JOURNAL OF ALGEBRA 199, 379᎐403Ž. 1998 ARTICLE NO. JA977198
Symmetry Groups of Boolean Functions and Constructions of Permutation Groups
Andrzej Kisielewicz*
Institute of Mathematics, Uni¨ersity of Wrocław, Wrocław, Poland View metadata, citation and similar papers at core.ac.uk brought to you by CORE Communicated by Walter Feit provided by Elsevier - Publisher Connector Received July 10, 1996
In this paper we deal with the symmetry group SfŽ.of a boolean function f on n-variables, that is, the set of all permutations on n elements which leave f invariant. The main problem is that of concrete representation: which permutation groups on n elements can be represented as G s SfŽ.for some n-ary boolean function f. Following P. Clote and E. Kranakiswx 6 we consider, more generally, groups represented in such a way by k-valued boolean functionsŽ i.e., functions on a two-element set with k possible values. and call such permutation groups k-representable Ž.k G 2 . The starting point of this paper is a false statement in one of the theorems ofwx 6 that every k-representable permutation group is 2-represen- table for all k G 2. We show that there exists a 3-representable permutation group that is not 2-representable. A natural question arises whether there are other examples like this. This question turns out to be not easy and leads us to considering general constructions of permutation groups and investigating their properties. The main result of the present paper may be summarized as follows: using known groups and ‘‘standard’’ constructions no other example like that above can be constructed. Nevertheless, we conjecture that there are k q 1-representable permutation groups that are not k-representable for all k G 2. If this is true, then this would open an interesting avenue toward investigating and classifying finite permutation groups. ᮊ 1998 Academic Press
1. INTRODUCTION
Let f s fxŽ.1,..., xn be a function of n variables. The set Sf Ž.of all permutations in the symmetric group Sn such that
fxŽ.Ž.12,x,..., xnsfxŽ1.,xŽ2.,..., xŽn.
is a subgroup of Sn called the symmetry group of f.
* Supported by KBN Grant 2 P301 027 07. E-mail address: [email protected].
379
0021-8693r98 $25.00 Copyright ᮊ 1998 by Academic Press All rights of reproduction in any form reserved. 380 ANDRZEJ KISIELEWICZ
Quite surprisingly, symmetry groups of functions has received some attention only recently in connection with applications in computer science on one hand, and universal algebra on the other hand. Symmetry groups of boolean functions appear naturally in the study of parallel complexity of formal languagesŽ for details and further references seewx 6 ; cf. also wx 5. . More precise knowledge on symmetry groups of algebraic operations is needed in investigation of sizes of clones and free spectra of algebrasŽ see, e.g.,wx 7, 11, 20.Ž . It seems that symmetries of functions especially those of boolean functions. are worthy of study not only from a purely algebraic or combinatorial point of view, but as well as a tool to provide possibly new insight into the permutation group theory. In this paper we deal with the symmetry groups of boolean functions, n i.e., those of the form f: 2 ª 2, where 2 s Ä40, 1 . This leads us naturally to considering also symmetry groups of k-valued boolean function f: n 2ªk, where k s Ä40, 1, . . . , k y 1. A permutation group G : Sn of the form G s SfŽ.for some k-valued boolean function f s fxŽ.12,x,..., xnwill be called representable by a k-¨alued boolean function, or briefly, k-representable. A group G is said to be representable if it is k-representable for some k G 2. The main question we are interested in is which subgroups of Sn are representable by boolean functions, i.e., 2-representableŽ inwx 6 such groups were called strongly representable.. Generally, it is not difficult to see that every abstract group is isomor- phic to some representable groupŽ cf.wx 6, Theorem 11. . What we need howeverŽ what is needed in problems in computer science and universal algebra mentioned above. is an answer to the following more concrete problem: Given n ) 1, what groups are representable by functions in n variables? In Section 2 we recall basic facts to use later in the proofs and show a counterexample to the last statement in the representation theorem for- mulated by Clote and Kranakis. Inwx 6, Theorem 13 they claim that every representable subgroup of Sn is 2-representable. Yet, the proof contains a gap. On page 572, in line 3 from the bottom, the claim that SgŽ.is properly contained in ShŽ.has no ground. It seems that the authors have just overlooked the possibility that while SgŽ.must be different from Sh Ž., it does not need to be contained in the latter at all. This possibility makes the problem here, in fact, much more complicated than it would seem at first sight. After finding the gap I was convinced that the theorem is simply false. Yet, for a long time I could not find any counterexample. Finally, I have discovered this quite surprising example: The subgroup D of S4 of even symmetries of the square is 3-representable, but not 2-representable Ž.Theorem 2.3 . SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 381
This example is surprising not only because it is so simple, but also because it seems quite exceptional. I did not succeed in generalizing it in any way. Moreover, so far, I found no other example like this; all other permutation groups I have checked are 2-representable or not repre- sentable at all. The next sections are devoted to investigating this very problem. Here, it turns out that the problem in questionŽ and probably many similar problems in permutation groups. requires a quite new approach. Representability and k-representability are properties that are invariant under isomorphism of permutation groups, but not under isomorphism of abstract groups. Therefore, it seems, that the most advanced parts of finite permutation group theory, like the O’Nan᎐Scott theorem for finite primi- tive permutation groupsŽ see, e.g.,wx 12, 4. will have rather little application here. Also more related to this question, the study of the abstract group of automorphismswx 2, 21 , the study of permutation groups through invariant relationswx 14, 19 , the theory of V-ringswx 10, 13 , or the classifications of boolean functionswx 9 seem to have no tools that would be suitable for our question. In this situation, we check first the simplest classes of permutation groups and then try to make a transition to more complex classes. A natural attempt is, of course, trying to reduce our problem to transitive and, if possible, to primitive permutation groups. Yet, at this point, it turns out that ‘‘reduction results’’ considered and used so far, while satisfactory from the abstract group point of view, are useless for us, since they involve loss of information about how the permutation groups involved act, which is the most essential for our purposeŽ seewx 8, p. 63; 18, p. 13 ; also cf. remarks about transitive constituents inwx 4, p. 3. . Also the chapter on constructions of permutation groups inwx 14 mentions no other construc- tion but those used in the abstract group theory. Therefore, in Section 4 we define suitable construction and formulate necessary reduction results. Among them we distinguish the most natural and transparent constructions and show that the class of 2-representable groups is closed on them. More generally, we show that no new repre- sentable group, which is not 2-representable may be constructed using these natural constructions and ‘‘known’’ permutation groups. The only exception is the construction that shows how the above mentioned group D is built up from smaller groups: this construction preserves 2-represen- tability except for the case of D itself. This shows that D is indeed exceptional. An intriguing question whether there are other examples of repre- sentable but not 2-representable groups remains open. We conjecture that there are k q 1-representable permutation groups that are not k-repre- sentable for all k G 2, since both in our research and analyzing the 382 ANDRZEJ KISIELEWICZ
Clote᎐Kranakis proof we found no indication that this may fail to hold. Other open questions are suggested at the end of the paper. In Section 6 we show how two other important results fromwx 6 which depended on the last statement of their representation theorem can be suitably mended, and give a new, simpler proof of one of them. We suggest also an application of our results to finding subgroups of a given permuta- tion group. Our terminology agrees generally with that ofwx 18 . The main difference that should be observed by the reader is that in this paper we read products of permutations from right to left. Two permutation groups isomorphic as permutation groups are considered, basically, as identical. In particular, we are interested in representatives of conjugate classes of subgroups of Sn rather than in particular groupsŽ as those representing really different kinds of symmetry on n-element objects. . Consequently, the set V of elements being permuted is always identified with V s Ä41, 2, . . . , n , unless otherwise stated. The symmetric and alternating groups on V are denoted simply by Snnand A . The identity permutation in Snis denoted just by 1, if no confusion can arise. For specific boolean n-tuples we adopt the convention to write them as strings of 0’s and 1’sŽ. without commas . In this notation, 0ii and 1 denote i consecutive 0’s or 1’s, respectively, and round brackets are used to denote orbits or blocks.
