1 Introduction

Ternary heaps (introduced by Prüfer [10]) have interesting applications to projective geometry [1], affine geometry [2], theory of nets (webs), theory of knots and even to the differential geometry [7]. A special case of ternary heaps defined on a group were considered by Certaine [3]. Vakerelov considered in [13] the affine geometry defined by ternary groups that are heaps satisfying some additional conditions. S.A. Rusakov extended this concept to the case of n-ary groups (see [11, 12]). His research was continued by his students and co-workers (see for example [8, 9]). Dudek noted in [5] that the proofs of the results obtained by Rusakov and his co-workers can be much shorter and clearer if we use other much simpler algebras called flocks instead of n-ary groups (see also [6]).

In this short note, we continue this line of research. We give some short and simple proofs of some important results previously proven for affine geometry induced by an n-ary group.

2 Preliminaries

We use standard terminology and notation (cf. [4, 12]). We just recall that an n-ary group (n-group) (G, f) is semiabelian if \( f ({x},a,\ldots ,a,y)= f(x, {\mathop {a}\limits ^{(n-2)}},y) =f(y, {\mathop {a}\limits ^{(n-2)}},x) \) for all \(x,y\in G\) and any fixed \(a\in G\).

Starting with the observation that in the affine geometry of the plane \({\mathbb {R}}^2\) intervals \(\overline{AB}\) and \(\overline{CD}\) are parallel if \(\overrightarrow{CA\;}=\overrightarrow{DB\;}\) we say two intervals (a, b) and (c, d) of a group \((G,+)\) are equal if \(a-c=b-d\), i.e. \(a-c+d=b\). Consequently, two intervals (a, b) and (c, d) of a ternary group (G, f) derived from the group \((G,+)\) are parallel if and only if \(f(a,\overline{c},d)=b\). In this case we also say that the points a, b, c, d form a parallelogram \(\langle a,c,d,b\rangle \).

Rusakov (cf. [11] or [12]) extended this concept to n-groups in his way that points a, b, c, d of an n-group (G, f) form a parallelogram if and only if

$$\begin{aligned} f\left( f\left( a,c^{[-2]},\!{\mathop {c}\limits ^{(n-2)}}\right) ,\!{\mathop {c}\limits ^{(n-2)}}\!,d\right) =b, \end{aligned}$$

where \(c^{[-2]}\) is an element such that \(f(f(c,c^{[-2]},\!{\mathop {c}\limits ^{(n-2)}}),\!{\mathop {c}\limits ^{(n-1)}})=c\).

Putting \(f(x^{[-2]},\!{\mathop {x}\limits ^{(n-1)}})=\overline{x}\) we can rewrite the above definition in the form

\( \langle a,c,d,b\,\rangle \;\;\;\text {forms a parallelogram if and only if}\;\;\;[a,\overline{c},d]=b,\)

where \([a,\overline{c},d]=f(a,\overline{c},\!{\mathop {c}\limits ^{(n-3)}},d)\), or using the free covering group for an n-group (G, f)

\( \langle a,c,d,b\,\rangle \;\;\;\text {forms a parallelogram if and only if}\;\;\;a\cdot c^{-1}\cdot d=b. \)

The operation \([x,y,z]=x\cdot y^{-1}\cdot z\) is idempotent and satisfies the following para-associative law:

$$\begin{aligned}{}[[x,y,z],u,w]=[x,[u,z,y],w]=[x,y,[z,u,w]]. \end{aligned}$$
(1)

Any ternary groupoid \((G,[\;\;])\) satisfying (1) is called para-associative or a semiheap (cf. [7]). A para-associative quasigroup is called a flock.

A ternary groupoid \((G,[\;\;])\) is semiabelian if \([x,y,z]=[z,y,x]\) holds for all \(x,y,z\in G\). A semiabelian flock is a ternary group and conversely, any semiabelian ternary group is a flock. But there are flocks that are not ternary groups and ternary groups that are not flocks.

If a is a fixed element of a flock \((G,[\;\;])\), then \((G.\cdot ),\) where \(x\cdot y=[x,a,y],\) is a group called the retract of \((G,[\;\;])\) and denoted by \(\textrm{ret}_a(G,[\;\;])\).

All retracts of a given flock are isomorphic (cf. [5]). A flock is a ternary group if and only if it has a commutative retract.

