Translatability and translatable semigroups

Definition 2.1

A finite groupoid Q is called k-translatable, where 1⩽k<n, if its multiplication table is obtained by the following rule: If the first row of the multiplication table is a₁,a₂,…,a_n, then the q-th row is obtained from the (q−1)-st row by taking the last k entries in the (q−1)−st row and inserting them as the first k entries of the q-th row and by taking the first n−k entries of the (q−1)-st row and inserting them as the last n−k entries of the q-th row, where q∈{2,3,…,n}. Then the (ordered) sequence a₁,a₂,…,a_n is called a k-translatable sequence of Q with respect to the ordering 1,2,…,n. A groupoid is called translatable if it has a k-translatable sequence for some k∈{1,2,…,n−1}.

Example 2.2

Consider the following groupoids:

The first table determines a 3-translatable semigroup isomorphic to the additive group Z₄. The second table says that after a change of ordering this semigroup is not translatable. The last table determines an idempotent 2-translatable groupoid which is not a semigroup.

Lemma 2.3

For a fixed ordering a left cancellative groupoid may be k-translatable for only one value of k.

Proof

By renumeration of elements we can assume that a groupoid Q of order n has a natural ordering 1,2,…,n. If the first row of the multiplication table of such groupoid has the form a₁,a₂,…,a_n, then all a_i are different. The first element of the second row is equal to a_n−k+1 if a groupoid is k-translatable, or a_n−t+1 if it is t-translatable. So, 1⋅(n−k+1)=a_n−k+1=2⋅1=a_n−t+1=1⋅(n−t+1), which, by left cancellativity, implies n−k+1=n−t+1, and consequently k=t.

Lemma 2.4

A k-translatable groupoid containing at least one left cancellable element is left cancellative.

Proof

Indeed, let i be the left cancellable element of a k-translatable groupoid Q. Then, i-th row of the multiplication table of this groupoid contains different elements. By k-translability other rows also contain different elements. So, for all j,s,t∈Q from j⋅s=j⋅t it follows that s=t. Hence, Q is left cancellative.

Lemma 2.5

Let a₁,a₂,…,a_n be the first row of the multiplication table of a groupoid Q of order n. Then Q is k-translatable if and only if for all i,j∈Q the following (equivalent) conditions are satisfied:

1. i⋅j=a_{[(i−1)(n−k)+j]}n=a_[k−ki+j]n,

2. i⋅j=[i+1]_n⋅[j+k]_n,

3. i⋅[j−k]_n=[i+1]_n⋅j.

Proof

From the definition of a k-translatable groupoid it is clear that a groupoid is k-translatable if and only if i⋅j=[i+1]_n⋅[j+k]_n, which implies i⋅[j−k]_n=[i+1]_n⋅j.

Assuming that i⋅[j−k]_n=[i+1]_n⋅j, we prove by induction on i that i⋅j=a_[k−ki+j]n. First note that since the first row of the Cayley table is a₁,a₂,…,a_n, it follows by the definition of k-translatable that, in every other row, a_i is always followed by a_[i+1]n and preceded by a_[i−1]n. For i=1, i⋅j=1⋅j=a_j=a_[k−k1+j]n. Assume that i⋅j=a_[k−ki+j]n. Then, [i+1]_n⋅j=i⋅[j−k]_n=a_{[i−ki+j−k]}n=a_{[k−k(i+1)+j]}n. Hence, by induction, i⋅j=a_[k−ki+j]n. Assuming i⋅j=a_[k−ki+j]n, clearly [i+1]_n⋅[j+k]_n=a_{[k−k(i=1)+j+k]}n=a_[k−ki+j]n=i⋅j.

Lemma 2.6

For a k-translatable left cancellative groupoid Q of order n the following statements are valid:

1. a_i=a_j if and only if i=j,

2. Q is idempotent if and only if i=a_[k−ki+i]n for all i∈Q,

3. if Q is idempotent then i⋅j=t if and only if [j−ki]_n=[t−kt]_n for all i,j,t∈Q,

4. Q is elastic if and only if [i+ki]_n=[(j⋅i)+k(i⋅j)]_n for all i,j∈Q,

5. Q is strongly elastic if and only if [i+ki]_n=[(j⋅i)+k(i⋅j)]_n and [i+kj]_n=[(i⋅j)+k(i⋅j)]_n for all i,j∈Q,

6. Q is bookend if and only if i=a_{[k−k(j⋅i)+(i⋅j)]}n for all i,j∈Q,

7. Q is left distributive if and only if [(i⋅j)+k(s⋅i)]_n=[(s⋅j)+ks]_n for all i,j,s∈Q,

8. Q is right distributive if and only if [s+k(i⋅s)]_n=[(j⋅s)+k(i⋅j)]_n for all i,j,s∈Q,

9. Q is medial if and only if [(w⋅z)+k(i⋅w)]_n=[(j⋅z)+k(i⋅j)]_n for all i,j,w,z∈Q,

10. Q is alterable if and only if [j+kw]_n=[z+ki]_n implies that [w+kz]_n=[i+kj]_n for all i,j,w,z∈Q,

11. Q is commutative if and only if k=n−1,

12. Q is associative if and only if [i+kj]_n=[(s⋅i)+k(j⋅s)]_n for all i,j,s∈Q.

Proof

(i)−(xii) follow from Lemma 2.5, the fact that left cancellation implies that a_x=a_y if and only if x=y and the definitions of idempotent, elastic, strongly elastic, bookend, etc. For example, in (x), Q is alterable if and only if i⋅j=w⋅z implies j⋅w=z⋅i. Using Lemma 2.5(i), this is valid if and only if a_[k−ki+j]n=a_[k−kw+z]n implies a_[k−kj+w]n=a_[k−kz+i]n. Left cancellation then implies the required result.

Lemma 2.7

The sequence a₁,a₂,…,a_nis a k-translatable sequence of Q with respect to the ordering 1,2,…,n if and only if a_k,a_k+1,…,a_n,a₁,…,a_k−1 is a k-translatable sequence with respect to the ordering n,1,2,…,n−1.

