ON THE NUMBER OF SUBGROUPS OF A GIVEN EXPONENT IN A FINITE ABELIAN GROUP. Marius Tărnăuceanu and László Tóth

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1 PUBLICATIONS DE L INSTITUT MATHÉMATIQUE Nouvelle série, tome 101(115) (2017), htts://doi.org/ /pim t ON THE NUMBER OF SUBGROUPS OF A GIVEN EXPONENT IN A FINITE ABELIAN GROUP Marius Tărnăuceanu and László Tóth Abstract. This aer deals with the number of subgrous of a given exonent in a finite abelian grou. Exlicit formulas are obtained in the case of rank two and rank three abelian grous. An asymtotic formula is also resented. 1. Introduction One of the most imortant roblems of combinatorial abelian grou theory is to determine the number of subgrous of a finite abelian grou. This toic has enjoyed a constant evolution starting with the first half of the 20 th century. Since a finite abelian grou is a direct roduct of abelian -grous, the above counting roblem can be reduced to -grous. Formulas which give the number of subgrous of tye µ of a finite -grou of tye λ were established by Delsarte [4], Djubjuk [5] and Yeh [15]. An excellent survey on this subject together with connections to symmetric functions was written by Butler [2] in Another way to find the total number of subgrous of finite abelian -grous was described by Bhowmik [1] by using divisor functions of matrices. By invoking different arguments, formulas in the case of rank two -grous were obtained by Călugăreanu [3], Tărnăuceanu [11, 12], Hamejs, Holighaus, Tóth, Wiesmeyr [6], Tóth [14] and for rank three -grous by Hamejs, Tóth [7], Oh [8]. Note that the aers [6, 7, 14] include also direct formulas for the grous Z m Z n and Z m Z n Z r, resectively, where m, n, r N := {1, 2,... } are arbitrary. The urose of the current aer is to count the number of subgrous of a given exonent in a finite abelian -grou. Exlicit formulas are obtained for rank two and rank three -grous. The numbers of subgrous of exonent, resectively 2 in an arbitrary -grou are also considered. We rove that if two finite abelian grous have the same number of subgrous of any exonent, then they are isomorhic. We also deduce comact formulas for the number of subgrous of a given exonent of the grou Z m Z n, where m, n N are arbitrary. Furthermore, 2010 Mathematics Subject Classification: Primary 20K01; Secondary 20K27, 11N37. Key words and hrases: finite Abelian grou, subgrou, exonent. Communicated by Zoran Petrović. 121

2 122 TĂRNĂUCEANU AND TÓTH we obtain an exact formula for the sum of exonents of the subgrous of Z m Z n, and an asymtotic formula for the arithmetic means of exonents of the subgrous of Z n Z n. For the roofs we use two different aroaches. The first one is based on the known formula for the number of subgrous of a given tye in an abelian - grou, given in terms of gaussian coefficients (Theorem 2.1). The second method, alicable only for rank two grous, uses the reresentation of the subgrous of Z m Z n obtained by the second author in [14] (Theorem 3.1). Most of our notation is standard and usually will not be reeated here. For basic notions and results on grou theory we refer the reader to [10]. 2. First aroach Let G be a finite abelian grou of order n and G = G 1 G 2 G m be the rimary decomosition of G, where G i is a i -grou (i = 1, 2,..., m). For every divisor d = α1 1 α αm m of n we denote E d(g) = {H G : ex(h) = d}. Since the subgrous H of G are of tye H = H 1 H 2 H m with H i G i (i = 1, 2,..., m), we infer that m (2.1) E d (G) = E α i (G i ). i i=1 Equality (2.1) shows that the roblem of counting the number of subgrous of exonent d in G is reduced to -grous. So, in this section we will assume that G is a finite abelian -grou, that is a grou of tye Z λ 1 Z λ 2 Z λ k with λ 1 λ 2 λ k 1. In this case we will say that G is of tye λ, where λ is the artition (λ 1, λ 2,..., λ k, 0,... ), and we will denote it by G λ. We recall the following well known result, which gives the number of subgrous of tye µ of G λ (see [4, 5, 15]). Theorem 2.1. For every artition µ λ (i.e., µ i λ i for every i N ) the number of subgrous of tye µ in G λ is α λ (µ; ) = ( ) ai b (ai bi)bi+1 i+1, b i b i+1 i 1 where λ = (a 1, a 2,..., a λ1, 0,... ), µ = (b 1, b 2,..., b µ1, 0,... ) are the artitions conjugate to λ and µ, resectively, and ( ) n n i=1 = (i 1) k k i=1 (i 1) n k i=1 (i 1) is the gaussian binomial coefficient (it is understood that m i=1 (i 1) = 1 for m = 0). By using Theorem 2.1 a way to comute the number of subgrous of exonent i in G λ can be inferred, namely E i(g λ ) = α λ (µ; ) (i = 0, 1,..., λ 1 ), µ λ, µ 1=i

