PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 109–120 DOI: 10.2298/PIM141208013V

GEOMETRY OF PENTAGONAL QUASIGROUPS

Stipe Vidak

Abstract. Pentagonal quasigroups are IM-quasigroups in which the additional identity of pentagonality holds. Motivated by the example C(q), where q is a solution of the equation q^{4} − 3q^{3} + 4q^{2} − 2q + 1 = 0, some basic geometric concepts are introduced and studied in a general pentagonal quasigroup. Such concepts are parallelogram, midpoint of a segment, regular pentagon and regular decagon. Some theorems of Euclidean plane which use these concepts are stated and proved in pentagonal quasigroups.

1. Introduction

A quasigroup (Q, ·) is a groupoid in which each of the equations a · x = b and y · a = b has a unique solution for given a, b ∈ Q. A quasigroup (Q, ·) is called IM-quasigroup if it satisfies the identities of idempotency and mediality:

(1.1)  aa = a
ab · cd= ac · bd. (1.2)

The immediate consequences of these identities are the identities known as elasticity, left distributivity and right distributivity:

ab · a= a · ba, (1.3)
a · bc= ab · ac, (1.4)
ab · c= ac · bc. (1.5)

Definition 1.1. Pentagonal quasigroup is an IM-quasigroup (Q, ·) in which the identity of pentagonality holds:

(1.6)  (ab · a)b · a = b.

In pentagonal quasigroups, along with pentagonality and the identities which are valid in any IM-quasigroup, some other identities hold. They are stated in the following theorem (see [2]).

2010 Mathematics Subject Classification: 20N05.

Key words and phrases: IM-quasigroup, parallelogram, midpoint, regular pentagon, regular decagon.

Communicated by Žarko Mijajlović.

Theorem 1.1. In every IM-quasigroup (Q, ·) identity (1.6) and the identities

(ab · a)c · a= bc · b, (1.7)

(ab · a)a · a= ba · b, (1.8)

ab · (ba · a)a= b (1.9)

are mutually equivalent and they imply the identity

(1.10)  a(b · (ba · a)a) · a = b

for every a, b, c ∈ Q.

Example 1.1. The basic example of the pentagonal quasigroup is C(q) = (ℂ, ∗), where ∗ is defined by a ∗ b = (1 − q)a + qb and q is a solution of the equation

(1.11)  q^{4} − 3q^{3} + 4q^{2} − 2q + 1 = 0.

The Equation (1.11) has four complex solutions:

q_{1,2} = \frac{1}{4}( 3 + \sqrt{5} ± i\sqrt{2(5 + \sqrt{5})} ) , q_{3,4} = \frac{1}{4}( 3 − \sqrt{5} ± i\sqrt{2(5 − \sqrt{5})} ) .

The most general example of the pentagonal quasigroup is given by the representation theorem proved in [2].

Theorem 1.2. Pentagonal quasigroup (Q, ·) exists if and only if there exists an Abelian group (Q, +) with an automorphism φ that satisfies

(1.12)  φ^{4} − 3φ^{3} + 4φ^{2} − 2φ + 1 = 0

and a · b = a + φ(b − a) for all a, b ∈ Q.

The example C(q) motivates the introduction of many geometric concepts in pentagonal quasigroups. We can regard elements of the set ℂ as points of the Euclidean plane. For any two different points a, b ∈ ℂ the equality a ∗ b = (1 − q)a + qb can be written in the form

\frac{a ∗ b − a}{b − a} = \frac{q − 0}{1 − 0} .

That means that the points a, b and a∗b are vertices of a triangle directly similar to the triangle with vertices 0, 1 and q. Each q_{i}, i = 1, 2, 3, 4 gives a certain type of triangle and we get so called characteristic triangles for pentagonal quasigroups. In C(q_{1}) the point a ∗ b is the third vertex of the regular pentagon determined by its adjacent vertices a and b. Any identity in the pentagonal quasigroup C(q) = (ℂ, ∗) can be interpreted as a theorem of the Euclidean geometry. Some figures which show that can be found in [2].

