# The Number of Chains of Subgroups in the Lattice of Subgroups of the Dicyclic Group

Authors: Ju-Mok Oh; Yunjae Kim; Kyung-Won Hwang
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We give an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dicyclic groupB4n of order 4n by finding its generating function
of multivariables.


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## Body

## 1. Introduction

Throughout this paper, all groups are assumed to be finite. Thelattice of subgroups of a given group G is the lattice (L(G),≤) where L(G) is the set of all subgroups of G and the partial order ≤ is the set inclusion. In this lattice (L(G),≤), achain of subgroups of G is a subset of L(G) linearly ordered by set inclusion. A chain of subgroups of G is called G-rooted (or rooted) if it contains G. Otherwise, it is called unrooted.The problem of counting chains of subgroups of a given groupG has received attention by researchers with related to classifying fuzzy subgroups of G under a certain type of equivalence relation. Some works have been done on the particular families of finite abelian groups (e.g., see [1–4]). As a step of this problem toward non-abelian groups, the first author [5] has found an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dihedral group D2n of order 2n where n is an arbitrary positive integer. As a continuation of this work, we give an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dicyclic group B4n of order 4n by finding its generating function of multivariables where n is an arbitrary integer.


## 2. Preliminaries

Given a groupG, let 𝒞(G), 𝒰(G), and ℛ(G) be the collection of chains of subgroups of G, of unrooted chains of subgroups of G, and of G-rooted chains of subgroups of G, respectively. Let C(G):=|𝒞(G)|, U(G):=|𝒰(G)|, and R(G):=|ℛ(G)|.The following simple observation is useful for enumerating chains of subgroups of a given group.Proposition 2.1.
LetG be a finite group. Then R(G)=U(G)+1 and C(G)=R(G)+U(G)=2R(G)-1.For a fixed positive integerk, we define a function λ as follows:
(2.1)λ(xk)≔1-2xk,λ(xk,xk-1,…,xj)≔λ(xk,xk-1,…,xj+1)-(1+λ(xk,xk-1,…,xj+1))xj
for any j=k-1,k-2,…,1.Proposition 2.2 (see [5]).
Letℤn be the cyclic group of order
(2.2)n=p1β1p2β2⋯pkβk,
where p1,…,pk are distinct prime numbers and β1,…,βk are positive integers. Then the number R(ℤn) of rooted chains of subgroups in the lattice of subgroups of ℤn is the coefficient of x1β1x2β2⋯xkβk of
(2.3)ϕβ1,…,βk(xk,…,x1)=1λ(xk,…,x1).Letℤ be the set of all integer numbers. Given distinct positive integers i1,…,it, we define a function
(2.4)πi1⋯it:Zk⟼Zk,(x1,…,xk)⟼(y1,…,yk),
where
(2.5)yl={xl,ifl≠ij∀j=1,…,t,xl-1,l=ijforsomejsuchthat1≤j≤t.Most of our notations are standard and for undefined group theoretical terminologies we refer the reader to [6, 7]. For a general theory of solving a recurrence relation using a generating function, we refer the reader to [8, 9].


## 3. The Number of Chains of Subgroups of the Dicyclic GroupB4n

Throughout the section, we assume that(3.1)n:=p1β1p2β2⋯pkβk,
is a positive integer, where p1,…,pk are distinct prime numbers and β1,…,βk are nonnegative integers and the dicyclic group B4n of order 4n is defined by the following presentation:
(3.2)B4n:=〈a,b∣a2n=e,b2=an,bab-1=a-1〉,
where e is the identity element.By the elementary group theory, the following is wellknown.Lemma 3.1.
The dicyclic groupB4n has an index 2 subgroup 〈a〉, which is isomorphic to ℤ2n, and has pi index pi subgroups
(3.3)〈api,b〉,〈api,ab〉,…,〈api,api-1b〉,
which are isomorphic to the dicyclic group B4n/pi of order 4n/pi where i=1,2,…,k.Lemma 3.2.
(1) For anyi=1,2,…,k,
(3.4)〈api,arb〉∩〈api,asb〉=〈api〉≅Z2n/pi,
where 0≤r<s≤pi-1.
(2) For any distinct prime factorspi1,pi2,…,pit of n,
(3.5)〈api1,ar1b〉∩〈api2,ar2b〉∩⋯∩〈apit,artb〉≅B4n/pi1⋯pit,
where r1,…,rt are nonnegative integers.Proof.