2. BASIC FACTS
We start from an observation, which we shall use later, that the family of permutation groups representable by boolean functions contains, in partic- ular, all automorphism groups of graphs.
THEOREM 2.1. If a permutation group G consists of all automorphisms of a graph D s Ž.V, E , then G is representable by a Ž.2-¨alued boolean function. In consequence, e¨ery abstract group is isomorphic to the symmetry group of a Ž.2-¨alued boolean function.
ŽThe second part of this theorem is due to P. Clote and E. Kranakiswx 6 . It follows from their sharper result that every subgroup G of Sn is isomorphic to a 2-representable subgroup H of Snиlog n..
Proof. Assuming V s Ä41, 2, . . . , n , we choose the n-tuple Ž.x1,..., xn with xijs x s 1 and x rs 0 for r / i, j to represent the edge ij g E.We define an n-ary boolean function f by putting f s 1 for those n-tuples SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 383
that represent edges in E, and f s 0, otherwise. It is clear that permuta- tions leaving f invariant are precisely the automorphisms of graph D. Hence, the first claim. The second claim follows now immediately from the well-known fact that every abstract group is isomorphic to the automorphism group of a graphŽ see, e.g.,wx 21. . The theorem above does not hold for directed graphs. For instance,
groups Cnngenerated by a cycle s Ž.1, 2, . . . , n for n s 3, 4, 5 are exam- ples of automorphism groups of directed graphs that, as we will see later, are not k-representable for any k.
To check whether a group G : Sn is representable we have to check first how G acts on the set 2n Ž.of arguments of an n-ary boolean function . The natural action here is given by
x s Ž.Žx1,..., xnªx s xŽ1., xŽ2.,..., xŽn. .. G n The basic role in this study plays the set of orbits x s Äx : g G4in 2 , which we shall denote OGŽ.. n Indeed, if G s SfŽ.for a Žk-valued . boolean function f: 2 ª k, then: Ž.1 all n-tuples in every orbit of OG Ž .must have the same value under f, and n Ž.2 for every permutation f G there must be x g 2 such thatŽ. a xand x are in different orbits of OGŽ., and Ž. b these orbits have different values under f. Hence, in order to find a boolean function whose symmetry group is G it is sufficientŽ. and necessary to find a function F: OG Ž.ªksuch that conditionŽ. 2 above Ž with f replaced by F .is satisfied. The reason why in conditionŽ. 2 two steps Ž. a and Ž. b are distinguished is that a failure in stepŽ. a shows quite a different cause of nonrepresentabil- ity of G than that in stepŽ. b . n Namely, if there is a permutation g G such that for every x g 2 , x and x are always in the same orbit of OGŽ., then the group GЈ generated by G and has precisely the same orbits in 2n as G alone. Therefore, every function f on 2n invariant under G is invariant under GЈ, as well.
Hence, G cannot be representable, at all. Consequently, if G : Sn is representable, then it must be maximal among the subgroups of Sn with the same set OGŽ.of orbits in 2n. Conversely, if G is such a maximal subgroup, then it is k-representable for k s ⌰Ž.G , where ⌰ Ž.G denotes the number of different orbits in OGŽ.. This is so, because a function F: OGŽ.ªk assigning pairwise distinct values to elements of OGŽ., in view of maximality of G, obviously satisfies conditionŽ. 2 . 384 ANDRZEJ KISIELEWICZ
Clearly, maximality with regard to OGŽ.may be replaced by maximality with regard to the number ⌰Ž.G , since if GЈ = G, then every orbit in OGŽ.is contained in some orbit in OG ŽЈ .. The number ⌰ Ž.G can be computed easily by means of the cycle indexŽ seewx 6 , where also an application of this fact is given to show in a precise sense that almost all boolean functions have a trivial symmetry group. . Moreover, we may establish a more precise bound on k, independent of G. To this end, we observe first that OGŽ.consists of n q 1 natural le¨els whose members are n-tuples with the same number i of 1’s. Obviously, for n every x g 2 and every g G, x and x are always in the same level.Ž As a matter of fact, the levels are just orbits OSŽ.n of the full symmetric group..Ž The set of orbits in OG.belonging to the ith level will be denoted
OGiŽ.. As a result, we do not need to care about assigning different values to orbits at different levels, and therefore k n q 1 Žthe cardinality of s ž/?@nq1r2 the largest level.Ž suffices in every case. ?@x denotes the largest integer not exceeding x.. We summarize our considerations in the following.
THEOREM 2.2. A permutation group G : Sn is representable if and only if it is a maximal subgroup of Sn among those ha¨ing the same number ⌰Ž.Gof n orbits in 2 . In such a case G is representable by a k-¨alued boolean function with k n q 1 . F ž/?@nq1r2
In particular, it follows from this theorem that the alternating group An is not representable for any n ) 2, since as it is not difficult to check, ⌰Ž.Annsnq1s⌰ Ž.S. We close this section by pointing out the mentioned counterexample to the representation theorem inwx 6 .
THEOREM 2.3. Let D : S41 be a group generated by s Ž.Ž.1, 2 3, 4 and 2 sŽ.Ž.1, 3 2, 4 . Then D is 3-representable, but not 2-representable.
Proof. There are two more elements in D: 3 s Ž.Ž.1, 4 2, 3 and the identity 0. Next, it is not difficult to check that there are ⌰Ž.D s 7 orbits in ODŽ., and only level OD2Ž.consists of more than one orbit. Namely, we have OD2Ž.sÄŽ.Ž..Ž.1100, 0011 , 1010 , 0101 , 1001, 01104 . First, we show that D is indeed 3-representable. To this end we have to 4 show that D s SfŽ., where f is a function on 2 assigning different values to the orbits in OD2Ž., or in other words, that the subgroup P : S4 of all permutations preserving these orbits equals D.
The first orbit in OD2Ž.shows thatÄŽ.Ž. 1, 2 , 3, 44 forms a block system in P. Similarly, the next orbit shows thatÄŽ.Ž. 1, 3 , 2, 44 is a block system. It follows that P contains no transposition, and now it is easy to see that P s D. SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 385
Suppose, in turn, that D s ShŽ.for some two-valued boolean function h. It follows that two orbits of OD2Ž.have the same value under h, say, the first one and the second one. Yet, in such a case, h is invariant under sŽ.2, 3 , and in consequence, g Sh Ž., a contradiction. The other two cases are symmetrical. Whence, D is not 2-representable.
