Extensions of the Golden Hypergroup by FiniteAbelian Groups

Extensions of the Golden Hypergroup by Finite Abelian Groups

著者 KAWAKAMI Satoshi, KAWASAKI Ken‑ichiroh, YAMANAKA Satoe

journal or

publication title

奈良教育大学紀要. 自然科学

volume 57

number 2

page range 1‑10

year 2008‑10‑31

URL http://hdl.handle.net/10105/710

奈良教育大学紀要 第57巻 第

2

号（自然）平成20年

1 Bull. Nara Univ. Educ., Vol. 57, No.2 ( Nat. ) , 2008

Extensions of the Golden Hypergroup by Finite Abelian Groups

Satoshi KAWAKAMI, Ken-ichiroh KAWASAKI ^* and Satoe YAMANAKA

（Department of Mathematics, Nara University of Education, Nara 630-8528, Japan）

（Received May 7, 2008 ）

Abstract

The purpose of this paper is to investigate extension problem for the category of finite com- mutative hypergroups. In fact, we determine all extensions of the Golden hypergroup by finite abelian groups. (AMS Subject Classification : 43A62, 20N20.)

Key Words : hypergroup, extension, group

* Supported in part by Grant-in-Aid for Scientific Research (C)

♯

20540043.

1. Introduction

Let H and L be finite commutative hypergroups.

A finite commutative hypergroup K is called an exten- sion of L by H if the sequence :

1 H K L 1

is exact, i.e. if the quotient hypergroup K/H is isomor- phic to L. Here, the notions of subhypergroup, quo- tient hypergroup and isomorphism between hyper- groups are taken from Bloom-Heyer

⁽¹⁾

and Wildberger

⁽¹⁴⁾

, a source from which all the elementary knowledge needed in the sequel will be taken.

In our previous paper(10) all extensions K of hyper- groups of order two by finite abelian groups H are determined. In the present paper we will report that all extensions K of the Golden hypergroup L by finite abelian groups H are determined including non-splitting case. Wildberger

⁽¹⁵⁾

determined all hypergroup struc- tures of order three in 2002. In his work, he pointed that the Golden hypergroup was in an interesting posi- tion among strong hypergroups of order three. This is a motivation that we consider extensions of the Golden hypergroup. Hypergroup structure of order four is not yet determined. Such hypergroups arising from exten- sions are determined in our paper (6). Therefore it is important to determine hypergroup structure of low order which is greater than 3. On the other hand in our

paper (3), we studied the relationship between splitting extensions and extensions arising from fields. In the present paper we also discuss splitting extensions and weaken the notion of splitting of extension.

In the present paper we determine all extensions K of the Golden hypergroup L={ r

0

, r

1

, r

2

} by finite abelian groups H as described in the following theorem.

Let K be a commutative hypergroup extension of the Golden hypergroup L by a finite abelian group H, which means that there exists a hypergroup homomor- phism ϕ from K onto L such that Kerϕ = H. Let H( r

1

) be the stability group of H at s

₀

∈ S := ϕ

⁻¹

( r

1

) and H( r

2

) be the stability group of H at t

₀

∈ T := ϕ

⁻¹

( r

2

). Let ω ( r

i

) denote the normalized Haar measure of H( r

i

) (i = 1, 2).

Then we have S = {hs

0

: h ∈ H} and T = {kt

0

: k ∈ H}.

When s

^＊₀

= hs

₀

and t

^＊₀

= kt

₀

for some h, k ∈ H, s

^＊₀

s

₀

= ω ( r

1

) + c

₁

t

₀

, t

^＊₀

t

₀

= ω ( r

2

) + c

₂

s

₀

, and s

₀

t

₀

= c

^＊₁

s

₀

+ c

^＊₂

t

₀

for c

1

, c

2

∈ M

¹

( H) such thatω ( r

1

)ω ( r

2

) c

_i

= c

_i

(i=1, 2).

Moreover, c

^＊1

= c

1

k

^＊

, c

^＊2

= c

2

h

^＊

, c

²1

=ω( r

1

) ω ( r

2

)k, c

²₂

=ω( r

1

)ω ( r

2

)h.

All extensions K of L by H are characterized in this way.

Such an extension K is denoted by K ( s

₀

, t

₀

, h, k, c

₁

, c

₂

).

Then two extensions K (s

₀

, t

0

, h, k, c

₁

, c

2

) and K (u

₀

, v

0

, h

1

, k

₁

, d

₁

, d

2

) of L by H are mutually equivalent as extensions if and only if there exist b

₁

, b

₂

∈ H such that u

₀

= b

^＊₁

s

₀

, v

₀

= b

^＊₂

t

₀

, d

₁

= b

₂

c

₁

, d

₂

= b

₁

c

₂

, ω ( r

1

)h

₁

=ω ( r

1

)b

²₁

h, and ω ( r

2

)k

₁

=

( r

2

)b

²₂

k.

Moreover, the extension K = K ( s

₀

, t

0

, h, k, c

₁

, c

2

) is

1 2 1

2 1

2 1

2 1

2 1 2

ω

weakly splitting if and only if there exist b

1

, b

2

∈ H such that c

1

= ω( r

1

)ω ( r

2

) b

₂

, c

2

= ω ( r

1

)ω ( r

2

) b

₁

, ω ( r

1

) h = ω ( r

1

)b

²₁

, and ω ( r

2

) k = ω ( r

2

)b

₂²

. K is splitting if and only if K is weakly splitting and H( r

1

) = H( r

2

).

2. Preliminaries

We recall some notions and facts on finite commu- tative hypergroups from Bloom-Heyer s book

⁽¹⁾

and Wildberger

⁽¹⁴⁾

. K := (K, A) is called a finite commutative hypergroup if the following conditions (1) 〜 (6) are satis- fied.

(1) A is a

＊

-algebra over with the unit c

₀

. (2) K = { c

₀

, c

₁

, … , c

_n

} is a linear basis of A.

(3) K

^＊

= K.

