CC BY
33
NANOSYSTEMS: PHYSICS, CHEMISTRY, MATHEMATICS, 2016, 7 (5), P. 888-892
Characterization of the normal subgroups of finite index for the group representation of a Cayley tree
F. H. Haydarov National University of Uzbekistan, Tashkent, Uzbekistan haydarov_imc@mail.ru
DOI 10.17586/2220-8054-2016-7-5-888-892
In this paper we give a characterization of normal subgroups for the group representation of the Cayley tree. Keywords: Cayley tree, G^-group, normal subgroup, homomorphism, epimorphism.
Received: 4 August 2016 Revised: 29 August 2016
1. Introduction
The method used for the description of Gibbs measures on Cayley trees is the method of Markov random field theory and recurrent equations of this theory, however, the modern theory of Gibbs measures on trees uses new tools such as group theory, information flows on trees, node-weighted random walks, contour methods on trees, nonlinear analysis. A very recently published book [1] discusses all above-mentioned methods for describing Gibbs measures on trees. In the configuration of 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. There are many works devoted to several kind of partitions of groups (lattices) (see e.g. [1-5,7]).
One of the central problems in the theory of Gibbs measures is to study periodic Gibbs measures corresponding to a given Hamiltonian. For any normal subgroups H of the group Gk, we define H-periodic Gibbs measures.
In Chapter 1 of [1] several normal subgroups were constructed for the group representation of the Cayley tree. In [6], we found full description of normal subgroups of index four and six for the group. In this paper, we continue this investigation and construct all normal subgroups of index eight and ten for the group representation of the Cayley tree.
Cayley tree. A Cayley tree (Bethe lattice) rk of order k > 1 is an infinite homogeneous tree, i.e., a graph without cycles, such that exactly k + 1 edges originate from each vertex. Let rk = (V, L) where V is the set of vertices and L that of edges (arcs).
A group representation of the Cayley tree. Let Gk be a free product of k +1 cyclic groups of the second order with generators a1,a2, ...ak+1, respectively.
A one to one correspondence is known to exist between the set of vertices V of the Cayley tree rk and the group Gk (see [1]).
To obtain this correspondence, we fix an arbitrary element x0 g V and let it correspond to the unit element e of the group Gk. Using ai,..., ak+i, we numerate the nearest-neighbors of element e, moving by positive direction. Next, we give the numeration for the nearest-neighbors of each ai,i = 1,...,k + 1 by aiaj,j = 1,...,k + 1. Since all ai have the common neighbor e, we give to it aiai = a2 = e. Other neighbors are numerated starting from aiai by the positive direction. We numerate the set of all the nearest-neighbors of each aiaj by words aiajaq, q =1,..., k + 1, starting from aiajaj = ai by the positive direction. Iterating this argument, one gets a one-to-one correspondence between the set of vertices V of the Cayley tree rk and the group Gk.
Any(minimally represented) element x g Gk has the following form: x = ai1 ai2...ain, where 1 < im < k + 1, m = 1,..., n. The number n is called the length of the word x and is denoted by l(x). The number of letters ai, i = 1,..., k +1, that enter the non-contractible representation of the word x is denoted by wx(ai).
Proposition 1. [8] Let f be homomorphism of the group Gk with the kernel H. Then H is a normal subgroup of the group Gk and f(Gk) ~ Gk/H, (where Gk/H is a quotient group) i.e., the index |Gk : H | coincides with the order |f (Gk )| of the group f(Gk).
Let H be a normal subgroup of a group G. The natural homomorphism g from G onto the quotient group G/H is given by the formula g(a) = aH for all a g G. Then, Ker f = H.
Definition 1. Let M1,M2, ..,Mm be some sets and Mi = Mj, for i = j. We call the intersection nm=1Mi contractible if there exists io(1 < io < m) such that:
nZiMi = (ni=-11Mi) n (nm=i0+1Mi).
