On the subgroups and normal subgroups of the group representation of the Cayley tree

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by  F. H. Haydarov

ON THE SUBGROUPS AND NORMAL SUBGROUPS OF THE GROUP
REPRESENTATION OF THE CAYLEY TREE.
F. H. HAYDAROV
6
Abstract. In this paper, we give a characterization of the normal subgroups of index
1
0 2s(2n+1), s∈{1,2}, n∈Nandofthesubgroupsofindexthreeofthegrouprepresentation
2 of the Cayley tree.

n
a
J Key words. G - group, subgroup, normal subgroup, homomorphism, epimorphism.
k
2
1
AMS Subject Classification: 20B07, 20E06.

]
R
G
1. Introduction
.
h
t In the theory of groups there are very important unsolved problems, most of which
a
m arise in solving of problems of natural sciences as physics, biology etc. In particular, if
configuration of a physical system is located on a lattice (in our case on the graph of a
[

group) then the configuration can be considered as a function defined on the lattice. Usu-
1
v ally, more important configuration (functions) are the periodic ones. It is well-known that
1 if the lattice has a group representation then periodicity of a function can be defined by a
9
given subgroup of the representation. More precisely, if a subgroup, say H, is given, then
2
3 one can define a H- periodic function as a function, which has a constant value (depending
0
only on the coset) on each (right or left) coset of H. So the periodicity is related to a
.
1 special partition of the group (that presents the lattice on which our physical system is
0
located). There are many works devoted to several kinds of partitions of groups (lattices)
6
1 (see e.g. [1], [3], [5], [7]).
:
v
Xi One of the central problems in the theory of Gibbs measures is to study periodic
Gibbs measures corresponding to a given Hamiltonian of model. For any subgroup H of
r
a
the group G we define H-periodic Gibbs measures. To find new periodic and weakly peri-
k
odic Gibbs measures one usually needs to find new subgroups of the group representation
of the Cayley tree. In Chapter 1 of [5] it is given a one to one correspondence between the
set of vertices V of the Cayley tree Γk and the group G and it is given a full description
k
of subgroups of index two and there are also constructed several normal subgroups of the
group G . But there wasn’t a full description of normal subgroups of finite index (without
k
index two) and to the best of our knowledge until now there wasn’t any description of a not
normal subgroup of finite index of the group representation of the Cayley tree. In [4] it is
given a full description of normal subgroups of indices 4 and 6 for the group representation
of the Cayley tree. In this paper we continue this investigation and construct all normal
subgroups of index 2s(2n+1), s ∈ {1,2}, n ∈ N and all subgroups of index three for the
1
2 F. H. HAYDAROV
group representation of the Cayley tree.
Cayley tree and its group representation. A Cayley tree (Bethe lattice) Γk of or-
der k ≥ 1 is an infinite homogeneous tree, i.e., a graph without cycles, such that exactly
k + 1 edges originate from each vertex. Let Γk = (V,L) where V is the set of vertices
and L that of edges (arcs). Let G be a free product of k + 1 cyclic groups of the sec-
k
ond order with generators a ,a ,...a , respectively. It is known that there exists a one
1 2 k+1
to one correspondence between the set of vertices V of the Cayley tree Γk and the group
G . To give this correspondence we fix an arbitrary element x ∈ V and let it correspond
k 0
to the unit element e of the group G . Using a ,...,a we label the nearest-neighbors
k 1 k+1
of element e, moving in positive direction. Now we’ll label the nearest-neighbors of each
a ,i = 1,...,k + 1 by a a ,j = 1,...,k + 1. Since all a have the common neighbor e we
i i j i
have a a = a2 = e. Other neighbors are labeled starting from a a in positive direction.
i i i i i
We label the set of all the nearest-neighbors of each a a by words a a a ,q = 1,...,k +1