(4) c

_i

c

_j

= n

^k_ij

c

_k

, where n

^k_ij

is a non-negative real number such that

c

^＊_i

= c

_j

n

_ij⁰

＞ 0, i.e. c

^＊_i

≠ c

_j

n

_ij⁰

= 0.

(5) n

^k_ij

= 1 for any i, j.

(6) c

_i

c

_j

= c

_j

c

_i

for any i, j.

The weight of an element c

_i

∈ K is defined by w(c

_i

) := (n

⁰_ij

)

⁻¹

where c

_j

= c

^＊_i

, and the total weight of K is given by w(K) := Σ

ⁿⁱ⁼⁰

^w(c

ⁱ

^).

Let M

¹

(K) denote the set of probability measures on K, i.e.

M

¹

(K) := {c = a

_k

c

_k

: a

_k

0 (k = 0, 1, … , n), a

_k

= 1}.

For c = Σ

^nk=0

^a

^k

^c

^k

^∈ A(K), support of c is defined by supp(c) := {c

_k

: a

_k

≠0. k = 0, 1, … , n}.

Let ω (K) denote the normalized Haar measure of K which is given by

ω(K) = c

_k

.

For a finite commutative hypergroup K, a complex valued function χ on K is called a character of K if

(1) χ (c

₀

) = 1, (2) χ ( c

^＊_i

) = χ ( c

_i

) , (3) χ (c

_i

) χ (c

_j

) = n

^k_ij

χ (c

_k

).

The set K ˆ of all characters of K is called the dual of K which is not necessarily a hypergroup in general. A finite hypergroup K is called a strong hypergroup if the dual K ˆ of K is also a hypergroup.

Let L = { r

0

, r

1

, r

2

} be the Golden hypergroup which is determined by r

¹²

= r

0

+ r

2

, r

2

2

= r

0

+ r

1

, r

1

r

2

= r

1

+ r

2

and H = { h

₀

, h

₁

, … , h

_n

} a finite commutative hyper-

group. Then, a hypergroup join H ∨ L of H by L is defined by

H ∨ L = { h

₀

, h

₁

, … , h

_n

, s

₀

, t

₀

}, h

_k

s

₀

= s

₀

(k = 0, 1, … , n), s

₀²

= ω (H) + t

₀

,

h

_k

t

₀

= t

₀

(k = 0, 1, … , n), t

₀²

= ω (H) + s

₀

.

3. Extensions of the Golden hypergroup

Let L = { r

0

, r

1

, r

2

} be the Golden hypergroup where r

0

is the unit of L. The hypergroup structure of L is determined by

r

²1

= r

0

+ r

2

, r

^＊1

= r

1

,

r

²2

= r

0

+ r

1

, r

^＊2

= r

2

, r

1

r

2

= r

1

+ r

2

.

Let H = {h

₀

, h

₁

, … , h

_n

} be a finite abelian group where h

₀

is the unit of H.

We investigate the structure of extensions K of L by H. Let ϕ be a homomorphism from K onto L such that Ker ϕ = H, where H is assumed to be a subhyper- group of K . Then K is written as the disjoint union of H = ϕ

⁻¹

( r

0

), S := ϕ

⁻¹

( r

1

), and T := ϕ

⁻¹

( r

2

). Let H( r

1

) and H( r

2

) denote the stability group of H at s

₀

∈ S and t

₀

∈ T respectively, i.e.

H( r

1

) = {h ∈ H : hs

₀

= s

₀

}, H( r

2

) = {h ∈ H : ht

0

= t

0

}.

We note that H( r

1

) does not depend on the choice of s

0

∈ S but only on S and H( r

2

) also depends only on T.

Proposition 1. For each s ∈ S and t ∈ T, there exist h and k ∈ H such that s = hs

₀

and t = kt

₀

.

Proof. For s ∈ S, there exists h ∈ H such that h ∈ supp(s

^＊₀

s) because ϕ (s

^＊₀

s)= r

^＊1

r

1

. Hence we see that h

0

= h

^＊

h ∈ supp((hs

₀

)

^＊

s). This implies that s ∈ supp(hs

₀

). Then supp(h

^＊

s ) ⊂ supp(h

^＊

hs

₀

) = supp(h

₀

s

₀

) = supp(s

₀

) = {s

₀

}.

Hence we see that h

^＊

s = s

₀

, namely s = hs

₀

. In a similar way, we have the same conclusion for t ∈ T .

[Q.E.D.]

Let ω (H

₀

) denote the normalized Haar measure of a subgroup H

₀

of H. The next lemma is useful for our arguments hereafter.

Lemma 2. For a subgroup H

₀

of H, if c ∈ M

¹

(H), supp(c) ⊂ H

₀

and ω (H

₀

)c = c, then we have c =ω (H

₀

).

1 2 1 2 1 2 1 2

1 2 1 2

1 2 1

2

1 2 1

2

1 2

1 2 1

2 1 2 1

2 1 2

n

Σ

k=0

w(c

_k

) w(K)

n

Σ

k=0 n

Σ

_k=0

n

Σ

k=0

n

Σ

_k=0

Σ

_k=0ⁿ

Proof. For c ∈ M

¹

(H) and supp(c) ⊂ H

₀

, we can write c = a

_k

h

_k

where a

_k

= 1. Then, we have c = ω (H

₀

)c = a

_k

ω (H

₀

)h

_k

= a

_k

ω (H

₀

) = ( ^a

k

) ω (H

₀

) =ω (H

₀

).

Hence we get the desired conclusion.

[Q.E.D.]

Let ω( r

1

) denote the normalized Haar measure of H( r

1

) and ω( r

2

) denote the normalized Haar measure of H( r

2

).

Proposition 3. For s

₀

∈ S and t

₀

∈ T, s

^＊₀

= hs

₀

and t

^＊₀

= kt

₀

hold for some h ∈ H and k ∈ H. Then we have s

^＊₀

s

₀

= ω ( r

1

) + c

₁

t

₀

, t

^＊0

t

₀

= ω ( r

2

) + c

₂

s

₀

, s

0

t

₀

= c

₃

s

₀

+ c

₄

t

₀

where c

_i

∈ M

¹

(H) (i = 1, 2, 3, 4) such that c

^＊₁

k = c

1

and c

^＊2

h

= c

₂

, and ω ( r

1

) ω ( r

2

)c

_i

= c

_i

( i = 1, 2, 3, 4). Moreover we have c

²₁

= ω ( r

1

) ω ( r

2

)k, c

²₂

=ω ( r

1

) ω ( r

2

)h, c

₃

= c

^＊₁

, and c

₄

= c

^＊₂

.