Here we define and study some basic geometric concepts in pentagonal quasigroups. We give the definition of the parallelogram and the midpoint of a segment and give an example of a quasigroup in which segments can have multiple midpoints. We give definitions of the regular pentagon and the regular decagon and their centres. In the last section we use defined concepts to state and prove two well known theorems of Euclidean plane in pentagonal quasigroups.

Figure 1. Definition of parallelogram

2. Parallelogram and midpoint of a segment

Definition 2.1. Let (Q, ·) be a quasigroup. Elements of Q are called points. A pair of points a and b is called a segment and is denoted by {a, b}. Cyclic n-tuple of points a_{1}, a_{2}, …, a_{n} is called n-gon and is denoted by (a_{1}, a_{2}, …, a_{n}).

Given four elements a, b, c, d of a medial quasigroup (Q, ·) the concept of parallelogram was defined in [5]. However, it is necessary to define many nontrivial geometric concepts in a medial quasigroup in order to define it properly. If we observe medial quasigroups with the additional identity of idempotency, IM-quasigroups, the definition of parallelogram becomes much more elegant (see [6]). In many subclasses of IM-quasigroups [7, 8] this definition can become even simpler due to an additional identity, which determines that subclass. In pentagonal quasigroups the definition is motivated by Figure 1.

Definition 2.2. Let (Q, ·) be a pentagonal quasigroup. We say that a, b, c, d ∈ Q form a parallelogram and we denote it by Par(a, b, c, d) if

(2.1)  d = ((ba · c)a · c)a.

Definition 2.3. Let Q be a set and let P ⊂ Q^{4} be a relation. Structure (Q, P) is called parallelogram space if:

i) For every a, b, c ∈ Q there is only one d ∈ Q such that P (a, b, c, d) holds. ii) If (e, f, g, h) is any cyclic permutation of (a, b, c, d) or (d, c, b, a), then

P (a, b, c, d) implies P (e, f, g, h), for every a, b, c, d ∈ Q. iii) For every a, b, c, d, e, f ∈ Q P (a, b, c, d) and P (c, d, e, f) imply P (a, b, f, e).

Our goal is to show that the structure (Q, Par), where Q is a pentagonal quasigroup and Par ⊂ Q^{4} a quaternary relation defined by (2.1) is a parallelogram space.

Proposition 2.1. If (Q, ·) is a pentagonal quasigroup and a, b, c, d ∈ Q, then Par(a, b, c, d) ⇒ P ar(b, c, d, a).

Proof. Par(a, b, c, d) is equivalent to d = ((ba · c)a · c)a and we have to prove the identity

a = ((cb · d)b · d)b.

Using (1.6) we get a = (ba · b)a · b. Right cancellation by b gives

(ba · b) · a = (cb · d)b · d.

If we apply (1.5) on the left hand side and plug in d = ((ba·c)a·c)a on the right-hand side, we get

(ba · a) · ba = (cb · d)b · ((ba · c)a · c)a.

Using (1.2) and cancelling from the right by ba, we need to prove

ba · a = (cb · d)((ba · c)a · c).

Now we have:

(cb · d)((ba · c)a · c) =^{(1.2)} (cb · (ba · c)a) · dc =^{(1.2)} (c(ba · c) · ba) · dc

=^{(1.4)} (c(ba · c) · ba)d · (c(ba · c) · ba)c

=^{(1.3),(1.6)} (c(ba · c) · ba)d · ba.