i j i j q
starting from a a a = a by the positive direction. Iterating this argument one gets a one-
i j j i
to-onecorrespondence between theset of vertices V of the Cayley tree Γk andthe groupG .
k
Any(minimal represented) element x ∈ G has the following form: x = a a ...a ,
k i1 i2 in
where 1 ≤ i ≤ k+1,m = 1,...,n. The number n is called the length of the word x and is
m
denoted by l(x). The number of letters a ,i = 1,...,k +1, that enter the non-contractible
i
representation of the word x is denoted by w (a ).
x i
The following result is well-known in group theory. If ϕ is a homomorphism of a
group G with the kernel H then H is a normal subgroup of the group G and ϕ(G) ≃ G/H,
(where G/H is the quotient group) so the index |G : H| coincides with the order |ϕ(G)| of
the group ϕ(G).
Usually we define natural homomorphism g from G onto the quotient group G/H by
the formula g(a) = aH for all a ∈ G. Then Kerϕ = H.
Definition 1. Let M ,M ,...,M be some sets and M 6= M , for i 6= j. We call the
1 2 m i j
intersection ∩m M contractible if there exists i (1 ≤ i ≤ m) such that
i=1 i 0 0
∩m M = ∩i0−1M ∩ ∩m M .
i=1 i i=1 i i=i0+1 i
Let N = {1,...,k+1}. The follo(cid:0)wing Pro(cid:1)pos(cid:0)ition descri(cid:1)bes several normal subgroups
k
of G .
k
Put
H = x ∈ G | ω (a ) is even , A ⊂ N . (1.1)
A k x i k
( )
i∈A
X
Proposition 1. [5] For any ∅ =6 A ⊆ N , the set H ⊂ G satisfies the following properties:
k A k
(a) H is a normal subgroup and |G : H | = 2;
A k A
(b) H 6= H , for all A 6= B ⊆ N ;
A B k
(c) Let A ,A ,...,A ⊆ N . If ∩m H is non-contractible, then it is a normal subgroup of
1 2 m k i=1 Ai
index 2m.
Theorem 1. [5]
SUBGROUPS AND NORMAL SUBGROUPS 3
1. The group G does not have normal subgroups of odd index (6= 1).
k
2. The group G has normal subgroups of arbitrary even index.
k
2. Subgroups and normal subgroups of finite index.
2.1. Normal subgroups of index 2s(2n+1), s ∈ {1,2}, n ∈ N.
Definition 2. A commutative group G of order 2n is called K −group if G is generated
2n
by free product of n elements c ,i = 1,n satisfying the relations o(c ) = 2, i = 1,n,
i i
Proposition 2. Let ϕ be a homomorphism of the group G onto a finite commutative group
k
G. Then ϕ(G ) is isomorphic to K for some i ∈ N.
k 2i
Proof. Let (G,∗) be a commutative group of order n and ϕ : G → G be an epimorphism.
k
We will first show that n ∈ {2i| i ∈ N}. Suppose n does not belong to {2i| i ∈ N}.
Then there exist m > 1, m,s ∈ N ∪ {0} such that n = 2sm, where m is odd. Since
ϕ : G → G is an epimorphism there exist distinct elements ϕ(a ), p = 1,s. Clearly,
k jp
< ϕ(a ),ϕ(a )...,ϕ(a ) > is a subgroup of G and | < ϕ(a ),ϕ(a )...,ϕ(a ) > | = 2s.
j1 j2 js j1 j2 js
Since m > 1 there exist at least one element a ∈ G such that ϕ(a ) 6= ϕ(a ), i ∈
j0 k j0 ji
{1,2,...,s}. Hence < ϕ(a ),ϕ(a )...,ϕ(a ) > is a subgroup of G. Then by Lagrange’s
j0 j1 js
theorem |G| : 2s+1 which contradicts our assumption that m is odd. Thus n ∈ {2i| i ∈ N}.
If n = 2q then G is equal to < ϕ(a ),ϕ(a )...,ϕ(a ) > We define the following mapping
j1 j2 jq
γ(a ) = c ∈ K for all i ∈ {1,2,...,q}. It is easy to check that this mapping is a one to one
ji i q
correspondence from G to K . This completes the proof. (cid:3)
q
Let A ,A ,...,A ⊂ N , m ∈ N and ∩m H is non-contractible. Then we denote by
1 2 m k i=1 Ai
ℜ the following set
ℜ = {∩m H | A ,A ,...,A ⊂ N , m ∈ N}.
i=1 Ai 1 2 m k
Theorem 2. Let ϕ be a homomorphism from G to a finite commutative group. Then there
k
exists an element H of ℜ such that Kerϕ ≃ H and conversely.
Proof. We give a one to one correspondence between {Kerϕ| ϕ(G ) ≃ G ∈ {K | i ∈ N}}
k 2i
andℜ.LetϕbeahomomorphismfromG toafinitecommutativegroupoforderp. Thenby
k
Proposition 2 the number p belongs to the set {2i| i ∈ N} and if p = 2n then ϕ : G → K
k 2n
is an isomorphism. For any nonempty sets A ,A ,...,A ⊂ N and a ∈ G , i ∈ N we
1 2 n k i k k
define the following homomorphism
4 F. H. HAYDAROV
c , if i ∈ A \(A ∪A ∪...∪A ), j = 1,n
j j 2 3 n
. . . . . . . .