Proof. One can take h, k ∈ H such that s

^＊0

= hs

0

and t

^＊0

= kt

0

by Proposition 1 because s

^＊0

∈ S and t

^＊0

∈ T by the relations r

^＊1

= r

1

and r

^＊2

= r

2

. It is easy to see that s

^＊₀

s

₀

is written as

s

^＊₀

s

₀

= c

₀

+ c

₁

t

₀

for some c

0

, c

1

∈ M

¹

(H) .

First, we show the equality c

₀

=ω ( r

1

). The fact ω ( r

1

)s

₀

= s

₀

implies that ω ( r

1

)c

₀

= c

₀

and ω ( r

1

)c

₁

= c

₁

. We suppose that h′ ∈ / ^H( r

1

). Then h′s

₀

≠s

0

holds so that (h′s

₀

)

^＊

≠s

^＊0

. Then h

₀

∈ / supp((h′s

₀

)

^＊

s

₀

) by the axiom of hypergroup. Since (h′s

₀

)

^＊

s

₀

= (h′)

^＊

c

₀

+ (h′)

^＊

c

₁

t

₀

, h

₀

∈ / supp((h′)

^＊

c

₀

). Therefore h′

∈ / ^supp(c

0

). Hence we see that supp(c

₀

) ⊂ H( r

1

). By Lemma 2, we get c

₀

= ω ( r

1

). By the fact that ω ( r

1

)s

₀

= s

0

and ω ( r

2

) t

₀

= t

₀

, we see that ω ( r

1

)ω ( r

2

)c

₁

= c

1

. By the equality:

(s

^＊₀

s

₀

)

^＊

= ( ω ( r

1

))

^＊

+ c

^＊₁

t

^＊₀

= ω ( r

1

)+ c

^＊₁

kt

₀

and (s

^＊₀

s

₀

)

^＊

= s

^＊₀

s

₀

, we get c

^＊₁

k = c

₁

. In a similar way to the above, we have t

^＊0

t

₀

= ω( r

2

) + c

₂

s

₀

where ω ( r

1

)ω ( r

2

)c

₂

= c

₂

and c

^＊2

h = c

2

. It is easy to see that s

0

t

₀

= c

₃

s

₀

+ c

₄

t

₀

where ω ( r

1

)ω ( r

2

)c

₃

= c

3

andω ( r

1

)ω ( r

2

)c

₄

= c

4

.

Next, we show the equation c

²₁

= ω ( r

1

) ω ( r

2

)k, c

²₂

= ω ( r

1

) ω ( r

2

) h, c

₃

= c

^＊₁

and c

₄

= c

^＊₂

. We have s

²₀

= ω ( r

1

)h

^＊

+ c

₁

h

^＊

t

₀

, t

²0

= ω( r

2

)k

^＊

+ c

₂

k

^＊

s

₀

, and s

0

t

₀

= c

₃

s

₀

+ c

₄

t

₀

. It is easy to see by simple calculations that

(s

²₀

)t

₀

= c

₁

h

^＊

k

^＊

+ c

₁

c

₂

h

^＊

k

^＊

s

₀

+ ω ( r

1

) ω ( r

2

)h

^＊

t

₀

, s

₀

(s

₀

t

₀

) = c

₃

h

^＊

+ c

₃

c

₄

s

₀

+ (c

₁

c

₃

h

^＊

+ c

²₄

)t

₀

. By the associativity: (s

²₀

) t

₀

= s

₀

(s

₀

t

₀

), we have 2 ω ( r

1

)ω ( r

2

)h

^＊

= c

₁

c

₃

h

^＊

+ c

²₄

and c

₃

= c

₁

k

^＊

= c

^＊₁

. In a similar way, by the associativity: s

₀

(t

²₀

) = (s

₀

t

₀

)t

₀

, we have c

₄

= c

₂

h

^＊

= c

^＊₂

. By these relations, we have 2 ω ( r

1

)ω r (

2

) = c

²₁

k

^＊

+ c

²2

h

^＊

. This fact implies that supp( ω( r

1

)ω( r

2

)) = supp(c

²₁

k

^＊

) ∪ supp(c

²₂

h

^＊

).

Hence we see that supp( c

²₁

k

^＊

) ⊂ H( r

1

)H( r

2

) and supp(c

²₂

h

^＊

)

⊂ H( r

1

) H( r

2

). Applying Lemma 2, we have c

²₁

k

^＊

= ω ( r

1

) ω ( r

2

)

and c

²₂

h

^＊

= ω ( r

1

) ω ( r

2

). Therefore, we get c

²₁

=ω ( r

1

) ω ( r

2

)k and c

²₂

= ω ( r

1

) ω ( r

2

)h.

[Q.E.D.]

Remark If K is an extension of the Golden hyper- group L by a finite abelian group H, we can reformulate Proposition 3 as follows.

(0) K is the disjoint union of H = ϕ

^−１

( r

0

), S= ϕ

^−１

( r

1

), and T = ϕ

^−１

( r

2

), and s

₀

∈ S, t

0

∈ T.

(1) s

^＊₀

= hs

₀

and t

^＊₀

= kt

₀

for h, k∈ H . (2) s

²₀

= ω ( r

1

)h

^＊

+ c

₁

h

^＊

t

₀

for c

₁

∈ M

¹

(H).

(3) t

²₀

= ω ( r

2

)k

^＊

+ c

₂

k

^＊

s

₀

for c

2

∈ M

¹

(H).

(4) s

₀

t

₀

= c

^＊₁

s

₀

+ c

^＊₂

t

₀

.

(5) ω ( r

1

) ω ( r

2

)c

₁

= c

₁

and ω ( r

1

) ω ( r

2

)c

₂

= c

₂

. (6) c

^＊₁

= c

₁

k

^＊

and c

^＊₂

= c

₂

h

^＊

.

