An arithmetic method of counting the subgroups of a finite abelian group

Transcription

1 arxiv: v1 [mathgr] 21 May 2018 An arithmetic method of counting the subgroups of a finite abelian group Marius Tărnăuceanu October 1, 2010 Abstract The main goal of this paper is to apply the arithmetic method developed in our previous paper [13] to determine the number of some types of subgroups of finite abelian groups MSC 2000): Primary 20K01; Secondary 20K27, 11C20, 15A36 Key words: finite abelian groups, number of subgroups, number of cyclic subgroups, number of elements, matrices of integers 1 Introduction One of the most important problems of the combinatorial abelian group theoryistodeterminethenumberofsubgroupsofafiniteabeliangroup This topic has enjoyed a constant evolution starting with the first half of the 20 th century Since a finite abelian group is a direct product of abelian p-groups, the above counting problem is reduced to p-groups Formulas which give the numberofsubgroupsoftypeµofafinitep-groupoftypeλwereestablishedby SDelsartesee[7]), PEDjubjusee[8])andYYehsee[15]) Anexcellent survey on this subject together with connections to symmetric functions was writtenbymlbutlersee[5])in1994 Anotherwaytofindthetotalnumber of subgroups of finite abelian p-groups is presented in [6] and applied for ran two p-groups, as well as for elementary abelian p-groups Also, remind here the paper [1] which gives an explicit formula for the number of subgroups in a finite abelian p-group by using divisor functions of matrices 1

2 The starting point for our discussion is given by the paper [13] see also Section I2 of [14]), where we introduced and studied the concept of fundamental group lattice, that is the subgroup lattice of a finite abelian group These lattices were successfully used to solve the problem of existence and uniqueness of a finite abelian group whose subgroup lattice is isomorphic to a fixed lattice see Proposition 28 21) of [13]) Some steps in finding the total number of subgroups of several particular finite abelian groups have been made in Section 22 of [13], too Thepurposeofthecurrentpaperistoextendtheabovestudy, byapplying the fundamental group lattices in counting some different types of subgroups of finite abelian groups Explicit formulas are obtained for the number of subgroups of a given order in a finite abelian p-group of ran 2, improving Proposition 29 22) of [13], and for the number of maximal subgroups and cyclic subgroups of a given order of arbitrary finite abelian groups The number of elements of a prescribed order in such a group will be also found The paper is organized as follows In Section 2 we recall the notion of fundamental group lattice and its basic properties Section 3 deals with the number of subgroups of finite abelian groups In Section 4 the precise expressions for the number of cyclic subgroups, as well as for the number of elements of a given order in a finite abelian group will be determined In the final section some conclusions and further research directions are indicated Most of our notation is standard and will usually not be repeated here Basic definitions and results on lattices respectively on groups) can be found in [9] respectively in [12]) For subgroup lattice concepts we refer the reader to [10] and [14] 2 Fundamental group lattices Let G be an abelian group of order n and LG) be the subgroup lattice of G By the fundamental theorem of finitely generated abelian groups, there exist uniquely determined by G) the numbers IN, d 1,d 2,,d IN\{0,1} satisfying d 1 d 2 d, d 1 d 2 d = n and G = i=1z di This decomposition of a finite abelian group into a direct product of cyclic groups together with the formof subgroups of Z see Lemma 21, 21, [13]) 2

3 leads us to the concept of fundamental group lattice, defined in the following manner: Let 1beanaturalnumber Then, foreachd 1,d 2,,d ) IN\{0,1}), weconsider theset L ;d1,d 2,,d ) consisting ofall matrices A = a ij ) M Z) which have the following properties: i) a ij = 0, for any i > j, ii) 0 a 1j,a 2j,,a j 1j < a jj, for any j = 1,, iii) 1) a 11 d 1, a ) 12 2) a 22 d 2,d 1, a 11 a 23 3) a 33 d 3,d 2,d 1 a 22 a 12 a 13 a 22 a 23 ), a 22 a 11 a 1 ) a d,d 1,d 2 a 1 1 d 1 a 2 1 a 2 a 1 1 a 1,, a 1 1 a 2 2 a 12 a 13 a 1 a 22 a 23 a a ) 1, a 1 1 a 2 2 a 11 where by x 1,x 2,,x m ) we denote the greatest common divisor of the numbers x 1,x 2,,x m Z On the set L ;d1,d 2,,d ) we introduce the next partial ordering relation denoted by ), as follows: for A = a ij ), B = b ij ) L ;d1,d 2,,d ), put A B if and only if the relations 1) b 11 a 11, a 11 a 12 b 2) 11 b 12 ) b 22 a 22,, b 11 3) b 33 a 33, a 22 a 23 b 22 b 23, b 22 ), b 22 b 11 a 11 a 12 a 13 b 11 b 12 b 13 0 b 22 b 23 3

