example 3.5.38. Consider the transducerCbelow. This is a transducer of order 3, in particular, it is a conjugate of the single state transducer which can be identified with the permutation(0, 1, 2).
q0 q1 q2 0|1 2|0 1|2 0|0 1|2 2|1 1|1 0|2 2|0
This transducer, as noted before, is bi-synchronizing at the second level. The level 3 dual has 27 nodes and so we shall not give this below. However utilising either the AAA package or the Automgrp package [41] in GAP [25], together with (in AutomGrp) the function “MinimizationOfAutomaton( )” which returns anω-equivalent transducer, applied to the third
power of the dual transducer, we get the following result:
a0
a1 a2
q0|q0, q1|q0, q2|q0
q0|q1, q1|q1,q2|q1
q0|q2,q1|q2, q2|q2
Figure 3.11: The level 3 dual ofC.
Since the original transducerChas order 3 we can see from its level 3 dual above that the states in the core will be cyclic rotations of(q0,q1,q2)all of which are locally identity.
We illustrate another example below, but now with an element of order 2.
Example 3.5.39. Consider the transducer of order two given below constructed based on an example in [10]. q0 q1 0|0 1|1 2|2 0|1 1|0 2|2
Figure 3.12: An element of order 2
This transducer is synchronizing on the first level. We give the dual below.
0 1 2 q1|q0 q0|q0 q1|q0 q0|q0 q0|q1, q1|q1
Figure 3.13: The level 1 dual.
that producesq0for all inputs. The states in the core of the square of the original transducer will be(q0,q0)and(q1,q1).
For a transducer of finite order,A, as above, we have the following result about the semigroup
hA∨i+.
Theorem 3.5.40. LetA∈Penbe synchronizing at levelk. Suppose that the semigrouphAi+is finite with
j∈N1the maximum of the minimal synchronizing levels of the elements ofhAi+, thenA∨j = (A∨)jis a zero inhA∨i+, the semigroup generated byA∨.
Proof. It suffices to show thatA∨j is a right zero of the semigroup since the semigrouphA∨i+is commutative.
Our strategy shall be to to show that for any stateΓofA∨j and any statexofA∨, that the state
Γ xofA∨j+1isω-equivalent to a state ofAj∨. To this end letΓ ∈Xjnbe a word of lengthj+1. By Lemma 3.5.31 and Remark 3.5.37 there is a pair of wordsW1,Γ1andW2,Γ1 such that any input read fromΓ1has output of the formW1(W2)lvforl∈Nandva prefix ofW2, otherwise the output is a prefix ofW1. Letγ∈Xnj be the lengthjsuffix ofΓ1. Observe that the outputs of the stateγof
A∨j must also all be of the formW1(W2)lvforl∈Nandva prefix ofW2, otherwise the output is a prefix ofW1and the output depends only on the length of the input word. Therefore we must have thatΓ andγareω-equivalent.
On the other hand, given a wordγ∈Xjn, then a similar argument demonstrates that the state
xγfor anyx∈Xjnisω-equivalent toγ.
The next result observes that Lemma 3.5.36 gives a complete characterisation of elements of
Hnwith finite order.
Proposition 3.5.41. LetAbe an element ofPen and supposeAis synchronizing at levelk. Then the semigrouphAi+generated byAis finite if and only if there is somem∈Nsuch that the following hold:
(i) A∨mis a zero of the semigrouphA∨i,
(ii) A∨misω-equivalent to a transducer withrcomponentsDi16i6r. For each componentDithere is a fixed pair of wordswi,1,wi,2(in the states ofA) associated toDisuch that whenever we read any input from a state in theDi, the output is of the formwi,1wli,2vforl∈Nandva prefix ofwi,2
or has the fromufor some prefixUofwi,1. Moreover the output depends only on which state in the componentDiwe begin processing inputs.
Proof. ⇒: This direction follows from Lemma 3.5.36, Remark 3.5.37 and Theorem 3.5.40.
⇐: Assume thatA∨m hasr components and letwi,1 andwi,2 1 6 i 6 rbe the pair of words in the states ofAassociated with each componentDi. To see that the semigrouphAi+is