3. CONSTRUCTING PERMUTATION GROUPS
Searching for another example like that in Theorem 2.3, we first observe that the most common classes of permutation groups are representable by 2-valued boolean functions. The following observations are due to P. Clote and E. Kranakiswx 6, Sect. 4 . For all n G 2 we have:
Ž.1 Symmetric group. Sn s Sf Ž ., where f is any constant boolean functionŽ. on n variables .
Ž.2 Identity group. In s Sf Ž ., where f s 1 for all n-tuples of the i n i form 1 0 y , and f s 0, otherwise.
Ž.3 Reflection group. Rn s ² :, where Ž.i s n q 1 y i is a reflec- tion permutation. A boolean function representing Rn is that assigning 1 to all n-tuples of the form 1i 0 nyi or 0 i1nyi.
Ž.4 Dihedral group Dn Žsymmetries of a regular n-gon, generated by reflection permutation and a cyclic permutation nns Ž..1, 2, . . . , n . D is representable by a boolean function, since it is the automorphism group of n the corresponding n-vertex circulant graph C . Dn s SfŽ.for f assign- ing 1 to all n-tuples representing the edges of C n; see the proof of Theorem 2.1.
Ž.5 Cycle group. Cnns ² :, where nis the cyclic permutation above. Cn is representable by a boolean function for n / 3, 4, 5. To see it, it is enough to modify the function f representing the dihedral group above in such a way that reflection no longer leaves f invariant. For example, put n f s 1 for all n-tuples representing the edges of C , and in addition, for all ny4 n-tuples x , where x s 11010 and g Cn. Otherwise, put f s 0. Then, for n ) 5, only permutations of Cn leave f invariant.Ž For n s 2, C22s S ..
It is important to note that Cn for n s 3, 4, 5 are not representable at all, neither by a boolean function, nor by a k-valued boolean function for any k. Indeed, for n s 3 we have C33s A , and for n s 4, 5 it is easy to compute ⌰Ž.C4455s 6 s ⌰ Ž.D , and ⌰ Ž.C s 8 s ⌰ Ž.D . Now we would like to show that other, more complex permutation groups constructed from those considered so far are also representable by boolean functions. Of course, we are interested only in constructions 386 ANDRZEJ KISIELEWICZ
whose ingredients and results are permutation groups acting on concrete sets. The simplest and most natural from this point of view is a direct product
of two permutations groups G : Snmand H : S acting on disjoint sets. We denote it just by G = H, since no abstract form of direct product is considered here. The set G = H acts on is usually identified with Ä41, 2, . . . , n q m Ži.e., Sm is assumed to act on Ä4n q 1,...,nqm; note that inwx 14 this construction is called the direct sum of permutation groups. . The following theorem is a generalization of Lemma 16 inwx 6 .
THEOREM 3.1. If G : Snm and H : S are k-representable for some k G 2 Ž. n,mG1, then G = H : Sisrnqm -representable for e¨ery r satisfying 2 r y r G k; in particular, G = Hisk-representable, as well. On the other hand, if one of G or H is not representable, then G = H is not representable, either. Proof. Let G s SgŽ.and H s Sh Ž., where g and h are k-valued functions on n and m boolean variables, respectively. Without loss of generality we may assume that m F n. Moreover, we assume that k different values of g and h are taken from the Cartesian product P s 2 Ä4Ä41, 2, . . . , r = 1, 2, . . . , r y 1 . This is possible, since r y r G k. Denoting by 12and first and second projection operations on P, respectively, we define an r-valued function f on 2nqm as gx ,ifzx0mn for some x 2 , x 0,1,nn ¡ 1Ž.Ž. s g / mnnn 2Ž.gxŽ.,ifzsx1 for some x g 2 , x / 0,1, Ž.hyŽ.,ifz0nmmmyfor some y 2 , y / 0,1, fzŽ.s~1 s g nmmm 2Ž.hyŽ.,ifzs1yfor some y g 2 , y / 0,1, nm r,ifzs10 , ¢0, otherwise. We show in two steps that SfŽ.sG=H. n m At first, if z12s xy and z s x y for some x g 2 , y g 2 , and g G, then obviously fzŽ.12sfz Ž.Žsince gx Ž.sgx Ž ... Similarly, if z1s xy nm and z2s xy for some x g 2 , y g 2 and g H, then fzŽ.12sfz Ž..It follows that SfŽ.=G=H. To prove the converse inclusion we first note that SfŽ.:Snm=S. nm Indeed, it follows easily from the fact that fuŽ.srfor u s 1 0 , and fzŽ.-rfor every z / u with 1 occurring precisely n times in z. This is so, because the only case in the definition of f, apart from the last one, when
z may have precisely n 1’s is the second one, while the values of 2 are less than r. Now, let s Ž. , f G = H, say, f G Žfor f H the argument is n analogous. . Then, there is x g 2 such that gxŽ./gx Ž .and, obviously, SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 387
nn x / 0 , 1 . Hence, either 11ŽŽ..gx / ŽŽgx..or 2 Žgx Ž ../ 2 Žgx Ž ... mm It follows that fzŽ./fz Ž .for either z s x0orzsx1 . Consequently, SfŽ.sG=H. For the second statement observe first that ⌰Ž.Ž.Ž.G = H s ⌰ G ⌰ H . Now, if, for example, G is not representable, then in view of Theorem 2.2, there is some GЈ : Sn such that ⌰Ž.GЈ s ⌰ Ž.G , GЈ = G, and GЈ / G.In view of our observation, ⌰Ž.Ž.GЈ = H s ⌰ G = H . Using Theorem 2.2 again we infer that G = H is not representable. Theorem 3.1 shows that we cannot expect to find a new representable group which is not 2-representable using the operation of direct product: neither starting from 2-representable groups, nor using nonrepresentable groups. Instead, the second part of this theorem shows how to produce new nonrepresentable groups: every direct product having a nonrepre- sentable factor is such a one. 2 In many cases, the bound r y r G k in Theorem 3.1 can be improved by one. In particular, using similar arguments as those in Theorem 3.1 one can prove that the class of all 2-representable groups together with the 3-representable group D from Theorem 2.3 is closed under the operation of direct product, so even using D as a factor of direct product we cannot expect to produce another such a group.
4. INTRANSITIVE PRODUCT
We would like to get a result stronger than Theorem 3.1, which would allow us to restrict our search to transitive groups. To this end we need a full description of how intransitive groups are constructed from transitive ones. From the abstract group theory point of view, a sufficient information is that an intransitive permutation group is a subdirect product of its transi- tive constituents. We need to have a full description how the action of the resulting group is composed from the actions of components.