4 a 1 1 a 1 b ) 1 1 b 1 b a,, b 1 1 a 2 2 a 2 1 a 2 b 2 2 b 2 1 b 2 0 b 1 1 b 1,, b 1 1 b 2 2 a 11 a 12 a 1 b 11 b 12 b b 1 ) b 1 1 b 2 2 b 11 hold Then L ;d1,d 2,,d ), ) is a complete modular lattice, which is called a fundamental group lattice of degree Moreover, from Proposition 22, 21, [13], we now that L ;d1,d 2,,d ) is isomorphic to LG) and so the problem of counting the subgroups of G can be translated into an arithmetic problem: finding the number of elements of L ;d1,d 2,,d ) On the other hand, if n = p n 1 1 pn 2 2 pnm m is the decomposition of n as a product of prime factors and G = m i=1g i is the corresponding primary decomposition of G, then it is well-nown that we have m LG) = i=1lg i ) The above lattice isomorphism shows that LG) = m LG i ), i=1 therefore our counting problem is reduced to p-groups In this way, we need to investigate only fundamental group lattices of type L ;p α 1,p α 2,,p α ), where p is a prime and 1 α 1 α 2 α 4

5 Concerning these lattices, the following elementary remars will be very useful: a) The order of the subgroup of A = a ij ) L ;p α 1,p α 2,,p α ) is i=1z p α i corresponding to the matrix α i p i=1 a ii i=1 b) The subgroup of i=1z p α i corresponding to the matrix A = a ij ) L ;p α 1,p α 2,,p α ) iscyclic if andonlyif< 0 1,0 2,,ā ) > < 0 1,0 2,,ā 1 1 1,ā 1 ) > < ā11ā2 12,,ā 1 ) >,where, for every i = 1,, we denote by x i the image of an element x Z through the canonical homomorphism: Z Z p α i c) If A = a ij ) is an element of L ;p α 1,p α 2,,p α ), then the linear system A x 1 p α 1 x 2 = p α 2 x p α admits solutions in Z 3 The number of subgroups of a finite abelian group As we have seen in the previous section, in order to determine the number of subgroups of finite abelian groups it suffices to reduce the study to p- groups and our problem is equivalent to the counting of elements of the fundamental group lattice L ;p α 1,p α 2,,p α ) This consists of all matrices of integers A = a ij ) i,j=1, satisfying the conditions: 5

6 i) a ij = 0, for any i > j, ) ii) 0 a 1j,a 2j,,a j 1j < a jj, for any j = 1,, iii) 1) a 11 p α 1, 2) a 22 p α 2,p α a ) 12 1, a 11 3) a 33 p α 3,p α a 23 2,p α 1 a 22 ) a p α,p α 1 a 12 a 13 a 22 a 23 ), a 22 a 11 a 1 a 1 1,p α 2 p α 1 a 2 1 a 2 a 1 1 a 1,, a 1 1 a 2 2 a 12 a 13 a 1 a 22 a 23 a a 1 ) a 1 1 a 2 2 a 11 An explicit formula for L ;p α 1,p α 2,,p α ), and consequently for L i=1z p α i), can be easily obtained in the particular case α 1 = α 2 = = α = 1 see Proposition 212, 22, [13]) Proposition 31 For α {0,1,,}, the number of all subgroups of order p α in the finite elementary abelian p-group Z p is 1 if α = 0 or α =, and 2 if 1 α 1 In particular, the total 1 i 1 <i 2 <<i α p i 1+i 2 ++i α αα+1) 1 number of subgroups of Z p is 2+ α=1 1 i 1 <i 2 <<i α p i 1+i 2 ++i α αα+1) 2 In the general case, our method gives an immediate result in counting 6