Thus, all flocks of orders \(p, p^{2} \) and pq, where \( p>q \) are prime integers such that \( (p-1,q)=1\), are ternary groups. A minimal flock which is not a ternary group has six elements and is defined on the symmetric group \({\mathbb {S}}_3\).

Retracts of isomorphic flocks are isomorphic. The converse statement is not true. For example, the two flocks \(({\mathbb {Z}}_2,[\;\;]_0)\) and \(({{\mathbb {Z}}_2,[\;\;]_{1})}\), where \([x,y,z]_k=(x+y+z+k)(\textrm{mod}\,2)\), are not isomorphic (the first is idempotent, the second has no idempotents), but their retracts are isomorphic.

We are going to need the following two theorems proved in [5].

Theorem 2.1

A para-associative groupoid \((G,[\;\;])\) with the unary operation \(\;\bar{}: x\rightarrow \overline{x}\) is a flock if and only if (at least) one of the identities

$$\begin{aligned}{} & {} [y,\overline{x},x]=[x,\overline{x},y]=y, \end{aligned}$$
(2)
$$\begin{aligned}{} & {} [y,x,\overline{x}]=[x,\overline{x},y]=y, \end{aligned}$$
(3)
$$\begin{aligned}{} & {} [y,x,\overline{x}]=[\overline{x},x,y]=y, \end{aligned}$$
(4)
$$\begin{aligned}{} & {} [y,\overline{x},x]=[\overline{x},x,y]=y. \end{aligned}$$
(5)

is satisfied.

Theorem 2.2

In every flock :  (a) \([\overline{x},x,x]=[x,\overline{x},x]=[x,x,\overline{x}]=x,\) (b) \(x=\overline{\overline{x}},\) (c) \(\overline{[x,y,z]}=[\overline{x},\overline{y},\overline{z}].\)

3 Geometry of flocks

The relation \( \equiv \) defined on the set of all intervals of a flock \((G,[\;\;])\) by

$$\begin{aligned} (a,b)\equiv (c,d)\quad \text {if and only if}\quad [a,\overline{c},d]=b \end{aligned}$$
(6)

is an equivalence.

Equivalence classes of this relation can be interpreted as vectors, i.e. \(\overrightarrow{ab}=\{(c,d)\,|\, (c,d)\equiv (a,b)\}\). The set of such defined vectors is denoted by V(G). Intervals belonging to the same equivalence class are parallel.

Another concept of parallel intervals defined by flocks was proposed in [5] (see also [4]). In [5], following Rusakov’s definition for n-groups, two intervals (a, b) and (c, d) of a flock \((G,[\;\;])\) are parallel if \([a,\overline{b},d]=c\). Intervals parallel in the Rusakov’s sense may not be parallel in the sense of (6).

Parallel intervals (a, b), (c, d) define a parallelogram \(\langle a,c,d,b\rangle \). In other words, four points \(a,b,c,d\in G\) form a parallelogram

$$\begin{aligned} \langle a,b,c,d\,\rangle \quad \text {if and only if}\quad [a,\overline{b},c]=d. \end{aligned}$$
(7)

From the fact that \( \langle a,b,c,d\,\rangle \) is a parallelogram it does not follow that \( \langle a,d,c,b\rangle \) and \( \langle b,c,d,a\rangle \) are parallelograms, i.e. \((a,b)\equiv (d,c)\) does not imply \((a,d)\equiv (b,c)\). As an illustration of this situation we can take the flock \((S_{3},[\;\;])\), where \(S_3\) is the symmetric group with \([x,y,z]=xy^{-1}z\), and the parallelogram determined by \( a={ 1\;2\;3 \atopwithdelims ()2\;1\;3 }\), \( b={ 1\;2\;3 \atopwithdelims ()3\;2\;1 }\), \( c={ 1\;2\;3 \atopwithdelims ()3\;1\;2 }\), \( d={ 1\;2\;3 \atopwithdelims ()2\;3\;1 }\). In this flock vector \(\overrightarrow{ab\;}\) in our sense is not identical with vector \(\overrightarrow{ab\;}\) in the Rusakov’s sense.