Let us now prove (c(ba · c) · ba)d = ((ba · (ba · a)) · ba)(ba · a). We have:

(c(ba · c) · ba)d= (c(ba · c) · ba) · ((ba · c)a · c)a

=^{(1.2)} (c(ba · c) · ((ba · c)a · c))(ba · a)

=^{(1.3)} ((c · ba)c · ((ba · c)a · c))(ba · a)

=^{(1.5)} ((c · ba) · (ba · c)a)c · (ba · a)

=^{(1.2),(1.3)} ((c · ba)c · (ba · a))c · (ba · a)

=^{(1.7)} ((ba · (ba · a)) · ba)(ba · a).

In the end using (1.6) we have ((ba · (ba · a)) · ba)(ba · a) · ba = ba · a. Hence a = ((cb · d)b · d)b, i.e. Par(b, c, d, a). □

Proposition 2.2. If (Q, ·) is a pentagonal quasigroup and a, b, c, d ∈ Q, then Par(a, b, c, d) ⇒ Par(c, b, a, d).

Proof. Since Par(a, b, c, d) ⇔ d = ((ba · c)a · c)a, we have:
d = ((ba · c)a · c)a =^{(1.6)} ((ba · c)a · c)((ca · c)a · c)

=^{(1.5)} ((ba · c)a · (ca · c)a)c =^{(1.5)} ((ba · c)(ca · c) · a)c

=^{(1.5)} ((ba · ca)c · a)c =^{(1.5)} ((bc · a)c · a)c,

which is equivalent to Par(c, b, a, d). □

Lemma 2.1. If (Q, ·) is a pentagonal quasigroup and a, b, c, d ∈ Q, then

Par(a, b, c, d) ⇔ ∃ p, q ∈ Q such thatpa = qb and pd = qc.

Proof. Let us put p = (ab · a)b and q = b. Then by (1.1) and (1.6) pa = qb holds. If we show that

(ab · a)b · ((ba · c)a · c)a = bc,

we will show that Par(a, b, c, d) implies pd = qc with the above defined p and q. In fact, proving that we will prove the other implication as well, because with such defined p and q we have pa = qb and pd = qc, and we must show d = ((ba · c)a · c)a. When we show

(ab · a)b · ((ba · c)a · c)a = bc,

we see that ((ba · c)a · c)a satisfies the equation pd = qc with given p and qc, and since in a quasigroup such solution must be unique, d must be equal to ((ba·c)a·c)a.

Let us now show that (ab · a)b · ((ba · c)a · c)a = bc. We have:

(ab · a)b · ((ba · c)a · c)a =^{(1.4)} ((ab · a)b · ((ba · c)a · c)) · ((ab · a)b · a)

=^{(1.6)} ((ab · a)b · ((ba · c)a · c))b

=^{(1.2)} (((ab · a) · (ba · c)a) · bc)b

=^{(1.5)} ((ab · (ba · c))a · bc)b =^{(1.2)} (((a · ba) · bc)a · bc)b

=^{(1.3)} (((ab · a) · bc)a · bc)b =^{(1.7)} ((b · bc)b · bc)b =^{(1.6)} bc. □

Proposition 2.3. If (Q, ·) is a pentagonal quasigroup and a, b, c, d, e, f ∈ Q, then

Par(a, b, c, d), Par(c, d, e, f) ⇒ Par(a, b, f, e).

Proof. By Lemma 2.1 we have

Par(a, b, c, d) ⇔ ∃ p_{1}, q_{1} ∈ Q such that p_{1}a = q_{1}b and p_{1}d = q_{1}c,
Par(c, d, e, f) ⇔ ∃ p_{2}, q_{2} ∈ Q such that p_{2}c = q_{2}d and p_{2}f = q_{2}e,

wherefrom we conclude that p_{1} = q_{2} and q_{1} = p_{2}. By Lemma 2.1 we have

Par(a, b, f, e) ⇔ ∃ p_{3}, q_{3} ∈ Q such that p_{3}a = q_{3}b and p_{3}e = q_{3}f.