 c c , if i ∈ (A ∩A )\(A ∪...A ∪A ∪...
 j1 j2 j1 j2 1 j1−1 j1+1


φ (a ) =  ...∪Aj2−1 ∪Aj2+1 ∪...∪An), 1 ≤ j1 < j2 ≤ n
A1A2...An i  . . . . . . . .


c ...c c ...c , if i ∈ (A ∩...∩A ∩A ...∩A )\A , j = 1,n
1 j−1 j+1 n 1 j−1 j+1 n j



 c1c2...cn, if i ∈ (A1 ∩A2 ∩...∩An)



 e, if i ∈ Nk \(A1 ∪A2 ∪...∪An).



If i ∈ ∅ then we’ll accept that there is no index i ∈ Nk satisfying the required condition.
For x ∈ G if ϕ (x) = e the number of a , i = 1,n of a appearing in the word x
k A1A2...An i i
must be even. Therefore
Kerϕ = H ∩H ∩...∩H .
A1A2...An A1 A2 An
Thus the following equality holds (up to isomorphism)
{Kerϕ| ϕ(G ) ≃ G ∈ {K | i ∈ N}} = ℜ.
k 2i
(cid:3)
The group G has finitely generators of the order two. Assume that r is the minimal
number of such generators of the group G and without loss of generality we can take these
generators to be b ,b ,...b . Let e be the identity element of the group G. We define a
1 2 r 1
homomorphism from G onto G. Let Ξ = {A , A ,...,A } be a partition of N \A , 0 ≤
k n 1 2 n k 0
|A | ≤ k+1−n. Then we consider the homomorphism u : {a ,a ,...,a } → {e ,b ...,b }
0 n 1 2 k+1 1 1 m
given by
e , if x = a ,i ∈ A
1 i 0
u (x) = (2.1)
n
( bj, if x = ai,i ∈ Aj,j = 1,n.
For b ∈ G we denote by R [b ,b ,...,b ] a representation of the word b by generators
b 1 2 m
b ,b ,...,b , r ≤ m. Define the homomorphism γ : G → G by the formula
1 2 r n
e , if x = e
1 1
γ (x) =  b , if x = b ,i = 1,r (2.2)
n i i


