For simplicity we assume that all groupoids considered in this note are finite and have the form *Q*={1,2,…,*n*} with the *natural ordering* 1,2,…,*n*, which is always possible by renumeration of elements. Moreover, instead of *i*≡*j*(mod*n*) we will write [*i*]_{n}=[*j*]_{n}. Additionally, in calculations of modulo *n*, we assume that 0=*n*.

#### 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 *a*_{1},*a*_{2},…,*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 *a*_{1},*a*_{2},…,*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}.

It is important to note that a *k*-translatable sequence of a groupoid *Q* depends on the ordering of the elements in the multiplication table of *Q*. A groupoid may be *k*-translatable for one ordering but not for another.

#### Example 2.2

*Consider the following groupoids*:

$$\begin{array}{ccccc}\cdot & 1& 2& 3& 4\\ 1& 1& 2& 3& 4\\ 2& 2& 3& 4& 1\\ 3& 3& 4& 1& 2\\ 4& 4& 1& 2& 3\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{ccccc}\cdot & 1& 3& 4& 2\\ 1& 1& 3& 4& 2\\ 3& 3& 1& 2& 4\\ 4& 4& 2& 3& 1\\ 2& 2& 4& 1& 3\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{ccccc}\cdot & 1& 2& 3& 4\\ 1& 1& 4& 3& 2\\ 2& 3& 2& 1& 4\\ 3& 1& 4& 3& 2\\ 4& 3& 2& 1& 4\end{array}$$

The first table determines a 3-translatable semigroup isomorphic to the additive group Z_{4}. 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.

On the other hand, a groupoid with the operation *x*⋅*y*=*a*, where *a* is fixed, is a *k*-translatable semigroup for each *k*, but the semigroup (*Q*,⋅) with the operation

$$x\cdot y=\left\{\begin{array}{ll}1& \mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}x+y\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n},\\ 2& \mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}x+y\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{d}\mathrm{d}\end{array}\right.$$

is *k*-translatable for each odd *k*<*n*, if *n* is even, and *k*-translatable for each even *k*, if *n* is odd. All these groupoids are easily proved to be semigroups.

#### 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 *a*_{1},*a*_{2},…,*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*.

Obviously, for another ordering it may be *k*-translatable for some other value of *k*, but in some cases the change of ordering preserves *k*-translatability.

We start with the following simple observation.

#### 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.

The proofs of the following two Lemmas are straightforward and are omitted.

#### Lemma 2.5

Let *a*_{1},*a*_{2},…,*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:

*i*⋅*j*=*a*_{[(i−1)(n−k)+j]}*n*=*a*_{[k−ki+j]}*n*,

*i*⋅*j*=[*i*+1]_{n}⋅[*j*+*k*]_{n},

*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 *a*_{1},*a*_{2},…,*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*.

Recall that a groupoid *Q* is called

*alterable* if *i*⋅*j*=*w*⋅*z* implies *j*⋅*w*=*z*⋅*i*,

*bookend* if and only if (*j*⋅*i*)⋅(*i*⋅*j*)=*i*,

*paramedial* if and only if (*i*⋅*j*)⋅(*w*⋅*z*)=(*z*⋅*j*)⋅(*w*⋅*i*),

*strongly* elastic if and only if *i*⋅(*j*⋅*i*)=(*i*⋅*j*)⋅*i*=(*j*⋅*i*)⋅*j*,

*left* modular if and only if (*i*⋅*j*)⋅*z*=(*z*⋅*j*)⋅*i*,

*right* modular if and only if *i*⋅(*j*⋅*z*)=*z*⋅(*j*⋅*i*)

for all *i*,*j*,*w*,*z*∈*Q*.