Proof

(⇒): The new ordering can be written as 1_′,…,n_′, where i_′=[i−1]_n. Then, since Q is k-translatable with respect to the ordering 1,2,…,n, by Lemma 2.5(ii), [i_′+1]_n⋅[j_′+k]_n=i⋅[j−1+k]_n=[i−1]_n⋅[j−1]_n=i_′⋅j_′. Hence, Q is k-translatable with respect to the ordering n,1,2,…,n−1, with k-translatable sequence n⋅n,n⋅1,n⋅2,…,n⋅(n−1). That is, using Lemma 2.5(i), the k-translatable sequence is a_k,a_k+1,…,a_n,a₁,a₂,…,a_k−1.

(⇐): The order n,1,2,…,n−1 can be written as 1_′,2_′,…,n_′, where i_′=[i−1]_n for all i∈{1,2,…,n}. The order 1,2,…,n can be written as i=[i_′+1]_n. Then, by Lemma 2.5(ii), we have [i+1]_n⋅[j+k]_n=[i+2]_′n⋅[j+k+1]_′n=[i+1]_′n⋅[j+1]_′n=i⋅j and so 1⋅1,1⋅2,…,1⋅n is a k-translatable sequence with respect to the order 1,2,…,n. But a_k,a_k+1,…,a_n,a₁,…,a_k−1 is the k-translatable sequence with respect to the order 1_′,2_′,…,n_′. Let’s call this sequence b₁,b₂,…,b_n. Note, then, that b_i=a_[i−1+k]n. So we have 1⋅j=[i+1]_′n⋅[j+1]_′n=2_′⋅[j+1]_′n=1_′⋅[j+1−k]_′n=b_[j+1−k]n=a_j. This implies that the sequence a₁,a₂,…,a_n is a k-translatable sequence with respect to the order 1,2,…,n.

Corollary 2.18.8

A k-translatable groupoid of order n has at least nk-translatable sequences, each with respect to a different order.

Proposition 2.9

An alterable, right solvable and right distributive groupoid is an idempotent quasigroup.

Proof

A right solvable groupoid Q is left cancellative. Right distributivity and left cancellativity give idempotency.

Then the right solvability means that for all z,w∈Q there are uniquely determined elements q,q_′∈Q such that z⋅q=w and w⋅q_′=z⋅w. Then, using alterability, w=w⋅w=q_′⋅z. The element q_′ is uniquely determined. Indeed, if w=q_′⋅z=q_′′⋅z, then, by alterability, z⋅q_′′=z⋅q_′. This implies q_′=q_′′. Hence Q is also left solvable and, hence, it is a quasigroup.

Lemma 2.10

For fixed k≠1 and n there is an idempotent k-translatable groupoid of order n. It is left cancellative.

Proof

First observe that an idempotent groupoid cannot be k-translatable for k=1. Let k>1. Then, by Lemma 2.5, in an idempotent k-translatable groupoid Q, for all i∈Q we have i=a_[k−ki+i]n. Since Q is idempotent, the first row of the multiplication table of this groupoid contains the distinct elements a₁,a_[2−k]n,a_[3−2k]n,…,a_[n+k(1−n)]n, with i appearing in the [k−ki+i]_n-column. So, by Lemma 2, this groupoid is left cancellative. By k-translatability, other rows are uniquely determined by the first.

Lemma 2.11

An idempotent k-translatable groupoid that is alterable, strongly elastic or bookend is cancellative, i.e., it is a quasigroup.

Proof

By Lemma 2.10 such groupoid is left cancellative. Let i⋅j=s⋅j. Then obviously s=[i+t]_n for some t∈{1,2,…,n}. Since a_[k−ki+j]n=i⋅j=s⋅j=a_[k−ks+j]n, [k−ki+j]_n=[k−ks+j]_n=[k−ki−kt+j]_n and so [kt]_n=0. But then, since Q is idempotent, t⋅t=t=a_[k−kt+t]n=a_[k+t]n=1⋅[k+t]_n=n⋅t. If Q is alterable, by definition, i⋅j=w⋅z implies j⋅w=z⋅i. Therefore, t⋅n=t⋅t=t. If Q is strongly elastic then, by definition, (i⋅j)⋅i=(j⋅i)⋅j. Therefore t=n⋅t=t⋅t=t⋅(n⋅t)=(n⋅t)⋅n=t⋅n. In either case, left cancellation implies t=n. If Q is bookend then, by definition, (i⋅j)⋅(j⋅i)=j. So, t=t⋅t=n⋅t=(n⋅t)⋅(t⋅n)=t⋅(t⋅n) and left cancellation implies t=n. Hence, if Q is alterable, strongly elastic or bookend, t=n and so s=[i+t]_n=[i+n]_n=i. Therefore, Q is also right cancellative, and consequently, it is a quasigroup.

Theorem 2.12

Idempotent k-translatable groupoids of the same order are isomorphic.

Proof

Suppose that (Q,⋅) and (S,∗) are idempotent k-translatable groupoids of order n. Then Q={1,2,…,n} has a k-translatable sequence a₁,a₂,…,a_n with respect to the ordering 1,2,…,n, and S={c₁,c₂,…,c_n} has a k-translatable sequence b₁,b₂,…,b_n with respect to the ordering c₁,c₂,…,c_n. Then, by Lemma 2.5 (i), we have i⋅j=a_[k−ki+j]n and c_i∗c_j=b_[k−ki+j]n. Since both groupoids are idempotent, i=a_[k−ki+i]n and c_i=b_[k−ki+i]n for all elements of Q and S.

Define a mapping φ:Q⟶S as follows: φ(i)=c_i (for all i∈Q). Then φ is clearly a bijection. For i,j∈Q, we have t=i⋅j=a_[k−ki+j]n=a_[k−kt+t]n, and since Q is left cancellative (Lemma 2.10), by Lemma 2.6, we obtain [k−ki+j]_n=[k−kt+t]_n. Thus φ(i⋅j)=c_i⋅j=c_t=b_[k−kt+t]n=b_[k−ki+j]n=c_i∗c_j=φ(i)∗φ(j) and so φ is an isomorphism.

Proposition 2.13

An idempotent, k-translatable groupoid Q of order n is right distributive if and only if it is left distributive.