(1) To the contrary suppose that(3.6)〈api,arb〉∩〈api,asb〉≠〈api〉.
Then apiu+rb=apiv+sb for some integers u and v. This implies pi∣s-r. Since 0≤r<s≤pi-1, we have s=r, a contradiction.
(2) We only give its proof whent=2. The general case can be proved by the inductive process. Let
(3.7)K:=〈api1,ar1b〉∩〈api2,ar2b〉.
Clearly, api1pi2∈K. Since gcd(pi1,pi2)=1, there exist integers u and v such that pi1u+pi2v=1. Note that api1(-u(r1-r2))+r1b=api1(-u(r1-r2))ar1b∈〈api1,ar1b〉. On the other hand,
(3.8)api1(-u(r1-r2))+r1b=a-pi1u(r1-r2)+r1b=api2v(r1-r2)-(r1-r2)+r1bsincepi1u+pi2v=1=api2v(r1-r2)+r2b∈〈api2,ar2b〉.
Considering the order of K, one can see that K=〈api1pi2,api1(-u(r1-r2))+r1b〉. Since
(3.9)(api1pi2)4n/pi1pi2=e,(api1(-u(r1-r2))+r1b)2=b2=an=(api1pi2)n/pi1pi2,(api1(-u(r1-r2))+r1b)(api1pi2)(api1(-u(r1-r2))+r1b)-1=(api1pi2)-1,
we have K≅B4n/pi1pi2.By Lemma3.1, we have
(3.10)U(B4n)=C(〈a〉≅Z2n)⋃i=0k‍⋃j=0pi-1‍C(〈api,ajb〉≅B4n/pi).
Using the inclusion-exclusion principle and Lemma 3.2, one can see that the number U(B4n) has the following form:
(3.11)U(B4n)=C(Z2n)+∑1≤i1<⋯<it≤k,1≤t≤kzi1,…,itC(Z2n/pi1⋯pit)+∑1≤i1<⋯<it≤k,1≤t≤kbi1,…,itC(B4n/pi1⋯pit)
for suitable integers zi1,…,it and bi1,…,it. In the following, we determine the numbers zi1,…,it and bi1,…,it explicitly.Lemma 3.3.
(1)bi1,i2,…,it=(-1)t+1pi1pi2⋯pit.
(2)zi1,i2,…,it=(-1)tpi1pi2⋯pit.Proof.
(1) Clearlybi1=(-1)1+1pi1=pi1 for any i1=1,…,k. For any integer t≥2, one can see by Lemma 3.2 that among intersections of the subgroups of the right-hand side of (3.10), the group isomorphic to B4n/pi1pi2⋯pit only appears in t-intersection of the subgroups
(3.12)〈api1,aj1b〉,〈api2,aj2b〉,…,〈apit,ajtb〉,
where 0≤jr≤pir-1 and 1≤r≤t. Since there are (pi11)(pi21)⋯(pit1)=pi1pi2⋯pit such choices, we have bi1,i2,…,it=(-1)t+1pi1pi2⋯pit.
(2) By Lemma3.2, one can see that among intersections of the subgroups of the right-hand side of (3.10), the group isomorphic to ℤ2n/pi1pi2⋯pit only appears one of the following two forms:
(3.13)〈a〉∩〈api1,aj1b〉∩〈api2,aj2b〉∩⋯∩〈apit,ajtb〉,〈api1,aj1b〉∩〈api2,aj2b〉∩⋯∩〈apit,ajtb〉,
where 0≤jr≤pir-1 and 1≤r≤t, and each subgroup type in the first form must appear at least once, and it can appear more than once, while each subgroup type in the second form must appear at least once, and one of the subgroup types must appear more than once. Let γ be the number of the groups isomorphic to ℤ2n/pi1pi2⋯pit obtained from the first form, and let δ be the number of the groups isomorphic to ℤ2n/pi1pi2⋯pit obtained from the second form. Then clearly zi1,i2,⋯,it=γ+δ. Note that
(3.14)γ=∑k=0pi1+⋯+pit-t‍(-1)t+2+k∑j1+⋯+jt=t+k,1≤jr≤pir,1≤r≤k∏r=1t‍(pirjr)=∑k≥0‍∑j1+⋯+jt=t+k,1≤jr≤pir,1≤r≤t(-1)t+k∏r=1t‍(pirjr)=∑1≤jr≤pir,1≤r≤t‍(-1)j1+⋯+jt∏r=1t‍(pirjr)=∏r=1t‍∑1≤jr≤pir‍(-1)j1+⋯+jt(pirjr)=(-1)t.