(7) c

²₁

= ω ( r

1

)ω ( r

2

)k and c

²2

= ω ( r

1

)ω ( r

2

)h.

We remark that it is easy to check that these con- ditions assure that K is a commutative hypergroup which is an extension of L by H. Hence we see that all extensions K of L by H are determined in this way by

s

₀

∈ S, t

₀

∈ T, h, k ∈ H, c

₁

, c

2

∈ M

¹

(H) satisfying the above conditions (1) 〜 (7). Therefore we denote such an extension K by K = K (s

₀

, t

0

, h, k, c

1

, c

2

) .

Let K

₁

= H ∪ S

₁

∪ T

₁

and K

₂

= H ∪ S

₂

∪ T

₂

be two extensions of L by H and ϕ

1

(resp. ϕ

2

) be a canonical quotient mapping from K

1

(resp. K

₂

) onto the Golden hypergroup L. Then K

1

is called to be equivalent to K

2

as extensions if there exists a hypergroup isomorphism ψ from K

₁

onto K

₂

such that ψ (h) = h for all h ∈ H and ϕ

2○

ψ = ϕ

1

.

When we take u

0

∈ S, v

0

∈ T, h

1

, k

1

∈ H, and d

1

, d

2

∈ M

¹

(H) satisfying the above conditions (1) 〜 (7), we have another extension K (u

₀

, v

₀

, h

₁

, k

₁

, d

₁

, d

₂

) of L by H.

Proposition 4. Two extensions K (s

₀

, t

0

, h, k, c

1

,c

2

) and K ( u

₀

, v

0

, h

1

, k

1

, d

1

, d

2

) of L by H are mutually equiva- lent as extensions if and only if there exist b

₁

, b

₂

∈ H such that u

₀

= b

^＊₁

s

₀

, v

₀

= b

^＊₂

t

₀

, d

₁

= b

₂

c

₁

, d

₂

= b

₁

c

₂

, ω ( r

1

)h

₁

= ω ( r

1

)b

²₁

h, and ω( r

2

)k

₁

= ω ( r

2

)b

²₂

k .

Proof. Suppose that K

1

= K( s

₀

, t

₀

, h, k, c

₁

, c

₂

) is equiv- alent to K

₂

= K (u

₀

, v

₀

, h

₁

, k

₁

, d

₁

, d

₂

). Then it is easy to see that both stability groups of H in K

₁

and K

₂

at s

₀

and u

₀

coincide and both stability groups of H at t

0

and v

0

also coincide. Hence we may assume that ϕ

2

−1

( r

1

) = ϕ

1

−1

( r

1

) = S and ϕ

2

−1

( r

2

) = ϕ

1

−1

( r

2

) = T. For u

₀

∈ S and v

₀

∈ T, there exist b

₁

and b

₂

∈ H such that u

₀

= b

^＊₁

s

₀

and v

₀

= b

^＊₂

t

₀

respectively by Proposition 1. By the relation that s

^＊₀

= hs

0

and u

^＊0

= h

1

u

₀

, we get

h

₁

s

₀

= b

²₁

hs

₀

.

Hence we have ω ( r

1

)h

₁

= ω ( r

1

)b

²₁

h. In a similar way, we

1 2 1 2

1 4 1

4 1 4

1 2 1

4 1

4

1 2 1 2 1

2 1

2 1

2

1 2

1 2 1 2 1

2 1

2

1 2 1 2 1

2 1

2 1 2 1

2

1 2 1 2

1 2 1 2 1

2 1

2 1

2 1

2

h

Σ

k∈H0

Σ

hk∈H0

Σ

hk∈H0

Σ

k

Σ

hk∈H0

Extensions of the Golden Hypergroup by Finite Abelian Groups 3

1 2

1 2 1

2

1

2

also obtain ω ( r

2

)k

₁

= ω ( r

2

)b

²₂

k. Since u

^＊0

u

₀

= s

^＊0

s

₀

, compar- ing coefficients of t

0

of s

^＊0

s

₀

and u

^＊0

u

₀

, we get d

1

= b

2

c

₁

. In a similar way, we see that d

₂

= b

₁

c

₂

.

Conversely, if there exists b

₁

, b

₂

∈ H such that u

₀

= b

^＊₁

s

₀

, v

0

= b

^＊2

t

₀

, d

₁

= b

2

c

₁

, d

2

= b

1

c

₂

, ω ( r

1

)h

₁

= ω ( r

1

)b

²₁

h, and

( r

2

)k

₁

=ω ( r

2

)b

²₂

k, it is easy to check that K (s

₀

, t

0

, h, k, c

1

, c

₂

) is equivalent to K (u

₀

, v

₀

, h

₁

, k

₁

, d

₁

, d

₂

).

[Q.E.D.]

Let K be an extension of L by a finite abelian group H. If there exists injective mapping φ from L into K such that

(0) ϕ ( φ ( r )) = r for r ∈ L.

(1) φ (e

_L

) = e

_K

andφ ( r

^＊

) =φ( r )

^＊

,

(2) The set H( r ) = { h ∈ H : h φ ( r ) =φ r ( )} is a subgroup of H,

(3) φ ( r

i

)φ ( r

j

) =φ( r

i

r

j

)ω ( r

i

)ω ( r

j

),

(4) ω ( r

i

)ω ( r

j

)ω ( r ) = ω ( r

i

)ω ( r

j

) if r ∈ supp( r

i

r

j

), (5) K = H φ (L), and H ∩φ (L) = {e

_K

},

then we call that the extension K of L by H splits or K is a splitting extension in our paper (3).

Definition of weakly splitting. We call the extension K of L by H weakly splitting if the condi- tions (1), (2), (3), (5) are satisfied.

Proposition 5. The extension K = K(s

₀

, t

₀

, h, k, c

1

, c

2

) is weakly splitting if and only if there exist b

1

, b

2

∈ H such that c

₁

= ω ( r

1

) ω ( r

2

)b

₂

, c

₂

= ω ( r

1

) ω ( r

2

)b

₁

, ω ( r

1

)h = ω ( r

1

)b

²₁

, and ω ( r

2

)k = ω ( r

2

)b

²₂

. Moreover, K is splitting if and only if K is weakly splitting and H( r

1

) = H( r

2

).