Two points a and c of a flock \((G,[\;\;])\) are symmetric with respect to the point \(x\in G\) if \(\overrightarrow{ax\;}=\overrightarrow{xc\;}\), i.e. if \((a,x)\equiv (x,c)\). In other words, two points a and c are symmetric if there is a point \(x\in G\) such that \([a,\overline{x},c]=x\). In this case \([x,\overline{a},x]=c\). Thus, the mapping \(S_x:G\rightarrow G\) defined by \(S_x(a)=[x,\overline{a},x]\), is the symmetry with respect to the point x. Obviously, \(S_x(a)=c\) if and only if \(S_x(c)=a\). In addition, \(S_xS_x(a)=a\) for all \(a,x\in G\).

Below, we prove several elementary properties of parallelograms defined by flocks. These results were previously proved for n-groups. The proofs for n-groups are long and hard to read. For flocks, they are much shorter and simpler.

Proposition 3.1

If at least one tetragon \( \langle a,b,c,d\,\rangle ,\) \( \langle b,a,d,c\rangle ,\) \( \langle c,d,a,b\rangle \) or \( \langle d,c,b,a\rangle \) is a parallelogram,  then the other three remaining tetragons are also parallelograms.

Proof

Let \( \langle a,b,c,d\,\rangle \) be a parallelogram. Then, \( [a,\overline{b},c]=d\). Thus, by Theorem 2.1

$$\begin{aligned}{}[b,\overline{a},d]=[b,\overline{a},[a,\overline{b},c]]=[[b,\overline{a},a],\overline{b},c]=c , \end{aligned}$$

which proves that \( \langle b,a,d,c\rangle \) is a parallelogram. In the similar way

$$\begin{aligned}{}[c,\overline{d},a]=[[b,\overline{a},d],\overline{d},a]=[b,\overline{a},[d,\overline{d},a]]=b . \end{aligned}$$

Hence, also \(\langle c,d,a,b\rangle \) is a parallelogram. Analogously,

$$\begin{aligned}{}[d,\overline{c},b]=[d,\overline{c},[c,\overline{d},a]]=[[d,\overline{c},c],\overline{d},a]=a \end{aligned}$$

shows that \(\langle d,c,b,a\rangle \) is a parallelogram, too.\(\square \)

Corollary 3.2

\((a,b)\equiv (d,c)\;\Leftrightarrow \; (b,a)\equiv (c,d)\).

Thus, the following fact is obvious.

Proposition 3.3

A flock in which \(\langle a,b,c,d\,\rangle \) is a parallelogram implies that \(\langle a,d,c,b\rangle \) is a parallelogram is semiabelian. Conversely,  in a semiabelian flock,  if \(\langle a,b,c,d\rangle \) is a parallelogram,  then \(\langle a,d,c,b\rangle \) is a parallelogram.

Observe that for every vector \(\overrightarrow{p}\in V(G)\) and every point \(a\in G\) there is only one point \(c\in G\) such that \(\overrightarrow{p}=\overrightarrow{ac}\). Indeed, if \(\overrightarrow{p}=\overrightarrow{uv}=\overrightarrow{ac}\), then \((u,v)\equiv (a,c)\), i.e. \([u,\overline{a},c]=v\). Since for fixed u, v, a the last equation has only one solution c, the point c is uniquely determined.

The addition of vectors determined by a flock \((G,[\;\;])\) is defined by

$$\begin{aligned} \overrightarrow{ab\;}+\overrightarrow{cd\;}=\overrightarrow{a[b,\overline{c},d]\;}. \end{aligned}$$
(8)

This addition is well defined. Indeed, if \(\overrightarrow{p\;}=\overrightarrow{ab\;}=\overrightarrow{xy\;}\) and \(\overrightarrow{q\;}=\overrightarrow{cd\;}=\overrightarrow{zv\;}\). Then, also \(\overrightarrow{cd\;}=\overrightarrow{be\;}\) for some \(e\in G\). Thus, \(\overrightarrow{p\;}+\overrightarrow{q\;}=\overrightarrow{ab\;}+\overrightarrow{be\;}=\overrightarrow{ae\;}\) and, on the other side, \(\overrightarrow{p\;}+\overrightarrow{q\;}=\overrightarrow{xy\;}+\overrightarrow{zv\;}=\overrightarrow{x[y,\overline{z},v]}.\) But \(\overrightarrow{ab\;}=\overrightarrow{xy\;}\Leftrightarrow [a,\overline{x},y]=b\), whence multiplying by \(\overline{z},v\) we obtain \([[a,\overline{x},y],\overline{z},v]=[b,\overline{z},v]\). Since \(\overrightarrow{be\;}=\overrightarrow{zv\;}\) means that \([b,\overline{z},v]=e\), the last equation has the form \([[a,\overline{x},y],\overline{z},v]=e\), i.e. \([a,\overline{x},[y,\overline{z},v]]=e\). This implies \(\overrightarrow{ae\;}=\overrightarrow{x[y,\overline{z},v]\;}\). Hence, the addition of vectors is well defined.

