P. G. KONTOROVICH, S. G. IVANOV, G. P. KONDRASHOV

DISTRIBUTIVE PAIRS OF ELEMENTS IN A LATTICE

(Presented by Academician A. I. Mal’tsev, 22 VIII 1964)

§ 1. A.d.-pairs and m.d.-pairs of elements in a lattice

Let \(L\) be a lattice, \(a, b \ldots\) its elements, and let \(a \cup b\), \(a \cap b\) denote, respectively, the lattice sum and intersection. A pair \((a,b)\) of elements of \(L\) will be called an a.d.-pair (additively distributive pair) if
\[ (a \cup b)\cap x=(a\cap x)\cup(b\cap x) \]
for every \(x \in L\). The pair \((a,b)\) will be called an m.d.-pair (multiplicatively distributive pair) if
\[ (a\cap b)\cup x=(a\cup x)\cap(b\cup x) \]
for every \(x \in L\). The pair \((a,b)\) will be called a strict a.d.-pair (respectively, a strict m.d.-pair) if it is not an m.d.-pair (respectively, an a.d.-pair). In the familiar five-element nonmodular lattice \(\{0,1,a,b,c\}\), \(a>b\), \(a\cup c=b\cup c=1\), \(a\cap c=b\cap c=0\), the pair \((a,c)\) is a strict a.d.-pair, and \((b,c)\) is a strict m.d.-pair. A pair \((a,b)\) that is simultaneously an a.d.-pair and an m.d.-pair will be called an a.m.d.-pair. An ordered pair of elements \((a,b)\) will always be an a.m.d.-pair; such a pair will be called trivial.

In the literature an a.d.-pair is usually called a distributive pair. M.d.-pairs and the relations between the a.d.- and m.d.-properties for pairs have almost not been considered. Note, however, the following result, due to J. von Neumann \(\left({}^{1}\right.\), p. 95):

In a modular lattice, the sublattice generated by the elements \(x,y,z\) is distributive if and only if
\[ (x\cup y)\cap z=(x\cap z)\cup(y\cap z). \]

It follows from this, in particular, that in a modular lattice there are no strict a.d.-pairs or m.d.-pairs.

In the present note we consider some relations between the a.d.- and m.d.-properties for pairs in a lattice and applications of the theory of such pairs to group lattices. In particular, we indicate an important class of groups characterized by the absence of a.d.-pairs, and we prove that in a periodic group with the normalizer condition there are no strict a.d.- and m.d.-pairs.

Lemma 1. \((a,b)\) is an a.d.-pair in \(L\) if and only if it is an a.d.-pair in the principal ideal \((a\cup b]=\{x\mid x\leq a\cup b\}\).

Dually: \((a,b)\) is an m.d.-pair in \(L\) if and only if it is an m.d.-pair in the dual principal ideal \([a\cap b)=\{x\mid x\geq a\cap b\}\).

Theorem 1. An a.d.-pair \((a,b)\) in \((a\cup b]\) will be an m.d.-pair in the same ideal if and only if one of the following conditions is satisfied:

\(1^\circ.\) The pairs \((b,x)\) and \((a,y)\) are a.d.-pairs in \((a\cup b]\), where
\[ x=(a\cap b)\cup(a\cap u), \qquad y=(a\cap b)\cup(b\cap v); \]
\(u\), \(v\) independently range over the ideal \((a\cup b]\).

\(2^\circ.\) In \((a\cup b]\) there do not exist elements \(x,y\) such that either
\[ a\cap b<x<y\leq a \]
and the elements \(b,x,y,b\cup x=b\cup y,a\cap b\) form a five-element nonmodular sublattice, or
\[ a\cap b<x<y\leq b \]
and the elements \(a,x,y,a\cup x=a\cup y,a\cap b\) form a five-element nonmodular sublattice.

\(3^\circ.\) For every \(x\in(a\cup b]\) one has:
\[ a\cap(b\cup x)=(a\cap b)\cup(a\cap x), \tag{1} \]
\[ b\cap(a\cup x)=(b\cap a)\cup(b\cap x). \tag{2} \]

In other words, the relation
\[ (a\cup b)\cap x=(a\cap x)\cup(b\cap x) \]
admits all possible permutations of the elements \(a,b,x\).

Remark. In \(3^0\) one may consider \(x\) varying in the intervals \([a \cap b, a]\), \([a \cap b, b]\), respectively, for (1) and (2).

Theorem \(1'\). An m.d.-pair \((a,b)\) in \([a \cap b)\) will then and only then be an a.d.-pair in the same ideal when one of the conditions dual to the conditions of Theorem 1 is satisfied.

Theorem 2. If \((a,b)\) is an a.d.-pair in \(L\), then \((a \cap x, b \cap x)\) is also an a.d.-pair in \(L\) for every \(x \in L\).

The dual assertion for an m.d.-pair is also valid.

Lemma 2. If \((a,b)\) is an a.d.-pair in \(L\), then every pair \((a_1,b_1)\) such that \(a_1 \ge a\), \(b_1 \ge b\), \(a \cup b = a_1 \cup b_1\), is an a.d.-pair in \(L\).