On the other hand,
(3.15)δ=∑k=0pi1+⋯+pit-t-1‍(-1)t+2+k∑j1+⋯+jt=t+1+k,1≤jr≤pir,1≤r≤t∏r=1t‍(pirjr)=∑k=0pi1+⋯+pit-t-1‍(-1)t+2+k∑j1+⋯+jt=t+1+k,1≤jr≤pir,1≤r≤t∏r=1t‍(pirjr)+(-1)t+1∑j1+⋯+jt=t,1≤jr≤pir,1≤r≤t∏r=1t‍(pirjr)-(-1)t+1∑j1+⋯+jt=t,1≤jr≤pir,1≤r≤t∏r=1t‍(pirjr)=∑k=0pi1+⋯+pit-t‍(-1)t+1+k∑j1+⋯+jt=t+k,1≤jr≤pjr,1≤r≤t∏r=1t‍(pirjr)-(-1)t+1∑j1+⋯+jt=t,1≤jr≤pir,1≤r≤t∏r=1t‍(pirjr)=(-1)t-(-1)t+1pi1⋯pit.
Therefore, we have zi1,i2,…,it=(-1)tpi1⋯pit.By Proposition2.1 and Lemma 3.3, (3.11) becomes
(3.16)R(B4n)=2R(Z2n)+2∑1≤i1<⋯<it≤k,1≤t≤k(-1)tpi1⋯pitR(Z2n/pi1⋯pit)+2∑1≤i1<⋯<it≤k,1≤t≤k(-1)t+1pi1⋯pitR(B4n/pi1⋯pit).
Let aβ1,…,βk:=R(B4n) and let bβ1,…,βk:=R(ℤ2n). Then (3.16) becomes
(3.17)aβ1,…,βk=2bβ1,…,βk+2∑1≤i1<⋯<it≤k,1≤t≤k(-1)tpi1⋯pitbπi1⋯it(β1,…,βk)+2∑1≤i1<⋯<it≤k,1≤t≤k(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk).Throughout the remaining part of the section, we solve the recurrence relation of (3.17) by using generating function technique. From now on, we allow each βi to be zero for computational convenience.Let(3.18)ψβ1,…,βk(xk,xk-1,…,xj):=∑βj=0∞‍⋯∑βk-1=0∞‍∑βk=0∞‍aβ1,…,βkxkβkxk-1βk-1⋯xjβj,ϕβ1,…,βk(xk,xk-1,…,xj):=∑βj=0∞‍⋯∑βk-1=0∞‍∑βk=0∞‍bβ1,…,βkxkβkxk-1βk-1⋯xjβj,
where j=k,k-1,…,1.For a fixed integern=p1β1p2β2⋯pkβk such that p1,…,pk are distinct prime numbers and β1,…,βk are non-negative integers, we define a function μ as follows.
(3.19)μ(xk)≔1-2pkxk,μ(xk,…,xj)≔μ(xk,…,xj+1)-(1+μ(xk,…,xj+1))pjxj
for any j=k-1,k-2,…,1.Lemma 3.4.
Letk be a positive integer. If k=1, then
(3.20)μ(x1)ψβ1(x1)=(1+μ(x1))ϕβ1(x1).
If k≥2, then
(3.21)μ(xk,…,xj)ψβ1,…,βk(xk,…,xj)=(1+μ(xk,…,xj))×[ϕβ1,…,βk(xk,…,xj)+∑1≤i1<⋯<it≤j-1,1≤t≤j-1(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj)+∑1≤i1<⋯<it≤j-1,1≤t≤j-1(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk,…,xj)]
for any j=k,k-1,…,2.Proof.
Assume first thatk=1. Then (3.17) with k=1 gives us that
(3.22)aβ1=2bβ1+2p1aβ1-1-2p1bβ1-1.
Taking ∑β1=1∞‍ to both sides of (3.22), we have
(3.23)(1-2p1x1)ψβ1(x1)=(2-2p1x1)ϕβ1(x1)
because a0=R(B4p10)=R(ℤ22)=22 and b0=R(ℤ2p10)=R(ℤ2)=2 by a direct computation.