3 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT which is a olynomial in with integer coefficients. If i 1 and k 2 are arbitrary, then the olynomial E i(g λ ) can not be given exlicitly, but we will do this in some articular cases. Namely, we will consider the following cases: k {2, 3} and i 1 arbitrary, i {1, 2} and k 2 arbitrary. We remark first that since H E i (G λ ) H G λ and H E i (G λ ) H is of tye Z i Z i Z λr Z λ k, where r = min{j : λ j < i}, we have E i(g λ ) = E i(z i Z i Z λr Z λ k ) with the convention that the direct roduct Z λr Z λ k is trivial for i λ k. Our first result gives a recise exression for E i(g λ ) in the case k = 2. Proosition 2.1. Let G λ = Z λ 1 Z λ 2 with λ 1 λ 2 1. Then { i+1 + i 2 1, 1 i λ 2, E i(g λ ) = λ , λ 2 < i λ 1. Proof. We have λ = (λ 1, λ 2, 0,... ) and consequently λ = (2, 2,..., 2, 1, 1,..., 1, 0,... ), where the number of 2 s is λ 2 and the number of 1 s is λ 1 λ 2. Then i E i(g λ ) = N i,j, where N i,j denotes the number of subgrous of tye (i, j) in G λ, or equivalently in Z i Z i if i λ 2 or in Z i Z λ 2 if λ 2 < i λ 1. In the first case we obtain N i,j = ( + 1) i j 1 for j = 0, 1,..., i 1, and N i,i = 1. Therefore j=0 i 1 E i(z i Z i) = 1 + ( + 1) i j 1 = i+1 + i 2. 1 j=0 In the second case we obtain N i,j = λ2 j for j = 0, 1,..., λ 2. Therefore This comletes the roof. λ 2 E i(z i Z λ 2 ) = λ2 j = λ j=0 Examle 2.1. We have 1, i = 0, + 2, i = 1, E i(z 4 Z 2) = , i = 2, , i = 3 or i = 4. In articular, by summing all quantities E i(g λ ) (i = 0, 1,..., λ 1 ), we obtain the total number of subgrous of G λ (see also [11, Pro. 2.9], [12, Th. 3.3]).

4 124 TĂRNĂUCEANU AND TÓTH Corollary 2.1. The total number of subgrous of G λ = Z λ 1 Z λ 2 where λ 1 λ 2 1, is (2.2) 1 ( 1) 2 [ (λ1 λ 2 +1) λ2+2 (λ 1 λ 2 1) λ2+1 (λ 1 +λ 2 +3)+(λ 1 +λ 2 + 1) ]. Examle 2.2. The total number of subgrous of Z 4 Z 2 is Now consider the case of rank three -grous. We need the following lemma. Lemma 2.1. Let N i,j,l denote the number of subgrous of tye (i, j, l) (i j l 0) in G λ = Z λ 1 Z λ 2 Z λ 3 with λ 1 λ 2 λ 3 1. Then 2i 2l 3 ( + 1)( ), l < j < i λ 3, 2(i l 1) ( ), l < j = i λ 3 or l = j < i λ 3, 1, l = j = i λ 3, λ3+i 2l 2 ( + 1) 2, l < j λ 3 < i λ 2, λ3+i 2j 1 ( + 1), l = j λ 3 < i λ 2, N i,j,l = 2λ3+i j 2l 1 ( + 1), l λ 3 < j < i λ 2, 2(λ3 l), l λ 3 < j = i λ 2, λ2+2λ3 j 2l, l λ 3 < j λ 2 < i λ 1, λ2+λ3 2l 1 ( + 1), l < j λ 3 λ 2 < i λ 1, λ2+λ3 2l, l = j λ 3 λ 2 < i λ 1. Proof. We use Theorem 2.1. We distinguish the following three cases: I. If i λ 3, then N i,j,l is the number of subgrous of tye (i, j, l) in Z i Z i Z i. Here we need to consider λ = (i, i, i, 0,...) with λ = (3, 3,..., 3, 0... ), where the number of 3 s is i, and µ = (i, j, l, 0,... ) with µ = (3, 3,..., 3, 2, 2,..., 2, 1, 1,..., 1, 0,... ), where the number of 3 s is l, the number of 2 s is j l, the number of 1 s is i j. In the subcase l < j < i λ 3 we deduce N i,j,l = ( 2 ) j l 1 ( + 1)( 2 ) i j 1 ( ) = 2i 2l 3 ( + 1)( ). The subcases l < j = i λ 3, l = j < i λ 3 and l = j = i λ 3 are treated similar. II. If λ 3 < i λ 2, then N i,j,l is the number of subgrous of tye (i, j, l) in Z i Z i Z λ 3. We consider λ = (i, i, λ 3, 0,... ) with λ = (3, 3,..., 3, 2, 2,..., 2, 0... ) where the number of 3 s is λ 3, the number of 2 s is i λ 3, and µ, µ like in the case I. For examle, in the subcase l = j λ 3 < i λ 2 we obtain N i,j,l = ( 2 ) λ3 j i λ3 1 ( + 1) = λ3+i 2j 1 ( + 1). III. If λ 2 < i λ 1, then N i,j,l is the number of subgrous of tye (i, j, l) in Z i Z λ 2 Z λ 3. We consider λ = (i, λ 2, λ 3, 0,... ) with λ = (3, 3,..., 3, 2, 2,..., 2, 1, 1,..., 1, 0... ),

