iullimra HHHi ffiS DNHH m m B3M m BBS HnRMa cSJwx kHSKx fflHHBfiH wSSfta Ea HHwhB LIBRARY OF THE UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAICN 5/0.84 no. 59S-6O0 cop. 2. >/tA V T J 6aS ** ^a n uiucdcs-R-73-597 /K*JXt SUBGROUPS OF THE GROUP G n by Carol Shigeko Abe Edwards October, 1973 DEPARTMENT OF COMPUTER SCIENCE UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN URBANA, ILLINOIS Digitized by the Internet Archive in 2013 http://archive.org/details/subgroupsofgroup597edwa UIUCDCS-R-73-597 SUBGROUPS OF THE GROUP G n by Carol Shigeko Abe Edwards October, 1973 Department of Computer Science University of Illinois at Urbana-Champaign Urbana, Illinois 6l801 Ill ACKNOWLEDGMENT The author is grateful to Professor Franz E. Hohn without whose encouragement, wise counsel, and generous patience graduate study and this paper would never have "been completed. The author would also like to thank Professor Wilson M. Zaring for his encouragement over the years and The National Science Foundation for support during parts of her graduate work. IV TABLE OF CONTENTS CHAPTER PAGE I INTRODUCTION: STATEMENT AND HISTORY OF THE PROBLEM. 1 II SUBGROUPS DETERMINED BY SETS OF BOOLEAN FUNCTIONS 18 III ASSOCIATED LATTICE STRUCTURES 56 IV CONCLUSION 62 LIST OF REFERENCES 6k VITA 68 LIST OF SYMBOLS G Group of permutations and complementations on n independent variables S Symmetric group on n letters C Group of complementations on n independent variables B The set of Boolean functions of n independent variables n cp,\|/,7 Elements of G ti Elements of S 1 n I Elements of C n f.g Elements of B ,& n < . . . > Subgroup generated by . . . c Is a subset of © Addition modulo 2 x. x. (uncomplemented variable) x. x. (complemented variable) N_ Set of elements of G which leave the Boolean function f fixed f n F Set of Boolean functions of n variables left individually fixed by the elements of H, H being a subgroup of G M-, Set of elements of G which leave the elements of F u individually n „. n n fixed N H Set of elements of G which leave F TT setwise fixed n H VI G^ Set of orbit functions for H H F A set of Boolean functions M_ Subgroup of G which leaves each function of F fixed P. Partition of the minterms associated with function f . 1 1 P Partition of the minterms associated with cp e G cp n k [ P. Partition which is the product of the partitions P, , P p , •••> P v i=l o Orbit function for H; orbit of H m, n. Minterms (in decimal notation) 1 m Complementary minterm to m (in decimal notation) m Image under cp of the minterm m The Boolean function zero L The Boolean function one < Is a subgroup of 4 Is a normal subgroup of U Set union n Set intersection ^ Set of all F„ where H < G H n \ Set of all M^ where H < G \ Set of all N^ where H < G V Disjunction of minterms; sup in & A Inf in & U Sup in ^ n Inf in 1\ VI 1 g Less than or equal to; relation on ^ C Relation on 171 CHAPTER I INTRODUCTION: STATEMENT AND HISTORY OF THE PROBLEM The group of permutations and complementations of n independent variables, denoted G in this thesis, is important in switching and automata theory and has been studied from a number of points of view. Geometrically- this group is the group or rotations of the n-cube, and its members permute the set of Boolean functions by permuting or comple- menting the variables of the functions. Elementary and general results about the n-cube and Boolean functions may be found in a number of books, including Harrison [14 L McCluskey [28], and Miller [29]. The classification of Boolean functions by group actions was of interest to many researchers in the 1950' s and 1960's. Two Boolean functions of n variables are said to be equivalent with respect to G if there exists a transformation in G which carries one function into the n other. In his 1953 paper, Slepian [43] obtained results about the number of equivalence classes of functions with respect to G by applying results of Polya [33],' [34]. Slepian also pointed out that such equivalent functions have the same physical networks, provided both a variable and its complement are available as inputs. As a result, in design problems only representatives of equivalence classes need to be considered, thus reducing the number of functions which have to be checked. How enormous this reduction is can be found in calculations for three and four variables by the staff of the Harvard Computational Laboratory [19 ]. There are 22 equivalence classes of 3-variable Boolean functions while the number of 2 5 such functions is 2 = 256; there are 1+02 equivalence classes for the 2 = 65.536 Boolean functions of 1+ variables. There is no known formula which gives the number of equivalence classes, but the number can be found for any value of n by using a generating function. Improved generating functions have been developed by Ninomiya [3IL Lorens [25], Harrison [11+], and Jackson and Ankerlin [20]. Equivalence classes with respect to G have also been helpful in the study of codes. Two binary codes are said to be equivalent if one can be transformed into the other by a transformation in G . Results of such * n studies include those of Hamming [12], Gilbert [9], Harper [13], Lindsey [21+], and Steiglitz and Bernstein [1+1+]. Further classification of Boolean functions, primarily by considering groups which have G as a subgroup, has been done by Ninomiya [32], Rouche [37] , Golomb [10], Harrison [16], [IT], and Stone and Jackson [1+5]. Function complementation and affine transformations are two operations considered in addition to those in G . Invertible Boolean functions, those n ' for which every output is the result of one and only one input, were investigated by Loresn [25] and Harrison [15 ]• The symmetric group on n letters, S , is a subgroup of G and was the group used by early researchers on the problem of invar iance of Boolean functions. Shannon [1+0] first recognized the possibility of complementing some of the variables before testing for symmetry. He showed that Boolean functions which are invariant under any permutations of their variables, that is, functions which are symmetric, can be realized by more economical circuits than the best series-par allel relay circuit. This property for partially symmetric functions, functions which are fixed under all permutations of some proper subset of their variables, and symmetric functions was further explored by Shannon [1+2], Washburn [1+7], Keister, Ritchie, and Washburn [22], and Epstein [7]. Techniques for detecting and counting symmetric and partially symmetric functions were developed by many individuals, including Povarov [35], [36], Caldwell [3], Lee [23], Marcus [26], Sagolivich [38], Elspas [6], Kautz [21], Choudhury and Basu [1+], Arnold and Harrison [1], Mukhopadhyay [30], Sheng [1+2], and Dietmeyer and Schneider [5]. McCluskey [27] presented a method for determining the subgroup of G which leaves a given Boolean function invariant. In Forbes [8; p. 1+] the concept of symmetry is extended to say that "a Boolean function possesses some symmetry if it is invariant when acted upon by some nonidentity element of G ." In her work (p. 5), Forbes answers the more general question: "if a function is known to be fixed by a cyclic subgroup of G , but nothing else is known about the function, can it be shown that it is also fixed by some larger subgroup of G ?" She determines the largest subgroup of G which fixes individually the set of functions which are invariant under application of elements of a cyclic subgroup of G . In this thesis this problem is extended to determining, for an arbitrary subgroup H of G , the largest subgroup of G which fixes every function that H fixes. In addition, we also determine the largest subgroup of G which leaves setwise invariant the set of functions fixed by an arbitrary subgroup H of G . Some interesting properties of these maximal subgroups are established, as are some properties of associated lattices. We begin by