Proof. Suppose that the extension K is given by K (s

₀

, t

₀

, h, k, c

₁

, c

₂

). We assume that φ ( r

0

) = h

₀

,φ ( r

1

) = s

₀

,

( r

2

) = t

₀

. Then we have s

^＊₀

= s

₀

and t

^＊₀

= t

₀

by weakly split- ting condition (1). This implies that we can assume that h = h

0

and k = k

0

so that c

1

= c

2

=ω( r

1

)ω ( r

2

). Since weakly splitting extensions are equivalent to this extension K = K (s

₀

, t

₀

, h

₀

, k

₀

, c

₁

, c

₂

), we get the desired conclusion applying Proposition 4.

By the structure equations (2) and (3) as described in Remark combined with splitting condition (4), we get

( r

1

) = ω ( r

2

), i.e. H( r

1

) = H( r

2

).

[Q.E.D.]

Theorem. Let K be a commutative hypergroup extension of the Golden hypergroup L = { r

0

, r

1

, r

2

} by a finite abelian group H, which means that there exists a hypergroup homomorphism ϕ from K onto L such that Ker ϕ = H. Let H( r

1

) be the stability group of H at s

₀

∈ S

= ϕ

⁻¹

( r

1

) and H( r

2

) be the stability group of H at t

₀

∈ T =

ϕ

⁻¹

( r

2

). Let ω( r

i

) denote the normalized Haar measure of H( r

i

) ( i =1, 2 ).

(1) Then we have S = {hs

₀

: h ∈ H} and T = {kt

₀

: k ∈ H}.

When s

^＊₀

= hs

₀

and t

^＊₀

= kt

₀

for some h, k ∈ H, we have s

^＊₀

s

₀

= ω( r

1

) + c

₁

t

₀

, t

^＊₀

t

₀

= ω ( r

2

) + c

₂

s

₀

, and s

0

t

₀

= c

^＊₁

s

₀

+ c

^＊₂

t

₀

for c

1

, c

2

∈ M

¹

(H) such thatω ( r

1

)ω ( r

2

)c

_i

= c

_i

(i =1, 2).

Moreover, c

^＊₁

= c

₁

k

^＊

, c

^＊₂

= c

₂

h

^＊

, c

²₁

= ω ( r

1

) ω ( r

2

) k, and c

²₂

= ω ( r

1

) ω ( r

2

)h.

(2) All extensions K of L by H are characterized in this way, so that we denote such an extension K by K (s

₀

, t

0

, h, k, c

1

, c

2

).

Two extensions K (s

₀

, t

₀

, h, k, c

₁

, c

₂

) and K ( u

₀

, v

₀

, h

₁

, k

₁

, d

₁

, d

₂

) of L by H are mutually equivalent as extensions if and only if there exists b

1

, b

₂

∈H such that u

0

= b

^＊1

s

₀

, v

0

= b

^＊2

t

₀

, d

1

= b

2

c

₁

, d

2

= b

1

c

₂

, ω ( r

1

)h

₁

= ω ( r

1

)b

²₁

h, and ω ( r

2

)k

₁

= ω ( r

2

)b

²₂

k.

(3) Moreover, the extension K = K (s

₀

, t

₀

, h, k, c

₁

, c

₂

) is weakly splitting if and only if there exist b

₁

, b

₂

∈ H such that c

₁

= ω ( r

1

)ω ( r

2

)b

₂

, c

2

= ω ( r

1

)ω ( r

2

)b

₁

, ω ( r

1

)h = ω ( r

1

)b

²₁

, and ω ( r

2

)k

= ω ( r

2

)b

²₂

. The extension K is splitting if and only if K is weakly splitting and H( r

1

) = H( r

2

).

Proof. These statements follow immediately from Proposition 1, 2, 3, 4, and 5 so that we omit the details.

[Q.E.D.]

4. Applications and Examples

Under these preparations we calculate all exten- sions K of the Golden hypergroup L by concrete abelian groups H =

₂

,

₃

,

₄

,

₅

, and

₆

. We denote the order of K by│ K │ .

Model 1. H =

₂

= {h

₀

, h

₁

}, h

²₁

= h

₀

.

(1) Case of│ K │ = 6, i.e. H( r

1

) = {h

₀

}, H( r

2

) = {h

₀

}, and K

⁶

= H × L.

K

⁶

= {h

0

, h

1

, s

0

, s

1

, t

0

, t

1

}, s

1

= h

1

s

₀

, t

1

= h

1

t

₀

,

s

^＊₀

= s

₀

, s

^＊₁

= s

₁

, t

^＊₀

= t

₀

, t

^＊₁

= t

₁

, s

²₀

= h

₀

+ t

₀

, t

₀²

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

.

(2) Case of │ K│ = 5 .

(2-a) When H ( r

1

) = H, H( r

2

) = {h

0

}, i.e.

K

⁵_a

= {h

₀

, h

₁

, s

₀

, t

₀

, t

₁

}, t

₁

= h

₁

t

₀

.

(2-a-1) K = K

⁵a1

(s

^＊₀

= s

₀

, t

^＊₀

= t

₀

, t

^＊₁

= t

₁

) which is character- ized by

s

²₀

= h

₀

+ h

₁

+ t

₀

+ t

₁

, t

²₀

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

.

(2-a-2) K = K

⁵a2

(s

^＊₀

= s

₀

, t

^＊₀

= t

₁

, t

^＊₁

= t

₀

) which is character- ized by

s

²₀

= h

₀

+ h

₁

+ t

₀

+ t

₁

, t

²₀

= h

₁

+ s

₀

, s

₀

t

₀

= ¹ ₂ s

₀

+ ¹ ₄ t

₀

+ ¹ ₄ t

₁

.