From now on, we assume thatk≥2. We prove (3.21) by double induction on k and j. Equation (3.17) with k=2 gives us that
(3.24)aβ1,β2=2bβ1,β2-2p1bβ1-1,β2-2p2bβ1,β2-1+2p1p2bβ1-1,β2-1+2p1aβ1-1,β2+2p2aβ1,β2-1-2p1p2aβ1-1,β2-1.
Taking ∑β2=1∞‍x2β2 of both sides of (3.24), we have
(3.25)(1-2p2x2)ψβ1,β2(x2)=(2-2p2x2)[p1ψβ1-1,β2(x2)+ϕβ1,β2(x2)-p1ϕβ1-1,β2(x2)]
because aβ1,0=aβ1 and bβ1,0=bβ1 by the definition, and
(3.26)aβ1,0-2bβ1,0-2p1aβ1,0+2p1bβ1-1,0=0
by (3.17) with k=1. That is,
(3.27)μ(x2)ψβ1,β2(x2)=(1+μ(x2))[ϕβ1,β2(x2)-p1ϕβ1-1,β2(x2)+p1ψβ1-1,β2(x2)].
Thus (3.21) holds for k=2.
Assume now that (3.21) holds from 2 to k-1 and consider the case for k. Note that the last two terms of the right-hand side of (3.17) can be divided into three terms, respectively, as follows:
(3.28)2∑1≤i1<⋯<it≤k,1≤t≤k(-1)tpi1⋯pitbπi1⋯it(β1,…,βk)=-2pkbβ1,…,βk-1,βk-1-2pk∑1≤i1<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitbπi1⋯it(β1,…,βk-1,βk-1)+2∑1≤i1<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitbπi1⋯it(β1,…,βk),2∑1≤i1<⋯<it≤k,1≤t≤k(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk)=2pkaβ1,…,βk-1,βk-1-2pk∑1≤i1<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk-1,βk-1)+2∑1≤i1<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk).
Taking ∑βk=1∞‍xkβk of both sides of (3.17) and using (3.28), one can see that
(3.29)(1-2pkxk)ψβ1,…,βk(xk)=(2-2pkxk)×[ϕβ1,…,βk(xk)+∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk-1,βk)(xk)+∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk)]+aβ1,…,βk-1,0-2bβ1,…,βk-1,0-2∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitbπi1⋯it(β1,…,βk-1,0)-2∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk-1,0).
Further since
(3.30)aβ1,…,βk-1,0-2bβ1,…,βk-1,0-2∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitbπi1⋯it(β1,…,βk-1,0)-2∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitaπi1⋯it(β1,…,βk-1,0)=0
by (3.17), we have
(3.31)(1-2pkxk)ψβ1,…,βk(xk)=(2-2pkxk)×[ϕβ1,…,βk(xk)+∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk-1,βk)(xk)+‍∑1≤i<⋯<it≤k-1,1≤t≤k-1(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk)].
Thus (3.21) holds for j=k. Assume that (3.21) holds from k to j and consider the case for j-1. Note that the last two terms of the right-hand side of (3.21) can be divided into three terms, respectively, as follows:
(3.32)∑1≤i1<⋯<it≤j-1,1≤t≤j-1(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj)=-pj-1ϕβ1,…,βj-2,βj-1-1,βj,…,βk(xk,…,xj)-pj-1∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βj-2,βj-1-1,βj,…,βk)(xk,…,xj)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj),(3.33)∑1≤i1<⋯<it≤j-1,1≤t≤j-1(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk,…,xj)=pj-1ψβ1,…,βj-2,βj-1-1,βj,…,βk(xk,…,xj)-pj-1∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βj-2,βj-1-1,βj,…,βk)(xk,…,xj)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj).
Taking ∑βj-1=1∞‍xj-1βj-1 of both sides of (3.21), we have
(3.34)μ(xk,…,xj,xj-1)ψβ1,…,βk(xj-1,…,xk)=(1+μ(xk,…,xj,xj-1))×[ϕβ1,…,βk(xj-1,…,xk)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj,xj-1)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk,…,xj,xj-1)]+μ(xk,…,xj)ψβ1,…,βj-2,0,βj,…,βk(xk,…,xj)-(1+μ(xk,…,xj))∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βj-2,0,βj,…,βk)(xk,…,xj)-(1+μ(xk,…,xj))∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βj-2,0,βj,…,βk)(xk,…,xj).