Such defined addition is associative because for \(\overrightarrow{p\;}=\overrightarrow{ab\;}, \ \overrightarrow{q\;}=\overrightarrow{cd\;}, \ \overrightarrow{r\;}=\overrightarrow{ef\;}\), we have

\((\overrightarrow{p\;}+\overrightarrow{q\;})+\overrightarrow{r\;}=\overrightarrow{a[b, \overline{c},d]\;}+\overrightarrow{r\;}=\overrightarrow{a[[b,\overline{c},d],\overline{e},f]\;}\)

\(\overrightarrow{p\;}+(\overrightarrow{q\;}+\overrightarrow{r\;})=\overrightarrow{p\;} +\overrightarrow{c[d,\overline{e},f]\;}=\overrightarrow{a[b,\overline{c},[d,\overline{e},f]]\;}= \overrightarrow{a[[b,\overline{c},d],\overline{e},f]\;}.\)

Moreover, \(\overrightarrow{aa\;}\) is the zero vector, and \(-\overrightarrow{ab\;}=\overrightarrow{ba\;}\) for all \(a,b\in G\). Consequently, \((V(G),+)\) is a group.

Such defined addition of vectors is a generalization of Rusakov’s addition of vectors defined on semiabelian n-groups. For vectors of a semiabelian n-group (G, f) Rusakov defined (cf. [11] or [12]) the addition of vectors by the formula

$$\begin{aligned} \overrightarrow{ab\;}+\overrightarrow{cd\;}=\overrightarrow{ax\;}=\overrightarrow{zd\;}, \end{aligned}$$
(9)

where \(x=f(b,c^{[-2]},\!{\mathop {c}\limits ^{(2n-4)}}\!,d)\) and \(z=f(c,b^{[-2]},\!{\mathop {b}\limits ^{(2n-4)}}\!,a)\). For flocks, this formula has the form

$$\begin{aligned} \overrightarrow{ab\;}+\overrightarrow{cd\;}=\overrightarrow{a[b,\overline{c},d]\;}=\overrightarrow{[c,\overline{b},a]d\;}. \end{aligned}$$

Since \(\,\overrightarrow{a[b,\overline{c},d]\;}=\overrightarrow{[c,\overline{b},a]d\;}\) is equivalent to \([a,[\overline{c},b,\overline{a}],d]=[b,\overline{c},d]\), which is valid for all flocks, in the Russakov’s definition of addition of vectors the assumption that an n-group is semiabelian can be omitted. Consequently, for flocks the addition of vectors can be defined by (8) or by

$$\begin{aligned} \overrightarrow{ab\;}+\overrightarrow{cd\;}=\overrightarrow{[c,\overline{b},a]d\;}. \end{aligned}$$

Proposition 3.4

The flock \((G,[\;\;])\) is semiabelian if and only if for all \(\overrightarrow{p},\overrightarrow{q}\in V(G),\) we have \(\overrightarrow{p}+\overrightarrow{q}=\overrightarrow{q}+\overrightarrow{p}\).

Proof

Let \(\overrightarrow{p}=\overrightarrow{xy}\). Then, for every \(\overrightarrow{q}\in V(G)\) there is \(z\in G\) such that \(\overrightarrow{q}=\overrightarrow{yz}\). Similarly, for \(\overrightarrow{p}\) there is \(v\in G\) such that \(\overrightarrow{p}=\overrightarrow{zv}\). Consequently, \(\overrightarrow{p}+\overrightarrow{q}=\overrightarrow{xz\;}\) and \(\overrightarrow{q}+\overrightarrow{p}=\overrightarrow{yv\;}\). If \(\overrightarrow{p}+\overrightarrow{q}=\overrightarrow{q}+\overrightarrow{p}\), then \((x,z)\equiv (y,v)\), i.e. \([x,\overline{y},v]=z\). Since \(\overrightarrow{p}=\overrightarrow{xy\;}=\overrightarrow{zv\;}\), we also have \([x,\overline{z},v]y\). But then \(\langle x,z,v,y\rangle \) is a parallelogram. By Proposition 3.1 also \(\langle v,y,x,z\rangle \) is a parallelogram. Thus, \([v,\overline{y},x]=z\). Therefore, \([x,\overline{y},v]=[v,\overline{y},x]\) for all \(x,y,v\in G\). The retract \(\textrm{ret}_{\overline{y}}(G,[\;\;])\) is commutative. But retracts of \(G,[\;\;])\) are isomorphic (cf. [5]), so all retracts of \((G,[\;\;])\) are commutative. Consequently, for all \(x,y,z\in G\) we have \([x,y,z]=[z,y,x]\). Hence, the flock \((G,[\;\;])\) is semiabelian.