1 2 1 2 1 4 1 4 1 4 1 4

1 4 1 4 1 2

1 2 1 2 1

4 1 4 1 4 1 4

1 2 1 2 1

2

1 2 1 2 1 2

1 2 1

2 1

2 1

2 1

1 2

ω 2

ω

φ

(2-b) When H( r

1

) ={h

₀

}, H( r

2

)= H, in a similar way, we have K

⁵b1

and K

⁵b2

.

(3) Case of│ K│ = 4, i.e. H( r

1

) = H, H( r

2

)= H

K

⁴

= H ∨ L = {h

0

, h

1

, s

0

, t

0

} which is the join of H by L and characterized by

s

^＊₀

= s

₀

, t

^＊₀

= t

₀

, s

²₀

= h

₀

+ h

₁

+ t

₀

, t

²₀

= h

₀

+ h

₁

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

.

Next, we consider the dual of this model. Let K ＾

a1⁵

= { χ

₀

, χ

₁

, χ

₂

, χ

₃

, χ

₄

}, be the dual of K

⁵a1

. The character table of K

⁵a1

is as follows.

Hence the structure equations of the dual K ＾

a1⁵

of K

⁵a1

are given in the following way.

χ

1

χ

1

= χ

0

+ χ

2

, χ

1

χ

2

= χ

1

+ χ

2

, χ

2

χ

2

= χ

0

+ χ

1

, χ

1

χ

3

= χ

3

+ χ

4

, χ

1

χ

4

= χ

3

+ χ

4

, χ

2

χ

3

= χ

3

+ χ

4

, χ

2

χ

4

= χ

3

+ χ

4

, χ

3

χ

3

= χ

4

χ

4

= χ

0

+ χ

1

+ χ

2

, χ

3

χ

4

= χ

1

+ χ

2

.

By this fact we see that K

⁵a1

is a strong hypergroup.

In a similar way, it is easy to check that K

⁵a2

, K

⁵b1

, and K

⁵b2

are also strong. It is well known that H × L and H ∨ L are strong.

Remark 1.

1. K is a splitting extension of L by H if and only if K = K

⁶

= H × L or K

⁴

= H ∨ L.

2. K is a weakly splitting extension of L by H if and only if K = K

⁶

= H × L, K

⁴

= H ∨ L, K

⁵a1

, or K

⁵b1

.

3. All extensions as above are strong.

Model 2. H=

₃

= {h

₀

, h

₁

, h

₂

}, h

³₁

= h

₀

, h

²₁

= h

₂

, h

²₂

= h

₁

, h

^＊₁

= h

₂

, h

^＊₂

= h

₁

.

(1) Case of│ K │ = 9 , i.e. H( r

1

) = {h

0

}, H( r

2

) = {h

0

}.

K

⁹

= {h

0

, h

1

, h

2

, s

1

, s

2

, s

3

, t

0

, t

1

, t

2

}, s

_k

= h

_k

s

₀

(k = 0, 1, 2), t

_j

= h

_j

t

₀

( j = 0, 1, 2).

(1-a) K = K

⁹_a

= H × L (s

^＊₀

= s

₀

, s

^＊₁

= s

₂

, s

^＊₂

= s

₁

, t

^＊₀

= t

₀

, t

^＊₁

= t

₂

, t

^＊₂

= t

₁

) which is characterized by

s

²₀

= h

₀

+ t

₀

, t

²₀

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

. (1-b) K = K

⁹b

(s

^＊₀

= s

₁

, s

^＊₁

= s

₀

, s

^＊₂

= s

₂

, t

^＊₀

= t

₀

, t

^＊₁

= t

₂

, t

^＊₂

= t

₁

) which

is characterized by

s

²₀

= h

₂

+ t

₂

, t

²₀

= h

₀

+ s

₂

, s

₀

t

₀

= s

₀

+ t

₁

. (1-c) K = K

⁹c

(s

^＊₀

= s

2

, s

^＊1

= s

1

, s

^＊2

= s

0

, t

^＊0

= t

0

, t

^＊1

= t

2

, t

^＊₂

= t

1

) which

is characterized by

s

²₀

= h

₁

+ t

₁

, t

²₀

= h

₀

+ s

₁

, s

₀

t

₀

= s

₀

+ t

₂

. (1-d) K = K

⁹d

(s

^＊₀

= s

1

, s

^＊1

= s

0

, s

^＊2

= s

2

, t

^＊₀

= t

1

, t

^＊1

= t

0

, t

^＊2

= t

2

) which

is characterized by

s

²₀

= h

₂

+ t

₁

, t

²₀

= h

₂

+ s

₁

, s

₀

t

₀

= s

₁

+ t

₁

. (1-e) K = K

⁹_e

(s

^＊₀

= s

₂

, s

^＊₁

= s

₁

, s

^＊₂

= s

₀

, t

^＊₀

= t

₁

, t

^＊₁

= t

₀

, t

^＊₂

= t

₂

) which is

characterized by

s

²₀

= h

₁

+ t

₀

, t

²₀

= h

₂

+ s

₀

, s

₀

t

₀

= s

₁

+ t

₂

. (1-f) K = K

_f⁹

(s

^＊₀

= s

₂

, s

^＊₁

= s

₁

, s

^＊₂

= s

₀

, t

^＊₀

= t

₂

, t

^＊₁

= t

₁

, t

^＊₂

= t

₀

) which is

characterized by

s

²₀

= h

₁

+ t

₂

, t

²₀

= h

₁

+ s

₂

, s

₀

t

₀

= s

₂

+ t

₂

. (2) Case of│ K│＝7.

(2-a) When H( r

1

) = H, H( r

2

) = {h

₀

}, i.e.

K

_a⁷

= {h

₀

, h

₁

, h

₂

, s

₀

, t

₀

, t

₁

, t

₂

} t

_j

= h

_j

t

₀

( j = 0, 1, 2).

(2-a-1) K = K

⁷a1

( s

^＊₀

= s

0

, t

^＊₀

= t

0

, t

^＊₁

= t

2

, t

^＊2

= t

1

) which is char- acterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ t

₀

+ t

₁

+ t

₂

, t

₀²

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

.