 R [b ,...,b ], if x = b , i ∈/ {1,...,r}.
bi 1 r i
Put 

(p) 
H (G) = {x ∈ G | l(γ (u (x))) : 2p}, 2 ≤ n ≤ k −1. (2.3)
Ξn k n n
Let γ (u (x))) = x˜. We introduce the following equivalence relation on the set G :
n n k
x ∼ y if x˜ = y˜. It’s easy to check this relation is reflexive, symmetric and transitive.
SUBGROUPS AND NORMAL SUBGROUPS 5
Let (G,∗) be a group of order n, generated by two elements of order 2, and let ℑ be
n
the set of all such groups.
Proposition 3. For the group G the following equality holds
k
{Kerϕ| ϕ : G → G ∈ ℑ is an epimorphism} =
k 2n
(n)
= {H (G)| B ,B is a partition of the set N \B , 0 ≤ |B | ≤ k −1}.
B0B1B2 1 2 k 0 0
Proof. For an arbitrary group G ∈ ℑ we give a one to one correspondence between the
2n
two given sets. Let e be the identity element of G and B ,B be a partition of N \B ,
1 1 2 k 0
where B ⊂ N , 0 ≤ |B | ≤ k −1. Then we define the homomorphism ϕ : G → G
0 k 0 B0B1B2 k
by the formula
b , if i ∈ B
1 1
ϕ (a ) = (2.4)
B0B1B2 i
( b , if i ∈ B .
2 2
There is only one such a homomorphism (corresponding to B ,B ,B ). It’s clearly x ∈
0 1 2
Kerϕ if and only if x˜ is equal to e . Hence it is sufficient to show that if y ∈
B0B1B2 1
(n)
H (G) then y˜ = e . Suppose that there exist y ∈ G such that l(y˜) ≥ 2n. Let
B0B1B2 1 k
y˜ = b b ...b , q ≥ 2n and S = {b ,b b ,...,b b ...b }. Since S ⊆ G there exist x ,x ∈
i1 i2 iq i1 i1 i2 i1 i2 iq 1 2
S such that x = x which contradicts the fact that y˜ is a non-contractible. Thus we
1 2
(n)
have proved that l(y˜) < 2n. Since y ∈ H (G) the number l(y˜) must be divisible
B0B1B2
(n)
by 2n. Thus for any y ∈ H (G) we have y˜ = e . Hence for the group G we get
B0B1B2 1
Kerϕ = H(n) (G). (cid:3)
B0B1B2 B0B1B2
Let us denote by ℵ the following set
n
(n)
{H (G)| B ,B is a partition of the set N \B , 0 ≤ |B | ≤ k −1, |G| = 2n}.
B0B1B2 1 2 k 0 0
Theorem 3. Any normal subgroup of index 2n with n = 2i(2s+1), i = {0,1}, s ∈ N has
(n)
the form H (G), |G| = 2n i.e.,
B0B1B2
ℵ = {H| H ⊳G , |G : H| = 2n}
n k k
Proof. We will first prove that ℵ ⊆ {H| H ⊳G , |G : H| = 2n}. Let G be a group with
n k k
|G| = 2nandB ,B apartitionofthesetN \B , 0 ≤ |B | ≤ k−1.Forx = a a ...a ∈ G
1 2 k 0 0 i1 i2 in k
it’s sufficient to show that x−1H(n) (G) x ⊆ H(n) (G). We have (as in the proof of
B0B1B2 B0B1B2
(n)
Proposition 3) if y ∈ H (G) then y˜ = e , where e is the identity element of G.
B0B1B2 1 1
Now we take any element z from the set x−1H(n) (G) x. Then z = x−1h x for some
B0B1B2
(n)
h ∈ H (G). We have
B0B1B2
z˜= γ (v (z)) = γ v (x−1h x) = γ v (x−1)v (h)v (x) =
n n n n n n n n
= γ v (x−1) γ (v (h))γ(cid:0) (v (x)) =(cid:1)(γ (v(cid:0)(x)))−1γ (v (h))γ(cid:1) (v (x)).