Let G11, H : Smand G22, H : Srbe permutation groups such that H1 and H21are normal subgroups of G and G2, respectively. Suppose, in addition, that factor groups G11rH and G 22rH are isomorphic and : G11rHªG 22rHis the isomorphism mapping. Then, by Ž.G11rH = Ž.G22rHwe denote the subset of Smr= S consisting of all permutations Ž.,such that ŽH12 .sH. It is easy to see that Ž.G11rH = Ž.G22rHis a subgroup of Smr= S . We shall call it the intransiti¨e product of permutation groups with respect to isomorphism , assuming always that all necessary conditions listed above are satisfied. Via this product every intransitive group can be obtained from transitive ones. The following resulat is implicit inwx 8, Theorem 5.5.1, p. 63 . 388 ANDRZEJ KISIELEWICZ
THEOREM 4.1. Let G : Sn be an intransiti¨e group on a set ⍀ s ⌽ j ⌿, ⌽ ⌿ where ⌽ and ⌿ are disjoint fixed blocks of G. Let G12s G and G s G be respecti¨e constituents of G, and H12, H subgroups induced in G1 and G 2, respecti¨ely, by those permutations which lea¨e each point of ⌿ or ⌽ indi¨idually fixed. Then H12 and H are normal subgroups of G12 and G , respecti¨ely, and GsŽ.Ž.G11rH=G22rH with the isomorphism defined by Ž.H1s ⌿ ⌿ H2 ,where is any permutation in G inducing in G1 and is the permutation induced in G2 by . The intransitive product is a good tool to construct further nonrepre- sentable groups.
THEOREM 4.2. Let G s Ž.Ž.G11rH =G22rH : Snm= S . If one of H1 or H2 is not representable, then G is not representable.
Proof. Suppose that H1 is not representable. Then, there is H : Sn with ⌰Ž.H s ⌰ ŽH11 ., H > H , and H / H 1. It follows that there is g n HyH11preserving orbits of OHŽ., i.e., such that for every x g 2 there is g H1 such that x s x . Suppose now, to the contrary, that G is representable and consider the
permutation Ž. ,1 ;Snm=S , where 1 is the identity permutation. For nm every z s xyŽ x g 2 , y g 2 .we have
Ž ,1. Ž,1. z sx ysx ysz .
It follows that Ž , 1 . preserves the orbits of Ž. , 1 , and since Ž. ,1 gG,it preserves the orbits of G. Now, since G is representable, Ž. ,1 gG.It follows from the definition of = that g H1, a contradiction.
An analogous claim for G12and G is not true. Later we will see that even if both G12and G are not representable, Ž.Ž.G11rH =G22rH may be representable. At this moment, we do not know in general whether the intransitive product preserves representability andror 2-representability. It seems to depend heavily on the isomorphism . This suggests we consider special cases, where is given in a simple and obvious form. One of them is just
the direct product defined and considered earlierŽ the case when G11s H and G22s H and is the trivial mapping on one-element sets. . We consider two more natural cases. Let H be a normal subgroup of a group G. Then, product Ž.GrH = Ž.GrH, where is the identity mapping on GrH, will be called the Ž2. parallel power of G by H and denoted Ž.GrH . It is easy to see that 2 Ž2. Ž2.2 HsH=His a normal subgroup of Ž.GrH , and Ž.GrH rH is SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 389
isomorphic in a natural way with GrH. Using this, we define also by Žn. Žn-.1 induction the nth parallel power: Ž.GrH s Ž.GrH = Ž.GrH , where is the natural isomorphism mapping. If H s Ä41 we will write simply GŽn.. We prove
THEOREM 4.3. Let G : Sbeakn -representable group, and let H be a normal subgroup. If H is representableŽ. i.e., r-representable for some r , then Žm. the parallel powerŽ. GrHisk-representable for all m ) 1.
Proof. We show the proof for m s 2. For m ) 2 the proof is analo- gous. 2 n Ž2. Our aim is to construct a function h: 2 ª k such that ShŽ.s ŽGrH . . n Let A, B g OHŽ.be two orbits of H in 2 . Denote AB s Äxy: x g A, 2 n Ž2. ygB4 :2. Clearly, AB is contained entirely in one orbit of Ž.GrH in 2 n Ž2. 2, since H = H : Ž.GrH . So, to define h we need only to assign suitable values from k to the sets of the form AB. We use the notation hABŽ.for the common value of all elements z g AB and define h in three steps. If A g OH10Ž.and B g OH Ž.then we put hAB Ž .s1. If A g OH0 Ž. 2n and B g OH1Ž.then we put hABŽ.s0. This defines h for all z g 2 having precisely one occurrence of 1. It is easy to note that it follows from
this part of the definition that ShŽ.:Snn=S, i.e., every permutation in ShŽ.is of the form Ž. , s . In the second step of the definition we wish to make sure that ShŽ.:G =G. To this end, let SfŽ.sG, where f is a k-valued boolean function. If n BsŽ1.Žsthe unique orbit in OHnŽ ..then we put hABŽ.Ž.sfA, and if n AsŽ1.Ž then we put hAB.sfBŽ.Ž. Now, if , .g ShŽ.and, for exmaple, f G, then there is X g OHŽ.such that fX Ž./fX Ž .. Con- n sequently, hXBŽ./hX Ž B.Žfor B s 1. , a contradiction. We have used here a simple fact that if H is a normal subgroup of G, then X g OHŽ.for every X g OHŽ.. In the last step for every A, B g OHiŽ.,1FiFuvnr2 , we put hABŽ.s1 whenever A s B, and h s 0, otherwise. Ž2. Ž2. Now, first observe that obviously Ž.GrH : ShŽ., since ŽGrH . con- sists of all permutations Ž. , g G = G such that s for some g H. To prove the converse, let g ShŽ.. From what we have observed so far sŽ.,for some , g G. Suppose, to the contrary, that f H. 1 1 Then, using the fact that Žy , y .Ž.g Sh, we infer that Ž.Ž.␦,1 gSh, 1 where ␦ s y and 1 is the identity. Moreover, ␦ f H. ␦ Whence, there is A g OHiŽ.such that A / A . We may restrict 1 F i uv Fnr2 , since levels Oinand O yiare symmetrical with regard to action of 390 ANDRZEJ KISIELEWICZ the permutation group. In view of our definition, we have hAAŽ.s1, ␦ while hAAŽ .Žs0, contradicting the fact that ␦,1.gShŽ.. This com- pletes the proof.
Note that the restriction 1 F i F uvnr2 used in the proof above makes Ž2. room for an improvement on the rank k of representability of Ž.GrH , similar to that in Theorem 3.1. We leave it to the reader.
In general, the action of the product Ž.Ž.G11rH =G22rH depends on isomorphism and may not be easy to visualize. There is still a case, when this action is relatively transparent, namely, when Gi is a cyclic extension of Hi. Even if we do not know precisely what is the isomorphism in such a case, we are able to prove the following.