(2-a-2) K = K

⁷a2

( s

^＊₀

= s

0

, t

^＊₀

= t

1

, t

^＊₁

= t

0

, t

^＊2

= t

2

) which is char- acterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ t

₀

+ t

₁

+ t

₂

, t

₀²

= h

₂

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

.

(2-a-3) K = K

a3⁷

( s

^＊₀

= s

₀

, t

^＊₀

= t

₂

, t

^＊₁

= t

₁

, t

^＊₂

= t

₀

) which is char- acterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ t

₀

+ t

₁

+ t

₂

, t

₀²

= h

₁

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

.

(2-b) When H( r

1

) = {h

₀

}, H( r

2

) = H, in a similar way,we have K

⁷b1

, K

⁷b2

, and K

b3⁷

.

(3) Case of│ K│ = 5, i.e. H( r

1

) = H, H( r

2

) = H.

K

⁵

= H ∨ L = {h

₀

, h

₁

, h

₂

, s

₀

, t

₀

} which is the join of H by L and characterized by

1 6 1 6 1 6 1 2 1

2 1 2

1 6 1 6 1 6 1 6 1 6 1 6

1 6 1 6 1 6 1 2 1

2 1 2

1 6 1 6 1 6 1 6 1 6 1 6

1 6 1 6 1 6 1 2 1

2 1 2

1 6 1 6 1 6 1 6 1 6 1 6

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

5−√ 5 10 5 + √ 5

10

3 + √ 5 10 3−√ 5

10 2 5

3 + √ 5 8 5−√ 5

8

5−√ 5 8 3 + √ 5

8

3−√ 5 8 5 + √ 5

8

5 + √ 5 8 3−√ 5

8 1 2 1 2

1 2 1 2 1

2 1 2

1 2 1 2 1

2

1 4 1 4 1 2 1 4 1 4

Extensions of the Golden Hypergroup by Finite Abelian Groups 5

h

0

χ₀

χ₁

χ₂

χ₃

χ₄

h

1

s

0

t

0

t

1

1 1 1

−1＋√

5 4

1 1

1 1

1 1

1

−

1

1

0

−1

0

−1−√

5 4

−1−√

5 4

−1−√

5 4

−1＋√

5 4 1

√

2

1

√

2 1

√

2

−1＋√

5 4

− 

1

√

2

− 

s

^＊₀

= s

₀

, t

^＊₀

= t

₀

, s

₀²

= h

₀

+ h

₁

+ h

₂

+ t

₀

, t

₀²

= h

₀

+ h

₁

+ h

₂

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

. Remark 2.

1. We remark that H × L = K

⁹a

K

⁹b

K

⁹c

K

⁹d

K

⁹e

K

⁹f

, K

⁷a1

K

⁷a2

K

⁷a3

, and K

⁷b1

K

⁷b2

K

⁷b3

, as extensions of L by H.

2. K is a splitting extension of L by H if and only if K H × L or K

⁵

= H ∨ L.

3. K is a weakly splitting extension of L by H if and only if K H × L, K

⁵

= H ∨ L, K

a1⁷

, or K

⁷b1

. Model 3. H =

4

= {h

0

, h

1

, h

₂

, h

3

}, h

⁴₁

= h

₀

, h

^k₁

= h

_k

( k = 1, 2, 3), h

²₂

= h

₀

, h

²₃

= h

₂

, h

^＊₁

= h

3

, h

^＊₂

= h

2

.

(1) Case of│ K │ = 12 , i.e. H( r

1

) = {h

₀

}, H( r

2

) = {h

₀

}.

K

¹²

= {h

₀

, h

₁

, h

₂

, h

₃

, s

₀

, s

₁

, s

₂

, s

₃

, t

₀

, t

₁

, t

₂

, t

₃

}, s

_k

= h

_k

s

₀

(k = 0, 1, 2, 3), t

_j

= h

_j

t

₀

( j = 0, 1, 2, 3).

(1-a) K = K

¹²a

= H × L ( s

^＊₀

= s

0

, s

^＊1

= s

3

, s

^＊2

= s

2

, s

^＊3

= s

1

, t

^＊0

= t

₀

, t

^＊₁

= t

₃

, t

^＊₂

= t

₂

, t

^＊₃

= t

₁

) which is characterized by

s

²₀

= h

₀

+ t

₀

, t

²₀

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

. (1-b) K = K

¹²b

(s

^＊₀

= s

2

, s

^＊1

= s

1

, s

^＊2

= s

0

, s

^＊3

= s

3

, t

^＊0

=t

0

, t

^＊1

= t

3

, t

^＊₂

=

t

₂

, t

^＊₃

= t

1

) which is characterized by

s

²₀

= h

₂

+ t

₂

, t

²₀

= h

₀

+ s

₁

, s

₀

t

₀

= s

₀

+ t

₃

. (1-c) K = K

¹²_c

( s

^＊₀

= s

₂

, s

^＊₁

= s

₁

, s

^＊₂

= s

₀

, s

^＊₃

= s

₃

, t

^＊₀

=t

₂

, t

^＊₁

= t

₁

,

t

^＊₂

= t

0

, t

^＊3

= t

3

) which is characterized by s

²₀

= h

₂

+ t

₃

, t

²0

= h

₂

+ s

₃

, s

0

t

₀

= s

₃

+ t

₃

. (2) Case of│ K │= 10.

(2-a) When H( r

1

) = {h

₀

, h

₂

}, H( r

2

) = {h

₀

}, i.e.

K

¹⁰a

= {h

0

, h

1

, h

2

, h

3

, s

0

, s

1

, t

0

, t

1

, t

2

, t

3

}, s

_k

= h

_k

s

₀

(k = 0, 1), t

_j

= h

_j

t

₀

( j = 0, 1, 2, 3).

(2-a-1) K = K

¹⁰a1

( s

^＊₀

= s

₀

, s

^＊₁

= s

₁

, t

^＊₀

= t

₀

, t

^＊₁

= t

₃

, t

^＊₂

= t

₂

, t

^＊₃

= t

₁

) which is characterized by

s

²₀

= h

₀

+ h

₂

+ t

₀

+ t

₂

,

t

²₀

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₂

.