Now, let (G, g) be a semiabelian flock and \(\overrightarrow{p},\overrightarrow{q}\in V(G)\) be any vectors. Then, as above \(\overrightarrow{p}=\overrightarrow{xz\;}=\overrightarrow{yv\;}\), \(\overrightarrow{q}={\overrightarrow{zy\;}}\), \(\overrightarrow{p}+\overrightarrow{q}=\overrightarrow{xy\;}\) and \(\overrightarrow{q}+\overrightarrow{p}=\overrightarrow{zv\;}.\) From \(\overrightarrow{xz\;}=\overrightarrow{y\;v}\) it follows that \(g(x,\overline{y},v)=z\). Since (G, g) is semiabelian, we also have \(g(v,\overline{y},x)=z\). So \(\langle v,y,x,z\rangle \) is a parallelogram. By Proposition 3.1 also \(\langle x,z,v,y\rangle \) is a parallelogram, i.e. \([x,\overline{z},v]=y\). Hence, \(\overrightarrow{xy\;}=\overrightarrow{zv\;}\), which shows \(\overrightarrow{p\;}+\overrightarrow{q\;}=\overrightarrow{q}+\overrightarrow{p}\). \(\square \)

The above result means that affine geometry induced by n-groups (or flocks) is not commutative. It is commutative only in the case when it is induced by a semiabelian n-group (respectively, by a flock that is a ternary group).

Further, for the sake of clarity, the expression [[x, y, z], u, v] will be written as [x, y, z, u, v].

Theorem 3.5

A flock \((G,[ \; ])\) is semiabelian if and only if

$$\begin{aligned} 2(\overrightarrow{xy\;}+\overrightarrow{zu\;})=2\overrightarrow{xy\;}+2\overrightarrow{zu\;} \end{aligned}$$

for any points \(x,y,z,u\in G\).

Proof

Since

$$\begin{aligned} 2(\overrightarrow{xy\;}+\overrightarrow{zu\;})=\overrightarrow{x[y,\overline{z},u,\overline{x},y,\overline{z},u]\;} \end{aligned}$$

and

$$\begin{aligned} 2\overrightarrow{xy\;}+2\overrightarrow{zu\;}=\overrightarrow{x[y,\overline{x},y,\overline{z},u,\overline{z},u]\;}, \end{aligned}$$

the equation given in the above theorem is equivalent to

$$\begin{aligned} \overrightarrow{x[y,\overline{z},u,\overline{x},y,\overline{z},u]\;}=\overrightarrow{x[y, \overline{x},y,\overline{z},u,\overline{z},u]\;}, \end{aligned}$$

i.e. to

$$\begin{aligned}{}[y,\overline{x},y,\overline{z},u,\overline{z},u]=[y,\overline{z},u,\overline{x},y,\overline{z},u]. \end{aligned}$$

The last equation is equivalent to

$$\begin{aligned}{}[y,\overline{x},y,\overline{z},u]=[y,\overline{z},u,\overline{x},y]. \end{aligned}$$

Substituting \(y=z\), we will see that this flock is semiabelian. \(\square \)

Theorem 3.6

A flock \((G,[\,\;\,])\) is semiabelian if and only if

$$\begin{aligned} \overrightarrow{[x,\overline{y},z]\,[u,\overline{v},w]\;}=\overrightarrow{xu\;}-\overrightarrow{yv\;}+\overrightarrow{zw\;} \end{aligned}$$
(10)

holds for all \(x,y,z,u,v,w\in G\).