The dual assertion for m.d.-pairs is also valid.

§ 2. Distributive pairs in the lattice of subgroups of a group

\(G\) is a group; \(a,b,\ldots\) are its elements; \(L(G)\) is the lattice of its subgroups; \(A,B,\ldots\) are subgroups of the group \(G\) (elements of \(L(G)\)); \(\cup,\cap\) are the lattice operations in \(L(G)\) (in what follows, for brevity, we shall speak of lattice operations in the group \(G\) itself).

Let us recall the following classical result of Ore ((2), p. 17):

Subgroups \(A\) and \(B\) form an a.d.-pair in the group \(G\) if and only if the orders with respect to \(A\) and \(B\) of every element \(x\) of \(A \cup B\) (but not of \(A\) and not of \(B\)) are finite and relatively prime.

Theorem 3. A pair \((A,B)\) of relatively prime subgroups will be an a.d.-pair or an m.d.-pair (and consequently also an a.m.d.-pair) in \(G = A \cup B\) if and only if \(G = A \times B\) is a direct product and the orders of the elements of \(A\) and \(B\) are finite and relatively prime.

Remark. For an a.d.-pair the theorem was proved in (2), p. 19.

Theorem 4. A periodic group \(G\) is free of nontrivial a.d.-pairs if and only if it is covered by its Sylow subgroups (i.e., every element of it has primary order).

Proof. If \(G\) has no nontrivial a.d.-pairs, then it contains no cyclic subgroups of order \(pq\), where \(p,q\) are distinct primes, as follows from the cited theorem of Ore. Consequently, \(G\) is covered by its Sylow subgroups. Conversely, if \(G\) is covered by its Sylow subgroups, then it is easy to see that there are no nontrivial a.d.-pairs, since Ore’s criterion is not satisfied.

Remark. Finite groups with a Sylow covering basis (s.b.-groups in the terminology of (3)) have been studied by many authors (Higman and others; see the bibliography in (3)). Theorem 4 gives a purely lattice-theoretic definition of this class of groups.

Lemma 3. Let \(G = A \cup B\) and let \(A \cap B = D\) be an invariant subgroup in \(G\). Then the pair \((A,B)\) is an m.d.-pair if and only if it is an a.d.-pair (and consequently also an a.m.d.-pair).

Proof. If \((A,B)\) is an a.d.-pair in \(G\), then \((A/D, B/D)\) is an a.d.-pair in the direct product
\[ G/D = A/D \times B/D. \]
But then \((A/D, B/D)\) is an a.m.d.-pair in \(G/D\) by Theorem 3. Hence it follows from Lemma 1 that \((A,B)\) is an m.d.-pair in \(G\).

If, however, \((A,B)\) is an m.d.-pair in \(G\), then \((A/D, B/D)\) is an m.d.-pair in \(G/D\), and consequently an a.m.d.-pair by Theorem 3. But then \((A,B)\) is an a.d.-pair in \(G\) by Ore’s theorem.

Theorem 5. A \(p\)-group with the normalizer condition, in particular a finite \(p\)-group, is free of nontrivial a.d.-pairs and m.d.-pairs.

Proof. That every \(p\)-group is free of nontrivial a.d.-pairs follows directly from Theorem 4. Suppose now that \((A,B)\) is an m.d.-pair. Then
\[ A \cap B = D \ne e. \]
By virtue of the normalizer condition, there exist \(A_1\) and \(B_1\) such that \(D \subset A_1 \subseteq A\), \(D \subset B_1 \subseteq B\), and \(D\) is invariant in \(A_1\) and \(B_1\). By the preceding lemma, \((A_1,B_1)\) will be a nontrivial a.d.-pair, which is impossible.

Lemma 4. In a periodic \(S\)-group (i.e., in a group that is the direct product of its Sylow subgroups), the intersection \((A \cap P, B \cap P)\) of an m.d.-pair \((A,B)\) with a \(p\)-Sylow subgroup \(P\) will be an m.d.-pair in \(P\).

Proof. Introduce the notation: \(P_A=P\cap A,\ P\cap B=P_B,\ D\cap P=P_D=P_A\cap P_B,\ D=A\cap B\). Let \(G=P\times H,\ A=P_A\times H_A,\ B=P_B\times H_B\). Consider a subgroup \(X\) of \(P\) for which \(P_D\subseteq X\) (see Lemma 1). Then
\[ X\subseteq (X\cup P_A)\cap (X\cup P_B)\subseteq P\cap (X\cup A)\cap (X\cup B)= \]
\[ =P\cap [X\cup (A\cap B)]=P\cap [X\cup (P_D\times H_D)]=P\cap (X\times H_D). \]
Since \((X,H_D)\) is a commuting pair, the modular identity is satisfied for it. Therefore
\[ P\cap (X\times H_D)=X\cup (P\cap H_D)=X, \]
since \(P\cap H_D=e\). Now the equality
\[ X\cup (P_A\cap P_B)=(X\cup P_A)\cap (X\cup P_B) \]
is easily obtained.