Proof

Let i⋅j=t, i⋅s=u, j⋅s=w. The right distributivity means that t⋅s=u⋅w. Then, by Lemmas 2.10 and 4, [s+ku]_n=[w+kt]_n,[t+ki]_n=[j+kt]_n, [u+ki]_n=[s+ku]_n and [s+kw]_n=[w+kj]_n. So, [s+ku]_n=[w+kt]_n=[u+ki]_n, and consequently, [−ki+w]_n=[−kt+u]_n. Thus i⋅w=t⋅u. Hence, Q is left distributive.

The converse statement can be proved analogously.

Theorem 2.14

An idempotent k-translatable groupoid of order n with (k,n)=1 is a quasigroup.

Proof

By Lemma 2.10 such groupoid is left cancellative. To prove that it is right cancellative consider an arbitrary column of its multiplication table. Suppose that it is j-th column. This column has the form a_j,a_[j−k]n,a_[j−2k]n,…, a_{[j−(n−1)k]}n. All these elements are different. In fact, if a_[j−sk]n=a_[j−tk]n for some s,t∈Q, then (s−t)k≡0(modn), which is possible only for s=t because (k,n)=1. So, in this groupoid s⋅j=t⋅j implies s=t for all j,s,t∈Q. Hence Q is a quasigroup.

Lemma 2.15

If a k-translatable groupoid of order n has a right cancellable element, then (k,n)=1.

Proof

Let t∈Q be a right cancellable element of a k-translatable groupoid Q. Then the mapping R_t(x)=x⋅t is one-to-one. So, it is a bijection. Hence all elements of the t-column are different. Thus a_[k−ki+t]n=i⋅t=j⋅t=a_[k−kj+t]n is equivalent to k(i−j)≡0(modn). The last, for all i,j∈Q, gives i=j only in the case when (k,n)=1.

Proposition 2.16

An idempotent, k-translatable groupoid Q of order n is elastic if and only if [(i⋅j)+(j⋅i)]_n=[i+j]_n holds for all i,j∈Q.

Proof

Since Q is idempotent, for all i,j∈Q we have i⋅j=(i⋅j)⋅(i⋅j), i.e.,

which together with elasticity and Lemma 2.6(iv), gives [j+k(i⋅j)−(i⋅j)]_n=[ki]_n=[(j⋅i)+k(i⋅j))−i]_n. This implies [(i⋅j)+(j⋅i)]_n=[i+j]_n.

Conversely, if [(i⋅j)+(j⋅i)]_n=[i+j]_n, then also [k(i⋅j)+k(j⋅i)]_n=[ki+kj]_n, and consequently

This gives the elasticity of Q.

Theorem 2.17

An alterable, left or right cancellative groupoid of order n may be k-translatable only for [k₂]_n=n−1.

Proof

Since in alterable groupoids left and right cancellativity are equivalent, we will consider only an alterable groupoid with left cancellativity.

Let Q be an alterable groupoid which is left cancellative and k-translatable. Then, by Lemma 2.5(iii), for all i,s∈{1,2,…,n} we have

for all i,s∈{1,2,…,n}.

Thus,

which, by alterability, gives

But n⋅n=n⋅[n+k−k]n=(2)[n+1]n⋅[n+k]n=[n+1]n⋅k, so

which, by alterability, implies k⋅k=[n−1]_n⋅[n+1]_n. Hence

Since Q is left cancellable, we obtain 0=[1+k₂]_n. Therefore k₂≡(−1)(modn).

Corollary 2.18.18

A k-translatable and left cancellative groupoid of order n is alterable if and only if [k₂]_n=n−1.

Proof

Let a left cancellative groupoid Q of order n be k-translatable. Assume that the first row of the multiplication table of this groupoid has the form a₁,a₂,…,a_n. Then all these elements are different.

If k₂≡−1(modn), then i⋅j=g⋅h, by Lemma 2.5(i), implies [k(g−i)]_n=[h−j]_n. Multiplying both sides of this equation by k gives [k₂(g−i)]_n=[k(h−j)]_n, i.e., [i−g]_n=[k(h−j)]_n. This implies j⋅g=h⋅i. Therefore Q is alterable.

The converse statement is a consequence of Theorem 3.

Theorem 3.1

Left unitary k-translatable groupoids of the same order are isomorphic.

Proof

Suppose that (Q,⋅) and (G,∗) are left unitary, k-translatable groupoids having k-translatable sequences 1,2,…,n and 1_′,2_′,…,n_′, with respect to the orderings 1,2,…,n and 1_′,2_′,…,n_′, respectively. Then i⋅j=[k−ki+j]_n and i_′∗j_′=[k−ki+j]_′n. Define φ:Q⟶G as φ(i)=i_′. Then φ is clearly a bijection and φ(i⋅j)=φ([k−ki+j]_n)=[k−ki+j]_′n=i_′∗j_′=φ(i)∗φ(j), so φ is an isomorphism.

Theorem 5.6

A k-translatable left unitary groupoid is medial, but it is not right distributive.

Proof

In a k-translatable left unitary groupoid Q of order n we have (x⋅y)⋅(z⋅w)=(x⋅z)⋅(y⋅w)⇔[k−k(kx+y)+(k−kz+w)]_n=[k−k(k−kx+z)+(k−ky+w)]_n⇔[k(k−kx)]_n=[k(k−kx)]_n. The last identity is always valid. So, this groupoid is medial.

If it is right distributive, then (x⋅y)⋅z=(x⋅z)⋅(y⋅z) for all x,y,z∈Q, or, equivalently, [k−k(k−kx+y)+z]_n=[k−k(k−kx+z)+(k−ky+z)]_n. This implies [k(z−1)]_n=0 for all z∈Q. Thus k=0, a contradiction. Hence, Q cannot be right distributive.

Lemma 2.5.3

A unitary groupoid of order n may be k-translatable only for k=n−1.

Proof

Indeed, i=i⋅1=[k−ki+1]_n for every i∈Q, which for i=2 gives 2=[1−k]_n. This is possible only for k=n−1.

Corollary 3.4

Groups of order n may be k-translatable only for k=n−1. Such groups are cyclic.