Note that
(3.35)μ(xk,…,xj)ψβ1,…,βj-2,0,βj,…,βk(xk,…,xj)-(1+μ(xk,…,xj))∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βj-2,0,βj,…,βk)(xk,…,xj)-(1+μ(xk,…,xj))∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βj-2,0,βj,…,βk)(xk,…,xj)=0
by induction hypothesis. Thus
(3.36)μ(xk,…,xj,xj-1)ψβ1,…,βk(xj-1,…,xk)=(1+μ(xk,…,xj,xj-1))×[ϕβ1,…,βk(xj-1,…,xk)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)tpi1⋯pitϕπi1⋯it(β1,…,βk)(xk,…,xj,xj-1)+∑1≤i1<⋯<it≤j-2,1≤t≤j-2(-1)t+1pi1⋯pitψπi1⋯it(β1,…,βk)(xk,…,xj,xj-1)].
Therefore, (3.21) holds for j-1.Equation (3.21) with j=2 gives us that
(3.37)μ(xk,…,x2)ψβ1,…,βk(xk,…,x2)=(1+μ(xk,…,x2))×[ϕβ1,…,βk(xk,…,x2)-p1ϕβ1-1,β2,…,βk(xk,…,x2)+p1ψβ1-1,β2,…,βk(xk,…,x2)].
Taking ∑β1=1∞‍x1β1 of both sides of (3.37), we get that
(3.38)μ(xk,…,x1)ψβ1,…,βk(xk,…,x2,x1)=(1+μ(xk,…,x1))ϕβ1,…,βk(xk,…,x2,x1)+μ(xk,…,x2)ψ0,β2,…,βk(xk,…,x2)-(1+μ(xk,…,x2))ϕ0,β2,…,βk(xk,…,x2).Lemma 3.5.
Ifk≥2, then
(3.39)μ(xk,…,x2)ψ0,β2,…,βk(xk,…,x2)=(1+μ(xk,…,x2))ϕ0,β2,…,βk(xk,…,x2).Proof.
Ifk=2, then since ψ0,β2(x2)=ψβ2(x2) and ϕ0,β2(x2)=ϕβ2(x2), the equation
(3.40)μ(x2)ψ0,β2(x2)=(1+μ(x2))ϕ0,β2(x2)
holds by (3.20). Assume now that (3.39) holds for k. Then by (3.38) we get that
(3.41)μ(xk,…,x1)ψβ1,…,βk(xk,…,x2,x1)=(1+μ(xk,…,x1))ϕβ1,…,βk(xk,…,x2,x1),
which implies that
(3.42)μ(xk+1,…,x2)ψ0,β2,…,βk+1(xk+1,…,x2)=(1+μ(xk+1,…,x2))ϕ0,β2,…,βk+1(xk+1,…,x2).
Thus (3.39) holds for k+1.By Lemmas3.4 and 3.5 and (3.38), we have
(3.43)μ(xk,…,x1)ψβ1,…,βk(xk,…,x1)=(1+μ(xk,…,x1))ϕβ1,…,βk(xk,…,x1).
We now need to find the function ϕβ1,…,βk(xk,…,x1) explicitly.Lemma 3.6.
Ifp1=2, then
(3.44)ϕβ1,…,βk(xk,…,x1)={2λ(x1),ifk=1,[1+1λ(xk,…,x2)]1λ(xk,…,x1),ifk≥2.
If pi≠2 for i=1,2,…,k, then
(3.45)ϕβ1,…,βk(xk,…,x1)=[1+1λ(xk,…,x1)]1λ(xk,…,x1).Proof.
We first assume thatp1=2. Then by Proposition 2.2,
(3.46)bβ1,β2,…,βk=R(Zp1β1+1p2β2p3β3⋯pkβk)
is the coefficient of x1β1+1x2β2x3β3⋯xkβk of
(3.47)1λ(xk,…,x1),
which implies that bβ1,β2,…,βk is the coefficient of x1β1x2β2x3β3⋯xkβk of
(3.48)2λ(x1)ifk=1,[1+1λ(xk,…,x2)]1λ(xk,…,x1)ifk≥2,
and hence by the definition of ϕ we get that
(3.49)ϕβ1,…,βk(xk,…,x1)={2λ(x1),ifk=1,[1+1λ(xk,…,x2)]1λ(xk,…,x1),ifk≥2.
Assume now thatpi≠2 for i=1,2,…,k. Since bβ1,…,βk=R(ℤ2p1β1p2β2⋯pkβk), by Proposition 2.2bβ1,…,βk is the coefficient of x11x2β1x3β2⋯xk+1βk of
(3.50)1λ(xk+1,…,x1).