(2-a-2) K = K

¹⁰a2

(s

^＊₀

= s

₁

, s

^＊₁

= s

₀

, t

^＊₀

= t

₂

, t

^＊₁

= t

₁

, t

^＊₂

= t

₀

, t

^＊₃

= t

₃

) which is characterized by

s

²₀

= h

₀

+ h

₂

+ t

₀

+ t

₂

,

t

²₀

= h

₂

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₂

.

(2-b) When H( r

1

) = {h

0

}, H( r

2

) = {h

0

, h

₂

}, in a similar way, we have K

¹⁰b1

and K

¹⁰b2

.

(3) Case of│ K │= 9.

(3-a) When H( r

1

) = H, H( r

2

) = {h

₀

}, i.e.

K

⁹a

= {h

0

, h

1

, h

2

, h

3

, s

0

, t

0

, t

1

, t

2

, t

3

}, t

_j

= h

_j

t

₀

( j= 0, 1, 2, 3).

(3-a-1) K = K

⁹a1

(s

^＊₀

= s

0

, t

^＊0

= t

0

, t

^＊1

= t

3

, t

^＊₂

= t

2

, t

^＊₃

= t

1

) which is characterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

+ t

₂

+ t

₃

, t

²₀

= h

₀

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

+ t

₃

. (3-a-2) K = K

a2⁹

(s

^＊₀

= s

₀

, t

^＊₀

=t

₁

, t

^＊₁

= t

₀

, t

^＊₂

= t

₃

, t

^＊₃

= t

₂

) which is

characterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

+ t

₂

+ t

₃

, t

²₀

= h

₃

＋ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

+ t

₃

. (3-a-3) K = K

a3⁹

( s

^＊₀

= s

0

, t

^＊0

=t

2

, t

^＊1

= t

1

, t

^＊2

= t

0

, t

^＊3

= t

3

) which is

characterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

+ t

₂

+ t

₃

, t

²₀

= h

₂

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

+ t

₃

. (3-a-4) K = K

a4⁹

( s

^＊₀

= s

0

, t

^＊0

= t

3

, t

^＊1

= t

2

, t

^＊2

= t

1

, t

^＊3

= t

0

) which is

characterized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

+ t

₂

+ t

₃

, t

²₀

= h

₁

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

+ t

₂

+ t

₃

. (3-b) When H( r

1

) = {h

₀

}, H( r

2

) = H, in a similar way,we

have K

b1⁹

, K

b2⁹

, K

b3⁹

, and K

b4⁹

.

(4) Case of│ K│＝ 8, i.e. H( r

1

) = {h

0

, h

₂

}, H( r

2

) = {h

0

, h

₂

}.

K

⁸

= {h

₀

, h

₁

, h

₂

, h

₃

, s

₀

, s

₁

, t

₀

, t

₁

}, s

_k

= h

_k

s

₀

( k= 0, 1 ), t

_j

= h

_j

t

₀

( j= 0, 1 ).

s

^＊₀

= s

₀

, s

^＊₁

= s

₁

, t

^＊₀

= t

₀

, t

^＊₁

= t

₁

, s

²₀

= h

₀

+ h

₂

+ t

₀

, t

²₀

= h

₀

+ h

₂

+ s

₀

, s

0

t

₀

= s

₀

+ t

₀

.

(5) Case of│ K │= 7.

(5-a) When H( r

1

) = H, H( r

2

) = {h

0

, h

2

}, i.e.

K

⁷a

= {h

0

, h

1

, h

2

, h

3

, s

0

, t

0

, t

1

}, t

_j

= h

_j

t

₀

( j = 0, 1).

(5-a-1) K = K

a1⁷

( s

^＊₀

= s

₀

, t

^＊₀

= t

₀

, t

^＊₁

= t

₁

) which is charac- terized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

, t

²₀

= h

₀

+ h

₂

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

.

(5-a-2) K = K

_a2⁷

(s

^＊₀

= s

₀

, t

^＊₀

= t

₁

, t

^＊₁

= t

₀

) which is character- ized by

s

²₀

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

+ t

₁

, t

²₀

= h

₁

+ h

₃

+ s

₀

, s

₀

t

₀

= s

₀

+ t

₀

+ t

₁

. (5-b) When H( r

1

) = {h

0

, h

2

}, H( r

2

) = H, in a similar way,

we have K

⁷b1

and K

b2⁷

.

(6) Case of│ K │= 6, i.e. H( r

1

) = H, H( r

2

) = H.

K

⁵

= H ∨ L = {h

₀

, h

₁

, h

₂

, h

₃

, s

₀

, t

₀

} which is the join of H by L and characterized by

s

^＊₀

= s

0

, t

^＊0

= t

0

, s

0²

= h

₀

+ h

₁

+ h

₂

+ h

₃

+ t

₀

, t

²₀

= ¹ ₈ h

₀

+ ¹ ₈ h

₁

+ ¹ ₈ h

₂

+ ¹ ₈ h

₃

+ ¹ ₂ s

₀

, s

₀

t

₀

= ¹ ₂ s

₀

+ ¹ ₂ t

₀

.

1 2 1 8 1 8 1 8 1 8

1 4 1 4 1 2 1

2 1 4 1 4

1 4 1 4 1 8 1 8 1 8 1 8

1 4 1 4 1 2 1

2 1 4 1 4

1 4 1 4 1 8 1 8 1 8 1 8

1 2 1 2 1

2 1 4 1 4

1 2 1 4 1 4

1 8 1 8 1 8 1 8 1 2 1

2 1 2

1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8

1 8 1 8 1 8 1 8 1 2 1 2 1

2 1 2

1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8

1 8 1 8 1 8 1 8 1 2 1

2 1 2

1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8

1 8 1 8 1 8 1 8 1 2 1

2 1 2

1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8

1 4 1 4 1 2 1

2 1 2

1 4 1 4 1 4 1 4

1 4 1 4 1 2 1

2 1 2

1 4 1 4 1 4 1 4

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 2 1 2 1 2

1 2 1 2 1

2 1 6 1 6 1 6

1

2

1

6

1

6

1

6