#### Lemma 2.6

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

*a*_{i}=*a*_{j} if and only if *i*=*j*,

*Q* is idempotent if and only if *i*=*a*_{[k−ki+i]}*n* for all *i*∈*Q*,

if *Q* is idempotent then *i*⋅*j*=*t* if and only if [*j*−*ki*]_{n}=[*t*−*kt*]_{n} for all *i*,*j*,*t*∈*Q*,

*Q* is elastic if and only if [*i*+*ki*]_{n}=[(*j*⋅*i*)+*k*(*i*⋅*j*)]_{n} for all *i*,*j*∈*Q*,

*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*,

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

*Q* is left distributive if and only if [(*i*⋅*j*)+*k*(*s*⋅*i*)]_{n}=[(*s*⋅*j*)+*ks*]_{n} for all *i*,*j*,*s*∈*Q*,

*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*,

*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*,

*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*,

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

*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 *a*_{1},*a*_{2},…,*a*_{n}is 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},*a*_{1},…,*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},*a*_{1},*a*_{2},…,*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},*a*_{1},…,*a*_{k−1} is the *k*-translatable sequence with respect to the order 1_{′},2_{′},…,*n*_{′}. Let’s call this sequence *b*_{1},*b*_{2},…,*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 *a*_{1},*a*_{2},…,*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.

A groupoid *Q* is called *right solvable* (*left solvable*) if for any {*a*,*b*}⊆*Q* there exists a unique *x*∈*G* such that *ax*=*b* (*xa*=*b*). It is a *quasigroup* if it is left and right solvable.

#### 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 *a*_{1},*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.

An idempotent *k*-translatable groupoid may not be right cancellative. It is not difficult to see that if the first row of the multiplication table of the 4-translatable groupoid *Q* of order 8 has form 1,6,3,8,5,2,7,4, then this groupoid is idempotent and left cancellative but it is not right cancellative.

For every odd *n* and every *k*>1 such that (*k*,*n*)=1 there is at least one idempotent *k*-translatable quasigroup. For even *n* there are no such quasigroups Theorem 4.1.

#### 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 *a*_{1},*a*_{2},…,*a*_{n} with respect to the ordering 1,2,…,*n*, and *S*={*c*_{1},*c*_{2},…,*c*_{n}} has a *k*-translatable sequence *b*_{1},*b*_{2},…,*b*_{n} with respect to the ordering *c*_{1},*c*_{2},…,*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(mod*n*), 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(mod*n*). 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.,

$$[-k(i\cdot j)+(i\cdot j){]}_{n}=[-ki+j{]}_{n},$$(1)

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

$$\begin{array}{rl}[i+ki{]}_{n}& =[i+k(i\cdot j)+k(j\cdot i)-kj{]}_{n}=[(k(j\cdot i)-kj+i)+k(i\cdot j){]}_{n}\\ & \stackrel{(1)}{=}[(j\cdot i)+k(i\cdot j){]}_{n}.\end{array}$$

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 [*k*_{2}]_{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

$$[i+1{]}_{n}\cdot s=i\cdot [n+s-k{]}_{n}=i\cdot [s-k{]}_{n}$$(2)

for all *i*,*s*∈{1,2,…,*n*}.

Thus,

$$(n-1)\cdot n=(n-1)\cdot [n+k-k{]}_{n}\stackrel{(2)}{=}n\cdot [n+k{]}_{n}=n\cdot k,$$

which, by alterability, gives

$$n\cdot n=k\cdot (n-1).$$(3)

But $n\cdot n=n\cdot [n+k-k{]}_{n}\stackrel{(2)}{=}[n+1{]}_{n}\cdot [n+k{]}_{n}=[n+1{]}_{n}\cdot k$, so

$$[n+1{]}_{n}\cdot k=k\cdot (n-1),$$

which, by alterability, implies *k*⋅*k*=[*n*−1]_{n}⋅[*n*+1]_{n}. Hence

$$k\cdot k=(n-1)\cdot [n+1{]}_{n}\stackrel{(2)}{=}n\cdot [n+1+k{]}_{n}=n\cdot (k+1)\stackrel{(2)}{=}k\cdot [k+1+{k}^{2}{]}_{n}.$$

Since *Q* is left cancellable, we obtain 0=[1+*k*_{2}]_{n}. Therefore *k*_{2}≡(−1)(mod*n*).

#### Corollary 2.18.18

A *k*-translatable and left cancellative groupoid of order *n* is alterable if and only if [*k*_{2}]_{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 *a*_{1},*a*_{2},…,*a*_{n}. Then all these elements are different.

If *k*_{2}≡−1(mod*n*), 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 [*k*_{2}(*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.

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