Let Nk = {1, ...,k + 1}. The following Proposition describes several normal subgroups of Gk. Put
Ha = < x € Gk | ^ux(ai) is even I , A c Nk. (1.1)
I ieA J
Proposition 2. [1] For any 0 = A C Nk, the set HA c Gk satisfies the following properties: (a) HA is a normal subgroup and |Gk : HA| = 2;
(b) Ha = Hb , for all A = B C Nk;
(c) Let A1,A2, ...,Am C Nk. If nm=1HA. is non-contractible, then it is a normal subgroup of index 2m. Theorem 1. [6]
1. The group Gk does not have normal subgroups of odd index (= 1).
2. The group Gk has a normal subgroups of arbitrary even index.
2. New normal subgroups of finite index 2.1. The case of index eight
Definition 2. A group G is called a dihedral group of degree 4 (i.e.,D4) if G is generated by two elements a and b satisfying the relations:
o(a) = 4, o(b) = 2, ba = a3b.
Definition 3. A group G is called a quaternion group (i.e., Q8) if G is generated by two elements a, b satisfying the relation:
o(a) = 4, a2 = b2, ba = a3b.
Remark 1. [8] D4 is not isomorph to Q8.
Definition 4. A commutative group G is called a Klein 8-group (i.e.,K8) if G is generated by three elements a, b and c satisfying the relations: o(a) = o(b) = o(c) = 2.
Proposition 3. [8] There exist (up to isomorphism) only two noncommutative nonisomorphic groups of order 8
Proposition 4. Let <p be a homomorphism of the group Gk onto a group G of order 8. Then, y>(Gk) is isomorph to either D4 or K8.
Proof. Case 1 Let <(Gk) be isomorph to any noncommutative group of order 8. By Proposition 1, <(Gk) is isomorph to either D4 or Q8. Let <(Gk) ~ Q8 and eq be an identity element of the group Q8. Then, eq = <(e) = <(a2) = (<(ai))2 where ai € Gk, i € Nk. Hence, for the order of <(ai), we have o(<(ai)) € {1, 2}. It is easy to check there are only two elements of the group Q8 which order of element less than two. This is contradict.
Case 2 Let <(Gk, *) be isomorph to any commutative group (G, *1) of order 8. Then, there exist distinct elements a,b € G such that o(a) = o(b) = 2. Let H = {e, a, b, ab}. It's easy to check that (H, *1) is a normal subgroup of the group (G, *1). For c € H we have H = cH (cH = c *1 H). Hence <(Gk, *) is isomorph to only one commutative group (cH U H, *1). Clearly (cH U H, *1) ~ K8. □
The group G has finit generators of the order two and r is the minimal number of such generators of the group G and without loss of generality, we can take these generators to be b1,b2,...br. Let e1 be an identity element of the group G. We define homomorphism from Gk onto G. Let En = {A1, A2, ...,An} be a partition of Nk\A0, 0 < |A0| < k +1 - n. Then, we consider homomorphism un : {a1 ,a2,..., ak+1} ^ {e1, b1..., bm} as
un(x) = ( ^ f x = ai,i€A0 . J- (2.1)
I bj, it x = ai,i € Aj,j = 1,n.
For b g G, we denote that Rb[b1, b2,..., bm] is a representation of the word b by generators b1, b2,..., br, r < m. We define the homomorphism Yn : G ^ G by the formula
ei, if x = ei
Yn(x) = { bi, if x = bi,i = 1,r (2.2)
Rbi [bi,...,br], if x = bi,i = 1,r
We set:
H^)(G) = {x e Gk\ l(Yn(uri(x))) is divisible by 2p}, 2 < n < k - 1. (2.3)
Let Yn(un(x))) = x. We introduce the following equivalence relation on the set Gk : x ~ y if x = y. This relation is readily confirmed to be reflexive, symmetric and transitive.
Proposition 5. Let En = {Ai, A2,..., Ar} be a partition of Nk\A0, 0 < \A0\ < k + 1 - n. Then H(G) is a normal subgroup of index 2p of the group Gk.