n n n n n n n n n n n n
(cid:0) (cid:1)
6 F. H. HAYDAROV
(n)
From γ (v (h)) = e we get z˜= e i.e., z ∈ H (G). Hence
n n 1 1 B0B1B2
H(n) (G) ∈ {H| H ⊳G , |G : H| = 2n}.
B0B1B2 k k
Now we’ll prove that {H| H ⊳ G , |G : H| = 2n} ⊆ ℵ . Let H ⊳ G , |G : H| = 2n.
k k n k k
We consider the natural homomorphism φ : G → G : H i.e. the homomorphism given by
k k
φ(x) = xH, x ∈ G . There exist e,b ,b ,...b such that φ : G → {H,b H,...,b H} is
k 1 2 2n−1 k 1 2n−1
an epimorphism. Let ({H,b H,...,b H},∗) = ℘, i.e., ℘ is the factor group. If we show
1 2n−1
that ℘ ∈ ℑ then the theorem will be proved. Assuming that ℘ ∈/ ℑ then there are
2n 2n
c ,c ,...,c ∈ ℘, q ≥ 3 such that ℘ =< c ,...,c > . Clearly < c ,c > is a subgroup of ℘
1 2 q 1 q 1 2
and | < c ,c > | is greater than three.
1 2
Case n=2s+1. By Lagrange’s theorem | < c ,c > | ∈ m| 2n ∈ N . If e is the
1 2 m 2
identity element of ℘ then from c2 = e we take | < c ,c > | = 2n. Hence < c ,c >= ℘
1 2 1 2 (cid:8) (cid:9) 1 2
but c ∈/< c ,c > . Thus ℘ ∈ ℑ .
3 1 2 2n
Case n=2(2s+1). FromLagrange’stheorem| < c ,c > | ∈ {4,2(2s+1),4(2s+1)}.
1 2
Let | < c ,c > | = 4. If the number four isn’t equal to one of these numbers | < c ,c > | or
1 2 1 3
| < c ,c > |thenwe’llchoosethatpairs. If| < c ,c > | = | < c ,c > | = | < c ,c > | = 4
2 3 1 2 1 3 2 3
then < c ,c ,c > is a group of order eight. Again by Lagrange’s theorem |℘| = 2n must
1 2 3
be divisible by 8 which is impossible.
Let | < c ,c > | = 2(2s+1). Then
1 2
< c ,c >= {e, c , c , c c , c c c , . . . , c c ...c } = A.
1 2 1 2 1 2 1 2 1 1 2 1
4s
It’s easy to check that
| {z }
c A∪A ⊆ ℘, c A∩A = ∅, |c A∪A| = |c A|+|A| = 2n = |℘|.
3 3 3 3
We then deduce that c A ∪ A = ℘ but we’ll show c c c ∈ ℘ does not belong to
3 3 1 3
c A ∪ A. Since c ,c ,c are generators we have c c c ∈/< c ,c > so c c c = xc with
3 1 2 3 3 1 3 1 2 3 1 3 3
x ∈< c ,c >. But x = c c ∈/< c ,c > .
1 2 3 1 1 2
If | < c ,c > | = 4(2l +1) then < c ,c >= ℘ but c ∈/< c ,c > . Hence ℘ ∈ ℑ .
1 2 1 2 3 1 2 2n
This completes the proof. (cid:3)
2.2. Subgroups of index three. In this section we will characterize the subgroups of
index three for the group representation of the Cayley tree.
Let (A , A ) be a partition of the set N \A , 0 ≤ |A | ≤ k−1. Put m be a minimal
1 2 k 0 0 j
element of A , j ∈ {1,2}. Then we consider the function u : {a ,a ,...,a } →
j A1A2 1 2 k+1
{e,a ...,a } given by
1 k+1
e, if x = a , i ∈ N \(A ∪A )
i k 1 2
u (x) =
A1A2
( a , if x = a , i ∈ A , j = 1,2.
mj i j
SUBGROUPS AND NORMAL SUBGROUPS 7
Define γ :< e,a ,a > → {e,a ,a } by the formula
m1 m2 m1 m2
e if x = e
a if x ∈ {a , a a }
 m1 m1 m2 m1