Proof

Let a flock (G, g) be semiabelian and

$$\begin{aligned} \overrightarrow{xu\;}-\overrightarrow{yv\;}+\overrightarrow{zw\;}=\overrightarrow{xu\;}+\overrightarrow{vy\;}+\overrightarrow{zw\;}= \overrightarrow{x[u,\overline{v},y]\;}+\overrightarrow{zw\;}=\overrightarrow{x[[u,\overline{v},y],\overline{z},w]\;}. \end{aligned}$$

Consider the quadrangle \(\langle [x,\overline{y},z],x,[u,\overline{v},y,\overline{z},w],[u,\overline{v},w]\rangle \). Since, as it is not difficult to verify, \( [[x,\overline{y},z],\overline{x},[u,\overline{v},y,\overline{z},w]]=[u,\overline{v},w] \), this quadrangle is a parallelogram. Thus, \(\overrightarrow{[x,\overline{y},z][u,\overline{v},w]\;}=\overrightarrow{x[u,\overline{v},y,\overline{z},w]\;}.\) This proves (10).

Conversely, if (10) holds for all \(x,y,z,u,v,w\in G\), then

$$\begin{aligned} \overrightarrow{[x,\overline{y},z]\,[u,\overline{v},w]\;}=\overrightarrow{xu\;} -\overrightarrow{yv\;}+\overrightarrow{zw\;}=\overrightarrow{x[u,\overline{v},y,\overline{z},w]\;}, \end{aligned}$$

i.e. \(\langle [x,\overline{y},z],x,[u,\overline{v},y,\overline{z},w],[u,\overline{v},w]\rangle \) is a parallelogram. Therefore,

$$\begin{aligned}{}[[x,\overline{y},z],\overline{x},[u,\overline{v},y,\overline{z},w]]=[u,\overline{v},w]. \end{aligned}$$
(11)

This for \(x=y=u\), \(z=w\) gives \([z,\overline{v},x]=[x,\overline{v},z]\). Hence, a flock satisfying (10) is semiabelian.

Since a semiabelian flock satisfies (11) Theorem 3.6 is proved. \(\square \)

Theorem 3.7

A flock \((G,[ \ ])\) is semiabelian if and only if

$$\begin{aligned} 2\overrightarrow{xy\;}=\overrightarrow{zu\;}+\overrightarrow{S_x(z)S_y(u)\;} \end{aligned}$$
(12)

for any points \(x,y,z,u\in G\).

Proof

Since \( 2\overrightarrow{xy\;}=\overrightarrow{xy\;}+\overrightarrow{xy\;}=\overrightarrow{x[y,\overline{x},y]\;} \) and

$$\begin{aligned} \overrightarrow{zu\;}+\overrightarrow{S_x(z)S_y(u)\;}=\overrightarrow{zu\;}+\overrightarrow{[x,\overline{z},x] [y,\overline{u},y]\;}=\overrightarrow{z[u,\overline{x},z,\overline{x},y,\overline{u},y]\;}, \end{aligned}$$

we have \(\overrightarrow{z[u,\overline{x},z,\overline{x},y,\overline{u},y]\;}=\overrightarrow{x[y,\overline{x},y]\;}\).

Thus, by (6), we obtain

$$\begin{aligned}{}[z,\overline{x},y,\overline{x},y]=[u,\overline{x},z,\overline{x},y,\overline{u},y], \end{aligned}$$

which for \(x=y\) gives \(z=[u,\overline{y},z,\overline{u},y]\). Whence, multiplying by \(\overline{y}\) and u, we get \([z,\overline{y},u]=[u,\overline{y},z]\).

Conversely, if a flock (G, g) is semiabelian, then

$$\begin{aligned}{}[x,\overline{z},[u,\overline{x},z,\overline{x},y,\overline{u},y]]=[y,\overline{x},y]. \end{aligned}$$

This, by (6), gives \(\overrightarrow{x[y,\overline{x},y]\;}=\overrightarrow{z[u,\overline{x},z,\overline{x},y,\overline{u},y]\;}\) and proves (12). \(\square \)

Theorem 3.8

A flock \((G,[ \ ])\) is semiabelian if and only if

$$\begin{aligned} \overrightarrow{ux\,}+\overrightarrow{S_x(u)y\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}=\overrightarrow{0} \end{aligned}$$
(13)

for any points \(x,y,z,u,w\in G\) such that \(\langle x,y,z,w\rangle \) is a parallelogram.