THEOREM 4.4. Let G1 : Sn, G2 : Sbekm -representable groups, and let H12, H be their normal subgroups, respecti¨ely, such that factor groups G11rH,G 22rH are cyclic groups of the same order r. If H12, H are representable, thenŽ.Ž. G11rH =G22rHisk-representableŽ for e¨ery iso- morphism of the factor groups.. Proof. The proof is similar to the proof of Theorem 4.3. The only essential differenceŽ. and the main difficulty is the third step in the definition of function h and therefore we concentrate on this point. nqm We define a function h: 2 ª k such that ShŽ.s ŽG11rH .= Ž.G22rH. If A g OH11Ž.and B g OH0 Ž. 2then we put hAB Ž .s1. If A g OH01Ž. and B g OH12Ž.then we put hABŽ.s0. Let G1s SfŽ. 1,G 2sSf Ž. 2, m where f12, f are k-valued boolean functions. If B s Ž1. then we put n hABŽ.sfA12 Ž., and if A s Ž1.Ž then we put hAB.ŽsfB.. From these parts of the definition it follows, as in the proof of Theorem 4.3, that
ShŽ.:G12=G. To complete the definition of h we consider the orbits of G12and G in nm 2and 2 , respectively. For every orbit X g OGiŽ.Ž11FiFuvnr2. we fix an orbit A s A0 g OHiŽ.111which is contained in X. Since G rH is a cyclic group of order r, there is ␥ g G11such that every coset of H in G1 u is of the form ␥ H1, where 0 F u - r. It is not difficult to see that
qy1 X A s D ¨ ¨s0
␥ ¨ Ž. is a disjoint sum of q orbits A¨ s A g OHi 1 , and q is a divisor of r. Similarly, for every orbit Y g OGjŽ.Ž11FjFuvmr2. we fix an orbit BsB0 gOHjŽ.2 which is contained in Y. Here, every coset of H22in G u is of the form H2 , where 0 F u - r. Moreover, we may choose such SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 391
that H21is the image of ␥ H under isomorphism . Then,
sy1 YB, sD¨ ¨s0
¨Ž. where B¨ s B g OHj 2 , and s is a divisor of r. Now, for every X and Y as above, let d s Ž.q, s be the greatest common Ž. Ž . divisor of q and s. We put hAB¨ u s1 whenever ¨ ' u mod d , and hs0, otherwise. Clearly, the permutation Ž.␥, preserves a so defined h. Since Ž.Ž.G11rH =G22rH consists of permutations of the form ¨¨ Ž␥12,.with 1g H 122, g H , and 0 F ¨ - r, it follows that Ž.Ž.Ž.G11rH=G22rH:Sh. ¨ u To prove the converse, let s Ž␥12,.Ž.gSh for some 1g H 1, 22gH,0F¨,u-r, and ¨ / u. In addition, we may assume that, for instance, ¨ ) 0. Then, taking a suitable power of , and multiplying the y1 y1 w result by Ž22, .Ž, we infer that s ␥,1.Ž.gSh for some g H1, 0-w-rwith 1 denoting the identity permutation. MoreoverŽ taking a suitable power again, if necessary. , we may assume that r s wp for some prime p. Now, for X and Y as above, Ž.AB00 sABw000. We have hABŽ.s1, and hAŽ.w B0 s0, unless w ' 0Ž. mod d . Thus to get the desired contra- diction it remains to show that there are X and Y such that the greatest divisor d assigned to them does not divides w s rrp. ␥w To this end, let C g OHŽ.1 be such that C and C are distinctŽ there w must be at least one such C, since ␥ f H11.. Let X be an orbit of G ␥w containing C Žand therefore C , as well. . The number q of orbits of H1 in X cannot be a divisor of w, since otherwise, applying ␥ to C we would ␥ w get C s C . On the other hand q is a divisor of r s wp. It follows that if pc is the highest power of p dividing w then pcq1 divides q. Similarly, w there is D g OHŽ.2 such that D and D are distinct, and the number s cq1 of orbits of H22in Y g OGŽ.containing D is also divisible by p . Hence, d s Ž.q, s does not divide w, as required. ŽBecause of the symmetry we can make the restriction i F uvnr2 and jFuvmr2 , which again makes room for some improvement on the rank of representability..
To conclude this section, note that the three special cases we dealt with covered most of the constructions of intransitive permutation groups from the transitive constituents we were able to visualize. The reader may check that it is not easy to give an example not covered by these cases. 392 ANDRZEJ KISIELEWICZ
5. IMPRIMITIVE PRODUCT
In this section we investigate from the same point of view the reduction of imprimitive groups to primitive ones. The well-known reduction result in this casewx 18, Proposition 7.2, p. 13 is far from being satisfactory for us. I have found nowhere the following details of this reduction, even in an implicit form. Let X be an mn-element set Ž.m, n ) 0 . In this section, it will be convenient to denote the elements of X as those of the m = n-matrix,
1 1 2 2 m m X s Ä4a1,...,an,a1,...,an,...,a1,...,an.
We will refer to m rows and n columns of the corresponding matrix i Xswajmx and identify S ns SXŽ.. Let XR be the maximal group permuting whole rows, i.e., XR is the set of those permutations g SXŽ.for which there is g Sm such that i Ži. Žajj. s a for all i, j. Note that, in other words, XRis the parallel Žn. power Sm whose n factors act in the parallel way on the columns of the matrix X.
Let XCCbe the maximal group fixing rows, i.e., X is the set of those permutations g SXŽ.for which there is a function Ž.i, j on indices of i i the matrix such that Ž.i, j g Snjfor every i and Ža .s aŽi,j.for all i, j. m In other words, XCnis the direct power S whose m factors act indepen- dently on the rows of the matrix X.
The subgroup W of SXŽ.generated by union of XRCand X is a well-known wreath product W s SmnX S , where S macts on the row indices and Sn acts on the column indices.Ž Inwx 14, p. 66 the first factor is referred to as acti¨e, and the second one as passi¨e; note that many authors write these factors in the reverse way.. It is easy to observe that W s XXRC,X C is normal in W, and XCRl X s Ä41 . In particular, W is a semidirect product of XRCand X with respect to the action XRCon X induced by conjugationŽ cf.wx 14, p. 66; 8, p. 88. . Every imprimitive permutation group on X for which the rows of X form a system of imprimitivity is a subgroup of W. The details are the following. Let a group G be a subsemigroup of a semigroup S. Then a function of G into S will be called a semihomomorphism if
1 Ž.sy Ž.Ž. for all , g G. We use this notion in a special case when S is the semigroup of all subsets of the wreath product W, and is a function of a subgroup G of XRCinto a set of left cosets of a subgroup of X . SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 393
THEOREM 5.1. Let G be a subgroup of XRC, H a subgroup of X , and a semihomomorphism of G into the set XCCrH of the left cosets of H in X . Then, the set P of all permutations such that g G and g Ž.forms a subgroup of the wreath product W s XXRC.Moreo¨er, Ž.1 s H, and HsXC lP,and H is a normal subgroup of P.
Proof. Let 12, g G, 1g Ž. 1, and 2g Ž. 2. Then Ž1122 .Ž. y1 sŽ.Ž12 2 122.Ž.g 12 11:P, which proves the first claim. Putting in the definition of semihomomorphism s 1 we get Ž.s Ž. Ž.1 , and since all the images are members of XCrH, it follows easily that Ž.1 s H s XC l P. To show that H is normal in P, let 1122g Ž., g Ž., and gŽ.12. Then, by the definition of semihomomorphism, y1 y1 2 1H 22HsH. It follows, in particular, that 2 1 22HsH. Comparing the left hand sides we get H22Hs 22H, and in conse- quence, H22s 22H.