γ(x) =  am2 if x ∈ {am2, am1am2}



 γ a a a ...γ(a a ) if x = a a a ...a , l(x) ≥ 3, i = 1,2
mi m3−i mi mi m3−i mi m3−i mi m3−i
 γ(cid:0)amiam3−iami...γ(am3−iami)(cid:1) if x = amiam3−iami...ami, l(x) ≥ 3, i = 1,2


For a partition(cid:0) (A , A ) of the set N \A(cid:1), 0 ≤ |A | ≤ k−1 we consider the following set.
1 2 k 0 0
Σ (G ) = {x ∈ G | γ(u (x)) = e}
A1A2 k k A1A2
Lemma 1. Let (A , A ) be a partition of the set N \A , 0 ≤ |A | ≤ k − 1. Then x ∈
1 2 k 0 0
(G ) if and only if the number l(u (x)) is divisible by 3.
A1A2 k A1A2
PProof. Let x = a a ...a ∈ G and l(u (x)) be odd (the even case is similar). Then we
i1 i2 in k A1A2
can write u (x) = a a ...a a a . γ(u (x)) is equal to
A1A2 mi m3−i mi m3−i mi A1A2
γ(a a ...a a a ) = γ a a ...a γ(a a ) = γ(a a ...a a ) = ...
mi m3−i mi m3−i mi mi m3−i mi m3−i mi mi m3−i m3−i mi
(cid:0) (cid:1) l(uA1A2(x))−3
We continue this process until the length of the word γ(uA1A2(x|)) will b{ezless tha}n three.
From γ(u (x)) = e we deduce that l(u (x)) is divisible by 3.
A1A2 A1A2
Conversely if l(u (x)) is divisible by 3 then x is generated by the elements
A1A2
a a a and a a a . Since γ(a a a ) = γ(a a a ) = e we get x ∈
m1 m2 m1 m2 m1 m2 m1 m2 m1 m2 m1 m2
(G ). (cid:3)
A1A2 k
PProposition 4. For the group G the following equality holds
k
{K| K is a subgroup of G of index 3} =
k
= {Σ (G )| A , A is a partition of N \A }.
A1A2 k 1 2 k 0
Proof. Let K be a subgroup of the group G with |G : K| = 3. Then there exist p,q ∈ N
k k k
such that |{K, a K, a K}| = 3. Put
p q
A = {i ∈ N | a ∈ K}, A = {i ∈ N | a a ∈ K}, A = {i ∈ N | a a ∈ K}
0 k i 1 k i p 2 k i q
From |G : K| = 3 we conclude that {A ,A } is a partition of N \ A . Let m be a
k 1 2 k 0 i
minimal element of A , i = 1,2. If we show (G ) is a subgroup of G (corre-
i A1A2 k k
sponding to K) then it’ll be given one to one correspondence between given sets. For
P
x = a a ...a ∈ (G ), y = a a ...a ∈ (G ) it is sufficient to show that
i1 i2 in A1A2 k j1 j2 jm A1A2 k
xy−1 ∈ (G ). Let u (x) = a a ....a , u (y) = a a ....a , where
A1A2 kP A1A2 mi m3−i Pms A1A2 mj m3−j mt
P
8 F. H. HAYDAROV
(i,j,s,t) ∈ {1,2}4. Since u is a homomorphism we have
A1A2
|l(u (x))−l(u (y−1))|, if s = t
l u (xy−1) = A1A2 A1A2
A1A2 ( l(uA1A2(x))+l(uA1A2(y−1)), if s 6= t.
(cid:0) (cid:1)
By Lemma 1 we see that l(u (x)) is divisible by 3 and l(u (y−1)) =
A1A2 A1A2
l (u (y))−1 is also divisible by 3, so l(u (xy−1)) is a multiple of 3, which shows
A1A2 A1A2
that xy−1 ∈ (G ). This completes the proof. (cid:3)
(cid:0) (cid:1)A1A2 k
P References
[1] Cohen,D.E, Lyndon,R.C., (1963), Free bases for normal subgroups of free groups,
Trans.Amer.Math.Soc., 108, pp. 526-537.
[2] D.S, Malik, J.N, Mordeson,M.K, Sen., (1997), Fundamentals of Abstract Algebra, McGraw-HillCom.
[3] Ganikhodjaev, N.N, Rozikov, U.A., (1997), Description of periodic extreme Gibbs measures of some
lattice model on the Cayley tree, Theor.Math.Phys. 111, pp. 480-486.
[4] U.A, Rozikov, F.H, Haydarov., (2014), Normal subgroups of finite index for the group represantation
of the Cayley tree, TWMS Jour.Pure.Appl.Math. 5, pp. 234-240.
[5] U.A, Rozikov., (2013) Gibbs measures on a Cayley tree, World Sci. Pub, Singapore.
[6] Kurosh, A.G., (1953) Group theory, Akademic Verlag, Berlin.
[7] Young,J.W.,(1927),Onthepartitionsofagroupandtheresultingclassification,Bull.Amer.Math.Soc.,
33, pp.453-461.
F.H.Haydarov, National University of Uzbekistan, Tashkent, Uzbekistan.
E-mail address: haydarov [email protected].