ON THE SUBGROUP LATTICE OF AN ABELIAN FINITE GROUP

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1 ON THE SUBGROUP LATTICE OF AN ABELIAN FINITE GROUP Marius Tărnăuceanu Faculty of Mathematics Al.I. Cuza University of Iaşi, Romania The aim of this paper is to give some connections between the structure of an abelian finite group and the structure of its subgroup lattice. 1 Preliminaries Let (G, +) be an abelian group. Then the set L(G) of subgroups of G is a modular and complete lattice. Moreover, we suppose that G is finite of order n. If L n is the divisors lattice of n, then the following function is well defined: ord : L(G) L n, ord(h) = H, for any H L(G), where by H we denote the order of the subgroup H. Main results Proposition 1. The following conditions are equivalent: (i) G is a cyclic group. (ii) ord is an one-to-one function. 1

2 (iii) ord is a homomorphism of the semilattice (L(G), ) into the semilattice (L n, (, )). (iv) ord is a homomorphism of the semilattice (L(G), +) into the semilattice (L n, [, ]). (v) ord is an isomorphism of lattices. Proof. (i) = (ii) Obvious. (ii) = (i) For any d L n let M d be the set of elements x G having the order d. Then the family {M d d L n } is a partition of G, therefore we have: (1) n = d L n M d. A set M d is nonempty if and only if there exists a cyclic subgroup H d of G having the order d. In this situation H d is the unique subgroup of G with the order d and we have: M d = {x G < x >= H d }. It results that M d = ϕ(d), where ϕ is the Euler function. Using the relation (1) and the identity n = d L n ϕ(d), we obtain that M d = ϕ(d), for any d L n. For d = n we have M d = ϕ(n) 1, so that G contains an element of order n. (i) = (iii) Let G 1, G be two subgroups of G, d i = G i, i = 1, and d = G 1 G. Since G 1 G is a subgroup of G i, i = 1,, we obtain that d/d i, i = 1,. If d is a divisor of d 1 and d, then d L n, so that there exists G L(G) with G = d. We have G G i, i = 1,, therefore G G 1 G. It results that d /d, thus d = (d 1, d ). (iii) = (ii) Let G 1, G be two subgroups of G such that ord(g 1 ) = ord(g ). From (iii) we obtain that ord(g 1 G ) = ord(g i ), i = 1,, therefore we have G 1 = G 1 G = G. (i) = (iv) Similarly with (i) = (iii). (iv) = (ii) Similarly with (iii) = (ii).

3 (i) = (v) Obvious. Next aim is to find necessary and sufficient conditions for L(G) in order to be a distributive lattice, respectively a complemented lattice in which every element has a unique complement. Lemma 1. If L(G) is a distributive lattice or a complemented lattice in which every element has a unique complement, then, for any H L(G), the lattice L(H) has the same properties. Proof. The first part of the assertion is obvious. We suppose that L(G) is a complemented lattice in which every element has a unique complement and let H 1 be a subgroup of H. Then H 1 L(G), thus there exists a unique subgroup H 1 L(G) such that H 1 H 1 = G. It results that H = G H = H 1 (H 1 H). If H 1 L(H) satisfies H 1 H 1 = H and K is the complement of H in G, then we have: and G = K H = (K H 1 ) (H 1 H) G = K H = (K H 1 ) H 1. Since the subgroup K H 1 has a unique complement in G, it follows that H 1 = H 1 H; hence, H 1 has a unique complement in H. Proposition. The following conditions are equivalent: (i) G is a cyclic group. (ii) L(G) is a distributive lattice. Proof. (i) = (ii) If G is a cyclic group, then, from Proposition 1, we have L(G) L n, thus L(G) is a distributive lattice. (ii) = (i) From the fundamental theorem on finitely generated abelian groups there exist (uniquely determined by G) the numbers d 1, d,..., d k IN\{0, 1} satisfying d 1 /d /.../d k and G k Z di. 3

4 We shall prove that k = 1. If we suppose that k, then let p be a prime divisor of d 1. Since d 1 /d, we obtain that G has a subgroup isomorphic to the group Z p Z p. Using Lemma 1, it is sufficiently to verify that L(Z p Z p ) is not a distributive lattice. If p =, then L(Z p Z p ) = L(Z Z ) M 3, which is not a distributive lattice. If p 3, then let H 1, H, H 3 be the subgroups of Z p Z p generated by the elements (ˆ1, ˆ), (ˆ1, ˆ0), respectively (ˆ0, ˆ1). It is easy to see that we have: H 1 H = H 1 H 3 = {(ˆ0, ˆ0)} H 1 + H = H 1 + H 3 = Z p Z p H = H 3, therefore L(Z p Z p ) is not a distributive lattice. Hence k = 1, i.e. G Z d1 is a cyclic group. Proposition 3. The following conditions are equivalent: (i) n = G is square free. (ii) L(G) is a complemented lattice in which every element has a unique complement. Proof. (i) = (ii) If n is square free, then L(G) L(Z n ) L n. Since L n is a complemented lattice in which every element has a unique complement (for any d L n there exists a unique element d L n, d = n d, such that (d, d) = 1 and [d, d] = n), L(G) has the same property. (ii) = (i) Let G 1 be the sum of all subgroups H L(G) which are simple groups. Then there exists a unique subgroup G 1 L(G) such that G 1 G 1 = G. We shall prove that G 1 = {0}. If we suppose that G 1 {0}, then there exists x G 1 \{0}. Using the Zorn s lemma, we obtain a maximal subgroup G of G 1 with property that x / G. From Lemma 1, there exists a unique subgroup G L(G 1 ) such that G G = G 1. It follows that G {0}. Let G 3 {0} be a subgroup of G. Then the inclusion G G G 3 implies that x G G 3. From the equality G 1 = G G 3 = G G it results that G 3 = G, thus G is a simple group. We obtain {0} G G 1 G 1, contrary to the fact that the sum G 1 + G 1 is direct. 4