The group P in the theorem above will be called the imprimiti¨e product of G by H with respect to semihomomorphism and will be denoted by P s G X X H. The group G will be usually identified with the corre- sponding subgroup of SmCacting on row indicies, while H : X will be m identified with the corresponding subgroup of the direct product Sn acting onÄ4 1, 2, . . . , mn . We show that every imprimitive permutation group P is of this form.
THEOREM 5.2. Let P be an imprimiti¨e permutation group on the matrix i X s waj x such that the rows of X form a system of imprimiti¨ity of P. Let H s P l XCR and G be the subgroup of X consisting of all those permuta- tions g XRC for which there is g X such that g P. By Ž.we denote the set of all permutations with this property. Then is a semihomo- morphism of G into the set XCCrH of all the left cosets of H in X and PsGXXH.
Proof. First we prove that is a function into XCrH. To this end suppose that g Ž.with g G. In particular, g XC and g P. Then g P for every g H. Hence, Ž.=H. To prove the equal- ity, suppose, in addition, that g Ž., that is, g XC and g P. Then y1 y1 y1 Ž.Ž.gP, and consequently, gP. It follows that gPlXC sH, and hence g H, as required. We show that for all 12, g G the condition Ž.12s y1 2122Ž. Ž.holds. Let 1g Ž. 1and 2g Ž. 2.Then Ž.Ž.11, 22gPand Ž.Ž.11 22s for some g XRCand g X . This is so, because P : W s XXRC. Moreover, since XCis a normal 394 ANDRZEJ KISIELEWICZ subgroup of W, we have also s 12gG. It follows that
y1 s2122 and g Ž.12. Since is a function into XCrH, Ž.12sH. Whence Ž. 12s y1 y1 y1 2 1 22H: 2 1H 22Hs 2Ž. 1 2 Ž. 2H. Since the choice of 111gŽ.sH was arbitrary, it follows that inclusion above is im- proper, thus showing that is an semihomomorphism and completing the proof. It is usually not easy to find and visualise an imprimitive product with a nontrivial semihomomorphism. A relatively easy example is the imprimi- Ž4. tive product D42X X C of dihedral group D44: S permuting rows by Ž4. the parallel power H s C28: S with the semihomomorphism gener- ated by Ž.Ž.s 1, 2 H for s Ž1, 2, 3, 4 . and Ž.sHfor s Ž.Ž.1, 4 2, 3 . We leave the reader to check that this definition is correct and D4 X X Ž4. C24is a subgroup of S X C2. Note that here, neither the image of D4by 4 is a subgroup of C2 , nor is a homomorphism. The simplest case of imprimitive product arises when is a constant function of G : XRCwhose only value is a subgroup H : X . It is easy to see that such a function is a semihomomorphism if and only if G is contained entirely in the normalizer NHŽ.of H in the wreath product
W s XXRC. Indeed, from the condition defining semihomomorphism, for 1 1 ,gG, we get H s y H H, and consequently, H = y H ,asre- quired. m Identifying G with a subgroup of Smn, and H with a subgroup of S we see that the condition with the normalizer NHŽ.formulated above is m Ž m. trivially satisfied in two cases: when H s K or H s K . In the first case, G X X H is a usual wreath product G X K. In the second case, we call G X X H the parallel wreath product and denote it briefly G XXK. ŽNote that G XXK acts on X just like the direct product G = K with G permuting rows and K permuting whole columns; cf.wx 14, p. 64 .. Generally, a similar result to that for intransitive products holds.
THEOREM 5.3. Let K s G X X H : SmnX S . If H is not representable, then K is not representable. For the usual wreath product it was observed inw 6, Theorem 15Ž. 1x . Proof. Suppose that H is not representable. Then, there is f H preserving orbits of OHŽ.. It follows that, in particular, g XC . Hence, by the definition of an imprimitive product, f K. Yet, since K = H, every orbit in OKŽ.is a union of orbits in OH Ž.. Therefore f K preserves orbits of OKŽ., which proves that K is not representable. SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 395
On the other hand, even if G is not representable, G X X H may be representable. An easy example to check is C33X S . For special cases distinguished above the following holds.
THEOREM 5.4. Let G : Smn and let H : Sbek-representable groups for some k G 2. Then both the standard and parallel wreath products G X H and G XXH are k-representable except for the case D s S22XXS .
Proof. Let G s SgŽ.and H s Sh Ž., where g, h are k-valued boolean functions. We may assume that m, n ) 1. A k-valued boolean function f such that SfŽ.sGXHis constructed as follows. mn n Let x s xx12... xmig2 with x g 2 for all 1 F i F m. n n If x jis 1 for some 1 F j F n and x s 0 for i / j, then we put fxŽ.s1. Otherwise, if there are precisely n occurrences of 1 in x, we put fxŽ.s0. This guarantees that Sf Ž.:SmnXS. nn If every xi is of the form 1 or 0Ž. and the previous case does not hold then we put fxŽ.sgx Ž.ˆˆ, where x is obtained from x by replacing each xi by a single 1 or 0, respectively. This guarantees that permutations in SfŽ. permuting whole rows are precisely those from G, provided that the orbit
OG1Ž.is handled, too, in some other way. To this aim, if x has precisely one occurrence of 1, we put fxŽ.sgx Ž., where x is obtained from x by n replacing xiiby 0, whenever x s 0 , and by 1, otherwise. This guarantees that SfŽ.:GXSn. n Now, if every xijis of the form 1 except for precisely one x Žwhich may nn be arbitrary, but other than 0 and 1.Ž , then we put fx.shxŽj .. This m m guarantees that permutations in Sn l SfŽ.are precisely those from H . Consequently, if we put fxŽ.s0 in all other cases, we get SfŽ.sGX H, as required in the first claim. To get SfŽ.sGXXH we modify f as follows. if there are precisely m occurrences of 1 in x, and moreover xijs x for all 1 F i, j F n Ži.e., 1 occurs once in each xiand always at the same place.Ž , then we change fx. m for fxŽ.s1. This guarantees that permutations in Sn l SfŽ.act in ‘‘parallel way’’ on rows, that is, are precisely those from H Ž m., providing that m is not a multiple of n Ži.e., the present case does not overlap the previous cases.Ž. . Then, Sf sGXXH, are required. If m is a multiple of n we may recover by putting fxŽ.s1 for only those x that have precisely m y 1 occurrences of 1 and all xi equal but one. This works well except for m s 2. Hence, our second claim is proved in all cases except for m s n s 2. The only nontrivial parallel wreath product in this case is just D s S2 XX S2 . From Theorem 2.3 we know that in this case our general claim is not true. 396 ANDRZEJ KISIELEWICZ
One sees easily that for larger m, n there is room in our proof for improvements like those in Theorem 3.1. In contrast, for m s n s 2it turned out to be too little room for our argument. Our result shows that the group D is indeed exceptional from the point of view of representabil- ity by boolean functions. It is the only permutation group that can be obtained by natural and simple construction from known groups, that is representable, but is not 2-representable. As in Section 4 we have also a result concerning imprimitive products with nontrivial semihomomorphism. This concerns a case of the following generalization of the parallel wreath product.