5 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT where the number of 3 s is λ 3, the number of 2 s is λ 2 λ 3, the number of 1 s is i λ 2, and µ, µ like in the case I. Proosition 2.2. Let G λ = Z λ 1 Z λ 2 Z λ 3 with λ 1 λ 2 λ 3 1. Then 2i 1 ((i+1) 4 +(i 1) 3 2 (i+2) i)+3 ( 2 1)( 1), 1 i λ 3, (λ E i(g λ ) = 3+1) λ 3 +i ( 2 1)(+1) 2 2λ ( 2 1)( 1), λ 3 < i λ 2, (λ 3+1) λ 2 +λ 3 +1 ( 2 1) 2λ ( 2 1)( 1), λ 2 < i λ 1. Proof. We have E i(g λ ) = 0 l j i N i,j,l and use Lemma 2.1. Case I. If i λ 3, then i 1 E i(g λ ) = N i,i,i + i 1 = 1 + i 1 j=l+1 i 1 j=l+1 i 1 i 1 N i,j,l + N i,i,l + N i,l,l i 1 2i 2l 3 ( + 1)( ) + 2 2(i l 1) ( ) = ( 2 1)( 1) ((i 1)2i+1 i 2i 1 + ) + 2( ) 2i = 2i 1 ((i + 1) 4 + (i 1) 3 2 (i + 2) i) + 3 ( 2, 1)( 1) by direct comutations. II. If λ 3 < i λ 2, then we have E i(g λ ) = = λ 3 λ 3 j=l+1 λ 3 λ 3 j=l+1 λ 3 N i,j,l + i 1 j=λ 3+1 λ 3 λ3+i 2l 2 ( + 1) 2 + λ 3 λ 3 N i,j,l + N i,i,l + i 1 j=λ 3+1 λ 3 λ 3 + 2(λ3 l) + λ3+i 2l 1 ( + 1) = λ 3 i 2 ( + 1) 2 + λ3 + i + i 1 λ3 ( 2 ( 2λ3+2 1) 1) N i,l,l 2λ3+i j 2l 1 ( + 1) i 2 ( 2λ3+2 (λ 3 + 1) 2 + λ 3 ) λ3 ( 1) 2 + (2λ3+2 1)( i λ3 1 1) ( 1) 2, which gives the above formula. III. Finally, if λ 2 < i λ 1, then E i(g λ ) = λ 3 λ 2 j=λ 3+1 λ 3 1 N i,j,l + λ 3 j=l+1 λ 3 N i,j,l + N i,l,l