Theorem 3.5

If a k-translatable groupoid Q of order n is left unitary then Q is

1. bookend if and only if [k₂]_n=[2k]_n=n−1,

2. elastic if and only if [k₂+k]_n=0,

3. left distributive if and only if [k₂]_n=0,

4. left modular if and only if k=n−1,

5. right modular if and only if [k₂]_n=1,

6. paramedial if and only if [k₂]_n=1.

Proof

1. Indeed, i=(j⋅i)⋅(i⋅j)⇔i=[k−k[k−kj+i]_n+[k−ki+j]_n]_n⇔[2k−k₂+(k₂+1)j−(2k+1)i]_n=0⇔[k₂+1]_n=[2k+1]_n=0.

2. i⋅(j⋅i)=(i⋅j)⋅i⇔[k−ki+(k−kj+i)]_n=[k−k(k−ki+j)+i]_n⇔[(k₂+k)i]_n=[k₂+k]_n⇔[k₂+k]_n=0.

In other cases the proof is similar.

Corollary 3.6

Let Q be a left unitary k-translatable groupoid. Then

1. Q is right modular if and only if it is paramedial,

2. if Q left modular, then it is right modular, elastic and paramedial,

3. Q is not strongly elastic.

Theorem 3.7

A k-translatable groupoid Q of order n is embeddable in a k-translatable groupoid B of order (t+1)n (t∈{1,2,3,…}) such that

1. if Q is left cancellative then B is left cancellative, and

2. if Q is left unitary then B is left unitary.

Proof

Let Q={1,2,…,n} have the k-translatable sequence a₁,…,a_n. Define B as follows: B={b_ij:1≤i≤n,1≤j≤t+1}, where b_i1=i for all i=1,2,…,n and B has (t+1)n distinct elements. Consider B with the following ordering: 1_′,2_′,…,((t+1)n)_′, where b_ij=((i−1)(t+1)+j)_′.

Let the first row c₁,c₂,…,c_(t+1)n of the multiplication table of B have the form

Then b_ij=c_{(i−1)(t+1)+j} for all j≠1 and a_i=c_{(i−1)(t+1)+1}.

The second row is obtained from the first by taking the last k entries and placing them as the first k entries of the second row and taking the first ((t+1)n−k) entries of the first row and placing them as the last ((t+1)n−k) entries of the second. In the same way we construct the (i+1)-th row from the i-th (for all i=2,3,…,(t+1)n−1). Clearly, such defined groupoid B is k-translatable. Also, if Q is left cancellative then B is left cancellative and if Q is left unitary then B is left unitary.

So, we only need to prove that the product ⋅ in Q is preserved in the new product, call it ∗, in B. In proving this, we do so using the facts that Q and B are k-translatable.

Since

for all i∈Q, for any i,j∈Q, by Lemma 2.5, we have

where x=[k−k(it+i−t)+jt+j−t]_(t+1)n=[k(t+1)−ki(t+1)+j(t+1)−t]_(t+1)n=([k−ki+j]_n−1)(t+1)+1.

But, on the other hand, using the fact that a_i=c_{(i−1)(t+1)+1},

Therefore, i⋅j=i∗j for all i,j∈Q. □

Theorem 3.1.1

If a left cancellative groupoid Q of order n is k-translatable, then its dual groupoid (Q,∗) is k_∗-translatable if and only if [kk_∗]_n=1.

Proof

(⇒): Suppose that (Q,⋅), where Q={1,2,…,n}, is k-translatable and the first row of its multiplication table has the form a₁,a₂,…,a_n. If its dual groupoid (Q,∗) is k_∗-translatable, then 1∗[1−k_∗]_n=[1−k_∗]_n⋅1=a_[k−k(1−k∗_)+1]n=a_[kk∗_+1]n. But on the other hand, 1∗[1−k_∗]_n=2∗1=1⋅2=a₂. So, a_[kk∗_+1]n=a₂. Left cancellability implies [kk_∗+1]_n=2, which gives [kk_∗]_n=1.

(⇐): Suppose that [kk_∗]_n=1. Then [i+1]_n∗[j+k_∗]_n=a_[k−k(j+k∗_)+(i+1)]n=a_[k−kj−kk∗_+i+1]n=a_[k−kj+i]n=i∗j. Hence, (Q,∗)is k_∗-translatable with respect to the ordering 1,2,…,n.

Corollary 4.2

A left cancellative, k-translatable groupoid (Q,⋅) of order n and its dual groupoid (Q,∗) are k-translatable, for the same value of k, if and only if [k₂]_n=1.

Corollary 4.3

A left unitary, k-translatable groupoid (Q,⋅) of order n and its dual groupoid (Q,∗) are k-translatable, for the same value of k, if and only if (Q,⋅) is paramedial.

Corollary 4.4

A dual groupoid of a left cancellative, k-translatable groupoid (Q,⋅) of order n is [n−tk]_n-translatable if and only if [tk₂]_n=n−1.

Corollary 4.5

A left cancellative k-translatable groupoid of order n is alterable if and only if its dual groupoid is (n−k)-translatable.

Proof

(⇒): Since, by Corollary 2.18, [k₂]_n=n−1 we have [k(n−k)]_n=[−k₂]_n=1. By Theorem 4.1, the dual of Q is (n−k)-translatable.

(⇐): Since Q is k-translatable then, by Theorem 4.1, [k(n−k)]_n=[−k₂]_n=1 . This implies [k₂]_n=n−1, which, by Corollary 2.18, shows that Q is alterable.

Theorem 5.6.1

A left cancellative, k-translatable groupoid Q of order n with a k-translatable sequence a₁,a₂,…,a_n is a semigroup if and only if [k₂+k]_n=0 and a_i=[i−k−ka_k]_n for all i=1,2,…,n.

Proof

(⇒): Setting x=y=n and z=[i−k]_n in (5) we obtain a_i=[i−k−ka_k]_n. Consequently, a_k=[−ka_k]_n=[a₁+k−1]_n, which implies [1−k−a₁]_n=[ka_k]_n=[ka₁+k₂−k]_n. Hence [1−k₂]_n=[a₁(k+1)]_n. Since, setting x=y=z=1 in (5) we obtain [a₁(k+1)]_n=[k+1]_n, the last equation gives [1−k₂]_n=[k+1]_n. Hence, [k₂+k]_n=0.