If we put p_{3} = p_{1} = q_{2} and q_{3} = q_{1} = p_{2}, then we immediately see that p_{3}a = q_{3}b and p_{3}e = q_{3}f, which implies Par(a, b, f, e). □

Theorem 2.1. If (Q, ·) is a pentagonal quasigroup, and Par ⊂ Q^{4} relation on Q defined by (2.1), then (Q, Par) is a parallelogram space.

Proof. Property i) is obvious.

Property ii) follows from Propositions 2.1 and 2.2, because all cyclic permutations of (a, b, c, d) and (d, c, b, a) can be achieved from (a, b, c, d) by successive applications of permutations from those propositions.

Property iii) follows from Proposition 2.3. □

The previous theorem proves that the parallelogram, defined in a pentagonal quasigroup, possesses all the properties of the parallelogram in medial and IMquasigroups stated and proved in [5] and [6]. The same holds for the concept of the midpoint of a segment whose definition uses the definition of the parallelogram.

Let (Q, ·) be a medial quasigroup. We say that m ∈ Q is the midpoint of the segment {a, b}, a, b ∈ Q if Par(a, m, b, m) holds. Using (2.1) the midpoint of a segment can be defined in a pentagonal quasigroup in the following way.

Definition 2.4. Let {a, b} be a segment in a pentagonal quasigroup (Q, ·) for a, b ∈ Q. We say that m ∈ Q is the midpoint of the segment {a, b} and denote it by M(a, m, b) if

(2.2)  m = ((ma · b)a · b)a

Applying Theorem 1.2 several times we get

((ma · b)a · b)a =(1 − φ)^{5}(m) + φ(1 − φ)^{4}(a) + φ(1 − φ)^{3}(b)
+ φ(1 − φ)^{2}(a) + φ(1 − φ)(b) + φ(a).

Plugging that into identity (2.2) and using (1.12) several times we get 2m = a + b, where + is the addition of the Abelian group (Q, +) associated to the pentagonal quasigroup (Q, ·).

Obviously, in the quasigroup C(q_{i}) = (ℂ, ∗), i = 1, 2, 3, 4, the midpoint m of any segment \{a, b\} is unique and can be expressed as m = \frac{a+b}{2} . However, that is not the case for all pentagonal quasigroups.

Example 2.1. Pentagonal quasigroup (Q_{16}, ·) is constructed from the group ℤ_{2}^{4} and the automorphism φ(x, y, z, w) = (w, x, y, z + w) satisfying (1.12).

Let a, b ∈ Q_{16} be such that a ≠ b. We notice that there exists no x ∈ Q_{16} such that M(a, x, b), while for every x ∈ Q_{16} M(a, x, a) holds. Thus the segment {a, a} has sixteen distinct midpoints in Q_{16}. The reason behind that lies in the Abelian group associated to the quasigroup (Q_{16}, ·) by Theorem 1.2, which is ℤ_{2}^{4}. Element x is midpoint of the segment {a, b} if and only if 2x = a + b holds. Since two different elements of ℤ_{2}^{4} that add up to the identity element do not exist, there exists no x ∈ Q_{16} such that M(a, x, b). Since every element of the group ℤ_{2}^{4} is of order 2, for every x ∈ Q_{16}, we have M(a, x, a).

The following theorem will be used repeatedly in the next section.

Theorem 2.2. In the pentagonal quasigroup (Q, ·) the identity

(2.3)  ((((ba · a)a · a)a · a)a · a)a = (ab · a)b

holds for every a, b ∈ Q.

Proof. Obviously Par(a, b, a, ((ba · a)a · a)a) holds. Using property ii) of the parallelogram space, Par(a, ((ba·a)a·a)a, a, b) holds as well. That gives the identity ((((ba·a)a·a)a·a)a·a)a·a = b. Applying (1.6) to the right-hand side and cancelling from the right by a gives the desired identity. □

3. Regular pentagon and regular decagon

Motivated by Figure 2 the following concept can be defined in a general pentagonal quasigroup.