6 126 TĂRNĂUCEANU AND TÓTH = λ 3 λ 2 j=λ 3+1 λ 3 1 λ2+2λ3 j 2l + = (2λ3+2 1)( λ2 λ3 1) ( 2 1)( 1) + λ2 λ3 2λ , 1 leading to the given result. λ 3 j=l+1 + λ 3 λ2+λ3 2l 1 ( + 1) + λ2+λ3 2l λ2 λ3+1 ( 2 1)( 1) λ 3 2λ3+2 (λ 3 + 1) 2λ3 + 1) Note that Proosition 2.2 is valid also in the case λ 3 = 0, when it reduces to Proosition 2.1. Examle 2.3. We have 1, i = 0, , i = 1, E i(z 4 Z 2 Z 2) = , i = 2, , i = 3 or i = 4. Corollary 2.2. The total number of subgrous of G λ = Z λ 1 Z λ 2 Z λ 3, where λ 1 λ 2 λ 3 1, is A ( 2 1) 2 ( 1), where A = (λ 3 + 1)(λ 1 λ 2 + 1) λ2+λ (λ 3 + 1) λ2+λ3+4 2(λ 3 + 1)(λ 1 λ 2 ) λ2+λ3+3 2(λ 3 + 1) λ2+λ3+2 + (λ 3 + 1)(λ 1 λ 2 1) λ2+λ3+1 (λ 1 + λ 2 λ 3 + 3) 2λ λ3+3 + (λ 1 + λ 2 λ 3 1) 2λ3+2 + (λ 1 + λ 2 + λ 3 + 5) (λ 1 + λ 2 + λ 3 + 1). Proof. The total number of subgrous of G λ is λ 1 i=0 E i(g λ ) = E i(g λ ) + E i(g λ ) + E i(g λ ) 0 i λ 3 λ 3<i λ 2 λ 2<i λ 1 and summing the quantities given in Proosition 2.2 we deduce the result. We remark that an equivalent formula to that given in Corollary 2.2 was obtained in [8, Cor. 2.2] by using different arguments. Corollary 2.2 is valid also in the case λ 3 = 0, when it reduces to Corollary 2.1. Examle 2.4. The total number of subgrous of Z 4 Z 2 Z 2 is

7 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT Next consider the number of subgrous of exonent (that is, the number of elementary abelian subgrous) in G λ, which equals the total number of nontrivial subgrous of (Z ) k. Since (Z ) k is a k dimensional linear sace overz, the number in question is exactly the total number of nonzero subsaces, which is, as wellknown, k ( k ) i=1 i (Galois number). So we have the following result. Proosition 2.3. Let G λ = Z λ 1 Z λ 2 Z λ k λ k 1. Then E (G λ ) = k ( k ) r=1 r. We give a direct roof of this formula based on Theorem 2.1. with λ 1 λ 2 Proof. We use Theorem 2.1 in the case λ = (1, 1,..., 1, 0,... ), where the number of 1 s is k, and µ = (1, 1,..., 1, 0,... ), where the number of 1 s is r with 1 r k. Here λ = (k, 0, 0,... ), µ = (r, 0, 0,... ) and obtain that the number of subgrous of tye µ is ( ) k r, while the number of subgrous of exonent 1 is exactly k ( k r=1 r). Examle 2.5. We have for any λ 1 λ 2 λ 3 λ 4 1, E (Z λ 1 Z λ 2 Z λ 3 Z λ 4 ) = Concerning the number of subgrous of exonent 2 in G λ, we have the following formula. Proosition 2.4. Let G λ = Z λ 1 Z λ 2 Z λ k with λ 1 λ 2 λ t > λ t+1 = = λ k = 1, where 0 t k is fixed (t = 0 if each λ j is 1 and t = k if each λ j is 2). Then E 2(G λ ) = ( ) ( ) k r t r(k r s) s r 1 r t 0 s k r which is zero (emty sum) for t = 0. Proof. Let t 1. Use Theorem 2.1 for λ = (2, 2,..., 2, 1, 1,..., 1, 0,... ), where the number of 2 s is t and the number of 1 s is k t and µ = (2, 2,..., 2, 1, 1,..., 1, 0,... ), where the number of 2 s is r and the number of 1 s is s with 1 r t, 0 s k r. Now λ = (k, t, 0,... ), µ = (r+s, r, 0,... ) and obtain that the number of subgrous of tye µ is r(k r s)( ) k r t ) s ( r and the number of subgrous of exonent 2 is deduced by summing over r and s. Examle 2.6. (k = 4, t = 2) We have E 2(Z 4 Z 2 Z Z ) = In what follows, let G λ = Z λ 1 Z λ 2 Z λ k and G κ = Z κ 1 Z κ 2 Z κ l be two finite abelian -grous, where λ 1 λ 2 λ k 1 and κ 1 κ 2 κ l 1. Assume that G λ and G κ have the same number of subgrous of exonent i for every i, i.e., E i(g λ ) = E i(g κ ) (i 0). Then,