defining the groups in question precisely and proving some results about G which are useful in later chapters and should also be of value in the further study of G . Let B be the set of Boolean functions of n independent variables { x, , x , ..., x }. The complement of x. will be denoted x. for i s 1, 2, . . . t n. In this study the word "function" will mean "Boolean function" unless otherwise specified. Let G be the group of permutations and complementations of n independent variables. G is the set of bij'ective mappings from the set f x, , x^, .... x , X-. , x^, .... x } onto itself where every cp in G is 1 1' 2 ' ' n' 1' 2 ' ' n } T n subject to the following conditions: For all i and j in {1, 2, . .., n} , 1. if cp(x. ) = x. then cp(x. ) = x., i J l 3 2. if cp(x. ) = x. then cp(x. ) = x.. l J i J Slepian [k3l pointed out that G is isomorphic to the hyperoctahedral group, the group of symmetries of the hyperoctahedron in Euclidean n- space, and stated and expanded results about this group that had been proved by Young [Jj-9] and Todd [>6]. Let S as usual denote the symmetric group on n letters. Using the cyclic notation for elements of this group, we can compute the action of an element ti of S on the variables and their complements, n Example . If r\ = (123), then T}( Xl ) = x g , tj^) = x^ tj(x 2 ) = x , n(x" 2 ) = xy T)(x ) = x ±f r](x ) = x ± , and t]( x .) = x ± , n(x ± ) = x\, for i = ) 4 , 5, ..., n. Let C be the group of complementations of n variables and let x. = x. and x. = x. . Then this group is conveniently represented by the set of n-tuples of zeros and ones under the operation "©" of coordinate- wise addition modulo 2. If the ith coordinate of an n-tuple is 0, this corresponds to leaving the ith variable x. unchanged. If the ith coordinate is 1, this corresponds to complementing x. . C is isomorphic to this group of n-tuples which is the direct product of n cyclic groups of order 2. It is well known that C is a normal subgroup of G so S is isomorphic n n n ^ to the quotient group G /C . It is also known that G is the semidirect product of C by S so G = S C = C S . It is easy to show that G is also * nnnnnnn n the wreath product of S by S . Lemma 1. If tj € S and | = (i, , i , .... i ) e C , then |ti = ti| ' where n 1' 2 n' n' V = (i ri(l)' Ste)' ""' S(n) } ' Proof : For j = 1, 2, ..., n, |T)(x ) = Oiv = nl'(x.). ■ J Theorem 1. G = S \ S . n 2 *■ n Proof: Since G = S C , we shall show that S C = S J S . By definition, nnn' n n 2 l n J ' s 2 I s = iy * T l I y '■ i 1 * 2 > •-• j n ) "*" S o and 1 £ S )* where for a in (1, 2) and b in (1, 2, ..., xi) , ( 7 *r])(a,b) = ^(a), t)0>)), 7^ = 7 (b) and Sp = {e, (12)} is the group of permutations of {1, 2}. Let be a mapping of S C to S n 1 S defined by: nn 2 L n J 0(r].(i 1 ,i 2 , ..., i n )) = 7 *tj, where r]eS n and (i^ ig, . .. , i Q ) eC n , and 7 is the mapping of {1, 2, ..., n} to S ? which maps j to e if i., = and j to (12) if i . = 1, for j = 1, 2, . . . , n. J 6 We shall show that is an isomorphism. Let y * t) be any element of S„ { S . Consider the n-tuple ( j, , j , . .., j ) where j = if y{\a) = e and j = 1 if ?(k) ■ (12), for k = 1, 2, . .., n. Then by definition of , ic ^T'dj-i* 3o> •••> J )) = 7* 1 )' Therefore, is a surjection, and since all sets involved are finite, $ is also a bisection. Now we show that is a homomorphism. Let (i_ 9 i , ..., i ) and (j-i > 3^t ••', J ) be elements of C and let T]_ and tj be elements of S . vu l' °2' ' °n n 1 2 n Let $(Ti 1 -(i 1 , i 2 , ..., i n )) = y ± *t) ± and O^-^, j 2 , ..., J n )) = 7 2 * V Then $(ti 1 - (i 1 , i gJ ..., 1^) • i] 2 • (j 1 , J* 2 , . .., j Q )) = $ (¥2' (i Ti 2 (l) ®h> \(2) ® h> ~> \(n) ® V ^ Lemma 1. This in turn is equal to y *" r l 1 T )p which is (y *t\ ){y< ? * y \ l? ) , where y maps {1, 2, ..., n] to S 2 by 7 (b) = y ' 7n > for a11 "b in {1, 2, . . . , n} . I n 2 (b) b The mappings in G can be extended to act upon n-tuples of variables by the following definition: Definition 1. For all cp in G , cp( X;L , x 2 , ..., x n ) = (cp( Xl ), cp(x 2 ), ..., cp(x n )). An element cp of G acts on a function f (x, , x„, . . . , x ) to give n 1' 2 7 n another function fcp(x, , x p , ..., x ) where fcp(x, , x„, ..., x ) = f(cp(x 1 , x 2 , ... , x n )). Example . If I \X^, Xg, X , X, ) = X-jX-jX x, V x,x 2 x x, and 9eG, with cp = Tjg where tj = (123 ) e S and g = (1,0,0,0) eC, , then s(x, , X~, X , X, ) - f(x^, X,->, X } X, J - x i x 2 X 3 X ^ V x l x 2 X 3 X ^ fTl( Xl , X 2 , X , X ) - f(x 2 , X , X-^ X ) - X^X-jX^ V XgX^X^ = x-.x x^x. V x, x„x^,x, , and l 2 3 i[ 123 k fcp(x 1 , x 2 , x , x ) = f(x 2 , X , X ± , X ) ^ P -L 4 ^ 2 -L 4- 1 2 3 ij. 1 2 3 4 In general, if cp e G and cp = rj| where T)eS and | = (i n , i_, .... i )e C , £. ^ -r n t ' n 1' 2' n n then X l X 2 X n fcp( Xl , x 2 , ..., x n ) =f(x T1 ^ 1 y Xti(2) , ..., ^ (n) ), where x^ = x and x, = x, for 1 ^ k ^ n. Also, fcp(x x , x 2 , ..., x Q ) = f(r)(|(x 1 , x 2 , ..., x n ))) and firiCx,, x , ..., x ) = f(A ( * } , x^), . x j(n) }- ,x 1' 2' ' n' v T](l) > t](2) ' ' T](n) y Theorem 2 . Each mapping in G (under the extended definition) permutes the set of minterms of the n variables. Proof: Let cp = n| be an element of G where iieS and T ' n n ^1 ^2 ^n I = (i n , i , .... i ) e C , and let f (x, , x , .... x ) = x n x . . .x b v 1' 2' ' n ' rr 1' 2' ' n 12 n be a minterm. Then fcp(x x , x 2 , . .., x n ) = f (cp(x x , x 2 , . .., x n )) = f(n^(x 1 , x 2 , ... , x n )) l l i 2 i n = f(x n(l)^ *n(2)' — ' X Tj(n) ) = X /, \ X_ /~\ ... X S,(l) ^(2) ••' X T](n) ' 1 1 ©3 1 i 2 ©j 2 i n ©j n and x /, s x nf2^ *** x n( ) is a minterm of the n variables since T) e S . n To show that cp is a surjection, define the minterm g(x, , x p , , x ) i l® J 'r 1 (l) i 2 ej T 1 (2) i n ej r 1 (n) _ as x, y x„ ... x . Then 12 n i l i 2 i gcp(x 1 , «g, ..., x n ) =g(x Tl(l) , Xt](2) , ..., x^ n) ) _ v J n(l) J n(2) d Tj(n) = X T)(1) X r}(2) '•• X T)(n) J 'l J 2 J 'n ~ X l X 2 * ' ' x n ' Since the set of minterms is finite, cp is also a bisection. ■ If S n denotes the entire group of permutations of the set of 2 n minterms of n variables, then by Theorem 2, it is clear that G can be embedded isomorphic ally as a subgroup of S n- Denote each minterm of n variables by the decimal name of the binary numeral formed by the exponents 1 — of the x. 's, where x. = x. and x. = x. , 1 ^ i g n. 1 11 1 l' Example. The minterm x n x x,x, x._ is denoted by 12, the decimal name for 1 ± d j 4 p 01100. An element of G can thus be written in the permutation notation of S n . 2 n Example . If cp = T]| e G where T) = (123 ) eS and {■ = (1,0,0) eC , then cp= (2 6570kl5/ = (° 2 3T54)(l6) in the two common notations for permutations. (There are eight minterms when n = 3. ) We order the set of 2 minterms by the usual order of the decimal 1 1 i 2 i names of the minterms. If m denotes the minterm x n x„ ... x n , where 12 n ' i e {0, 1}, k = 1, 2, . .., n, we use cp to denote the minterm which is the K. m X l X 2 X n product cp(x 1 ) cp(x 2 ) ... cp(x n ) There is an algorithm to convert an element of G from the S C - n n n notation to the usual cyclic notation in S n . The algorithm, which is outlined below, is readily adaptable for machine calculations. We begin with the following definition. Definition 2. Let tj e S and £ = (i_ , i_, ..., i ) e C . By ti(|) = n 1' 2' n n ^((i-Lj i 2 ' •"' 1 n) >) We sha11 mean the n- tuple 5' = (j_, j*, ..., j ), where j = i k n _1 (k) Example . If r\ = (123) eS, and | = (1,0,1,1) eC , then tj(0 = (1,1,0,1). The actual computation of t)(£) can be accomplished by permuting the elements of the string | according to rf . So in the example, the first coordinate moves to second place, the second coordinate moves to third place, the third coordinate moves to first place, and the fourth coordinate is fixed. 10 Lemma 2 . Let r\ e S n and let I - (i^ lg, . .., l fl ) and t; = (j^ J g , . . . , J n )eC n . Then T](| = T!