5 Hence G 1 = {0}, therefore G = G 1. Since G is finite, there exist H 1, H,..., H k L(G) such that H i is a simple group, for any i = 1, k and k G = H i. For each i {1,,..., k} let p i be a prime number with the property: H i Z pi. Using the fact that p i p j for i = j, we obtain k k G Z pi Z pi Z p1 p...p k, i.e. n = G is square free. Let Ab, Lat be the categories of abelian groups, respectively of lattices. We have a functor L : Ab Lat given by: a) for an abelian group G, L(G) is the lattice of subgroups of G; (b) for a homomorphism of groups f : G 1 G, L(f) : L(G 1 ) L(G ) is the homomorphism of lattices defined by L(f)(H 1 ) = f(h 1 ) for any H 1 L(G 1 ). Remark. L is an exact functor. Proposition 4. If f : G 1 G is an epimorphism of abelian groups and then: (i) L is a sublattice of L(G 1 ); L = {H 1 L(G 1 )/ ker f H 1 }, (ii) the function f : L L(G ), f(h 1 ) = L(f)(H 1 ) = f(h 1 ) for any H 1 L, is an isomorphism of lattices. Proof. (i) Obvious. (ii) It remains to prove only that f is one-to-one and onto. Let H 1, H 1 be two elements of L such that f(h 1 ) = f(h 1), i.e. f(h 1 ) = f(h 1). For any x H 1, we have f(x) f(h 1 ) = f(h 1), so that there exists y H 1 with f(x) = f(y). It results that f(x y) = 0, thus x y ker f. Since ker f H 1, we obtain x H 1; hence H 1 H 1. In the same way we can check the other inclusion; therefore f is one to one. 5

6 Let H be a subgroup of G. Then H 1 = f 1 (H ) L and, using the fact that f is onto, we obtain: therefore f is onto. f(h 1 ) = f(h 1 ) = (f f 1 )(H ) = H ; Remark. The functor L reflects the isomorphisms, i.e. if f : G 1 G is a homomorphism of abelian groups such that L(f) : L(G 1 ) L(G ) is an isomorphism of lattices, then f is an isomorphism of groups. Next aim is to study when, for two abelian groups G, G of the same order n, the isomorphism of lattice L(G) L(G ) implies the isomorphism of groups G G. In order to solve this problem, it is necessary to minutely study the k structure of the lattice L Z di, where di IN\{0, 1}, i = 1, k and d 1 /d /.../d k. We shall treat only the case k =, in the case k 3 the problem remaining open. Let π : Z Z Z d1 Z d be the function defined by π(x, y) = ( x, ŷ), for any (x, y) Z Z. π is an epimorphism of groups, therefore, by Proposition 4, we have the isomorphism of lattices: L (Z d1 Z d ) L d1,d, where L d1,d = {H L(Z Z ) ker π = d 1 Z d Z H}. It is a simple exercise to verify that: Lemma. L(Z Z ) = {H p,q,r = (p, q)z + (0, r)z p, q, r IN, q < r}. From the above results, we obtain that L(Z d1 Z d ) L d1,d = { ( = H p,q,r = ((p, q)z + (0, r)z p, q, r IN, q < r, p/d 1, r/ d, d )} 1q. p 6

7 Remark. 1) For two elements H p,q,r, H p,q,r L d 1,d the following conditions are equivalent: (i) H p,q,r = H p,q,r. (ii) p = p, r/(q q, r ), r /(q q, r). (iii) (p, q, r) = (p, q, r ). ) For two elements H p,q,r, H p,q,r L d 1,d the following conditions are equivalent: (i) H p,q,r H p,q,r (r,. q q pp ) (ii) p /p, r /. Let A n be the set {(d 1, d ) (IN\{0, 1}) d 1 /d, d 1 d = n} and g : A n IN be the function defined by g(d 1, d ) = card L(Z d1 Z d ) for any (d 1, d ) A n. If d 1 = p α 1 1 p α...p αs s, d = p β 1 1 p β...p β s s are the decompositions of d 1, respectively d, as a product of prime factors (i.e. p 1 = prime number, α i, β i IN, α i β i, i = 1, s and p i p j for i j), then we have the following result: Lemma 3. g(d 1, d ) = s 1 (p i 1) [(β i α i + 1)p α i+ i (β i α i 1)p α i+1 i (α i + β i + 3)p i + (α i + β i + 1)]. Corollary. We have: (i) card L(Z Z 4 ) = 8. (i) card L(Z Z ) = 5. (i) card L(Z p Z p ) = p + 3, where p is a prime number. 7

