ON THE SUBGROUP LATTICE OF AN ABELIAN FINITE GROUP

ON THE SUBGROUP LATTICE OF AN ABELIAN FINITE GROUP Marius Tărnăuceanu Faculty of Mathematics Al.I. Cuza University of Iaşi, Romania e-mail: mtarnauceanu@yahoo.com 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

(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).

(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

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

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

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

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

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

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 1 + 1 = 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 1 + 3 (α 1 + 1)x)p (α 1+1)x+1 1 (h 1 + 1 (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

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 i0 +1...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., 1967. [] Grätzer, G., General lattice theory, Academic Press, New York, 1978. [3] Suzuki, M., Group theory, I, II, Springer Verlag, Berlin, 1980, 1985. [4] Suzuki, M., Structure of a group and the structure of its lattice of subgroups, Springer Verlag, Berlin, 1956. [5] Ştefănescu, M., Introduction to group theory (Romanian), Editura Universităţii Al.I. Cuza, Iaşi, 1993. 10