Since
(3.51)1λ(xk+1,…,x1)=1λ(xk+1,…,x2)-(1+λ(xk+1,…,x2))x1=1λ(xk+1,…,x2)11-[1+(1/λ(xk+1,…,x2))]x1
by the definition, bβ1,…,βk is the coefficient of x2β1x3β2⋯xk+1βk of
(3.52)1λ(xk+1,…,x2)[1+1λ(xk+1,…,x2)].
By changing the variables x2,x3,…,xk+1 by x1,x2,…,xk, respectively, we get that bβ1,…,βk is the coefficient of x1β1x2β2⋯xkβk of
(3.53)1λ(xk,…,x1)[1+1λ(xk,…,x1)].
By the definition of ϕβ1,…,βk(xk,…,x1), we have
(3.54)ϕβ1,…,βk(xk,…,x1)=1λ(xk,…,x1)[1+1λ(xk,…,x1)].By Proposition2.1, (3.43), and Lemma 3.6, we have the following theorem.Theorem 3.7.
Let(3.55)n:=p1β1p2β2⋯pkβk,
be a positive integer such that p1,…,pk are distinct prime numbers and β1,…,βk are positive integers. Let
(3.56)B4n:=〈a,b∣a2n=e,b2=an,bab-1=a-1〉
be the dicyclic group of order 4n. Let R(B4n) be the number of rooted chains of subgroups in the lattice of subgroups of B4n.  (1)
Ifp1=2, then R(B4n) is the coefficient of x1β1x2β2⋯xkβk of
(3.57)ψβ1,…,βk(xk,…,x1)={[1+1μ(x1)]2λ(x1),ifk=1,[1+1μ(xk,…,x1)][1+1λ(xk,…,x2)]1λ(xk,…,x1),ifk≥2.(2)
Ifpi≠2 for i=1,2,…,k, then R(B4n) is the coefficient of x1β1x2β2⋯xkβk of
(3.58)ψβ1,…,βk(xk,…,x1)=[1+1μ(xk,…,x1)]1λ(xk,…,x1)[1+1λ(xk,…,x1)].
Furthermore, the numberC(B4n) of chains of subgroups in the lattice of subgroups of B4n is the coefficient of x1β1x2β2⋯xkβk of
(3.59)2ψβ1,…,βk(xk,…,x1)-∏i=1k‍11-xi.We now want to find the coefficient ofx1β1x2β2⋯xkβk of ψβ1,…,βk(xk,…,x1) explicitly. Since
(3.60)1μ(xk,…,x1)=1μ(xk,…,x2)11-[1+(1/μ(xk,…,x2))]p1x1,
by the definition, the coefficient of x1β1 of 1/μ(xk,…,x1) is
(3.61)1μ(xk,…,x2)[1+1μ(xk,…,x2)]β1p1β1=p1β1∑i1=0β1‍(β1i1)[1μ(xk,…,x2)]i1+1=p1β1∑i1=0β1‍(β1i1)[1μ(xk,…,x3)]i1+1[11-[1+(1/μ(xk,…,x3))]p2x2]i1+1.
Thus the coefficient of x1β1x2β2 of 1/μ(xk,…,x1) is
(3.62)p1β1p2β2∑i1=0β1‍(β1i1)(i1+β2β2)[1μ(xk,…,x3)]i1+1[1+1μ(xk,…,x3)]β2=p1β1p2β2∑i1=0β1‍∑i2=0β2‍(β1i1)(β2i2)(i1+β2β2)[1μ(xk,…,x3)]i1+i2+1.
Continuing this process, one can see that the coefficient of x1β1x2β2⋯xkβk of 1/μ(xk,…,x1) is
(3.63)2βkp1β1p2β2⋯pkβk∑i1=0β1‍∑i2=0β2‍⋯∑ik-1=0βk-1‍∏r=1k-1‍(βrir)(βr+1+∑m=1r‍imβr+1).