8 128 TĂRNĂUCEANU AND TÓTH λ 1 = κ 1. On the other hand, since the function f : N N, f(k) = k ( k ) i=1 i is one to one, by Proosition 2.3 we infer that k = l. Clearly, if k = 1 one obtains G λ = Gκ. The same thing can be also said for k = 2 and k = 3 by Proositions 2.1 and 2.2. Insired by these remarks, we state and rove the following result. Proosition 2.5. Two finite abelian -grous G λ and G κ are isomorhic if and only if they have the same number of subgrous of exonent i, for every i 0. Proof. Let λ = (λ 1, λ 2,..., λ k, 0,... ) and κ = (κ 1, κ 2,..., κ l, 0,... ), where λ k, κ l > 0, be artitions such that for -grous G λ and G κ one has E i(g λ ) = E i(g κ ), for every i 0. As noted before, λ 1 = κ 1 and k = l hold true. Let us define λ k+1 = κ k+1 = 0. We rove, by reverse induction on t k + 1 that λ t = κ t. So, let us assume that λ i = κ i for all k + 1 i t + 1, and rove that λ t = κ t. Suose to the contrary that w.l.o.g. λ t > κ t. Since λ 1 = κ 1, one has t > 1. Consider the artitions λ λt and κ λt defined with λ λt = (λ t,..., λ }{{} t, γ t+1,..., γ k, 0,... ), t κ λt = (λ t,..., λ }{{} t, κ s+1,..., κ t, γ t+1,..., γ k, 0,... ), s where γ i = λ i = κ i, for i t + 1, and s = max{j : κ j λ t }. Note that by our assumtion s < t, and since κ 1 = λ 1 λ t also s 1. From the definition of numbers α ω (µ; ), it is clear that for any three artitions µ, σ, τ, which satisfy µ σ τ, we have α σ (µ; ) α τ (µ; ). Using this remark, the fact that κ λt λ λt and α λ λ t (λ λt ; ) = 1, one has 1 + E λ t (G κ ) = 1 + E λ t (G κ λ t ) = 1 + α κ λt (µ; ) µ κ λ t, µ 1=λ t α λ λ t (λ λt ; ) + α λ λt (µ; ) α λ λt (µ; ) µ κ λ t, µ 1=λ t µ λ λ t, µ 1=λ t a contradiction. = E λ t (G λ λ t ) = E λ t (G λ ), Corollary 2.3. Two arbitrary finite abelian grous are isomorhic if and only if they have the same number of subgrous of any exonent. Finally, we note that another interesting roblem is to find the olynomial E i(g λ ) in the case k = Second aroach We need the following result giving the reresentation of subgrous of the grou Z m Z n. For every m, n N let { J m,n := (a, b, c, d, l) (N ) 5 : a m, b a, c n, d c, a b = c d,

9 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT l a ( b, gcd l, a ) } = 1. b Note that here gcd(b, d) lcm(a, c) = ad and gcd(b, d) lcm(a, c). For (a, b, c, d, l) J m,n define {( K a,b,c,d,l := i m a, iln c + j n ) } : 0 i a 1, 0 j d 1. d Theorem 3.1. [14, Th. 3.1] Let m, n N. i) The ma (a, b, c, d, l) K a,b,c,d,l is a bijection between the set J m,n and the set of subgrous of (Z m Z n, +). ii) The invariant factor decomosition of the subgrou K a,b,c,d,l is K a,b,c,d,l Z gcd(b,d) Z lcm(a,c). iii) The order of the subgrou K a,b,c,d,l is ad and its exonent is lcm(a, c). Let s E (m, n) stand for the number of subgrous of exonent E of the grou Z m Z n. Proosition 3.1. For every m, n N, E lcm(m, n) we have (3.1) s E (m, n) = gcd(i, j) (3.2) i m,j n lcm(i,j)=e = 1 E Proof. According to Theorem 3.1, s E (m, n) = a m c n b a d c i m,j n lcm(i,j)=e ij. a/b=c/d=e lcm(a,c)=e φ(e), where φ is Euler s totient function. This can be written (with m = ax, a = by, n = cz, c = dt) as s E (m, n) = φ(e) = φ(e) = bxe=m dze=n e lcm(b,d)=e i m j n lcm(i,j)=e e gcd(i,j) ix=m jz=n φ(e) = be=i de=j e lcm(b,d)=e i m j n lcm(i,j)=e which is (3.1). Formula (3.2) is its immediate consequence. gcd(i, j), Remark 3.1. Proosition 2.1 is a direct consequence of the above result. The total number s(m, n) of subgrous of the grou Z m Z n is (see [6, Th. 3], [14, Th. 4.1]) (3.3) s(m, n) = gcd(i, j) i m,j n