(0 © T)(£). Proof : tj(£ © = Ti(k 1 , kg, . .., k n ), where k = i + J tor I ■ 1, 2, . .. , n, (k, , k , , . . . ; k . ) n (i) ^(2) n (n) = (i i © J , , i i © J , , ..., i n © J , ) n (i) n (i) n (2) r 1 - 1 (2) n (n) n (n) = (i n » i -I i •• • , i i ) ?l(l) t 1 - 1 (2) T] (*) n (i) n (s) Ti _1 (n) = ii(6) © T)(0. ■ Theorem 3 . The representation of cp = tj| eG , where r\ e S and | = (i _, i p , . . . , i ) eC in S n >may be found by the following algorithm: Step 1 : Start the first cycle of the representation with minterm 0. Step 2 ; The image of under cp is the minterm whose associated n-tuple is i](£). Step 3 : If the preceding step leads to the minterm whose binary name is i-.j^.-.j then its image under cp is the minterm whose associated °1 2 u n ox n-tuple is the sum ^{{d^, 3 2 , ..., d n )) ©n(l). Step If. : Repeat Step 3 until the cycle is closed. Step ^ • Begin the next cycle with the first minterm which has not yet appeared. (The minterms are ordered by the usual ordering in their decimal names. ) Step 6 : Repeat Steps 3 and if. Step 7: Repeat Steps 5 and 6 until all minterms have been considered. 11 Step 8 : Eliminate the singleton cycles. Example . Before proceeding with the proof of Theorem 3> let us consider an example. Let cp = Tig = (125 ) (1,0,1) eS_C_. The conversion of cp into cyclic 3 3 notation is presented in the following table: Computations according to the algorithm Cyclic notation in S -, By Step 1, (0 By Step 2, (1,0,1) ^^ > (1,1,0) (06 By Step 3, (1,1,0) J—^U (0,1,1); and (065 (0,1,1) (1,1,0) = (1,0,1) By Step k, (1,0,1) ■ (l2 ^ > (1,1,0); and (065) (1.1.0) (1,1,0) = (0,0,0) By Step 5, (065) (1 By Step 6, (0,0,1) ^ 12 ^> (1,0,0); and (065)(12 (1,0,0) e (1,1,0) = (0,1,0) (0,1,0) (123) > (0,0,1); and (065)(l27 (0,0,1) e (1,1,0) = (1,1,1) (1.1.1) (123) > (1,1,1); and (065)(l2?) (1,1,1) © (1,1,0) = (0,0,1) By Step 7, (065) (127) (3 By Step 7, (0,1,1) J£^l> (1,0,1); and (065) (127) (3) (1,0,1) © (1,1,0) = (0,1,1) By step 7, (065) (127) (3) (^ 12 By Step 7, (1,0,0) (l23) > (0,1,0); and (065)(l27)(3)(U) (0,1,0) © (1,1,0) = (1,0,0) By Step 8, (065K127) In actual practice one need not write down all the computational details and the conversion goes very quickly. Proof of Theorem 3 - Let f represent the minterm 0, that is, f (x., , x„, ..., x ) =x n x_...x . Then v 1' 2' ' n 12 n in ig i fcp(x 1 , x 2 , ..., x n ) =f(x Tj(l) , ^ (2) , ..., x n(n) ) i-, ip i = x r](l) X T](2) ••' \(n) 1 -1 ± -1 i 1 12 n because for each k, k = 1, 2, . . . , n, x, = x / \ for some |, 1 ^ £ ^ n, and the exponent of x / \ is i , so £ = tj~ (k). This means that the image of minterm is the minterm with binary name i , i , . . . i , Td) n(2) vT(n) Let g represent the minterm whose binary name is j-,j Q ...j , that is, , s J 'l ^2 ^n g(x, , X^, . . . , X ) = X, X. . . .X &v 1' 2 n' 12 n Then gcp(x 1 , x 2 , ..., x n ) = g( N(i)' x n(2)' ■•■' \{n) ] 13 d l 1 u 2 2 n n = X ri(l) X n(2) •*• X T](n) J n ©In 3 -, ©in On © 1 n 1,^(1) rf^i) ^(2) n' 1 (2) , n (n) n _1 (n) 12 n by the reason in the preceding paragraph and by Lemma 2. I This idea of denoting elements of G as permutations of minterms will prove very useful in the next chapter. From this point on, unless other- wise stated, G will be considered as a subgroup of S with the appropriate notations. For convenience of notation, the Boolean functions of n variables are written as disjunctions of the decimal names of the minterms. Example. The function f (x, ,x ,x_,) = x.,x x^ V x,x x v V x,x x_. is — ^— — L d y A. d y X. d y ± d $ abbreviated to f = 1 V 2 V 6. The mappings in G , considered as permutations of minterms, can be extended to act upon Boolean functions by the following definition: Definition 3 - L e "t f =m V nu V ... V m, be a function of n variables with m. , 1 g i ^ k, the decimal name of the ith minterm, and let cp be an element of G . Then fcp = m,cp V mpCp V ... V m, cp. We now make the definitions which are necessary to state our problem precisely. Definition 1+ . An element cp of G fixes a function f if f = fcp. A- subgroup H of G fixes a function f if every cp in H fixes f. Definition 5 . For a function f, let N„ = {cp | cp e G and f = fcp). Defi nition 6. Let F be the set of n-variable functions fixed by the — — _ _ j-[ subgroup H of G , that is, F = {f | f eB and H fixes f ) . Ik Definition 7 . For any subgroup II of G , let NL = {cp | cp eG and f = f cp for every f in F u ) . n Theorem )| . For any subgroup H of G , M.. is a subgroup of G , and H is a n n n subgroup of K,. Proof ; Let cp and \|r be elements of K,. Clearly, cp^ is also in VL.. Now, f = fcp for every f eF , and so fcp" = f for every f eF . Thus, cp" e M^ and L is a subgroup of G . By definition of F , H fixes every function of F and so H is a subgroup of V As a result of Theorem \, 11. is the largest subgroup of G which fixes every function of F . 11 Definition 8 . For the subgroup H of G , It, is called the group of symmetries for the set of functions fixed by H . There is another group of G which is maximal in a sense and is of interest to us: Definition 9 - For any subgroup H of G , let N = {cp | cp e G and F = cp[F H J); that is, N 7T is the set of all elements of G which leave F v setwise invariant. ' H n H Clearly, HCJLCL, The fact that N is a subgroup of G and that it contains the normalizer of H in G will be proved in the next chapter. We now define the object which will be the basic tool in proving the results to be presented in Chapter II. Definition 10 . Let H be a subgroup of G . The orbit functions for H are the functions computed as follows: List all orbits of H, including the singletons. These orbits will be sets of minterms represented decimally. 15 The elements of H are represented as elements of S n . To each orbit there 2 corresponds an orbit function which is the disjunction of all the minterms in the orbit. Order the orbits according to the smallest minterm in the orbit. (The first orbit is the orbit containing the minterm 0; the second orbit contains the first minterm not in the first orbit; etc. ) The orbit functions are ordered according to the ordering of the orbits from which they are derived. Definition 11 . An orbit function for H is said to be in standard form if it is expressed as a disjunction of minterms, if the minterms of the orbit function are in increasing order from left to right, and if the minterms are distinct. Definition 12 . The length of an orbit of H is the number of minterms in the orbit. The length of an orbit function for H is the length of the orbit from which it is derived. Example. Consider the subgroup H = <(06)(l7), (03)(Vf)> of G_. The orbits 3 of H are: {0, 3, 6), {1, k, 7} , {2), and [5]. The orbit functions of H are: ^ = V 3 V 6, fg = 1 V 1^ V 7, f = 2, and f = 5. They are all written in standard form. The functions f and f have length 3 while f and f, have length 1. k Definition 13 . For a subgroup H of G , let C> denote the set of all orbit ^ — — — x\ n functions for H. Theorem 5 . For the subgroup H of G n _, F H is equal to the set containing the Boolean function C, the orbit functions for H, and all possible disjunctions of orbit functions for H. Proof : Since the function is fixed by every element of G , it is clearly in F u . It is also clear that all the orbit functions are in F TT . If H fixes a H 16 orbit functions f. and f., it obviously fixes f. V f . as well. Therefore, the set containing the function 0, the orbit functions for H, and all possible disjunctions of orbit functions for H is a subset of F„. n Suppose f e F and suppose that f is neither the function nor an orbit function for H. Let the minterms of f be m_ , itu, ..., m, in ascending order. Since H fixes f , M {cp(m. ) | cp g H) must be the set of i=l X all minterms of f. Now, for each i, 1 I i s k, the disjunction of the minterms in {cp(m. ) | cp e H) is an orbit function. Thus f is a disjunction of orbit functions for H. Hence F u is equal to the set containing the n function 0, the orbit functions for