In this section we describe a several ways of combining two elements ofePnandPem, into a single element ofPem+n. As we will see, this has interesting consequences when we restrict to the group
Hn.
LetA = hXn,QA,πA,λAi and B = hXm,QB,πB,λBibe elements of Pen andPem. It is a consequence of Proposition 3.4.6 that for all 16i6nan 16j6mthere is a state,qiofAandPj ofAandBrespectively such thatπA(i,qi) =qiandπB(j,pj) =pj.
Form a new transducerBe =h{n, . . . ,n+m−1},Q
e
B,πBe,λBeiwith input and output alphabet
{n, . . . ,n+m−1}such that the states ofBeare in bijective correspondence with the states ofB; we denote them byqe, whereqis a state ofB.
The transition and rewrite function ofBeare defined by the following rules fori,j∈Xm:
π e B(n+i,eq) =pe ⇐⇒ πB(i,q) =p λ e B(n+i,eq) =n+j ⇐⇒ λB(i,q) =j
Now we form a new transducer,AtB=hXn+m,QAtB,πAtB,λAtBias follows:
QAtB=QAtQBe; the states ofQAtransition exactly as inAfor all inputs inXnand the states
ofQ
e
Btransition as inBefor all inputs inXn+m\Xn. Finally for alli ∈Xn, and for anyeq∈QBe,
λAtB(i,qe) =i. An analogous condition holds inAtBfor the states ofAon inputs inX2n\Xn. We shall demonstrate this in Example 3.6.4.
Now further assume thatA∈PenandB∈Pemboth have finite order (notice that this implies
A∈HnandB∈Hm). We argue that the order ofAtBis at leastlcm(O(A),O(B))— the lowest common multiple of the orders ofAandB.
First we show thatAtBhas finite order.
Letkbe the maximal bi-synchronizing level ofAandB. We shall give two different proofs showing thatAtBis synchronizing; similar arguments show that(AtB)−1is also synchronizing at levelk+1.
Claim 3.6.1. AtBis bi-synchronizing at levelk+1
Proof 1. Observe that as soon we readi,i∈Xn, we must be processing from a state ofAand if we read ann+i,i∈Xnwe must be processing from a state ofBe.
LetΓ ∈Xk+12n a word of lengthk+1.
IfΓ ∈Xk+1n orΓ ∈{n, . . . , 2n−1}k+1, then the state ofAtBforced byΓ is the state ofAoreB forced byΓ. Since reading the first letter guarantees, by the observation in the first paragraph, that the active state is a state ofA(orBifΓ ∈{n, . . . , 2n−1}k+1), andAandBare bi-synchronizing at levelk.
Hence we need only consider the case thatΓ contains at least one letter fromXnand one letter fromX2n\Xn.
LetΓ =g0. . .gk. Letgi ∈Xn and assumeg0∈X2n\Xn(the other case follows by a similar argument) and suppose that 0< jandjis minimal bigger thanisuch thatgj∈X2n\Xn. By the observation in the first paragraph, regardless of the starting state, after processing theg0, the active state must be some state ofA. By the minimality ofj, after processinggj−1, the active state is still some state ofA. Now, notice that every state ofAwill readgjto a fixed stateeqgjofB. Therefore
regardless of the starting position, we always process the finalk−jinputs from the stateqgj.
To see that(AtB)−1is also synchronizing at levelk+1 observe that the states corresponding to statesA−1 in (AtB)−1 process words in X∗n exactly as A−1 does. Moreover all states of
(AtB)−1corresponding to states ofA−1read a fixed letterjin{n, . . . , 2n−1}to a unique state e
q−1j corresponding to the stateeq −1
j ofBe−1. Analogously for the states of(AtB)1corresponding to the states ofBe−1and elements of{n, . . . , 2n−1}∗and letters inXn. Therefore we may repeat the argument already given for(AtB)to show that(AtB)−1is synchronizing at levelk+1 also. Proof 2. We apply the Collapsing procedure (Construction 2.2.1). Since k is the maximal bi- synchronizing level ofAandB, then afterksteps of the procedure, the copies ofAandBhave both been reduced to singletons. Now since every input inXnis read intoAand every input in
Xn+m\Xnis read intoB, we only need at most one additional step to reduceAtBto a singleton. The result now follows. The same argument shows that (AtB)−1 is synchronizing at level
k+1.
We now free the symbolk. To show thatAtBhas finite order, we prove the following claim. Claim 3.6.2. Letkbe minimal such that bothA∨k andB∨k are the zero ofhA∨i+andhB∨i+respectively. ThenAtB∨k+1isω-equivalent to a disjoint union of cycles as in Proposition 3.5.41.