10 130 TĂRNĂUCEANU AND TÓTH and (3.1) shows the distribution of the number of subgrous according to their exonents. Formula (2.2) can be obtained also by using (3.3). Examle 3.1. The total number of subgrous ofz 12 Z 18 is s(12, 18) = 80 and we have s 1 (12, 18) = 1, s 2 (12, 18) = 4, s 3 (12, 18) = 5, s 4 (12, 18) = 3, s 6 (12, 18) = 20, s 9 (12, 18) = 4, s 12 (12, 18) = 15, s 18 (12, 18) = 16, s 36 (12, 18) = 12. Corollary 3.1 (m = n). For every n N and E n, (3.4) s E (n, n) = gcd(i, j), i E,j E gcd(e/i,e/j)=1 which equals the number of cyclic subgrous of the grou Z E Z E. The fact that s E (n, n) equals the number of cyclic subgrous of the grou Z E Z E, but without deriving formula (3.4) is [6, Th. 8], roved by using different arguments. Proof. In the case m = n, for every E n we have by (3.1), s E (n, n) = gcd(i, j) = gcd(i, j) = i n,j n lcm(i,j)=e ia=e,jb=e lcm(e/a,e/b)=e ia=e,jb=e gcd(a,b)=1 gcd(i, j), giving (3.4), which equals the number of cyclic subgrous of the grou Z E Z E by [6, eq. (16)]. Proosition 3.2. For every m, n N the sum of exonents of the subgrous of Z m Z n is σ(m)σ(n), where σ(k) = d k d. Proof. By (3.2) the sum of exonents of the subgrous of Z m Z n is Es E (m, n) = E 1 ij = ij E E lcm(m,n) E lcm(m,n) = i m i j n i m,j n lcm(i,j)=e j = σ(m)σ(n). i m,j n lcm(i,j) lcm(m,n) By Proosition 3.2, the arithmetic mean of exonents of the subgrous of Z m Z n is σ(m)σ(n)/s(m, n), where s(m, n) is given by (3.3). Now consider the case m = n. Let A(n) stand for the arithmetic mean of exonents of the subgrous of Z n Z n. We have (3.5) A(n) = σ(n)2 s(n), where s(n) := s(n, n). Recall that a function f : N C is said to be multilicative if f(nn ) = f(n)f(n ) whenever gcd(n, n ) = 1. It is well known that the sum-of-divisors function σ is multilicative. The function s(n) = i,j n gcd(i, j) is also multilicative,

11 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT as shown by the following direct roof: Let gcd(n, n ) = 1. Then s(nn ) = gcd(i, j) = gcd(aa, bb ) i,j nn = a,b n gcd(a, b) a,b n a,b n a,b n gcd(a, b ) = s(n)s(n ). We conclude that the function A given by (3.5) is multilicative and and for every rime ower ν, A( ν ) = ( ν+1 1) 2 ν+2 + ν+1 (2ν + 3) + 2ν + 1, cf. Corollary 2.1. Since the exonent of every subgrou of Z n Z n is a divisor of n, whence n, we deduce that A(n) n (n N ). For the function n f(n) := A(n)/n (0, 1] the series taken over the rimes 1 f() = 1 2 ( + 3) is convergent, and it follows from a theorem of H. Delange (see, e.g., [9]) that the function f has a nonzero mean value M given by 1 M := lim f(n) = x x n x = ( 1 1 ) f( ν ) ν ν=0 ) ( 1 1 ν=0 the roducts being over the rimes. We rove the following more exact result. Proosition 3.3. We have (3.6) A(n) = M 2 x2 + O(x log 3 x). n x ( ν+1 1) 2 2ν ( ν+2 + ν+1 (2ν + 3) + 2ν + 1). Proof. Let f(n) = d n g(d) (n N ), that is g = µ f in terms of the Dirichlet convolution, where µ is the Möbius function. Here g( ν ) = f( ν ) f( ν 1 ) for every rime ower ν (ν N ). Note that A( k ) = ( k i=0 i ) 2 k i=0 (2i + 1)k i (k 0), so if we denote S = ν 1 i=0 i and T = ν 1 i=0 (2i + 1)ν 1 i, we have g( ν ) = (S + 1) 2 ν (T + 2ν + 1) S2 ν 1 T = 2ST + T S2 (2ν + 1) ν T (T + 2ν + 1).