Suppose that G is a subgroup of SmRŽ.identified with a subgroup of X Žm. m and K s H is a subgroup of Sn . Suppose also that is a semihomo- Žm. Žm. m morphism of G into the set SnnrK. Note that S is a subgroup of S n Žm. containing K. Note also that every permutation in Sn Žcorresponding to permuting whole columns. commutes with every permutation in XR. Therefore the semihomomorphism condition reading in this case
Ž.s Ž.Ž. means that is a homomorphism. Moreover, as it is easy to check, the Žm. Žm. image Ž.G is contained in NKnnŽ.rK, where NKŽ.is the normalizer Žm. of K in Sn . Obviously, inducesŽ. and can be viewed as a homomor- phism of G into NHŽ.rH, where NH Ž.is the normalizer of H in Sn. Conversely, suppose we are given G : Smn, H : S , and a homomor- phism of G into NHŽ.rH. Then the homomorphism of XR into Ž m. Žm. Sn rH induced by is a semihomomorphism. The imprimitive product Žm. GXXHwill be called theŽ. general parallel wreath product with respect to homomorphism and denoted briefly G XX H. Note that G XX H is a subgroup of G XXSn all of whose elements Žwith Žm. g XRn, g S . satisfy s . We note that the construction of metacirculant graphs inwx 1 is based on permutation groups which are general parallel wreath products of cycle groups. The following result applies, in particular, for such groups, and more generally, for all general parallel wreath products with the first factor being a semiregular permutation group.
THEOREM 5.5. Let G : Smn, H : Sbek-representable groups for some kG2, and let be a homomorphism of G into NŽ. H rH. If the in¨erse y1 image Ž.H contains a subgroup Ga of G fixing a letter a g Ä41, 2, . . . , m then the general parallel wreath product G XX H is also k-representable, pro¨ided it is different from D s S22XXS . Proof. As in the proof of the previous theorem, we assume that GsSgŽ.and H s Sh Ž., where g and h are k-valued boolean function, SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 397
and m, n ) 1. For simplicity we assume also that a s 1. A k-valued boolean function f such that SfŽ.sGXX H is defined as follows. We start from the function f constructed as in the proof of Theorem 5.4
and such that SfŽ.sGXXSn. Ž It depends on the choice of the function hЈsatisfying Sn s GhŽ.Ј; we assume here that hЈŽ.x s 0 for all x.. mn n Now, we modify f. Let x s xx12... xmig2 with x g 2 for all n 1FiFm, and suppose that every xi s 1 except for x1 that is different nn from 1 and also from 0 . For every such x we put fxŽ.shx Ž1 .and fxŽ .Ž.sfxfor all g G XX H. Otherwise, f is left unchanged. First we show that f is well-defined. Suppose that x and y are like x ␥ ␦ described above, and x s y for some ␥, ␦ g G XX H. To the contrary, we suppose also that fxŽ./fy Ž.. Then there is g G XX H such that n x s y . Since xi / 1 if and only if i s 1 and y is of the same form, it follows that keeps the block of the first n lettersŽ the first row of the matrix X . fixed. By definition, we have s with g G˜˜and g Ž., where G is a subgroup of XR corresponding to G, and is the homomorphism of G˜ Žm.Žm. y1 into Sn rH induced by . The fact that Ž.H contains G1 is Žm. equivalent to that g H whenever fixes the first row. Consequently, if fixes the block of the first n letters, then fxŽ.s fyŽ .Žsfy.Ž.sfy, which is a contradiction proving that f is well- defined. Now, it is easy to see that G XX H:SfŽ.. To prove the equality, ˜ suppose that g SfŽ.:GXXSn with g G and f Ž.. We have Žm.Žm. Ž.s00H for some g Snand 0g G XXH. It follows that y1y1 00s g SfŽ.Žwe use the fact that the factors commute here. . It y1 follows that for s 0 and x as above, fxŽ.sfx Ž .Ž.. Since fxs Žm.Žm. hxŽ.1 , we infer that g H , and consequently, g 0 H , a contradic- tion.
6. APPLICATIONS
As the first application of our approach we give a new proof of the algebraic result contained in Theorem 17 fromwx 6 . Let G s wx be a permutation group generated by a single permutation . Let s 12...k be a decomposition of into k disjoint cycles of length l12, l ,...,lkG2, respectively. Then the group G is 2-representable if and only if the following conditions are satisfied for all 1 F i F k
lijs 3 « Ž.᭚j / i Ž.3 lijs 4 « Ž.᭚j / i Ž.gcdŽ. 4, l / 12Ž. 398 ANDRZEJ KISIELEWICZ and lijs 5 « Ž.᭚j / i Ž.5 The proof given inwx 6 is long and tedious. Moreover, most of it Ž.including the most complicated case is left to the reader. We give a simpler and more transparent proof based on one of our reduction theorems and on applying connection with the automorphism groups of graphs. THEOREM 6.1Ž P. Clote and E Kranakiswx 6. . Let G s wx be a permuta- tion group generated by a permutation which is a product of k disjoint cycles of length l12, l ,...,lkG2, respecti¨ely. The group G is 2-representable if for all s s 3, 4, 5 and 1 F i F k the equality li s s implies that there is j / i such that gcdŽ.lij, l / 1. Otherwise, the group G is not representable, at all. Proof. First suppose that the condition in the theorem does not hold. For example, suppose that l1 s 3 and l j is not divisible by 3, for j ) 1. m Then, for some m, s is a 3-element cycle, and G is a direct product of C3 s wx and a group GЈ acting on the remaining elements. In view of Theorem 3.1 and remarks concerning representability of cycle groups G is not representable. An analogous argument works for s s 4, 5. To prove the converse we show first the representability of G in some special cases. Let s 12...kbe the decomposition of into disjoint cycles. Ž k. First suppose that each i is a 3-element cycle. Then G s C3 . Suppose that G acts on the set V s Ä41, 2, . . . , 3k ŽŽi.e., i s 3i q 1, 3i q 2, 3i q 3.. and k G 2. We consider graph P s Ž.V, E , consisting of k circulants 3, i CsŽ.Ž3iq1, 3i q 2, 3i q 3,is0, 1, . . . , k y 1 here, we adopt the con- vention to denote circulant graphs as the corresponding cycles. joined by additional edgesŽŽ.. 3i q j,3 iq1 qj for all i - k y 1 and j s 1, 2, 3 Ž see Fig. 1Ž.. a . Clearly, the automorphism group AP Ž .of P contains G. From Theorem 2.1 we know that APŽ.sSf Ž.for some boolean function f: 3k 2 ª2, and from the proof of this theorem we know that fxŽ.s0, whenever the number of occurrences of 1 in x is other than two. We modify f to exclude some permutations from SfŽ.and obtain G. First, we 3k 3 put fxŽ.s1 for x s 100 y, 010 y, 001 y with y s 0y . This guarantees Žk. thatŽ 1, 2, 3 . is an orbit in Sf Ž., and consequently, that Sf Ž.:S3 . Moreover, we put fxŽ.s1 for x s 100010 y, 01001 y, 001100 y with y s 3ky6 Žk. 0 . This guarantees that SfŽ.:C3 , and consequently, SfŽ.sG,as required. Similarly, we prove that G s wx is 2-representable when each i is a Ž k. 4-element cycle or each i is a 5-element cycle, i.e., both G s C4 and Žk. G s C5 are 2-representable for all k G 2. SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 399 FIGURE 1 Suppose now, that s 12, where 1s Ž.1, 2, 3 and 2s Ž4, 5, . . . , 3Ž..rq1isa3r-element cycle with r G 2.