H, and all possible disjunctions of orbit functions for H. ■ Corollary 5.1 . For subgroups H and K of G , F = F if and only if H and K have identical orbits. Proof: It is obvious from Theorem 5 that F^ = F T _. if H and K have identical orbits. Suppose F TT = F T . and let o be an orbit function for H. Then o e F„ and o e F T ^. Suppose o is not an orbit function for K. Then o is the disjunction K of at least two distinct orbit functions for K. These two orbit functions must belong to F^ since F = F . Then o is not an orbit function for H. This contradiction shows that every orbit function for H is an orbit function for K. By a similar argument every orbit function for K is an orbit function for H. Since there is a one-to-one correspondence between orbits and orbit functions for a subgroup, H and K have identical orbits. ■ IT Before proceeding to the next chapter we mention the sizes of some of the sets we have defined. It is well-known that the set, B , of Boolean ' n' 2 n functions of n variables has 2 elements. The group of permutations and complementations, G , has 2 n -n! elements, and this has been embedded in the symmetric group S n which has 2 ! elements. If H has k orbits, then F u has p ti 2 elements in it, a result following immediately from Theorem 5- Although all the sets under consideration are finite, and one could determine 1YL and N by determining F and checking each element of G to see which ones fix every function of F H or which ones fix F as a set, respec- tively, the procedure would be most impractical because of the enormous sizes of the sets. In Chapter II an algorithm is presented which determines ML with considerably less computation. Results about N are also presented h. along with some of its relationships to M„. We conclude this chapter with an observation about G . n Theorem 6 . G is a transitive group. Proof : We must show that G has only one orbit, that is, we need to show that there is a minterm such that all other minterms can be obtained from it by applying all elements of G to it. Consider the minterm 0. The elements of G , written in the S C -notation, (l) (0,0,. . . ,0) , (l) (0,0,. . . ,0,1), (1) (0,0,. .. ,0,1,0), ..., (l)(l,l,. . . ,1), applied to minterm give all 2 n minterms . ■ 18 CHAPTER II SUBGROUPS DETERMINED BY SETS OF BOOLEAN FUNCTIONS In this chapter the main theorems of this thesis are proved. Through- out, minterms will be identified by their decimal names. We begin by determining the elements cp of G which fix a given function f : Theorem 7 . Let f e 33L and m, , m~, . . . , ni be the minterms of f. Let S„ be the set of all elements cp of G such that for each cycle of cp, including the trivial ones, the set of the minterms in the cycle is either a subset of {m.. , m p , . . . , m, } or disjoint from it. Then H_ = S . Proof : First, recall that N f = {cp | cp g G and f = f cp} . Let cp e S f . Partition {m, , nu, ..., itl ) according to the sets of minterms determined by the cycles of cp, that is, m. and m. are in the same block of the partition if and only if m. and m. are in the same cycle of cp. Since the cycle sets of cp are either subsets of {m , m , . . . , ni } or disjoint from it, cp leaves each block of the partition setwise invariant. This means that cp leaves {m-. , nu, ..., m,} setwise invariant, and so f cp = (m 1 V mp V . . . V m ) cp = f . Therefore cp e N„ and S is a subset of N f . Next let cp e N and let (n n p ...n ) be a cycle of cp such that [n^ n 2 , ..., np D {11^, m^ . . . , m^.) / 0- Since cp fixes f =m ) V m V ...V m, each n. , 1 ^ i g £, must be one of the m. 's, 19 1 I j Ik. Thus {n , n , ..., n ) is a subset of {m,, m , ..., m,} and cp e S-. The equality of N f and S f follows. ■ Lemma 3 . For f e B , L is a subgroup of G . Proof: If cp and cp are elements of N then it is clear from the definition of N_ that cp cp is also in N f . Then f = f cp from which it follows that fcp" 1 = t, and cp" 1 e N . ■ Corollary 7.1 . If F = {f ± , f g , . . . , f fc ) where f.eB^, i = 1, 2, ..., la, and M= {cp | cp e G and f cp = f . for i = 1, 2, . . . , k] then M= p) N , i=l i and ML, is a subgroup of G . Consider again the set of functions F. With each function f. = m. , V m.„ V ... V m. . we can associate a partition of the minterms 1 ll i2 lk. 1 P. = {m ±1 , m ..., m. k _; m m. k _ ..., ml 1 l+l 1+2 i2 Every element cp of G also determines a partition P of the minterms, each block of P being the set of minterms of a cycle of cp. Then by Theorem 7, f.cp = f. if and only if P IP.. This observation leads to 11 J cp 1 Corollary 7.2 . If F = {f , f , . . . , f } , f . e B , i = 1, 2, . . . , k, and if P. is the partition associated with f . , then ^ = {cp J cp e G n and P 1 JJ P.}. ^ KL k Proof : Every cp such that P ^ [ Pj_ satisfies the condition of Theorem 7 * i"=l for every f . } and every cp e G that satisfies the condition of Theorem 7 20 for every f . defines a partition P which is less than or equal to every k P. and hence is less than or equal to P. . ■ i ] ii x Theorem 8 . For a subgroup H of G n , 1VL = /) N = f| S_. f £ H fcO H Proof : Let cp be an element of ML. Then f = f cp for all f in F . If f is in 0.. then f is also in P„, so cp is an element of N„. Therefore, cp is in ti ti 1 f} N and Lis a subset of (~] N . f £ H feO H Let cp be an element of J '_ ML,. Then f = f cp for all f in 0„. This I € U-rj I n means that f = f cp for all f in F„ by Theorem 5 and the fact that every ti. element of G fixes the function 0. So, cp is in ML and ML. = \ \ N . f e H The second equality is obvious from Theorem 7- ' The first major theorem of this chapter determines M^. for any subgroup H of G . n Theorem 9 - Let H be a subgroup of G and cp an element of G . Then cp is in ML if and only if for each cycle of cp the set of minterms in the cycle is a subset of an orbit of H. Proof : Suppose that for each cycle of cp the set of minterms in the cycle is a subset of an orbit of H. Then cp is in S_ for every f in U by the I ti definition of S . Thus, cp is in f\ N„ and hence cp is a member of ML fcO H by Theorem 8. Conversely, let cp be in 1L and let (m,mp...m, ) be a cycle of cp. By Theorem 8, cp belongs to S for every f in • that is, for all f in 21 and for each cycle of cp, the set of minterms in the cycle is either a subset of the set of minterms in f or disjoint from it. Since the orbits of H exhaust the entire list of minterms there is an orbit of H which has a nonempty intersection with (m, , m p , . .., nO« This means that {jtl , m p , . . . , m, is a subset of this orbit of H. ■ Corollary $.1 . If (l) is the identity element of G then M ' , w = {(l)}. Corollary 9.2 . If H has exactly one orbit, then M H = G . In particular, VL = G . G n n Proof: By Theorem 6, G is transitive. ■ J ' n Corollary 9»3 » If H contains all cp in G for which the set of minterms of each cycle of cp is a subset of an orbit of H, then K, = H. Otherwise, M^H. Corollary 9.k - If H and K are subgroups of G that have identical orbits then Mp, = 1VL.. In particular, M. = 1VL. Corollary 9- 5 > If there is an element t in H such that for each cycle of \|r (including the trivial ones), the set of minterms in that cycle constitutes an orbit of H, then ]yL = M < > . My. has been completely determined for the case when H is a cyclic subgroup of G by Forbes [8]. Corollary 9»6 . For any subgroup H of G , F = F . Proof : By Corollary 5.1, if two subgroups of G have identical orbits then they fix the same set of functions. I 22 Although Theorem 9 and its corollaries completely determine M^ for an arbitrary subgroup H of G , the actual computation of M^ by these facts alone is impractical. We make some further observations which greatly reduce the number of cases which need to be checked and state an algorithm which is useful for hand calculations of NL for small n and can be programmed for machine computation in general. The key to the entire process lies in being able to recognize which elements of S _ are in G . 