(⇐): From a_i=[i−k−ka_k]_n we conclude a_k=[−ka_k]_n. Thus, [−ka_k]_n=[k₂a_k]_n. Now using the above and [k₂+k]_n=0 we obtain

which proves (5). Hence Q is a semigroup.

Corollary 5.2

A k-translatable semigroup is left cancellative if an only if it has a left neutral element.

Proof

If Q is a left cancellative and k-translatable semigroup, then by previous theorem a_i=[i−k−ka_k]_n and [−ka_k]_n=a_k. So, a_[−2ak_−1]n=[−k−1−a_k]_n and a_[−2ak_−1]n⋅a_j=a_t, where

Thus, a_[−2ak_−1]n is a left neutral element.

The converse statement is obvious. □

Corollary 5.3

A left unitary, k-translatable groupoid Q of order n is a semigroup if and only if [k+k₂]_n=0.

Corollary 5.4

A left unitary, k-translatable groupoid Q of order n=k+k₂ is a semigroup with the multiplication given by the formula

where i,j∈{1,2,…,k} and s,t∈{0,1,2,…,k}.

Proof

Observe first that all elements of Q can be written in the form i+sk, where i∈{1,2,…,k} and s∈{0,1,2,…,k}. Since Q is left unitary and k-translatable, x⋅y=[k−kx+y]_n for all x,y∈Q. Therefore (i+sk)⋅(j+tk)=[k−k(i+sk)+(j+tk)]_n=[j+(s+t−i+1)k]_n. It is straightforward to verify that this multiplication is associative.

Corollary 5.5

A groupoid Q of order n=k+k₂ with the multiplication defined by (6) is a left unitary, k-translatable semigroup.

Proof

Indeed, this multiplication is associative and 1 is its left neutral element. Also, [(i+1)+sk]_n⋅[j+(t+1)k]_n=[j+(t+s−i+1)k]_n=[i+sk]_n⋅[j+tk]_n, so Q is k-translatable.

Theorem 5.6

By re-ordering, any left cancellative k-translatable semigroup can be transformed into a left unitary k-translatable semigroup.

Proof

Let Q be a left cancellative k-translatable semigroup and let a₁,a₂,…,a_n be its k-translatable sequence with respect to the ordering 1,2,…,n. Consider a new ordering b₁,b₂,…,b_n of Q, where b_s=[−a_k−k+s−2]_n, s=1,2,…,n.

Using Theorem 5.1, it is not difficult to see that b_i⋅b_j=b_[j−ki+k]n. Thus b₁ is a left neutral element of Q and, by Lemma 2.5 (i), Q is k-translatable with respect to the ordering b₁,b₂,…,b_n.

Corollary 5.7

Left cancellative k-translatable semigroups of the same order are isomorphic.

Corollary 5.8

A left cancellative groupoid Q of order n with an (n−1)-translatable sequence of the form a_[i+1]n=[a_i+1]_n is a cyclic group.

Proof

An (n−1)-translatable groupoid is commutative. So, if it is left cancellative, then it also is right cancellative. Hence it is a quasigroup. Moreover, k=n−1 and a_[i+1]n=[a_i+1]_n imply [k+k₂]_n=0 and a_i=[i−k−ka_k]_n for all i=1,2,…,n. By Theorem 5.1 then, Q is a semigroup. Hence Q is a commutative group. By Corollary 3 it is cyclic.

Corollary 5.9

A left cancellative (n−1)-translatable semigroup of order n is isomorphic to the additive group Z_n.

Proof

By Theorem 5.1, a_i=[i+1+ak]_n. Therefore, a_[i+1]n=[(i+1)+1+a_k]_n=[a_i+1]_n. The result then follows from Corollary 2.18.

Proposition 5.10

There are no idempotent k-translatable semigroups of order n.

Proof

An idempotent k-translatable semigroup of order n is left cancellative. Thus, by Lemma 2.6 and Theorem 5.1, in such semigroup for all i∈Q should be i=a_[k−ki+i]n=[i−ki−ka_k]_n, i.e., [k(i+a_k)]_n=0, which is impossible.

Lemma 5.11

If Q is a left cancellative, k-translatable semigroup of order n, then

1. E(Q)={i∈Q:[k(i+a_k)]_n=0},

2. E(Q) is a subsemigroup of Q,

3. E(Q) equals the set of all left neutral elements of Q,

4. E(Q) is a right-zero semigroup.

Proof

1. By Lemma 2.6 and Theorem 5.1, we have

   i=i⋅i⇔i=a_[k−ki+i]n=[i−ki−ka_k]_n⇔[k(i+a_k)]_n=0.

2. If i,j∈E(Q), then (i⋅j)⋅(i⋅j)=[j−ki−ka_k]_n⋅[j−ki−ka_k]_n=[j−ki−ka_k−k(i+a_k)−k(j+a_k)]_n=[j−ki−ka_k]_n=i⋅j. So, i⋅j∈E(Q).

3. For every i∈E(Q) and j∈Q we have i⋅j=[j−ki−ka_k]_n=j.

4. This follows from (3).

Corollary 5.12

If 1 is a left neutral element of a left cancellative, k-translatable semigroup Q of order n, then E(Q)={j=[i+k(i−1)]_n:i∈Q}.

Corollary 5.13

Right cancellative semigroups of order n may be k-translatable only for k=n−1. Such semigroups are isomorphic to the additive groups Z_n.

Proof

Let Q be a right cancellative k-translatable semigroup of order n. Then the right cancellativity implies that each column of the multiplication table of Q consists of n distinct elements. Since Q is k-translatable, the first row of its multiplication table has n distinct elements, i.e., Q has a left cancellable element. Thus, by Lemma 2, it is left cancellative. So, Q is an associative quasigroup. Hence Q is a group. Corollary 3 ends the proof.