7 the maximal subgroups of i=1z p α i By the first remar of Section 2, such a subgroup corresponds to a matrix A = a ij ) L ;p α 1,p α 2,,p α ) satisfying a ii = p Then a ii = p for some i {1,2,,} and a jj = 1 for all i=1 j i From the condition ii) of ) we get a 1j = a 2j = = a j 1j = 0, for all j i Remar also that the condition iii) is satisfied and thus the elementsa 1i,a 2i,,a i 1i canbechosenarbitrarilyfromtheset{0,1,,p 1} Therefore we have p i 1 distinct solutions of the system ) Summing up these quantities for i = 1,, we determine the number of maximal subgroups of the finite abelian p-group i=1z p α i Proposition 32 The number of maximal subgroups of Z p α i is p 1 i=1 p 1 Next, we return to the problem of finding the total number of subgroups of i=1z p α i We shall apply our method for ran two abelian p-groups, ie when = 2 clearly, it can be extended in a natural way for an arbitrary ) Note also that the following theorem improves Proposition 29, 22, [13], by indicating the number of subgroups of a fixed order in such a group and by giving a proof founded on fundamental group lattices Theorem 33 For every 0 α α 1 +α 2, the number of all subgroups of order p α 1+α 2 α in the finite abelian p-group Z p α 1 Z p α 2 is: p α+1 1 p 1, if 0 α α 1 p α p 1, if α 1 α α 2 p α 1+α 2 α+1 1, if α 2 α α 1 +α 2 p 1 In particular, the total number of subgroups of Z p α 1 Z p α 2 is 1 p 1) 2 [ α2 α 1 +1)p α 1+2 α 2 α 1 1)p α 1+1 α 1 +α 2 +3)p+α 1 +α 2 +1) ] 7

8 Proof Let A = a ij ) be a solution of ) for = 2, corresponding to a subgroup of order p α 1+α 2 α In this situation, the condition iii) of ) becomes a 11 p α 1 and a 22 p α 2,p α a ) 1 12 a 11 Put a 11 = p i, where 0 i α 1 Then a 22 = p α i and so p α i p α 2,p α 1 i a 12 ), that is p α i p α 1 i p α 2 α 1 +i,a 12 ) If 0 α α 1, we must have i α and the above condition is satisfied by all a 12 < p α i So, one obtains p α i distinct solutionsof ),whichimpliesthatthenumberofsubgroupsoforderp α 1+α 2 α in Z p α 1 Z p α 2 is in this case 1) S 1 α) = α i=0 p α i = pα+1 1 p 1 Suppose now that α 1 α α 2 Then p α 1 α p α 2 α 1 +i,a 12 ) and thus a 12 can be any multiple of p α 1 α in the set {0,1,,p α i 1} It results p α 1 i distinct solutions of ) and the number of subgroups of order p α 1+α 2 α in Z p α 1 Z p α 2 is in this case α 1 2) S 2 α) = p α1 i = pα p 1 i=0 Finally, assume that α 2 α α 1 +α 2 We must have α 1 α α 2 α 1 +i and the number of distinct solutions of ) is again p α 1 i Thus the number of subgroups of order p α 1+α 2 α in Z p α 1 Z p α 2 is in this case 3) S 2 α) = α 1 p α1 i = pα1+α2 α+1 1 p 1 i=α α 2 By using the equalities 1), 2) and 3), one obtains the total number of subgroups of Z p α 1 Z p α 2, namely α 1 α=0 S 1 α)+ α 2 α=α 1 +1 S 2 α)+ α 1 +α 2 α=α 2 +1 S 3 α) = 1 p 1) 2 [ α2 α 1 +1)p α 1+2 α 2 α 1 1)p α 1+1 α 1 +α 2 +3)p+α 1 +α 2 +1)], 8

9 which completes our proof In the following let us denote by f p i,j) the number of all subgroups of the finite abelian p-group Z p i Z p j i j), determined in Theorem 33 Note that we have f p i,j) = 1 [ = j i+1)p i+2 j i 1)p i+1 i+j +3)p+i+j +1) ] = p 1) 2 = j i+1)p i +j i+3)p i 1 + +i+j 1)p+i+j +1) Put f p i,j) = f p j,i), for all i > j, and let n be a fixed positive integer and A p n) be the matrix f p i,j)) i,j=0,n Then A p n) induces a quadratic form n f p i,j)x i Y j Because i,j=0 by induction on n one easily obtains deta p n)=p 1)p n 1 deta p n 1), deta p n) = p 1) n p nn 1) 2, for any n 1 Hence, we have proved the next two corollaries Corollary 34 The quadratic form n i,j=0 A p n) is positive definite, for all n IN f p i,j)x i Y j induced by the matrix Corollary 35 All eigenvalues of the matrix A p n) are positive, for all n IN 4 The number of cyclic subgroups of a finite abelian group Another interesting application of fundamental group lattices not studied in [13]) is the counting of cyclic subgroups of finite abelian groups First of all, we obtain this number for a finite abelian p-group of ran 2 By the second remar of Section 2, the subgroup of Z p α 1 Z p α 2 determined by the 9