2 n n Definition 15 . Let m be a minterm of n variables. By the complementary minterm to m , we shall mean the minterm (2 n - l) - m and denote it by in. Theorem 10. Let cp be an element of S „. In order that cp belong to G the following conditions must be satisfied: 1. If cp maps the minterm m. to the minterm m. then cp must map minterm in. to minterm m., and, equivalently, 2. if cp maps minterm m. to minterm in. then cp must map minterm in. to minterm m . . 3 Proof : Let cp belong to G . Suppose that m.cp = m.. The binary notation of in. is obtained from the binary notation of m. by replacing each in m. by 1 and each 1 by 0. Since the means an uncomplemented variable and the 1 a complemented variable, by the definition of G , m.cp =21.. Since every in. = (2 - l) - m. the second condition is equivalent to the first. ■ 3 3 Example. The element (07) (1514.623) eS , satisfies the conditions of the 2 3 theorem and hence may be an element of G^ : is mapped to 7 and 7 is mapped 3 to 0, 1 is mapped to 5 and 6 is mapped to 2, 5 is mapped to k and 2 is mapped to 3, and k is mapped to 6 and 3 is mapped to 1. However, 23 (017562) (3I1) belongs to S but does not belong to G because 1 is mapped 23 3 to 7 but 6 is mapped to 2 instead of to 0. It should be noted that the conditions of Theorem 10 are not sufficient. An example illustrating this appears following Theorem 11. Corollary 10.1 . Let H be a subgroup of G . If {m, , m~, . . . , m, ) is an orbit of H then {m, , m^, . .., in, } is an orbit of H and either i\f ^2> • "f \) = {\> ™ 2 > '"' \* ° r {m 1 , n^, . .. , m^.3 n {t\, mg, ...,11^] = 0. Proof : Suppose {in, , EL, . .., El] ^ {m, , m~, ... , m, } . Then there is an m. in {m, , nip, . .., m,) such that in. / {m, , iru, . .., m,] . There is an orbit o of H such that m. e o. We shall show that {m, , SL, . . . , in, ) = o. Let in. e {EL, EL, . .. , in, } . By the definition of orbit and Theorem 10, there is a f.eH such that in. \|r . = EL. This means that in. and in. belong to the same orbit of H, namely o. Thus, {in , EL, ..., nl] £ o. Let m e o. If m = in. then m e {in, , in , . . . , m, } . If m ^ in. , there is a |eH such that m.ty = m. Thus, m.i|/ = in by Theorem 10 and in e {m, , iru, . . . , m, } , say in = m.. This means that m = in. e {EL , EL, . . . , m, } and o c {m f EL, . . . , in, ) . ■ Theorem 11. Let cp be an element of G and write cp in the two row notation, n ' 1 2 ... 2 n -l l cp cp cp cp /■ Parti tion the string cp^cp^ . .cp into pairs starting at the left. For each pair cp. cp. so obtained from a given string, the n-tuple for cp. differs from the n-tuple for cp. in exactly one coordinate, and in this same coordinate for i =1, 3, 5, ..., 2 n -l. 2k Proof: In the S C -notation, cp = t}£ for some r\ e S n and £ eC n . By- Theorem 3> 9-1 i s the minterm whose associated n-tuple is tj( (0,0,. . . ,0,1)) © ti(5)» Now tj( (0,0,. . . ,0,1)) has exactly one 1 so all the remaining coordinates are 0. Hence cp differs from cp in exactly one coordinate, say the £th coordinate, 1 ^ I % n. For i = 3, 5, 1, ••-, 2 -1, cp. = T] (n-tuple representing i) © T](£) and cp. = T) (n-tuple representing i-l) © n(i). The binary names of i and i-1, for i = 3* 5> It ••> 2 -1, differ in exactly one position, for i has a 1 as the nth coordinate and i-l has a as nth coordinate. Hence, x\ (n-tuple representing i) and r\ (n-tuple representing i-l) differ in exactly one coordinate, the £th coordinate—the same way that cp differs from cp, . ■ Example . The permutation cp = (036K17U) =1^ 2 ^ 1 5 1 e S V The conditions about complementary minterms in Theorem 10 hold for cp. Since 7-3=6-2=5-1=^-0= '4, the condition of Theorem 11 is met so cp may belong to G . On the other hand, the element of S j, : * = (0,9)(6,l5)(2,ll)(3,l0)a,13)(5A2) 1 2 3 k 5 6 7 a 9 10 11 12 13 ik 15 9 1 11 10 13 12 15 7 8 3 2 5 k ik 6 satisfies the conditions of Theorem 10 but does not belong to G. because 1-9/10-11. The element /o 1 2 3 ^ 5 6 7 8 9 10 11 12 13 14 15 7 \o 1 6 7 10 11 2 3 12 13 4 5 8 9 1J+ 15 = (2,6) (3,7) (4,10) (5,11) (8. 12) (9, 13) belongs to S , , and a quick check shows that 7 satisfies the conditions of 2 k 25 both Theorems 10 and 11. However, these conditions are not sufficient for membership in G, since y / G. . This situation is remedied by the following theorem. Theorem 12. Let cp e S n and write cp in the two-row notation 2 xi / 1 2 ... 2 n -l \ / ) . Then cp e G if and only if the following are true: ^ cp Q cp 1 cp 2 ... ( p 2n _ i / n 1. For i =0, 1, 2, . .., n-1, cp ; and cp differ in exactly one 2 ° coordinate in binary notation and cp - cp = ± 2 for some k , £ k £ n-1. Moreover, if 1 g £ g n-1, then k / k , for m = 0, 1, .... £-1. 2. If 1 S i g n-1, then for p = 2. 1 + 1, 2 l . 2, . . . , 2 i+1 - 1, 9 p - cp = (q> 2 , - cp Q ) + (cp^ - cp Q ). Example . Before proving Theorem 12, we give an example. Consider again, cp = (036X171O = (S 7 2 6 1 5 M in S y The conditions for complemen- tary minterms in Theorem 10 hold, and for this it is sufficient to check only the first half of the string. Thus (036)(l7^) belongs to G because 3 7-5=k/2-3=-l>6-3=3=^+-l. On the other hand, the element of S . : 2 4 (0, 9) (6, 15) (2, 11) (3A0) (4, 13) (5, 12) 1 2 3 h 5 6 7 8 9 10 11 12 13 Ik 15 9 1 11 10 13 12 15 7 8 3 2 5 '4 l!+ 6 satisfies the complementary minterm conditions, but does not belong to G. because 1 - 9 = -8, 11 - 9 = 2, and 10 - 9 = 1 / -8 + 2. Proof of Theorem 12 : We shall show first that any element cp of G must satisfy the given conditions. Considering cp in the S C notation, cp = T]| where r \eS> n and £ eC^. Let I e {0, 1, 2, ... , n-1). By Theorem 3, cp = the 26 minterm whose binary name is t](0 and cp - - the minterm whose binary name is T](n-tuple for 2*) © ^U)- Now > f](n-tuple for 2*) has a 1 in exactly one coordinate, the (n - k«)th coordinate, for some k., O^k^ n-1, with the remaining coordinates 0. If t)(£) has a as (n - k )th coordinate, then T)(n-tuple for 2*) © ^(0 and tj(|) differ in exactly one coordinate, the (n - k )th coordinate. If T)(|) has a 1 as (n - k^)th coordinate, the (n - k )th coordinate of T)(n-tuple for 2*) ©?]((;) is with the remaining coordinates identical to the corresponding coordinates of t)(|). Hence, in this case also, T](n- tuple for 2^) © n(l) and T](i) differ in exactly one coordinate, the (n - k )th coordinate. Therefore, cp and cp differ in k, exactly one coordinate in binary notation, and cp - cp = ± 2 , where 2 J! Ik ^ n-1. Since, moreover, cp is bijective, if J e {1, 2, ..., n-1) , k, d k for m = 0, 1, 2, ..., £-1. Thus any element cp of G^ satisfies condition 1. Let p e {2* + 1, 2^ + 2, ..., 2 i+1 - 1}. Then cp + cp - cp = the number whose binary name is 2 p-2 [(n(n-tuple for 2*) © n(§)) - n(S) + (n.(n- tuple for p - 2 £ ) © !}(§))]• Since T](n- tuple for 2*) has exactly one 1 among its coordinates, the- (n - k.)th coordinate, with the remaining coordinates 0, T](n- tuple for p-2*) must have a as (n - k )th coordinate. If T](|) has a as (n - k )th coordinate then [T](n- tuple for 2 l ) © l(i)] -.t)(|) = T](n- tuple for 2^). 27 In this case, T)(n- tuple for p - 2 ) © ?)(!) has (n - k.)th coordinate 0, so n(n-tuple for 2-0 + [r](n- tuple for p - 2 l ) © n(i)] = T)(n- tuple for 2. 