Proof

Let us start with the skew elements to \(S_yS_x(u)\) and \(S_zS_yS_x(u)\). Using Theorem 2.2, we obtain

$$\begin{aligned}{} & {} \overline{S_yS_x(u)\,}=\overline{[y,\overline{x},u,\overline{x},y]\,}=[\overline{y},x,\overline{u},x,\overline{y}], \\{} & {} \overline{S_zS_yS_x(u)}=\overline{[z,\overline{y},x,\overline{u},x,\overline{y},z]\,}=[\overline{z},y,\overline{x},u,\overline{x},y,\overline{z}]. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} \overrightarrow{ux\,}+\overrightarrow{S_x(u)y\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}\\{} & {} \quad \quad =\overrightarrow{u[x,\overline{S_x(u)},y]\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}\\{} & {} \quad \quad =\overrightarrow{u[x,[\overline{x},u,\overline{x}],y]\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}\\{} & {} \quad \quad =\overrightarrow{u[u,\overline{x},y]\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}\\{} & {} \quad \quad =\overrightarrow{u[x,\overline{y},z]\,}+\overrightarrow{S_zS_yS_x(u)w\,}\\{} & {} \quad \quad =\overrightarrow{u[x,\overline{y},z]\,}+\overrightarrow{[z,\overline{y},x,\overline{u},x,\overline{y},z]w\,}= \overrightarrow{u[u,\overline{x},y,\overline{z},w]\,}. \end{aligned}$$

Therefore,

$$\begin{aligned} \overrightarrow{ux\,}+\overrightarrow{S_x(u)y\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}= \overrightarrow{u[u,\overline{x},y,\overline{z},w]\,}. \end{aligned}$$

Since \(\langle x,y,z,w\rangle \) is a parallelogram, \(w=[x,\overline{y},z]\). Hence,

$$\begin{aligned} \overrightarrow{ux\,}+\overrightarrow{S_x(u)y\,}+\overrightarrow{S_yS_x(u)z\,}+\overrightarrow{S_zS_yS_x(u)w\,}= \overrightarrow{u[u,\overline{x},y,\overline{z},x,\overline{y},z]\,}. \end{aligned}$$

This means that (13) can be written in the form

$$\begin{aligned} \overrightarrow{u[u,\overline{x},y,\overline{z},x,\overline{y},z]\,}={\overrightarrow{0\,},} \end{aligned}$$

which is equivalent to

$$\begin{aligned} u=[u,\overline{x},y,\overline{z},x,\overline{y},z], \end{aligned}$$

i.e. to \([u,\overline{z},y]=[u,\overline{x},y,\overline{z},x]\). The last equation is satisfied only in a semiabelian flock. \(\square \)

Remark 3.9

In all the above results, the word “flock” can be replaced by “n-group”. Putting \([x,\overline{y},z]=f(f(x,y^{[-2]},\!{\mathop {y}\limits ^{(n-2)}}),\!{\mathop {y}\limits ^{(n-2)}}\!,z)\), where \((G,[\;\;])\) is a flock and (G, f) is an n-group, we obtain the short proof of these results for n-groups.

Remark 3.10

Another concept of vectors in flocks is presented in [4]. In this concept \(\overrightarrow{ab\;}=\overrightarrow{cd\;}\) if and only if \([a,\overline{b},d]=c\). The addition of vectors is defined by (9). The part \(=\overrightarrow{[c,\overline{b},a]d\;}\) should be omitted. It is not used in this article. This is a technical error not detected during the correction.

Yet another version was proposed by Kulazhenko (see for example [8]). In his version the addition of vectors is defined by (9) but \(\overrightarrow{ab\;}=\overrightarrow{cd\;}\) if and only if \([a,\overline{b},c]=d\). Then, by (9) and Theorem 2.2, \(\overrightarrow{a[b,\overline{c},d]\;}=\overrightarrow{[c,\overline{b},a]d\;}\) is equivalent to \([a,[\overline{b},c,\overline{d}],[c,\overline{b},a]]=d\). The last equation is not true because, for example, for the flock \([x,y^{-1},z]=xy^{-1}z\) induced by the group \((G,\cdot )\) implies \(d^{-1}=d\), which is not true. Thus, Lemma 4.1 in [8] (and other results of Sect.4 in [8] where this lemma is used) is wrong.