Corollary 5.14

A left cancellative, k-translatable semigroup of order n with k-translatable sequence a₁,a₂,…,a_n in which a_k=1 or a_k=n−1 is isomorphic to the group Z_n

Proof

If a_k=1 then by Theorem 5.1, a_k=1=[−ka_k]_n=[−k]_n. Hence k=n−1. If a_k=−1 then by Theorem 5.1 again, a_k=n−1=[−ka_k]_n=k. So, in both these cases the results follows from Corollary 2.18.85.

Theorem 5.6.15

A left cancellative, k-translatable semigroup of order n is a union of t disjoint copies of cyclic groups of order m, where m and t are the smallest positive integers such that [mk]_n=0 and [tm]_n=0.

Proof

For every i∈E(Q) consider the set Q_i=Q⋅i={x⋅i:x∈Q}. Then, by associativity and Lemma 10, (x⋅i)⋅(y⋅i)=x⋅(y⋅i)=x⋅[i−ky−ka_k]_n=[i−ky−kx−2ka_k]_n=y⋅(x⋅i)=(y⋅i)⋅(x⋅i). So, Q_i is a commutative semigroup contained in Q and i=i⋅i is its neutral element. Since (x⋅i)⋅(y⋅i)=i for y=[−x−2a_k]_n, this semigroup is a commutative group. Moreover, for every i∈E(Q), the mapping φ(x⋅i)=x⋅1 is an isomorphism between Q_i and Q₁. So, all groups Q_i are isomorphic to Q₁. Since Q is k-translatable, the group Q₁ contains elements of the form (t+1)⋅1=[1−kt]_n, t=0,1,…,m−1, where m is the smallest positive integer such that [mk]_n=0. It is not difficult to see that this group is generated by the element 2⋅1=[1−k]_n.

Groups Q_i are pairwise disjoint. If not, then for z∈Q_i∩Q_j we have z=x⋅i=y⋅j for some x,y∈Q. Whence, multiplying by i and j, and applying Lemma 10, we obtain z⋅i=x⋅i=y⋅i and z⋅j=x⋅j=y⋅j. Thus x⋅i=y⋅j=x⋅j, which by left cancellativity gives i=j. Hence Q_i=Q_j.

Obviously ⋃Q_i⊆Q, i∈E(Q). To prove the converse inclusion observe that, according to Theorem 5.6, in Q we can introduce a new ordering such that some element of E(Q) can be identified with 1. Then, by Corollary 2.18.86, [q+kq−k]_n∈E(Q) for each q∈Q. Thus, q⋅i=q for i=[q+kq−k]_n. Hence Q⊆⋃Q_i. So, Q is a union of t copies of Q₁, where t is the number of idempotents of Q. Since (1+wm)⋅j=k−k(1+wm)+j=j, the set L={1+wm:w=1,2,…} consists of left identities. Since t is the smallest positive integer such that [tm]_n=0, L={1+wm:w=0,1,…,t−1}. By Lemma 10, L=E(Q).

Corollary 5.16

A left cancellative, k-translatable semigroup of order n is a union of t disjoint pairwise isomorphic left ideals of order m, where m and t are the smallest positive integers such that [mk]_n=0 and [tm]_n=0.

Corollary 5.17

A left cancellative, k-translatable semigroup of order n=k₂+k is a union of k disjoint copies of cyclic groups isomorphic to Z_k+1.

Example 5.18

A left unitary 2-translatable semigroup of order 6 has two idempotents 1 and 4 and is a union of two cyclic groups isomorphic to the additive Z₃. On the other hand, a left unitary 3-translatable semigroup of order 6 has three idempotents 1,3 and 5 and is a union of three cyclic groups of order two.

Theorem 5.6.19

In left cancellative k-translatable semigroup every left and every right ideal is semiprime.

Proof

It is a consequence of our Theorem 5.15 and Theorem 3.1.3 from [5].

Theorem 5.6.20

A left cancellative k-translatable semigroup of order n is

1. medial,

2. anticommutative (i.e., nowhere commutative), if (1+k,n)=1,

3. conditionally commutative,

4. left commutative,

5. left and right regular,

6. regular,

7. intra-regular,

8. orthodox,

9. a Clifford right semigroup,

10. a Clifford left semigroup, if (k,n)=1.

Proof

1. This is a consequence of Theorem 5.6.

2. Indeed, i⋅j=j⋅i means that [j−ki−ka_k]_n=[i−kj−ka_k]_n, i.e., [(j−i)(1+k)]_n=0, which for (1+k,n)=1 gives i=j.

3. It is easy to see that i⋅j=j⋅i implies i⋅x⋅j=j⋅x⋅i for all x∈Q.

4. Since Q has a left neutral element, (1) implies i⋅j⋅x=j⋅i⋅x for all i,j,x∈Q.

5. x⋅(j⋅j)=j and (j⋅j)⋅y=j for each j∈Q and x=[−j−2ka_k]_n, y=[j+2kj+2ka_k]_n.

6. i=i⋅x⋅i for every i∈Q and x=[−i−2a_k]_n.

7. i=x⋅i⋅i⋅y for every i∈Q and x=[−2i−3a_k]_n, y=i.

8. It is a consequence of Lemma 10 and (6).

9. Q⋅i⊆i⋅Q for every i∈Q, because j⋅i=i⋅s for s=[i+k(i−j)]_n.

10. Analogously, i⋅Q⊆Q⋅i for every i∈Q, because i⋅j=s⋅i for s=[t(i−j)+i]_n, where [tk]_n=1. Such t exists only for (k,n)=1.

Theorem 5.6.21

A paramedial left cancellative k-translatable semigroup is a cyclic group.

Proof

Comparing Theorems 6 and 9 we can see that such semigroup is (n−1)-translatable and a_[i+1]n=[(i+1)+1+a_k]_n=[a_i+1]_n. Corollary 2.18 completes the proof.

Theorem 5.6.22

If the first row of the multiplication table of a naturally ordered k-translatable groupoid Q={1,2,…,n} has the form a₁,a₂,…,a_n, then Q is a semigroup with the multiplication i⋅j=a_j if and only if a_s=a_[s+k]n=a_as for any s∈Q.

Proof

(⇒): Indeed, (i⋅j)⋅s=a_s and i⋅(j⋅s)=a_as imply a_s=a_as for all s∈Q. From k-translability we obtain a_s=a_[s+k]n.

