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Finite Boolean Algebras and Subgroup Lattices Of Finite Abelian Groups[/FONT]
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Ryan Watkins[/FONT]
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Abilene Christian University[/FONT]
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Advisor: Jason Holland, Ph.D.[/FONT]
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Assistant Professor of Mathematics Abilene Christian University[/FONT]
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Abstract[/FONT]
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We use the fundamental theorem of finite abelian groups to give a characterization of when the subgroup lattice of a finite abelian group can be given the structure of a Boolean algebra. Arguments involving both complementation in a Boolean algebra as well as the distributivity of a lattice are used. Prerequisites include an elementary course in abstract algebra as well as familiarity with elementary lattice theory. [/FONT]
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Introduction[/FONT]
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We shall follow the notation in [4] and suggest this as a reference for unexplained terminology as well.
We consider only finite abelian groups in this paper. We use the notation Zn for the cyclic group of integers modulo n. The direct product of groups G and H is denoted by G ´ H. It is a fact that every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order. This result is important to our paper so we state it below as Theorem 1. For a proof, we refer the reader to [4], pages 200-203.
Theorem 1. Every finite abelian group G is isomorphic to a direct product of cyclic groups of the form
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Zp(1) ´ Zp(2) ´ Zp(3)´ ...´ Zp
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where the p(i)'s are powers of primes and the primes are not necessarily distinct. [/FONT]
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[FONT=Verdana, Arial, Helvetica, sans-serif] Boolean Algebras and Subgroup Lattices[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]A Boolean algebra is a lattice that is both distributive and complemented. Figure 1 shows the Hasse diagrams of two finite Boolean algebras.
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Figure 1[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]
For the lattice of divisors of an integer, the partial order is given by a < b if a divides b. It is easy to verify that the supremum of two elements a and b is given by the least common multiple of a and b. Similarly, the infimum of a and b is just the greatest common divisor of a and b. The lattice of divisors of an integer n has the same structure as the lattice of all subgroups of Zn. If H and G are subgroups of a given group, then the partial order in the subgroup lattice is given by H < G if H is a subgroup of G. The lattice operations are then as follows. The infimum of H and G is simply HÇG and the supremum of H and G is the smallest subgroup that contains both H and G (the subgroup generated by H and G). We denote the supremum of H and G by HÚG.
The divisor lattice of an integer n can be given the structure of a Boolean algebra if and only if n is square free (e.g. [1] page 182, number 7 or [4] page 322, number 2). One can immediately infer that the lattice of subgroups of Zn can be given the structure of a Boolean algebra if and only if n is square free. For the remainder of this paper, we shall demonstrate that the only finite abelian groups with Boolean subgroup lattices are those of type Zn where n is square free. We acknowledge that the latter result is a straightforward application of Theorem 1, yet we were unable to find this result in the literature. We draw inspiration from the following example.
Example. Let G = Z4´ Z3. This group is cyclic (see [3] Theorem 8.2) and therefore every subgroup can be described by a generator. We write = H if the element h generates H. The subgroups of G are
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{G, <(1,0)>, <(0,1)>, <(2,1)>, <(2,0)>, <(0,0)>}.
The Hasse diagram for the subgroup lattice of G is given in Figure 2.
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Figure 2
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Note that we can find elements in the lattice which do not have complements and therefore the lattice is not Boolean. We point out that one can also use the fact that the number of elements in the lattice is not a power of 2 (see [2] Chapter 3, Theorem 4). Inspired by the latter example, we state and prove the following lemma.
Lemma 2. Suppose that G = Zp(1) ´ Zp(2) ´ Zp(3)´ ...´ Zp where each p(i) is a power of a prime. If for some i, p(i) is not prime, then the subgroup lattice of G is not Boolean.
Proof. The case where there is only one summand in the product is easy and left for the reader. Suppose that p(i) is not a prime number and is a power of the prime number q. Let H denote the subgroup generated by the element with a q in the ith coordinate and zeros in all other coordinates. Let K be any subgroup such that HÇK = {(0,0,0,...,0)}. The supremum of H and K can never be all of G since the subgroup generated by H and K will not have all elements of Zp(i) in the ith coordinate. It follows that the subgroup lattice of G cannot be Boolean.
This does not take care of all possible cases in the class of finite abelian groups. We need another example.
Example. Let G = Z2´ Z2´ Z3. In the event that there are repeated primes, the group G is not cyclic (again, see Theorem 8.2 in [3]). One has to be careful about listing the subgroups. There are 8 cyclic subgroups:
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{<(0,0,0)>, <(1,0,0)>, <(0,1,0)>, <(0,0,1)>, <(1,1,0)>, <(1,0,1)>, <(0,1,1)>, <(1,1,1)>}.
The remaining subgroups which are not cyclic are the whole group and <(1,0,0), (0,1,0)>. The Hasse diagram for the subgroup lattice of G is given in Figure 3.
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Figure 3
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Note that we can find elements in the lattice which violate the distributive law. Again, the example motivates the following lemma.
Lemma 3. Suppose that G = Zp(1) ´ Zp(2) ´ Zp(3)´ ...´ Zp where each p(i) is a prime. Suppose further that p(i)=p(j) for some i not equal to j. Then the subgroup lattice of G is not Boolean.
Proof. Without loss of generality, assume that the first two groups in the product are the same. Let
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H = <(1,0,0,0,0,...)>, K = <(0,1,0,0,0,...)> and L = <(1,1,0,0,0,...)>. [/FONT] [FONT=Verdana, Arial, Helvetica, sans-serif]
We have that HÚ(KÇL) = H. However, (HÚK)Ç(HÚL) = <H,K>. The latter argument can be visualized if one labels the three circled elements on Figure 3 as H, K and L in that order. It follows that the subgroup lattice is not distributive and therefore not Boolean.
Combining Lemma 2 with Lemma 3 yields Theorem 4, which is our main result.
Theorem 4. If G is a finite abelian group, then G has a subgroup lattice which is a Boolean algebra if and only if G is isomorphic to Zn where n is a square free integer.
Proof. By Theorem 1, G = Zp(1) ´ Zp(2) ´ Zp(3)´ ...´ Zp where p(i) is a power of a prime for every i. By Lemma 2, all powers are 1. By Lemma 3, there are no repeated primes. We now apply Corollary 8.12 in [4] to get that G = Zn where n = p(1)´ p(2)´ ... ´ p. This is, of course, a square free integer.[/FONT]
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[FONT=Verdana, Arial, Helvetica, sans-serif] Extensions[/FONT]
It is known that the normal subgroups of any group form a modular lattice (see Theorem 11 in [2]. If a lattice is modular, then it is obviously distributive. Thus, when studying the structure of the subgroup lattice of nonabelian groups, one may first consider the lattice of normal subgroups. Perhaps a place to start would be subgroups of the permutation groups since by Cayley's Theorem, every finite group can be represented as a subgroup of a permutation group
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