8 Now we can prove the main result of this paper: Proposition 5. Let n be a natural number, s be the number of distinct prime divisors of n and G, G be two abelian groups of order n which have (corresponding to the fundamental theorem on finitely generated abelian groups) the decompositions: respectively G Z d1 Z d, G Z d 1 Z d. Then, for s {1, }, the isomorphism of lattice L(G) L(G ) implies the isomorphism of groups G G. Proof. If p 1, p,..., p s are the distinct prime divisors of n and n = p h 1 1 p h...p h s s, d 1 = p α 1 1 p α...p αs s, d = p β 1 1 p β...p β s s, d 1 = p α 1 1 p α...p α s s, d = p β 1 1 p β...p β s s are the decompositions of n, d 1, d, d 1, d as a product of prime factors (where h i, α i, β i, α i, β i IN, α i β i, α i β i, α i + β i = α i + β i = h i, i = 1, s), then we have: (1) g(d 1, d ) = g(d 1, d ). Since L(Z d1 Z d ) L(Z d 1 Z d ), there exists an isomorphism of lattice f : L d1,d L d 1,d. If we consider: (i) p 0 = 1, q 0 = 0, r 0 = d, then we have H 1,0,d H 1,0,d for any d IN, d/d, therefore H p 0,q 0,r 0 not ==f(h 1,0,d ) f(h 1,0,d ) ==H not p d,q d for any d IN, d/d ;,r d it results r d/r 0 for any d IN, d/d, thus the number of divisors of d is at most the number of divisors of r 0, so that (because r 0/d ) at most the number of divisors of d ; (ii) p 0 = d 1, q 0 = 0, r 0 = 1, then we have H d1,0,1 H e,0,1 for any e IN, e/d 1, therefore H p 0,q 0,r 0 not ==f(h d1,0,1) f(h e,0,1 ) ==H not p e,q e,r e for any e IN, e/d 1; 8

9 it results p e/p 0 for any e IN, e/d 1, thus the number of divisors of d 1 is at most the number of divisors of p 0, so that (because p 0/d 1) at most the number of divisors of d 1. Starting by the isomorphism of lattice f 1 : L d 1,d L d 1,d and making a similarly reasoning, we obtain the converses of the above inequalities. Thus, for i {1, }, the number of divisors of d i is equal with the number of divisors of d i, i.e. the following equalities hold: () s s (α i + 1) = (α i + 1) s s (β i + 1) = (β i + 1). If s = 1, then, from equalities (), we obtain α 1 = α 1 and β 1 = β 1, thus: G Z d1 Z d = Z d 1 Z d G. If s =, then, from the first of the equalities (), we obtain: α = u(α 1 + 1), α + 1 = 1 u (α + 1), where u Q. If we consider the function F : IR + IR + defined by F (x) = [(h (α 1 + 1)x)p (α 1+1)x+1 1 (h (α 1 + 1)x)p (α 1+1)x [( (h 1 + 3)p 1 + (h 1 + 1)] h + 3 α ) + 1 p α +1 x x ( h + 1 α ) ] + 1 p α +1 x (h + 3)p + (h + 1) x for any x IR +, then the equality (1) becomes: (3) F (u) = F (1). 1 We can suppose that u 1. On the interval [1, ) we have F < 0, so that F is an one to one function. Therefore the equality (3) implies u = 1. It follows that α 1 = α 1 and α = α. Since α 1 + β 1 = α 1 + β 1 = h 1 and 9

10 α + β = α + β = h, it results that β 1 = β 1 and β = β. Hence we have d 1 = d 1 and d = d, thus: G Z d1 Z d = Z d 1 Z d G. Remark. In the case when the number s of distinct prime divisors of n is arbitrary, the equalities (1) and () are not sufficient to obtain d 1 = d 1 and d = d. However, if the following conditions are satisfied: (i) h i = 1 for any i {1,,..., s} (i.e. n is square free) or (ii) h i = 1 for any i {1,,..., s}\{i 0 } (i.e. n has the form p 1...p i0 1p h i 0 i 0 p i p s ), then it is easy to see that the conclusion of Proposition 5 holds. References [1] Birkhoff, G., Lattice theory, Amer. Math. Soc., Providence, R.I., [] Grätzer, G., General lattice theory, Academic Press, New York, [3] Suzuki, M., Group theory, I, II, Springer Verlag, Berlin, 1980, [4] Suzuki, M., Structure of a group and the structure of its lattice of subgroups, Springer Verlag, Berlin, [5] Ştefănescu, M., Introduction to group theory (Romanian), Editura Universităţii Al.I. Cuza, Iaşi,