Ž Then G is, in fact, an intransi- tive product of C3 = C3rrH for suitably chosen H and .. Consider 3 graph Q s Ž.V, E , consisting of the circulant C s Ž.1, 2, 3 and the circu- 3r lant C s ŽŽ..4, 5, . . . , 3 r q 1 joined by additional edges ŽŽ..j,3 iq1 qj for all i - r and j s 1, 2, 3ŽŽ.. see Fig. 1 b for r s 2 . Clearly, the automor- phism group AQŽ.of Q satisfies G : AQŽ.:S33=Sr. Similarly, as in the previous case, let f be a boolean function such that SfŽ.sAQ Ž .. Using the argument for the representability of the cycle group Cn in Section 3 it is not difficult now to modify function f so that it does not preserve 3r reflections of C any longer, and consequently, SfŽ.sG. Similarly we prove that G s wx is 2-representable if is a product of disjoint cycles with 4 and 2 s elements or 5 and 5r elements, respectively, for s G 3 and r G 2. The only case we still need to consider is when s Ž.Ž.1, 2, 3, 4 5, 6 . 4 Here, let R be a graph consisting of two circulants C s Ž.1, 2, 3, 4 and 2 CsŽ.5, 6 joined by additional edgesŽ.Ž.Ž.Ž.ŽŽ.. 1, 5 , 2, 6 , 3, 5 , 4, 6 Fig. 1 c , and let f be the boolean function assigned for R by the proof of Theo- rem 2.1. We modify this function putting fxŽ.s1for xs 110010, 011001, 001110, 100101 and for x s 101010, 010101. Then still preserves f, while neither the reflectionŽ. 1, 3 nor the product Ž.Ž. 1, 3 5, 6 preserves so modified f. Consequently, SfŽ.sG. Now the proof proceeds by induction on k.If ks1, then is a cycle and the result follows from remarks in Section 3. If k G 2, then either G is 400 ANDRZEJ KISIELEWICZ one of the form considered above, and hence G is 2-representable, or else there is a cycle of length other than 3, 4, 5 and a permutation on the remaining elements such that s and itself generates a group that is 2-representable by induction hypothesis. In the latter case, by Theo- rem 4.1, G is an intransitive product of cyclic groups generated by and , respectivelyŽ. divided by certain subgroups . Using Theorem 4.4 we con- clude that G is 2-representable, which completes the proof. The theorem above shows that minimal subgroups of Sn are all 2- representable. Using a result ofwx 3 , in w 6, Theorem 14x it was proved also that with known exceptionsŽ. which are not representable maximal groups in Sn are also 2-representable. Since both the formulation of this result and the proof are not quite correct inwx 6Ž because of the false statement in the representation theorem. , we correct this here. Let G be a maximal subgroup of Snn. Then either ⌰Ž.G - ⌰ ŽS .or ⌰Ž.Gs⌰ ŽSni .snq1. In the former, there is a level OGŽ.consisting of more than one orbit. Then, if T g OGiŽ., the function f assigning 1 to all n-tuples in T, and 0, otherwise, satisfies obviously SfŽ.sG. Hence, G is 2-representable. In the latter, G is not representable at all. Moreover, G is set-transitive, i.e., for every two subsets S, SЈ ofÄ4 1, 2, . . . , n of the same cardinality there is a permutation g G such that Ž.S s SЈ. Fromwx 3 we know that in such a case G is either An or else G is a conjugate of one of the following groups: for n s 5, G5 s ÄŽ1, 2, 3, 4, 5 . , Ž 1, 3, 2, 5 .4 ; for n s 6, G6 s ÄŽ1, 2, 3, 4, 5 .Ž , 1, 2 .Ž 3, 5 .Ž , 1, 3, 4, 6, 5 .Ž , 1, 3, 2, 5 .4 ; for n s 9, G9 s ÄŽ1, 2, 5, 4, 6, 7, 3. , Ž 1, 5 . , Ž 2, 9 .Ž 4, 7 .Ž 6, 8 . , Ž 1, 2, 4 .Ž 7, 6, 5 .4. ŽŽG5 may be defined also as 1-dimensional, linear, affine group AGL1 5. over GFŽ.5,G6 as the group of linear transformations PGL2Ž.5 of the projective line over GFŽ.5 , and G9 as the group of semilinear transforma- tions P⌫L2Ž.8 of the projective line over GFŽ..8. Hence, we have the following. THEOREM 6.2Ž P. Clote and E. Kranakiswx 6. . All maximal subgroups of Sn are 2-representable except for the alternating groups An and for n s 5, 6, and 9, respecti¨ely, the conjugates of G56, G , and G 9 defined abo¨e. The results of this paper may be also used to improve algorithms for drawing the lattices of subgroups of a permutation group or listing conju- gacy types of subgroups of Sn. In this paper we do not deal with the algorithmic aspect of the question, so we give only as an example the SYMMETRY GROUPS OF BOOLEAN FUNCTIONS 401 FIG. 2. All conjugacy types of subgroups of S5. classification of subgroups of S5 Ž.see Fig. 2 . While it is a rather tedious task to make up such a list oneselfŽ checking that no conjugacy type is omitted.Ž , combining our results with a list of primitive groups like, e.g., that inwx 17. yields Fig. 2 almost immediately. In Fig. 2 a line going upward from a type X to a type Y means that every group of conjugacy type Y contains a subgroup of conjugacy type X. In notation we apply the convention to ignore the S1 factor in intransitive products. In such a way, analogous diagrams for S43, S , and S 2may be shown here as subdiagrams. Primitive groups are indicated by filled circles. To see how useful our results are the reader may wish to draw the analogous diagrams for S67and S . Using our results it is not difficult to check that the only nonrepre- sentable groups here, apart from Annand C for n ) 2, are AGL1Ž.5, Ž.S33rC=S 2and C 3= S 2. The remaining are 2-representable except for S22XXS , which is 3-representable. 402 ANDRZEJ KISIELEWICZ 7. PROBLEMS The first natural question to ask is whether there are other repre- sentable permutation groups that are not 2-representable, and if so, whether such groups are rather exceptional or there are many of them. Note, that if the latter is trueŽ. as we conjecture , then k-representability could serve as a certain measure of complexity of a permutation group. On the other hand, our approach opens a new area of research that, as we believe, may be quite fruitful. Other natural special cases of intransitive and imprimitive products should be considered. Similar questions as those considered here for symmetry groups of boolean functions should be asked for automorphism groups of graphs and other structures representing permutation groups. For example, my student W. 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