Definition 3.1. Let (a, b, c, d, e) be a pentagon in a pentagonal quasigroup (Q, ·). We say that (a, b, c, d, e) is a regular pentagon and denote it by RP(a, b, c, d, e) if ab = c, bc = d and cd = e hold.

We immediately note that any regular pentagon in a pentagonal quasigroup is uniquely determined by its pair of adjacent vertices. The remaining vertices can be expressed in terms of these two vertices using operation in the quasigroup.

Figure 2. Regular pentagon (a, b, c, d, e)

Theorem 3.1. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e ∈ Q such that RP(a, b, c, d, e) holds. Then d = ba · a and e = (ab · a)b hold.

Proof. We directly compute

d = bc = b · ab =^{(1.3)} ba · b, e = cd = ab · (ba · b) =^{(1.5)} (a · ba)b =^{(1.3)} (ab · a)b.□

Theorem 3.2. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e ∈ Q such that RP(a, b, c, d, e) holds. Then de = a and ea = b hold.

Proof. Using the previous theorem we compute

de = (ba · b) · (ab · a)b =^{(1.5)} (ba · (ab · a))b =^{(1.5)} (b · ab)a · b =^{(1.3)} (ba · b)a · b =^{(1.6)} a,

ea = (ab · a)b · a =^{(1.6)} b.  □

Next two corollaries are direct consequences of the previous theorem.

Corollary 3.1. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e ∈ Q. Then

RP(a, b, c, d, e) ⇔ ab = c, bc = d, cd = e, de = a, ea = b.

Corollary 3.2. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e ∈ Q. Then RP(a, b, c, d, e) implies RP(p, q, r, s, t), where (p, q, r, s, t) is any cyclic permutation of (a, b, c, d, e).

Figure 3. Centre of regular pentagon

Figure 3 motivates introduction of the following geometric concept in pentagonal quasigroups.

Definition 3.2. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e ∈ Q such that RP(a, b, c, d, e) holds. We say that o ∈ Q is the centre of the regular pentagon (a, b, c, d, e) if oa · b = o holds.

Theorem 3.3. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, o ∈ Q such that RP(a, b, c, d, e) holds and o is the centre of (a, b, c, d, e). Then ob · c = o, oc · d = o, od · e = o and oe · a = o hold.

Proof. We compute ob · c = ob · ab =^{(1.5)} oa · b = o. The other identities are proved analogously. □

The centre of a regular pentagon in pentagonal quasigroups does not have to exist as explained in the next example.

Table 1. Pentagonal quasigroup Q_{5}

\begin{matrix}· & 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 2 & 4 & 1 & 3 \\ 1 & 4 & 1 & 3 & 0 & 2 \\ 2 & 3 & 0 & 2 & 4 & 1 \\ 3 & 2 & 4 & 1 & 3 & 0 \\ 4 & 1 & 3 & 0 & 2 & 4 \end{matrix}

Example 3.1. The quasigroup Q_{5} (Table 1) is constructed from the Abelian group ℤ_{5} using the automorphism φ(x) = 2x which satisfies (1.12).

In (Q_{5}, ·) the statement RP(0, 1, 2, 3, 4) holds. The pentagon (0, 1, 2, 3, 4) is determined by its vertices 0 and 1. Now we want to try all possible candidates for the centre o of (0, 1, 2, 3, 4):

00 · 1 = 2 ≠ 0, 10 · 1 = 3 ≠ 1, 20 · 1 = 4 ≠ 2, 30 · 1 = 0 ≠ 3, 40 · 1 = 1 ≠ 4.

Hence, we conclude that the regular pentagon (0, 1, 2, 3, 4) in Q_{5} does not have a centre. Algebraic justification for that is the use of Theorem 1.2. If we apply a · b = a + φ(b − a) to the identity oa · b = o, we get successively:

(1 − φ)(oa) + φ(b) = o
(1 − φ)^{2}(o) + (1 − φ)φ(a) + φ(b)  = o
o − 2φ(o) + φ^{2}(o) − o = −φ(1 − φ)(a) − φ(b)
φ(2 · 1 − φ)(o) = φ(1 − φ)(a) + φ(b).