Similarly one can see that the coefficient of x1β1x2β2⋯xkβk of 1/λ(xk,…,x1) is
(3.64)2βk∑i1=0β1‍∑i2=0β2‍⋯∑ik-1=0βk-1‍∏r=1k-1‍(βrir)(βr+1+∑m=1r‍imβr+1),
the coefficient of x1β1x2β2⋯xkβk of [1+(1/λ(xk,…,x2))](1/λ(xk,…,x1)) is
(3.65)2βk∑i1=0β1+1‍∑i2=0β2‍∑i3=0β3‍⋯∑ik-1=0βk-1‍(β1+1i1)(β2+i1β2)∏r=2k-1‍(βrir)(βr+1+∑m=1r‍imβr+1)
and the coefficient of x1β1x2β2⋯xkβk of 1/λ(xk,…,x1)2 is
(3.66)(β1+1)2βk∑i1=0β1‍∑i2=0β2‍⋯∑ik-1=0βk-1‍∏r=1k-1‍(βrir)(βr+1+∑m=1r‍im+1βr+1).Therefore, one can have the following.Corollary 3.8.
Letn and B4n be the positive integer and the dicyclic group, respectively, defined in Theorem 3.7. Let R(B4n) be the number of rooted chains of subgroups in the lattice of subgroups of B4n.  (1)
Ifp1=2, then
(3.67)R(B4n)=2βk∑i1=0β1+1‍∑i2=0β2‍∑i3=0β3‍⋯∑ik-1=0βk-1‍(β1+1i1)(β2+i1β2)∏r=2k-1‍(βrir)(βr+1+∑m=1r‍imβr+1)+2βk∑j1=0β1‍∑j2=0β2‍⋯∑jk=0βk‍[[p1j1p2j2⋯pkjk∑i1=0j1‍∑i2=0j2‍⋯∑ik-1=0jk-1‍∏r=1k-1‍(jrir)(jr+1+∑m=1r‍imjr+1)]×[∑i1=0β1-j1+1‍∑i2=0β2-j2‍∑i3=0β3-j3‍⋯∑ik-1=0βk-1-jk-1‍(β1-j1+1i1)(β2-j2+i1β2-j2)×∏r=2k-1‍(βr-jrir)(βr+1-jr+1+∑m=1r‍imβr+1-jr+1)]],where if k=1, then R(B4·2β1)=22β1+2 and if k=2, then
(3.68)R(B4⋅2β1p2β2)=2β2∑i1=0β1+1‍(β1+1i1)(β2+i1β2)+2β2∑j1=0β1‍∑j2=0β2‍[[2j1p2j2∑i1=0j1‍(j1i1)(j2+i1j2)]×[∑i1=0β1-j1+1‍(β1-j1+1i1)(β2-j2+i1β2-j2)]].(2)
Ifpi≠2 for i=1,2,…,k, then
(3.69)R(B4n)=2βk∑i1=0β1‍∑i2=0β2‍⋯∑ik-1=0βk-1‍∏r=1k-1‍(βrir)(βr+1+∑m=1r‍imβr+1)+(β1+1)2βk∑i1=0β1‍∑i2=0β2‍⋯∑ik-1=0βk-1‍∏r=1k-1‍(βrir)(βr+1+1+∑m=1r‍imβr+1)+2βk∑j1=0β1‍∑j2=0β2‍⋯∑jk=0βk‍[[p1j1p2j2⋯pkjk∑i1=0j1‍∑i2=0j2‍⋯∑ik-1=0jk-1‍∏r=1k-1‍(jrir)(jr+1+∑m=1r‍imjr+1)]×[∑i1=0β1-j1‍∑i2=0β2-j2‍⋯∑ik-1=0βk-1-jk-1‍∏r=1k-1‍(βr-jrir)(βr+1-jr+1+∑m=1r‍imβr+1-jr+1)]]+2βk∑j1=0β1‍∑j2=0β2‍⋯∑jk=0βk‍[[p1j1p2j2⋯pkjk∑i1=0j1‍∑i2=0j2‍⋯∑ik-1=0jk-1‍∏r=1k-1‍(jrir)(jr+1+∑m=1r‍imjr+1)]×[(β1-j1+1)∑i1=0β1-j1‍∑i2=0β2-j2‍⋯∑ik-1=0βk-1-jk-1‍∏r=1k-1‍(βr-jrir)×(βr+1-jr+1+1+∑m=1r‍imβr+1-jr+1)]],where if k=1, then
(3.70)R(B4p1β1)=2β1+(β1+1)2β1+2β1∑j1=0β1‍p1j1+2β1∑j1=0β1‍p1j1(β1-j1+1)=2β1[β1+2+p1β1+1-1p1-1+p1β1+2-(β1+2)p1+β1+1(p1-1)2].



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*Source: 760246-2012-09-04.xml*