1 ) © Tj(n- tuple for p - 2^) © n(s) = T)(n- tuple for p) © n(£)> If T](l) has a 1 as (n - k )th coordinate then [ri(n- tuple for 2 1 ) © n(£)J - n(l) = - T](n-tuple for 2^), and T](n- tuple for p - 2^) © n(£) has a 1 as (n - k )th coordinate. Thus - T)(n- tuple for 2 l ) + [r](n-tuple for p - 2 £ ) © T)(|)] = T](n-tuple for 2^) © T)(n- tuple for p - 2*) © n(D = T)(n-tuple for p) © t)(£). Therefore, in either case, 9 - cp + cp = the number whose binary name is 2 £ p-2 r)(n-tuple for p) © T](|) = 9 , P and cp also satisfies condition 2. We now show conversely that if an element cp of S n satisfies 2 conditions 1 and 2, then cp is an element of G . We accomplish this by f inding T) in S and g in C such that cp = T]|. Since cp € S _. there is a n n o 28 minterm m, £ m ^ 2 n - 1, such that cp =0. Let £ = (n-tuple for m) = (i, , i^, .... i ). Clearly £eC . v 1' 2 n' J n We define an operator r\ as follows : By the first condition, for l = 0, 1, 2, . .., n-1, cp-cp=±2* for some k^, ^ k g a-1, and, if £ £ {1, 2, . . . , n-1] , k 7/ k for m - 0, 1, . . . , |-1. For each |, £ =0, 1, . .. , n-1, let T) (n- tuple for 2^) = n-tuple for 2 l = I n-tuple for cp - cp _ I . Reinterpreting t) as a permutation on n letters we see that, for each £, £ =0, 1, . .., n-1, T] maps (n - f) to (n - k ) because the n-tuple for 2^ has exactly one 1 and the n-tuple for 2 * has exactly one 1. Since the k.'s are all distinct, r\ is therefore a permutation of {1, 2, .., n}, that is, T) e S . Thus Tig e S C , and now we show that cp = Tig. 7 n n n' By Theorem 3> Hi 0*0 = minterm whose binary name is T) (n-tuple for m) © fl(g). But the n-tuple for m is g, and so Tjg(m) = 0. Therefore, cp = T]g(m). x m Suppose m = 2 for some £, ^ £ ^ n-1. Then by hypothesis the k n-tuple for (cp - cp ) = ± (n-tuple for 2 1L ) and, since cp =0, k n-tuple for cp = n-tuple for 2 . Therefore, n-tuple for cp = rj (n-tuple for 2 s -) = i\( n-tuple for m) = n(0. 29 If m = 2* + 1, 2* + 2, ... , or 2 i+1 - 1, with 1 g £ g n -l, then by condition 2, n-tuple for (cp^ - cp Q ) So, since cp = 0, m = n-tuple for (cp - cp + cp - cp ), 2 m-2 n-tuple for = n-tuple for (cp - cp + cp ) 2 m-2* = [n-tuple for (9 -

cp then cp - cp = 2 > 0. This means that the 2 2 n-tuple for cp has a 1 as the (n - k )th coordinate and the n-tuple for cp has a as the (n - k . )th coordinate, with the remaining coordinates identical. (Recall that by hypothesis cp and cp differ in exactly one 2 k^ coordinate in binary notation. ) Since the n-tuple for 2 has a 1 as its (n - k )th coordinate with the remaining coordinates 0, the n-tuple for kp 2 * contributes no carry digits when added to the n-tuple for cp . As a result, n-tuple for 2 + n-tuple for cp k i = n-tuple for 2 ® n-tuple for cp k £ If, on the other hand, cp < cp then cp - cp = - 2 < 0. Here, the n-tuple for cp has a as the (n - k )th coordinate and the n-tuple for cp 31 has a 1 as the (n - k )th coordinate, with the remaining coordinates identical. (Recall again that cp and cp differ in exactly one coordinate 2 in binary notation. ) Reasoning as above, the n- tuple for 2 requires no borrowing when subtracted from the n- tuple for cp , that is, - (n-tuple for 2 ) requires no carrying when added to the n-tuple for cp , and we have, k - (n-tuple for 2 ) + n-tuple for cp k ; = n-tuple for 2 © n-tuple for cp . Therefore, in either case, ( k * \ n-tuple for 2 + n-tuple for cp if cp >cp n-tuple for cp . = / -(n-tuple for 2 ) + n-tuple for cp if cp < q = n-tuple for 2 © n-tuple for cp = il (n-tuple for 2 l ) © T](|) = 111(2^). Hence, cp and tj| agree on all powers of 2. We have cp Q = t^(o), cp 1 = r\t(l), cp = n£(2) from the preceding argument. Now from condition 2, cp = cp - cp + cp , and so we have n-tuple for cp = n-tuple for (cp - cp + cp ) = 0) (n-tuple for 2) © *)(£)] - 1(0 + [t)( n-tuple for l) © tj(£)]. 52 Since t] (n-tuple for 2) has exactly one 1, the (n - k )th coordinate, and the remaining coordinates 0, and T](n- tuple for l) has exactly one 1, the (n - k )th coordinate, k ^ k , and the remaining coordinates 0, we have two cases to consider for t](£): the (n - k, )th coordinate of t](£) is either or 1. If the (n - k, )th coordinate of T](£) is 0, then [n (n-tuple for 2) © T)(|)] - n(S) = n(n- tuple for 2) © q(l) © n(l) = T](n- tuple for 2). Hence, n-tuple for cp = T](n- tuple for 2) 3 + [tj (n-tuple for l) © tj(£)] = tj (n-tuple for 2) © tj (n-tuple for l) © n(g) = T) (n-tuple for 3) © T)(|) = n-tuple for 111(3). If the (n - k, )th coordinate of n(g) is 1, then [t] (n-tuple for 2) © n(|)] - n(|) = - tj (n-tuple for 2). Therefore, n-tuple for cp = - n (n-tuple for 2) 3 33 + [tj (n-tuple for l) © tj(£)] = T] (n-tuple for 2) © T)(n- tuple for l) © tj(£) = T] (n-tuple for 3) tj(|) = n- tuple for ^(3). For 1 ^ H ^ n-1, we prove similarly that cp = T)|(p) if ir a 1 0+1 9 p e {2 + 1, 2 + 2, . . . , 2* - 1). Since p - 2* can be written as a sum of distinct powers of 2, we have p=2*+2+2^+...+2 s , where ^ £ < I -.<...< i < £ 1 < I and 1 ^ s ^ n-1. By repeated use of condition 2, n-tuple for cp = n-tuple for (cp - cp + cp - cp + . . . 2 2 1 -cp -cp + q> )• 2 s " 1 2 S Since cp and tj| agree on all powers of 2, n-tuple for cp = [t] (n-tuple for 2*) © ^(l)] -nU) + [t] (n-tuple for 2 1 ) © tj(|)] - T](|) + + [T) (n-tuple for 2 S_1 ) © tj(§)] - T)(£) + [t] (n-tuple for 2 S ) © tj(£)]. By the argument used to show that cp, = rjg (3 ) . we have 3 ft (n-tuple for 2 ) © T](|)] - T)(|) % T) (n-tuple for 2 £ ) if T}(|) has a as the (n - kjth coordinate I -^ (n-tuple for 2. 1 ) if T)(t) has a 1 as the (n - k )th coordinate and for j = 1, 2, . . . , s, [T)(n-tuple for 2 J) e tj(0] - tj(|) T] (n-tuple for 2 d ) if tj(|) has a as the (n - k )th coordinate J -il (n- tuple for 2 J ) if T](g) has a 1 as the (n - k )th coordinate Hence, n-tuple for cp = T](n-tuple for 2 ) © T)(n-tuple for 2 ^-) © ... T] (n-tuple for 2 ) © T)(|) Jl . . Js = T] (n- tuple for 2* + 2 + ... + 2 ) © ^(0 = T] (n-tuple for p) © T)(|) = n-tuple for T]£(p), and cp p = T]|(p). Therefore, cp = T]| and cp e G . Not only does the second half of Theorem 12 give a sufficient condition for membership in G but its proof gives an algorithm for converting an element of G from S notation to S C notation. We can now go easily from n 2 n n n B J one notation to the other. 35 Example. We know that cp = (065)(127) "belongs to G,. Since cp,_ = 0, we have 3 5 g = (n- tuple for 5) = (1,0,1). Since T](0,0,1) = | n- tuple for (cp 1 - cp Q ) | = | n- tuple for (2 - 6) | = n- tuple for k = (1,0,0), T] maps 3 "to 1. Since T](0,1,0) = | n- tuple for (cp 2 - cp Q ) | = n- tuple for (7-6) = n-tuple for 1 = (0,0,1), T] maps 2 to 3. Since T](1,0,0) = I n-tuple for (cp^ - cp Q )| = I n-tuple for (1+ - 6) | = n-tuple for 2 = (0,1,0), T] maps 1 to 2. So T] = (123), and (065)(127) is (123) (1,0,1) in S,C„. 3 3 Corollary 12. 1 . Let cp be an element of G and write cp in the two-row / 1 2 2 n -l N \ notation, . Then, with k as in Theorem 12, 1. for £ =0, 1, 2, ..., n-1, the n-tuple for cp differs from the 2 £ n-tuple for cp Q in exactly one coordinate, namely, the (n-k )th coordinate; 36 2. for p = 2 l + 2q where qe (1, 2, 3> •••> 2*" 1 - 1), the n- tuple for cp differs from the n- tuple for cc> in the (n - k )th coordinate and in x p T £ those coordinates in which the n-tuple for cp differs from the n-tuple for 2 q V Proof : The first conclusion restates the first part of Theorem 12. Suppose next that p = 2 s - + 2q with qe {1, 2, 3, ..., 2*' 1 - 1) and cp = r)£ with T) e S and g eC . By condition 2 of Theorem 12, we have (n-tuple for cp ) - (n-tuple for cp n ) = (n-tuple for cp ) - (n-tuple for cp ) + (n-tuple for cp ) - (n-tuple for cp ). The n-tuple for cp differs from the n-tuple for cp in the (n - k )th coordinate. Since 2q can be written as a sum of distinct powers of 2, with these powers less than £, the n-tuple for cp differs from the n-tuple for cp in coordinates other than the (n - k )th coordinate. So, the n-tuple for cp differs from the n-tuple for cp in the (n - k )th digit and in those digits in which the n-tuple for cp differs from the n-tuple for cp . B Corollary 12.2 . To verify that the conditions for Theorems 10, 11, and 12 hold for a given element of S _. it suffices to check the first half .of the two-row representation of the element. The preceding theorem and its corollaries permit us now to state a useful algorithm for computing the subgroup M^ determined by a given subgroup H of G . 37 Theorem 13 . Let H be an arbitrary subgroup of G . Then ]VL may be computed by the following algorithm. Step 1 : Find the orbits of H and order the orbits according to the smallest minterm in the orbit. We shall use cp to denote an element of S n . 