(⇐): First note that a_s=a_[s+k]n implies a_[s−k]n=a_[s−k+k]n=a_s. Thus, for any, i,j,s∈Q we have

which, by (4), means that Q is a semigroup. The rest is a consequence of Lemma 2.5 (i).

Example 5.23

Let Q={1,2,…,n} be naturally ordered set. Then

1. the sequences (12,12,9,10,9,12,12,12,9,10,9,12), (1,1,1,10,11,10,1,1,1,10,11,10) and (5,8,8,8,5,6,5,8,8,8,5,6) define three isomorphic 6-translatable semigroups of order n=12,

2. the sequences (1,3,3,1,3,3), (2,2,6,2,2,6) and (1,5,1,1,5,1) yield isomorphic 3-translatable semigroups of order 6.

3. the sequences (1,5,4,4,5,1,5,4,4,5), (4,3,3,4,10,4,3,3,4,10) and

   (1,2,8,2,1,1,2,8,2,1) define isomorphic 5-translatable semigroups of order n=10.

Corollary 5.24

A naturally ordered semigroup Q={1,2,…,n} with a k-translatable sequence a₁,a₂,…,a_n in which

1. a₁=a₂=n or

2. a₁=1 and a_j=j+1 for some j≠1

has multiplication defined by the formula i⋅j=a_j.

Proof

Let Q be a k-translatable semigroup. Then, by (4), for x=1, y=j and z=s we have

which for j=1 gives a_[k−ka1_+s]n=a_as. From this, for a₁=n, we obtain a_[s+k]n=a_as, whence we get a_s=a_a[s−k]n. Thus,

for a₂=n. So, for a₁=a₂=n the conditions of Theorem 5.22 are satisfied. Hence i⋅j=a_j for all i,j∈Q.

Now if a₁=1 and a_j=j+1 for some j≠1, then 1⋅1=a₁=1 and a_s=1⋅s=(1⋅1)⋅s=1⋅(1⋅s)=1⋅a_s=a_as. This together with (7) implies a_[k−kj+s]n=a_a[k−kj+s]n=a_[k−kaj_+s]n, which for s=[ka_j−k+t]_n gives a_[k(aj_−j)+s]n=a_t. But a_j−j=1 for some j≠1, so a_[k+t]n=a_t. This means that also in this case the condition of Theorem 5.22 are satisfied. Thus, also in this case i⋅j=a_jfor all i,j∈Q.

Corollary 5.25

A k-translatable groupoid Q={1,2,…,n} with an idempotent element j∈Q such that j=j⋅[j−1]_n is a semigroup if and only if for all s∈Q

1. j⋅[j−1+s−k]_n=j⋅[j−1+s]_n=j⋅[j−1+s+k]_n and

2. j⋅[j−1+s]_n=j⋅[(j−1)+(j⋅[j−1+s]_n)]_n.

Proof

First we introduce on Q a new ordering b₁,b₂,…,b_n starting with an idempotent j and such that b_s=[j−1+s]_n for all s∈Q. Then, by repeated application of Lemma 2.7, we can see that a₁,a₂,…,a_n, where a_s=j⋅b_s, is a k-translatable sequence with respect to this new ordering.

(⇒): Since Q is a k-translatable semigroup, for any s∈Q we have

From this, setting s=[t−k]_n we obtain a_t=a_[t−k]n, and so a_s=a_[s+k]n=a_[s−k]n. This proves (i) and, together with (4), implies a_s=a_as. The last is equivalent to (ii).

(⇐): (i) and (ii) imply a_s=a_[s+k]n=a_[s−k]n=a_as for all s∈Q. Therefore for i,j,s∈Q we have a_[k−ka[k−ki+j]n_+s]n=a_[k−ki+a[k−kj+s]n_]n and Q is a semigroup.

Corollary 5.26

A naturally ordered, k-translatable groupoid Q={1,2,…,n} with an idempotent element 1such that 1=1⋅n is a semigroup if and only if for all s∈Q we have 1⋅[s−k]_n=1⋅s=1⋅[s+k]_n=1⋅(1⋅s).

Lemma 6.11

Let t and x be positive integers. Then

1. [tx]_tn=t[x]_n,

2. [t(x−1)]_tn=t[[x]_n−1]_n,

3. if n=k+k₂ and k=tq, then [(k+(t−1)n)+(k+(t−1)n)₂]_tn=0.

Theorem 3.5.2

Let k=tq and [k+k₂]_tn=0. Suppose that (Q_i,∗_i), i=1,2,…,t, are pairwise disjoint left unitary k-translatable semigroup with ordering i₁,i₂,…,i_n and the cardinality n. Then Q=⋃Q_i, with ordering defined by i_r=t(r−1)+i (i=1,2,…,nandr=1,2,…,n), is a left unitary k-translatable semigroup, with respect to the product ∗ defined by

in which each (Q_i,∗) isomorphic to (Q_i,∗_i).

Furthermore, if we define a product ∗ on Q=⋃Q_i by (8), then there is an ordering on Q such that (Q,∗) is a left unitary k-translatable semigroup in which each (Q_i,∗) isomorphic to (Q_i,∗_i).

Proof

Note that [k+k₂]_tn=0implies [k+k₂]_n=0. Consider on Q=⋃Q_i the ordering i_r=t(r−1)+i and define the first row of the multiplication table of ∗ as a₁,a₂,…,a_tn, where a_i=i for all i=1,2,…,tn. We make the sequence a₁,a₂,…,a_tn into a k-translatable sequence in a natural way, as follows. The last k entries of the first row become the first k entries of the second row and the first (tn−k) entries of the first row become the last (tn−k) entries of the second row. Continue in this manner, making the last k entries of the i-th row become the first k entries of the (i+1)-th row and the first (tn−k) entries of the i-th row become the last (tn−k) entries of the (i+1)-th row. By definition, the resultant groupoid (Q,∗) is k-translatable, with 1₁=1 as a left identity element. Since [k+k₂]_tn=0, by Corollary 2.18.80, (Q,∗) is a left unitary k-translatable semigroup, with product r∗s=a_[k−kr+s]tn=[k−kr+s]_tn. Therefore, for all i_r,j_s∈Q we obtain

where w=[q−kr+k−qi+s]_n. Since, by Lemma 11, we have w=j[(q−kr+k−qi+s)]_n, the above proves (8) and shows that each Q_j is a left ideal of (Q,∗).