^p
Proof. For x = ai1 ai2 ...ain g Gk it's sufficient to show that x-iH(p)(G) x C H~\G). Suppose that there exist y G Gk such that l(y) > 2p. Let y = bi1 bi2...biq, q > 2p and S = {bi1 ,bi1 bi2,...,bi1 bi2...biq}. Since S C G there exist xi,x2 G S such that xi = x2. But this contradicts y, which is a non-contractible. Thus we have proved that l(y) < 2p. Hence, for any x G H(p)(G) we have x = ei. Next, we take any element z from the
set x~iH(p)(G) x. Then, z = x-ih x for some h G H^)(G). We have z = Yn(vn(z)) = Yn (vn(x-ih x)) = = Yn (vn(x-i)vn(h)vn(x)) = Yn (vn(x-i)) Yn (vn(h)) Yn (vn(x)). From Yn (vn(h)) = ei, we get z = ei i.e., z G H^)(G). This completes the proof. □
Since Ai, A2, A3 c Nk and n3=iHAi is non-contractible we denote the following set:
K = {nf=iHAi| Ai,A2,A3 C Nk}. Theorem 2. For the group Gk, the following statement is hold:
{H| H is a normal subgroup of Gk with Gk : H| = 8} =
= {H(b)oClc2 (D4)| Ci,C2 is a partition of Nk \ C0} U K. Proof. Let $ be a homomorphism with |Gk : Ker^ = 8. Then by Proposition 2 we have ^(Gk) ~ K8 and
HGk) - D4.
Let $ : Gk ^ K8 be an epimorphism. For any nonempty sets Ai,A2,A3 c Nk, we give a one to one correspondence between
operatornameKer$| $(Gk) — K8} and K. Let ai G Gk,i G Nk. We define following homomorphism corresponding to the set Ai,A2,A3:
' a, if i G Ai \ (A2 U A3)
b, if i G A2 \ (Ai U A3)
c, if i G A3 \ (Ai U A2)
ab, if i G (Ai n A2) \ (Ai n A2 n A3)
ac, if i G (Ai n A3) \ (Ai n A2 n A3) bc, if i G (A2 n A3) \ (Ai n A2 n A3) abc, if i G Ai n A2 n A3
&A1A2A3 (ai)
e, if i G Nk \ (Ai U A2 U A3).
If i G 0, then we'll accept that there is no index i g Nk for which that condition is not satisfied. It is easy to check Ker $a1a2a3 = HA1 n HA2 n HA3. Hence {Ker $| $(Gk) — K8} = K.
Now, we'll consider the case $(Gk) — D4. Let $ : Gk ^ D4 be epimorphisms. Put
Co = {i| $(ai) = e}, Ci = {i| $(ai) = b}, C2 = {i| $(ai) = ab}.
One can construct following homomorphism (corresponding to C0, C\, C2)
Фс0а1а2 (x) = <
e, if X = e
a, if x = b2bi a2, if X = b2bib2bi a3, if X = b2bib2b\b2b\
b, if x = bi ab, if X = b2 a2b, if X = b2 bi b2 a3b, if X = b2bib2bib2.
Immediately, we conclude Ker(0CoClC2) = H((4)CiC2(D4). We have constructed all homomorphisms 0 on the group Gk which |Gk : Ker = 8. Thus by Proposition 1, one gets:
{H| |Gfc : H| = 8} Ç [H^cc(D4)l Ci, C2 is a partition of Nk \ Co} U R.
By Proposition 2 and Proposition 5, we can easily see that:
R U HCC2 (D4)| Ci, C2 is a partition of Nk \ Co} Ç [H| G : H| = 8}.
The theorem is proved. □
Corollary 1. The number of all normal subgroups of index 8 for the group Gk is equal to: 8k+1 — 6 • 4k+1 + 3k+1 + 9 • 2k+1 — 5.
Proof. Number of elements of the set HA c Gk, 0 = A c Nk is 2k+1 — 1. Then |R| = (2k+l)(2k+1 — 2)(2k+2 — 3). Let C0 c Nk be a fixed set and |C0| = p. If Ci, C2 is a partition of Nk \ C0 then there are 2k-p+i — 2 ways to
choose the sets C1 and C2. Hence the cardinality of {H,
(4)
Co Ci C2
(D4)l C1,C2 is a partition of Nk \ C0} is equal to
(2k+1 - 2)C0+1 + (2k - 2)C'1+1 + ... + 2C'k-l = 3k+1 - 2k+2 + 1.
k— 1 _ qk + 1
)k+2
Since K and {HCgC\c2 (D4)}| C1,C2 c Nk are disjoint sets, the cardinality of the union of these sets is 8k+1 — 6 • 4k+1 +3k+1 + 9 • 2k+1 — 5. □
2.2. Case of index ten
Let the group Rw be generated by the permutations:
ni = (1, 2)(3,4)(5, 6), П2 = (2, 3)(4, 5).
Proposition 6. Let ip be a homomorphism of the group Gk onto a group G of order 10. Then, p(Gk) is isomorph to R10.