2 n Step 2 : Find the list all pairs cp cp that satisfy the following criteria: (l) For each i, i = 0, 1, • • • , 2 -1, are determined according to these criteria: (1) The minterm cp. . belongs to the same orbit as i - 1 and cp. belongs to the same orbit as i. 1 (2) In binary notation cp. , differs from cp. in exactly the same coordinate and in the same manner in this coordinate that cp„ differs from cp . This means that if the differing 58 coordinates are respectively a and b for cp and cp , then the differing coordinates are a and b for cp. , and cp. . l-l 1 (3) If i - 1 = 2 1 for some I ■ 1, 2, . .., n-2, then the n-tuple for cp. differs from the n-tuple for cp in exactly- one coordinate, the (n - k„)th coordinate, and k 4 k for i I T m m = 1, 2, . . . , 1-1. (k) If i - 1 = 2 £ + 2q where qe {1, 2, ... , 2 i_1 - 1), then the n-tuple for cp. differs from the n-tuple for cp in exactly the (n - k )th coordinate and in those coordinates in which the n-tuple for cp differs from the n-tuple for cp . Step k : Eliminate all half strings which are not parts of elements of G according to the test described in condition 2 of Theorem 12. n Step 5 ' Complete all remaining half strings by writing the complementary images. Step 6 : Convert all strings to cyclic notation. The set of all elements of G obtained by the above procedure constitute Mtj. Proof : The steps in this algorithm are justified by Theorems 9 and 11, Corollaries 12.2 and 12.1, Theorems 12 and 10, and Corollary 10.1. Clearly the mappings obtained are in ]VL. This procedure gives all of M. because we consider all possible elements of G which satisfy the conditions of Theorem 9' ■ Example . We now give an example illustrating this algorithm: Consider the subgroup 39 H = <(1,8)(5,10)(5,12)(7A4), (0,8,9A) (2,1X>,11,3) (4,12,13,5) (6,14 ,15,7> of G. . The orbits of H are the following and are ordered by their smallest members : (0,1,8,9), {2,3,10,11}, {4,5,12,13}, and (6,7,14,15). The possible images of are 0, 1, 8, and 9, and the possible images of 1 are also 0, 1, 8, and 9« The only compatible cp cp pairs are: 0,1 1 'n-tuple for differs from n- tuple for 1 in the 0,8 1 'n-tuple for differs from n- tuple for 8 in' the 1,0 ( 'n- tuple for 1 differs from n-tuple for in the 1,9 ( 'n-tuple for 1 differs from n-tuple for 9 in the 8,0 ( 'n-tuple for 8 differs from n-tuple for in the 8,9 ( 'n- tuple for 8 differs from n-tuple for 9 in the 9,1 ( 'n- tuple for 9 differs from n-tuple for l in the 9,8 ( 'n-tuple for 9 differs from n-tuple for 8 in the fourth coordinate) first coordinate) fourth coordinate) first coordinate) first coordinate) fourth coordinate) first coordinate) fourth coordinate), For the cp cp -pair 9, 1, the other cp. cp. pairs (not necessarily in order) which appear in the string are 8, 0; 10, 2; 11, 3; 12, k; 13, 5; 14, 6; and 15, 7 because the members of each of these ordered pairs differ in exactly the same way that 1 differs from 9 in binary notation. The image of 2 must be 2, 3, 10, or 11. But 11 is the only minterm among these whose n-tuple differs from the n-tuple for the image of in exactly one coordinate, namely, the third coordinate. Thus, the image of 2 is 11 and hence the image of 3 is 3« The image of k must be k, 5, 12, or 13, and 13 is the only one of these whose n-tuple differs from the n-tuple for the image of in exactly one coordinate, namely, the second coordinate. Thus, the image of k is 13, and therefore, the image of 5 has to be 5, with 13, 5 being the next pair. The image of 6 must be 6, 7> 14, or 15, and 15 is the only one among these which differs from the image of in the third and second coordinates in binary notation. So, the image of 6 is 15, and the image of 7 is 7* In this discussion there happens to be one choice at each step. However, if there is more than one possibility, all of them must be considered. When the remaining cp cp pairs have been treated similarly, the completed half table according to Step 3 is the following: 1 2 3 k 5 6 7 8 9 10 11 12 13 Ik 15 01231+567 8 2 10 ^ 12 6 14 10325476 1 9 3 11 5 13 7 15 8 10 2 12 4 14 6 B 9 10 11 12 13 14 15 9 l 11 3 13 5 15 7 9 8 11 10 13 12 15 Ik We now test these half strings according to the conditions of Theorem 12 and find that all of them may be completed to elements of G . For example, the second string satisfies Theorem 12: 8 - = 8 = 2? , 2 - = 2, 10 -0= 10 =8+ 2, k - = k = 2 2 , 12-0= 12 =8+4, 6-0=6 = 2 + k, and Ik - = 11+ = 10 + k. 41 The completed table is: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 8 2 10 4 12 6 14 1 9 3 11 5 13 7 15 1 3 2 5 4 7 6 9 8 11 10 13 12 15 14 1 9 3 11 5 13 7 15 8 2 10 4 12 6 14 8 o 10 2 12 4 14 6 9 l 11 3 13 5 15 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 9 1 11 3 13 5 15 7 8 10 2 12 4 14 6 9 8 11 10 13 12 15 14 1 3 2 5 4 7 6 In cyclic notation we have Mjj = {(1), (1,8)(3,10)(5,12)(7,14-), (0,1) (2, 3) (h, 5) (6, 7) (8, 9) (10, 11) (12, 13) (14,15), (0,l,9,8)(2,3,ll,10)(4,5,13,12)(6,7,15,lU, (o,8,9,l)(2,10,ll,3) (4,12,13,5) (6,14,15,7), (0,8)(1,9)(2,10)(3,H)(4,12)(5,13)(6,14)(7,15), (0,9) (2,11) (4,13) (6,15), (0,9)(1,8)(2,11)(3,10)(4,13)(5,12)(6,15)(7,1^)}. Since ]YL is the set of all elements of G which leave P pointwise invariant, once we have ]VL we can generate the collection of all subgroups of G which leave F pointwise invariant. Each subgroup in this collection is generated by the elements in a nonempty subset of 1VL and every nonempty subset of ]VL generates a subgroup which leaves F pointwise invariant. If 1VL contains k elements, there are at most 2-1 subgroups of G which leave F„ pointwise invariant. 1+2 Corollary 1 j. 1 . Let F ■ (f^, f^, . . . , f fc ) , f . e B , and let P., i = 1, 2, . .., k, be the corresponding partitions, and let P = i: P.. i=l X Then M^, is determined by employing the blocks of P as the orbits in the algorithm of Theorem 13. This generalizes the previously mentioned result of McCluskey [27]. We now proceed to study N , the largest subset of G which leaves the set of functions F^ setwise invariant. n Theorem 11+ . For a subgroup H of G , let be the set of all orbit functions for H and let cp be an element of GL. Then cp[0„] = 0„ if and only u xl xl if cp[F H ] = F H> so N H = {cp|cpeG n and cp[0 H ] = H } . Proof: Suppose cpCO^] = U . We shall first show that cp[F u ] is a subset of — — — ri n fi F„. Let f e F_. rl n Case 1: If f is the function then f cp = Ocp = 0, and so, fcp is in F„. ■ ' ft. Case 2: If f is a member of 0„, then fcp is also in 0^ by hypothesis. — — — ft ft Since TT c F fcp belongs to F . ft ~" ' ft ft Case 3 : If f is a disjunction of orbit functions, say f = o n V o„ V . . . V o. for o. in TT , i = 1, 2, .... k, for some k, then 12 k 1 Br fcp = o_cp V o p cp V ... V o cp. By hypothesis, o.cpeO for i = 1, 2, ..., k, and so fcp e F . a. Thus in every case f cp e F^ and cp[F u ] c F xl H xl Now we show that F„ S cpft?]. Let geF„. xl ft xl Case 1 : If g is the function then since Ocp = 0, OctppiV,]. Case 2: If g =o n Vo.V ... V o. where o. eO^ i = 1, 2, ... , k, ° 1 2 k l H' 7 ' then since cp[Ou] = Cy., for each o. , i = 1, 2, ... , k, there is an o. ' in n n 1 1 0^ such that o. 'cp = o.. Let f = o ' V o ' V . . . V o ' . Clearly f is in F TT h. 11 J. c. K. n and fcp = o 'cp V o 'q> V ... V o 'cp = o V o V . . . V = g. Therefore, F„ is a subset of cp[F u ], and we have cp[X T ] = F TT . xl xi xi H Conversely, suppose cpfF^.] = F^. We show first that cp[0 TT ] is a subset xi xi n of TJ . Let o e . Then since c j 1 ocp e F TT . Now ocp cannot be the h. xi n rl H function since cp permutes minterms and i 0^. So, ocp is either an orbit h. function or a disjunction of distinct orbit functions for H, say ocp = o, V o„ V ... V o. for o. e TT <= F TT , i =1, 2, .... k, for some k. 12 k l H — IT > > > > Since cp is bijective on F„, for each i, i =1, 2, ..., k, there is an f. in h. i F„ such that f.cp = o. , and so (f n V LV ... V f,)q> = H x l 12 k f cp V f_cp V . . . V f . cp = o n V o V . . . V o. . But ocp = o n V o n V . . . V o. 12^ k 1 2 k Y 1 2 k means that o = f , V f _ V . . . V f . . But o is an orbit function for H and 12 k £ F , so k = 1. This means that ocp is an orbit function, that is, n G n n Corollary 14.2 . If H and K are subgroups of G which have identical orbits, then N„ - 1L.. n J\ Corollary 14«3« For the subgroup H of G , if cp e II, then cp maps each orbit — — — — — ■ — — —— n H function for H to an orbit function for H of the same length. Corollary Ik-'K- For a subgroup H of G , L is a subgroup of G . 