Now, we define on Q_i new ordering by putting i_s=s for all s=1,2,…,n. Then r∗s=i_r∗i_s=i_x, where x=[k−kr+q−qi+s]_n=[k−kr+q−q(i+1)+s+q]_n. So, r∗s=i_[r+1]n∗i_[s+k]n=[r+1]_n∗[s+k]_n. By Lemma 2.5 (ii), (Q_i,∗) is k-translatable.

Since 0=[k+k₂]_tn=[t(q+tq₂)]_tn by Lemma 11 (a), [q+tq₂]_n=0. Therefore, [tq₂]_n=[−q]_n. If r=1−q(1−i), then we get [k−kr]_n=[k(1−r)]_n=[tq₂(1−i)]_n=[−q(1−i)]_n. Therefore, by (8), for r=1−q(1−i) we have i_r∗i_s=i_s. Thus i_r is a left neutral element of (Q_i,∗).

By Lemma 2.7, we can re-order (Q_i,∗) such that by Corollary 2.18.80, (Q_i,∗) is a left unitary k-translatable groupoid of order n. Then, since [k+k₂]_n=0, by Corollary 2.18.80, is a left unitary k-translatable semigroup. By Corollary 5.7 then, (Q_i,∗) is isomorphic to (Q_i,∗_i). This proves the first part of the Theorem.

To prove the second part we introduce on Q=⋃Q_i an ordering i_s=t(s−1)+i and consider a product ∗ defined by (8). Then,

where x=[(k−kr+q−q(i+1)+s+q)]_n=[(k−kr+q−qi+s)]_n. So, [i_r+1]_tn∗[j_s+k]_tn=j_x=i_r∗j_s. Thus (Q,∗) is k-translatable. Since, by (8), 1₁=1 is a left neutral element of (Q,∗) and [k+k₂]_tn=0, by Corollary 2.18.80 then, (Q,∗) is a left unitary k-translatable semigroup.

Analogously we can see that each (Q_i,∗) with natural ordering s=i_s is a left unitary k-translatable semigroup. By Corollary 5.7, semigroups (Q_i,∗) and (Q_i,∗_i) are isomorphic. This proves the second part of the Theorem.

Theorem 3.5.3

Let k=tq and n=k+k₂. Suppose that (Q_i,∗_i), i=1,2,…,t, are pairwise disjoint left unitary k-translatable semigroups with ordering i₁,i₂,…,i_n and the cardinality n. Then Q=⋃Q_i with the ordering i_r=t(r−1)+i is a left unitary (k+(t−1)n)-translatable semigroup with respect to the product ∗ defined by

in which each (Q_i,∗) isomorphic to (Q_i,∗_i).

Furthermore, if on Q=⋃Q_i a product ∗ is defined by (9), then Q can be ordered in such a way that (Q,∗) becomes a left unitary (k+(t−1)n)-translatable semigroup in which each (Q_i,∗) is isomorphic to (Q_i,∗_i).

Proof

Similarly as in the proof of Theorem 6.2, we can construct a multiplication table for (Q,∗) in this way that (Q,∗) becomes a left unitary (k+(t−1)n)-translatable semigroup. Then, according to the (k+(t−1)n)-translatability, Lemma 11 and k+(t−1)n=tk+(t−1)k₂, we have

where

because k+k₂=n.

This proves (9) and shows that each Q_j is a left ideal of (Q,∗). Next, similarly as in the previous proof, we can prove that each semigroup (Q_i,∗) has a left neutral element i_r, where r=q(i−1)+1, and is isomorphic to (Q_i,∗_i). This completes the proof of the first part of Theorem.

To prove the second part we introduce on Q=⋃Q_i an ordering i_s=t(s−1)+i and consider the product defined (9). Then,

where x=[(k−kr+(s−kq)+kqi)]_n=[(k−kr+s−kq(i−1))]_n.

So, [i_r+1]_tn∗[j_s+k+(t−1)n]_tn=j_x=i_r∗j_s, which means that (Q,∗) is a (k+(t−1)n)-translatable semigroup with a left neutral element 1₁=1. It is not difficult to see that each (Q_i,∗) with the ordering i_s=s is a k-translatable semigroup isomorphic to (Q_i,∗).

Example 6.4

Suppose that Q is a left unitary 8-translatable semigroup of order n=12. Since 8=2×4 and [8+8₂]₂₄=0, there is a left unitary 8-translatable semigroup G of order 24 that is a disjoint union of two copies of Q. It is a disjoint union of 8 copies of Z₃.

Example 6.5

Suppose that Q is a left unitary 8-translatable semigroup of order n=72. Then there is a left unitary 224-translatable semigroup of order 288 that is a disjoint union of 4 copies of Q. There is also a left unitary 512-translatable semigroup G of order 576 that is a disjoint union of 8 copies of Q. Also it is a disjoint union of 64 copies of Z₉.

Corollary 6.6

Let (Q,⋅) and (G,∘) be left unitary k-translatable semigroups of order n=k+k₂, with k=2q, Q={1,2,…,n}, G={g₁,g₂,…,g_n} and Q∩G=∅. Then B=Q∪G can be transformed into a left unitary (k+n)-translatable semigroup with product ∗ such that

1. (Q,⋅)=(Q,∗),

2. (G,∘) is isomorphic to (G,∗),

3. g_i∗g_j=g_{[k(1+q)−ki+j]}n,

4. i∗g_j=g_[k−ki+j]n for all i,j∈Q and

5. g_i∗j=[k(1+q)−ki+j]_n for all i,j∈Q.

Conversely, if n=k+k₂ and B=Q∪G with the ordering 1,g₁,2,g₂,…,n,g_n has the product ∗ defined by (i), (iii), (iv) and (v) then (B,∗) is a left unitary (k+n)-translatable semigroup satisfying (ii). Furthermore, Q∗G=G and G∗Q=Q.