12 132 TĂRNĂUCEANU AND TÓTH Since S T and T (2ν 1)S, we have g( ν ) < max{2st + T S2 (2ν 1), S 2 (2ν + 1) 2ST } ν T (T + 2ν + 1) 1 { } ST + T ν max T (T + 1), S2 (2ν 1) T 2 1 2ν 1 max{1, 2ν 1} = ν ν, valid for every rime ower ν (ν N ). Hence, g(n) τ(n 2 )/n for every n 1, where τ(k) stands for the number of ositive divisors of k. We deduce that g(d) g(d)(x/d + O(1)) f(n) = g(d) = 1 = n x de x d x e x/d d x ( g(d) = x d + O x ) ( g(d) + O d d=1 d>x d x ( = Mx + O x τ(d 2 ) ( ) τ(d 2 ) d 2 + O d d>x d x ) g(d) where in the main term the coefficient of x is M by Euler s roduct formula. It is known that n x τ(n2 ) = cx log 2 x + O(x log x) with a certain constant c and artial summation shows that n>x τ(n2 )/n 2 = O((log 2 x)/x), n x τ(n2 )/n = O(log 3 x). Therefore, (3.7) f(n) = Mx + O(log 3 x). n x Now (3.6) follows from (3.7) by artial summation. Formula (3.6) and its roof are similar to those of [13, Th ]. Acknowledgements. The authors thank the referee for very careful reading of the manuscrit, many useful comments and for the roof of Proosition 2.5. References 1. G. Bhowmik, Evaluation of divisor functions of matrices, Acta Arith. 74 (1996), M. L. Butler, Subgrou Lattices and Symmetric Functions, Mem. Am. Math. Soc. 112(539), G. Călugăreanu, The total number of subgrous of a finite abelian grou, Sci. Math. Jn. 60(1) (2004), S. Delsarte, Fonctions de Möbius sur les groues abeliens finis, Annals of Math. 49 (1948), P. E. Djubjuk, On the number of subgrous of a finite abelian grou, Izv. Akad. Nauk SSSR, Ser. Mat. 12 (1948), M. Hamejs, N. Holighaus, L. Tóth, C. Wiesmeyr, Reresenting and counting the subgrous of the grou Z m Z n, J. Numbers 2014, Article ID M. Hamejs, L. Tóth, On the subgrous of finite abelian grous of rank three, Ann. Univ. Sci. Buda. Rolando Eötvös, Sect. Comut. 39 (2013), ),

13 NUMBER OF SUBGROUPS OF A GIVEN EXPONENT J.-M. Oh, An exlicit formula for the number of subgrous of a finite abelian -grou u to rank 3, Commun. Korean Math. Soc. 28 (2013), A. G. Postnikov, Introduction to Analytic Number Theory, Transl. Math. Monogr. 68, American Mathematical Society, Providence, RI, M. Suzuki, Grou Theory, vol. I, II, Sringer-Verlag, Berlin, 1982, M. Tărnăuceanu, A new method of roving some classical theorems of abelian grous, Southeast Asian Bull. Math. 31(6) (2007), , An arithmetic method of counting the subgrous of a finite abelian grou, Bull. Math. Soc. Sci. Math. Roum., Nouv. Sér. 53(101) (2010), M. Tărnăuceanu, L. Tóth, Cyclicity degrees of finite grous, Acta Math. Hung. 145 (2015), L. Tóth, Subgrous of finite abelian grous having rank two via Goursat s lemma, Tatra Mt. Math. Publ. 59 (2014), Y. Yeh, On rime ower abelian grous, Bull. Am. Math. Soc. 54 (1948), Faculty of Mathematics (Received ) Al. I. Cuza University (Revised ) Iaşi Romania tarnauc@uaic.ro Deartment of Mathematics University of Pécs Pécs Hungary ltoth@gamma.ttk.te.hu