Proof. Let (G, *) be a group and |G| = 10. Suppose there exist an epimorphism from Gk onto G. It is easy to check that there are at least two elements a,b G Gk such that o(a) = o(b) = 2. If a * b = b * a, then (H, *) is a subgroup of the group (G, *), where H = {e, a, b, a*b}. Then, by Lagrange's theorem, |G| is divided by |H| but 10 is not divided by 4. Hence, a * b = b * a. We have {e,a,b,a * b,b * a} с G If G is generated by three elements, then there exist an element c such that c G {e, a, b, a * b, b*a}. Then, the set {e, a, b, a*b, b*a, c, c * a, c*b, c*a*b, c * b*a} must be equal to G. Since G is a group, we get a * b * a = b but from o(a) = 2 the last equality is equivalent to a * b = b * a. This is a contradiction. Hence, by Lagrange's theorem it is easy to see:
G = {e,a,b,a * b,b * a, a * b * a,b * a * b, a * b * a * b,b * a * b * a, a * b * a * b * a},
where o(a * b) = 5. Namely G ~ R10. This completes the proof. □
Theorem 3. For the group Gk, the following statement is holds:
{H| | H is a normal subgroup of Gk with |Gk : H| = 10} =
= {hB50)b1b (R1o)l B1, B2 is a partition of the set Nk \ Bo}.
Proof. Let $ be a homomorphism with |Gk : Ker $| = 10. By Proposition 6 $(Gk) — Ri0 and by Proposition 5 we can easily see:
{HbIb1b2(Rio)| Bi,B2 is a partition of the set Nk \ Bo} C {H| |Gk : H| = 10}. Let p : Gk ^ Ri0 be epimorphisms. We denote:
Bo = {i| p(ai) = e}, Bi = {i| p(ai) = a, B2 = {i| p(ai) = b}. Then, we can show this homomorphism (corresponding to Bi, B2, B3), i.e.,
e, if x = e
a, if x = bi
b, if x = b2
a * b, if x = bib2 b * a, if x = b2bi a * b * a, if x = bib2bi b * a * b, if x = b2bib2 a * b * a * b, if x = bib2bib2
b * a * b * a, if x = b2bib2bi a * b * a * b * a, if x = b^b^bi.
We have constructed all homomorphisms $ on the group Gk which |Gk : Ker $| = 10. Hence:
0B1B2
(x)
{Ker |Gfc : Ker ^ = 10} c {H(°o;BiB (Rio) | B1,B2 is a partition of the set Nk \ Bo}. By Proposition 1:
{H| |Gk : H| = 10} = {HB50)BlB2 (Rio)| Bi, B2 is a partition of the set Nk \ Bo}. The theorem is proved. □
Corollary 2. The number of all normal subgroups of index 10 for the group Gk is equal to 3k+1 — 2k+2 + 1. Proof. The proof of this Corollary is similar to proof of Corollary 1. □
Acknowledgements
I am deeply grateful to Professor U. A. Rozikov for the attention to my work. References
[1] U.A. Rozikov. Gibbs measures on a Cayley trees, World Sci. Pub, Singapore, 2013.
[2] Cohen D.E., Lyndon R.C. Free bases for normal subgroups of free groups, Trans. Amer. Math. Soc., 1963, 108, P. 526-537.
[3] Ganikhodjaev N.N., Rozikov U.A. Description of periodic extreme Gibbs measures of some lattice model on the Cayley tree, Theor. Math. Phys., 1997, 111, P. 480-486.
[4] G.I. Botirov. Functional equations for the Potts model with competing interactions on a Cayley tree. Nanosystems: physics, chemistry, mathematics, 2016, 7(3), P. 401-404.
[5] Young J.W. On the partitions of a group and the resulting classification, Bull. Amer. Math. Soc., 1927, 33, P. 453-461.
[6] U.A. Rozikov, F.H. Haydarov. Normal subgroups of finite index for the group represantation of the Cayley tree, TWMS Jour. Pure. Appl. Math., 2014, 5, P. 234-240.
[7] Kurosh A.G. Group theory, Akademic Verlag, Berlin, 1953.
[8] D.S. Malik, J.N. Mordeson, M.K. Sen. Fundamentals of Abstract Algebra, McGraw-Hill Com., 1997.