Theorem 6. A periodic group with the normalizer condition contains no proper a.d.-pairs and no proper m.d.-pairs.

Proof. Let \(G\) be a periodic group with the normalizer condition. As is known, \(G=\prod P_i\), where the \(P_i\) are Sylow \(p_i\)-subgroups, and \(\prod\) denotes the sign of the direct product.

1) Let \((A,B)\) be an m.d.-pair in \(G\). Put \(P_{iA}=P_i\cap A,\ P_{iB}=P_i\cap B\). By Theorem 5 and Lemma 4, either \(P_{iA}\subseteq P_{iB}\) or \(P_{iB}\subseteq P_{iA}\). Let
\[ H=\prod P_{iA},\quad P_{iA}\supseteq P_{iB}; \qquad F=\prod P_{jB},\quad P_{jB}\supset P_{jA}. \]
It is clear that \(A\cup B=H\cup F=H\times F\). By Lemma 1 and Theorem 3, \((H,F)\) is an a.d.-pair in \(G\). Therefore the orders \(m\) and \(n\) of an element from \([(H\times F)\setminus H]\setminus F\), respectively relative to \(H\) and \(F\), are finite and relatively prime. Now let \(x\in [(A\cup B)\setminus A]\setminus B\), and let the orders of the element \(x\) relative to \(A\) and \(B\) be respectively \(m_1\) and \(n_1\). From \(H\subseteq A,\ F\subseteq B\) it follows that \(x\in [(H\times F)\setminus H]\setminus F\), and \(m_1\) divides \(m\), \(n_1\) divides \(n\), i.e., \(m_1\) and \(n_1\) are also relatively prime. Consequently, \((A,B)\) is an a.d.-pair in \(G\).

2) Let \((A,B)\) be an a.d.-pair in \(G\). According to Theorem 2, \((P_{iA},P_{iB})\) is an a.d.-pair in \(G\), and hence all the more so in \(P_i\). By Theorem 5 it follows that either \(P_{iA}\subseteq P_{iB}\) or \(P_{iB}\subset P_{iA}\).

We now prove that \((A,B)\) is an m.d.-pair in \(G\). Form the following subgroups:
\[ H=\prod P_{iA},\ P_{iA}\supseteq P_{iB}; \qquad F=\prod P_{jB},\ P_{jB}\supset P_{jA}; \]
\[ H_1=\prod P_{iB},\ P_{iA}\supseteq P_{iB}; \qquad F_1=\prod P_{jA},\ P_{jB}\supset P_{jA}. \]
It is clear that \(A\cup B=H\times F,\ A=H\times F_1,\ B=H_1\times F\). Let \(\pi_2=\pi(F)\) be the set of all prime numbers dividing the orders of elements of \(F\), and let \(\pi_1\) be the totality of all the remaining prime numbers dividing the orders of elements of the group \(G\); in particular, \(\pi_1\supseteq \pi(H)\). Let \(R_1\) and \(R_2\) be respectively Sylow \(\pi_1\)- and \(\pi_2\)-subgroups of the group \(G\), so that \(G=R_1\times R_2\). Let \(Y\) be a subgroup of \(G\), and \(Y\supseteq A\cap B\) (see Lemma 1). It is not hard to see that
\[ Y\subseteq (Y\cup A)\cap (Y\cup B)= \]
\[ =\{[(Y\cap R_1)\cup H]\times [(Y\cap R_2)\cup F_1]\}\cap \{[(Y\cap R_1)\cup H_1]\times \]
\[ \times [(Y\cap R_2)\cup F]\}. \]
The pairs of direct factors standing in braces will be a.d.-pairs in \(G\). Taking into account that
\[ [(Y\cap R_1)\cup H]\cap [(Y\cap R_2)\cup F] = [(Y\cap R_1)\cup H_1]\cap [(Y\cap R_2)\cup F_1]=e \]
and that \(H_1\subseteq A\cap B,\ F_1\subseteq A\cap B\), we obtain, after some calculation,
\[ Y\cup (A\cap B)=(Y\cup A)\cap (Y\cup B), \]
as required.

Theorem 7. If \(A,B\) are quasinormal subgroups in a group \(G\) (i.e., they permute with every subgroup of \(G\)) and \((A,B)\) is an m.d.-pair in \(A\cup B\), then \((A,B)\) is an m.d.-pair in the whole group \(G\).

In the proof, the quasinormality of the subgroups and the modular identity for such subgroups are used.

Received
30 VII 1964

CITED LITERATURE

1. G. Szasz, Introduction to Lattice Theory, Budapest, 1963.
2. M. Suzuki, The Structure of Groups and the Structure of the Lattice of Their Subgroups, Moscow, 1960.
3. P. G. Kontorovich, A. S. Pekelis, A. I. Starostin, “Structural Questions in Group Theory,” Matem. zap., Sverdlovsk, 3, No. 1 (1961).