10 matrix A = a ij ) is cyclic if and only if < 0 1,ā 2 22 ) > < ā1 11,ā2 12 ) > This necessary and sufficient condition can be rewritten in the following manner Lemma 41 The subgroup of Z p α 1 Z p α 2 corresponding to the matrix A = a ij ) L 2;p α 1,p α 2) is cyclic if and only if a 22 = p α 2,p α a ) 12 1 a 11 Proof If < 0 1,ā 2 22 ) > < ā1 11,ā2 12 ) >, then we can choose an integer x such that 0 1,ā 2 22 ) = xā1 11,ā2 12 ) It results pα 1 xa 11 and p α 2 xa 12 a 22, therefore there exist y,z Z satisfying xa 11 = yp α 1 and xa 12 a 22 = zp α 2 These equalities imply that a 22 = zp α 2 +yp α a 12 1, which together with the a 11 condition 2) of iii) in ) show that a 22 = p α 2,p α a 12 1 a 11 Conversely, suppose that a 22 = p α 2,p α a 12 1 a 11 ) ) Then there are y,z Z with a 22 = zp α 2 + yp α a 12 1 Taing x = y pα1 Z, we easily obtain a 11 a ,ā 2 22 ) = xā1 11,ā2 12 ) and so < 01,ā 2 22 ) > is contained in < ā1 11,ā2 12 ) > By using the above lemma, the problem of finding the number of cyclic subgroups of Z p α 1 Z p α 2 reduces to an elementary arithmetic exercise Theorem 42 For every 0 α α 2, the number of cyclic subgroups of order p α in the finite abelian p-group Z p α 1 Z p α 2 is: 1, if α = 0 p α +p α 1, if 1 α α 1 p α 1, if α 1 < α α 2 In particular, the number of all cyclic subgroups of Z p α 1 Z p α 2 is 2+2p+ +2p α 1 1 +α 2 α 1 +1)p α 1 Proof Denote by g 2 p α) the number of cyclic subgroups of order pα in Z p α 1 Z p α 2 and let A=a ij ) L 2;p α 1,p α 2) be the matrix corresponding to 10

11 such a subgroup Then a 11 p α 1, a 22 = p α 2,p α a ) 12 1 Taing a 11 =p i with 0 i α 1, we obtain a 11 and a 11 a 22 =p α 1+α 2 α which implies that a 22 = p α 1+α 2 α i = p α 2,p α 1 i a 12 ) = p α 1 i p α 2 α 1 +i,a 12 ), 4) p α 2 α = p α 2 α 1 +i,a 12 ) Clearly, for α = 0 it results a 11 = p α 1, a 22 = p α 2, a 12 = 0, and thus 5) g 2 p0) = 1 For 1 α α 1 we must have α 1 α i If i = α 1 α, the condition 4) is equivalent to p α 2 α a 12, therefore a 12 can be chosen in p α ways If α 1 α+1 i, 4) is equivalent to p α 2 α a 12 and p α 2 α+1 a 12 Therearep α 1 i p α 1 i 1 elementsoftheset{0,1,,p α 1+α 2 α i }whichsatisfy the previous relations So, one obtains 6) g 2 p α) = pα + α 1 i=α 1 α+1 p α 1 i p α 1 i 1 ) = p α +p α 1, for 1 α α 1 Mention that if α 1 < α α 2, then the condition α 1 α i is satisfied by all i = 1,α 1, and hence α 1 7) gpα) 2 = p α 1 i p ) α 1 i 1 = p α 1, for α 1 < α α 2 i=0 Now, the equalities 5)-7) give us the total number of cyclic subgroups of Z p α 1 Z p α 2, namely α 1 1+ p α +p α 1) α 2 + p α 1 = α=1 = 1 p 1 α=α 1 +1 [ α2 α 1 +1)p α 1+1 α 2 α 1 1)p α 1 2 ] = = 2+2p+ +2p α 1 1 +α 2 α 1 +1)p α 1 11