Proof. First we consider the case whereΓ ∈Xk+1n orΓ ∈{n+1, . . . ,n+m−1}k+1. LetΓ =g0. . .gk. By the assumption that bothAandB(henceBe) have finite order, this implies that there is a state
qofA(orBe ifΓ ∈ {n+1, . . . ,n+m−1}k+1) such that the image ofΓ is in the setWq. This is because after readingg0we are enter a state ofA(oreBifΓ ∈{n+1, . . . ,n+m−1}k+1), and using the fact thatg1. . .gkbelongs to a cycle of states as in Proposition 3.5.41 inA∨k (orBe∨if
Γ ∈{n+1, . . . ,n+m−1}k+1). Moreover ifΓ ∈Xkn, then the image ofΓ through any state ofAtB
is also inXknand analogously ifΓ ∈{n+1, . . . ,n+m−1}k+1. Therefore the fact thatΓ belongs to such a cycle of states is a consequence of the fact thatAandBhave finite order.
Now we consider the case whereΓ contains a letter inXnand a letter inXn+m\Xn. Similarly to the proof of Claim 3.6.1, letΓ =g0g1. . .gk. Suppose that a letter fromXn(the other case being
analogous) occurs first and letjbe minimal such thatj >0 andgj ∈Xn+m\Xn. The proof of Claim 3.6.1 shows that the state ofAtBforced byΓdepends only on the suffixgjgj+1. . .gk. Let
qgjbe the state ofBe such thatπBe(gj,qgj) = Qgj. Since every state ofAacts as the identity on
Xn+m\Xn, it is the case that processingΓ from any state ofAtB, the first letter of the output is an element ofXnand thejth letter is the minimal element inXn+m\Xn and is in fact equal to
gj, moreover since we process the lengthj−ksuffix from the stateqgj, the lengthj−ksuffix is
independent of the state we begin processing from. However we can now repeat this argument to show that the output through any state of the set of images ofΓ, have the samej−ksuffix and thejthletter equal togj. It now follows from the observation above that the state ofAtBforced depends only on the lengthj−ksuffix, and by induction, thatΓ belongs to a cycle of states as in Proposition 3.5.41.
The above two paragraphs show that it is possible to decomposeAtBinto a disjoint union of cycles as in Proposition 3.5.41 and soAtBhas finite order.
The following alternative way of combining elements of Pen also has the property that combining finite order elements results in elements of finite order by mechanical substitutions in the arguments above.
LetA∈ePnandB∈Pembe as above and formBeas before. For each 16i6n, letpibe the state ofAsuch thatπA(i,pi) =pi, by the definition of the transformation (in the case whereA
andBhave finite order a permutation),A1, we haveλA(i,pi) =A1(i), likewise there is a stateqj ofBsuch thatλB(j,qj) =B1(j)for 16j6m.
We formA⊕Banalogously toAtB. The set of states, and the transition function,πA⊕Bare identical but we make some adjustments to the rewrite function. For any letteri∈Xn, and any stateqeofBe, we takeλA⊕B(i,qe) =A1(i), likewise for any lettern+j∈{n, . . .n+m−1}and any statepofA, we haveλA⊕B(j,p) =n+B1(j).
We have the following claim analogous to Claim 3.6.2 and proved in a similar way.
Claim 3.6.3. Letkbe minimal such that bothA∨k andB∨k are the zero ofhA∨i+andhB∨i+respectively. ThenA⊕B∨k+1isω-equivalent to a disjoint union of cycles as in Proposition 3.5.41.
The methods described above of combining elements ofPendo not exhaust all possibilities, for instance we could fix a state ofeBsuch that reading any letterXn+m\XnfromAgoes into this state, and likewise we could fix such a state ofBe. Similar arguments to those given above will show that these methods also give rise to elements of finite order whenever the initial elements have finite order, in fact we may prove versions of Claim 3.6.2 for each by making mechanical substitutions in the original proof.
We remark also that as there are new cycles of states introduced in(AtB)∨k+1and(A⊕B)∨k+1
that are not present inA∨k or B∨k it might be that the order ofAtB is strictly greater than
lcm(O(A),O(B))in some cases. We give an example below.
Example 3.6.4. Consider the elements of H3 and H2 =∼ Z/2Z of order three and order 2 respectively,
q0 q1 q2 p 1|2 2|0 0|1 2|2 0|1 1|0 1|1 0|2 2|0 0|1 1|0
Figure 3.22: Elements ofH3andH2.
we now combine them to give an element of order 6 inH5. Using any of the methods describe above will yield an element of order 6, we only illustrate one such method.
q0 q1 q2 p 1|2 2|0 0|1 3|3,4|4 2|2 0|1 1|0 3|3, 4|4 1|1 0|2 2|0 3|3,4|4 3|4 4|3 1|1 0|0 2|2
Figure 3.23: The result of combining the elements in Figure 3.22.