Since φ is an isomorphism, it is equivalent to

(2 · 1 − φ)(o) = (1 − φ)(a) + b.

The function 2 · 1 − φ = 0 is not invertible, so there exists no o ∈ Q_{5} which satisfies the previous equation.

Figure 4 gives motivation for the next definition.

Definition 3.3. We say that points a, b, c, d, e, f, g, h, i, j of the pentagonal quasigroup (Q, ·) form a regular decagon with the centre o ∈ Q if

ba · b = cb · c = dc · d = ed · e = fe · f = gf · g = hg · h = ih · i = ji · j = o.

We denote it by RD_{o}(a, b, c, d, e, f, g, h, i, j) or, if we want to omit the centre, by RD(a, b, c, d, e, f, g, h, i, j).

We note that any regular decagon in a pentagonal quasigroup is uniquely determined by its pair of adjacent vertices.

Lemma 3.1. Let (Q, ·) be a pentagonal quasigroup and a, b, o ∈ Q. Then ba · b = ois equivalent to b = (ao · o)o.

Proof. Multiplying the identity ba·b = o from the right by a and b respectively we get (ba · b)a · b = oa · b. Applying (1.6) we get a = oa · b and (1.9) finally gives b = (ao · o)o. □

Figure 4. Regular decagon (a, b, c, d, e, f, g, h, i, j) with the centre o

Theorem 3.4. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h, i, j, o ∈ Q. Then RD_{o}(a, b, c, d, e, f, g, h, i, j) implies j = co and j = (((ao·o)o·o)o·o)o.

Proof. Multiple use of Lemma 3.1 gives:

j = (io · o)o = ((ho · o)o · o)o · o = ((((go · o)o · o)o · o)o · o)o

=^{(2.3)} (og · o)g = (og · o) · (fo · o)o =^{(1.5)} (og · (fo · o))o =^{(1.2),(1.3)} ((of · o) · go)o

=^{(1.5)} (of · g)o · o = (of · (fo · o)o)o · o =^{(1.9)} fo · o = ((eo · o)o · o)o

= (((do · o)o · o)o · o)o · o = (((((co · o)o · o)o · o)o · o)o · o)o

=^{(1.4)} ((oc · o)c · o)o =^{(1.6)} co = (bo · o)o · o = (((ao · o)o · o)o · o)o.  □

Corollary 3.3. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h, i, j, o ∈ Q. Then RD_{o}(a, b, c, d, e, f, g, h, i, j) implies the following identities

a = do, b = eo, c = fo, d = go, e = ho, f = io, g = jo, h = ao, i = bo, j = co. Theorem 3.5. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h,

i, j, o ∈ Q. Then RD_{o}(a, b, c, d, e, f, g, h, i, j) implies aj · a = o.

Proof. Using Theorem 3.4 we get

aj · a = (a · (((ao · o)o · o)o · o)o)a =^{(1.6)} (((oa · o)a · o) · (((ao · o)o · o)o · o)o)a

=^{(1.5)} ((oa · o)a · (((ao · o)o · o)o · o))o · a =^{(1.2)} (((oa · o) · ((ao · o)o · o)o) · ao)o · a

=^{(1.5)} ((oa · ((ao · o)o · o))o · ao)o · a =^{(1.5)} ((oa · ((ao · o)o · o))a · o)o · a

=^{(1.4),(1.9)} ((a(oa · o) · a)o · o)a =^{(1.5),(1.6)} (ao · a)o · a =^{(1.6)} o.  □

Corollary 3.4. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h, i, j, o ∈ Q. Any nine of ten identities

ba · b= o, cb · c= o, dc · d= o, ed · e= o, fe · f = o,
gf · g = o, hg · h= o, ih · i = o, ji · j = o, aj · a = o

imply the remaining one.