12 and our proof is finished The above method can be used for an arbitrary > 2, too In order to do this we need to remar that where g 2 p α) = pα h 1 pα) p α 1 h 1 pα 1) p α p α 1, for all α 0, h 1 pα) = { p α, if 0 α α 1 p α 1, if α 1 α This equality extends to the general case in the following way Theorem 43 For every 1 α α, the number of cyclic subgroups of order p α in the finite abelian p-group i=1z p α i is g pα) = pα h 1 p α) p α 1 h 1 p α 1) p α p α 1, where h 1 p α) = p 1)α, if 0 α α 1 p 2)α+α 1, if α 1 α α 2 p α 1+α 2 ++α 1, if α 1 α Note that g p 0) = 1 and the number of all cyclic subgroups of i=1z p α i can be easily determined from Theorem 43 Since the numbers of cyclic subgroups and of elements of a given order in a finite abelian p-group are closely connected through the well-nown Euler s function ϕ), we also infer the following consequence of Theorem 43 Corollary 44 The number of all elements of order p α, 1 α α, in the finite abelian p-group i=1z p α i is g p α)ϕpα ) = g p α) p α p α 1) = p α h 1 p α) p α 1 h 1 p α 1) 12

13 As we have seen in Section 2, counting the subgroups of finite abelian groupscan bereduced to p-groups The same thing can bealso saidfor cyclic subgroups and for elements of a given order in an arbitrary finite abelian group G Suppose that p n 1 1 pn 2 2 is the decomposition of G as a product pnm m m ofprimefactorsandlet i=1g i bethecorrespondingprimarydecompositionof G Then every cyclic subgroup H of order p α 1 written as a direct product m 1 p α 2 2 p am m of G can be uniquely i=1h i, where H i is a cyclic subgroup of order p α i i of G i, i = 1,m This remar leads to the following result, that generalizes Theorem 43 and Corollary 44 Corollary 45 Under the previous hypotheses, for every α 1,α 2,,α m ) IN m with α i n i, i = 1,m, the number of cyclic subgroups respectively of elements) of order p α 1 1 p α 2 2 p αm m in G is m g i p i α i ) respectively m i=1 i=1 g i p i α i )ϕp α i i )), where i denotes the number of direct factors of G i, i = 1,m 5 Conclusions and further research All our previous results show that the arithmetic method introduced in [13] and applied in this paper can constitute an alternative way to study the subgroups of finite abelian groups Clearly, it can successfully be used in solving many computational problems in finite) abelian group theory These will surely be the subject of some further research Finally, we mention several open problems concerning this topic Problem 51 Extend Theorem 33, by indicating explicit formulas for the number of subgroups of a fixed order and for the total number of subgroups of a finite abelian p-group of an arbitrary ran 13

14 Problem 52 Let G be a finite abelian group Use the description of LG) given by the above arithmetic method and the well-nown description of AutG) to determine the characteristic subgroups of G Problem 53 By using the defining relations of a fundamental group lattice, create a computer algebra program that generates the subgroups of a finite abelian group Problem 54 Let G 1 be a finite abelian group and G 2 be a finite group such that G 1 = G 2 = n Denote π e G i ) = {oa) a G i } and, for every divisor d of n, let n i d) be the number of elements of order d in G i, i = 1,2 remar that the numbers n 1 d) are nown, by Corollary 45) Is it true that the conditions a) π e G 1 ) = π e G 2 ), b) n 1 d) = n 2 d), for all d, imply the group isomorphism G 1 = G2? In other words, study whether a finite abelian group is determined by the set of its element orders and by the numbers of elements of any fixed order) Acnowledgement The authors are grateful to the reviewers for their remars which improve the previous version of the paper References [1] Bhowmi, G, Evaluation of the divisor function of matrices, Acta Arith ), [2] Bhowmi, G, Ramaré, O, Average orders of multiplicative arithmetical functions of integer matrices, Acta Arith ), [3] Bhowmi, G, Ramaré, O, Algebra of matrix arithmetic, J Algebra ), [4] Birhoff, G, Subgroups of abelian groups, Proc Lond Math Soc , 1935),

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