Corollary 3.5. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h, i, j, o ∈ Q. Any nine of ten identities

ba · b= o, cb · c= o, dc · d= o, ed · e= o, fe · f = o,
gf · g = o, hg · h= o, ih · i = o, ji · j = o, aj · a = o

imply RD_{o}(a, b, c, d, e, f, g, h, i, j).

4. Two theorems of Euclidean plane

The following theorem was proved in [5].

Theorem 4.1. In a medial quasigroup any two of three statements Par(a, b, c, d), Par(e, f, g, h), Par(ae, bf, cg, dh) imply the remaining one.

As an illustration of the concepts introduced before we give versions of two theorems of Euclidean plane in pentagonal quasigroups.

Theorem 4.2. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, a_{1}, b_{1}, c_{1}, d_{1}, e_{1}, ∈ Q such that RP(a, b, c, d, e), M(a, a_{1}, b), M(b, b_{1}, c) hold. Then

RP(a_{1}, b_{1}, c_{1}, d_{1}, e_{1}) ⇒ M(c, c_{1}, d), M(d, d_{1}, e), M(e, e_{1}, a).

Proof. Statements M(a, a_{1}, b) and M(b, b_{1}, c) are respectively equivalent to Par(a, a_{1}, b, a_{1}) and Par(b, b_{1}, c, b_{1}). By Theorem 4.1 we have Par(ab, a_{1}b_{1}, bc, a_{1}b_{1}). Since we have RP(a, b, c, d, e), we get Par(c, a_{1}b_{1}, d, a_{1}b_{1}) or M(c, a_{1}b_{1}, d). The statement RP(a_{1}, b_{1}, c_{1}, d_{1}, e_{1}) gives a_{1}b_{1} = c_{1}. Hence we have M(c, c_{1}, d). The statements M(d, d_{1}, e) and M(e, e_{1}, a) are proved analogously. □

Theorem 4.3. Let (Q, ·) be a pentagonal quasigroup and a, b, c, d, e, f, g, h, i, j, ∈ Q. We have

RD(a, b, c, d, e, f, g, h, i, j) ⇒ RP(a, c, e, g, i), RP(b, d, f, h, j).

Proof. Let us denote by o the centre of the regular decagon (a, b, c, d, e, f, g, h, i, j). Since cb · c = o, by Lemma 3.1 we have c = (bo · o)o. Corollary 3.3 gives b = eo, wherefrom using (1.6) we get e = (ob · o)b. Also, we have ba · b = o, wherefrom using (1.6) we get ba = (bo · b)o. Using (1.3) and (1.10) on that we get a = ((bo · b)o · (((bo · b)o · b)b · b))b. Furthermore, we have:

a = ((bo · b)o · (((bo · b)o · b)b · b))b =^{(1.6)} ((bo · b)o · (ob · b))b

=^{(1.5)} ((bo · b)o · b) · (ob · b)b =^{(1.6)} o · (ob · b)b.

Now we compute:

ac = (o · (ob · b)b) · (bo · o)o =^{(1.4)} (o(ob · b) · ob) · (bo · o)o

=^{(1.5),(1.9)} (o(ob · b) · (bo · o)o)b =^{(1.4)} (((o · ob) · ob) · (bo · o)o)b

=^{(1.5),(1.9)} ((o · ob) · (bo · o)o)b · b =^{(1.4),(1.8),(1.9)} ((bo · b)b · b)b =^{(1.8)} (ob · o)b.

That shows ac = e. Identities ce = g and eg = i are proved analogously. Those three identities imply RP(a, c, e, g, i). The statement RP(b, d, f, h, j) is proved analogously. □

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Department of Mathematics (Received 17 12 2014) University of Zagreb Zagreb Croatia

    svidak@math.pmf.hr