1 n n n Proof: If cp n and cp are elements of K, then from the definition of N u , it is clear that cp-j_ cpp is also an element of N . Let cp be an element of N , and let o be an orbit function for H, that H is, o e TT . Then there is an orbit function o' in 0^ such that o'cp = o by Theorem 14. So, ocp" = o*. and cp" [0„] is contained in CL. n n Let o be an orbit function for H. Then ocp = o' for some o' in 0^. n So, o'cp" = o and 0„ is contained in cp [0,,]. Therefore, cp" [CL,] = 0^ h. rl rl n and cp" £ N u . ■ The subgroup N Ti of G is maximal in the sense that it is the largest ^ H n subgroup of G which leaves F setwise invariant. We now show how to compute U5 N "by modifying the procedure used to compute IVL. A parallel reading of the example following Theorem 15 will make the algorithm easier to follow. Theorem 15 . Let H he a subgroup of G . The subgroup IC may be computed by ____________ n ^ the following algorithm: Step 1 : Find all orbits of H. Step 2 : Partition the set of orbits of H according to length; that is, two orbits belong to the same block of the partition if and only if they have the same length. Order the blocks in the partition according to the smallest minterm that appears in the orbits belonging to the block. Call these distinct blocks C, , CL, .... C, . Within each block order the orbits 12 k according to the smallest minterm in each orbit. Call the orbits in block C. , o. , , o. , . . . , o . i ll i2 lk. i We shall use cp to denote an element of S n . 2 n Step 3 : Find and list all possible pairs cp cp according to the following: (1) C, is the block containing the orbit o which contains the minterm 0. (2) The possible images of are the minterms in the orbits in C, . List these minterms together with the names o n . of the 1 lj orbits to which they belong. (3) The possible images of 1 are the minterms in the orbits of the block containing 1. List these minterms with the names o. . of the orbits to which they belong. If 1 is in an orbit in C, then the restrictions of orbit-to-orbit mappings (orbits must map onto orbits of the same length) implied by a choice of 1*6 cp must be consistent with the restriction implied by the cp with which cp is to be paired. At this stage the pairs cp cp to be considered are those satisfying (3) for which cp differs from cp in exactly one digit in binary notation. Step k : Construct a table with column headings through 2 n - 1. First find the possible half-strings cp^ cp., . . . cp 1 p n-l Step 5 : F°r each listed pair cp cp and for each partial string cp Q cp_ L . . . cp i _ 1 cp i , 1 g i ^ 2 n ~ - 3, i odd, the possible pairs cp i+1 cp_. +2 are determined as follows : (1) cp. ., must differ from cp. _ in exactly the same ' l+l 1+2 coordinate and the same manner in this coordinate that cp differs from cp in binary notation. Choose cp from the from the set of minterms not previously chosen which appear in the orbits contained in the block containing the minterm i + 1. The orbit containing i + 2 and the orbit containing cp must be in the same block. The restrictions on orbit-to- orbit maps implied by these choices of cp and cp. _ must be consistent with those implied in the construction of cp Q cp 1 . . . cp._ x cp. . (2) If i + 1 = 2 l for some I = 1, 2, ... , n-2, then the n-tuple for cp. , must differ from the n-tuple for cp in exactly one coordinate, the (n - k )th coordinate, and k =/ k^, m = 0, 1, . . . , 1-1. kl (3) If i + 1 = 2* + 2q where qe{l, 2, 3, • • • , 2^ _1 - 1} , then the n- tuple for cp must differ from the n-tuple for cp in exactly the (n - k )th coordinate and in those coordinates in which the n-tuple for cp differs from the n-tuple for cp . Step 6 ; Eliminate all strings which are not parts of elements of G according to the test described in condition 2 of Theorem 12. Step 7 ' Complete all the remaining half- strings by writing the complementary images. Step 8 : Convert each string to cyclic notation. Proof of the validity of the algorithm : The steps in this theorem are justified by Theorem Ik, Corollary lk-3> Theorem 11, Corollary 12.1, Theorem 12, Corollary 12.2, Theorem 10, and Corollary 10.1. Since -N n consists of all elements of G which leave TT setwise invariant and, in n H ' the algorithm, we consider all possible mappings of an orbit to an orbit of equal length, we obtain all of N u . ■ rl Example . We now illustrate this algorithm by computing N for the subgroup H = <(0,9,3) (2,8,11) (If, 13,7) (6,12,15)> of G^: The orbits of H are: (0,3,9), {2,8,11}, {k,1,13), {6,12,15}, {1}, (5), {10}, [Ik). The blocks are: C 1 = {{0,3,9}, {2,8,11}, {k,lA3)> {6.12,15}}, and °11 °12 °13 °lk C 2 = {{1}, {5}, {10}, {14}}. °21 °22 °23 °2k The possible cp are as follows: for o to o : 0, 3, 9 for o i;l to o 12 : 2, 8, 11 for o ±1 to o • k, 1, 13 for o ±1 to o^: 6, 12, 15 . U8 The possible cp, are as follows for o 21 to o 21 : 1 for o 21 to o 22 : 5 for o to o • 10 for o to o 2) : 12f . The possible pairs cp cp are: for o u to o u , o 21 to o 21 : 0, 1 3, 1 9, 1 for o to o , o to o : none (by binary 11 11' 21 22 digit rule) for o xl to o lx , o 21 to o 23 ; none for o ±1 to o ±1 , o 21 to o ; none for o ±1 to o 12 , o 21 to o 21 : none for Qll to o 12 , o 21 to o 22 : none for o ±1 to o 12 , o 21 to o : 2, 10 8, 10 11, 10 for o ±1 to o 12 , o 21 to o 2V none for Gll to o iy o 21 to o 2r none for o 11 to o , o 21 to o 22 ; h, 5 1, 5 13, 5 for Qll to o iy o 21 to o 2y none for Dll to o 15 , o 21 to o 2V none for o-,-, to o-,j , o 21 to o 2 -, ; none '49 for o to o , Op to Op : none for o i:L to o , o gl to o • none for o i;L to o^, o 21 to o g ^: 6, 11+ 12, 11+ 15, 1»4 To illustrate how the following table is obtained, we explain the second row in the table in detail. For cp cp = 0, 1, the possible choices for cp_ cp according to binary digit differences alone are: 2, 3; Ij., 5; d 3 6, 7; 8, 9; 10, 11; 12, 13; and 1!+, 15. By the orbit length condition, these choices are reduced to 2, 3; 6, 7; 8, 9j and 12, 13. In view of the o to o and o to o restrictions under which cp cp was derived, the only choices for cp ? cp are 2, 3 and 8, 9. Both 2 and 8 differ from cp = in exactly one digit in binary notation. (The 2, 3 -c hoice appears in the first row of the table. ) Continuing with the second row of the table we note that for cp cp cp cp =0, 1, 8, 9, the possible choices for cp cp according to binary digit differences alone are: 2, 3; 1+, 5; 6, 7; 10, 11; 12, 13; and 11+, 15. By the orbit length condition, these choices are reduced to 2, 3; !+, 5; 6, 7; and 12, 13. In view of the o to o , o p to o , and o p to o restrictions under which cp cp cp_ cp was derived, the only choice for ±£- u J. c. 3 \ cp^ is h, 5 (o to o , o 22 to o 22 ). For cp cp cp 2 cp cp cp = 0, 1, 8, 9, k, 5, <|V must be 12 and cp must be 13. In binary notation 12 differs from cp = in exactly those coordinates in which cp_ = 8 and cp. = k differ from cp Q . . The choice cp^ cp = 12, 13 is consistent with the preceding restrictions. 50 The complete table, after adding complementary images, is the following: °11 t0 °11' °21 t0 °21' °12 to °12' °13 t0 °13' o„~ to o„, 0-, to o n . , 22 22' 14 14 °2 5 t0 °23' °24 t0 °2V °11 to °12' °21 t0 °23 °12 t0 °1V °13 t0 °lk> °22 tD W °lk t0 V °23 t<=> °21' °24 t0 °22* N l 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 8 9 4 5 12 13 2 3 10 11 6 7 14 15 3 1 2 7 4 5 4 11 9 10 8 15 13 14 12 3 1 11 9 7 5 15 13 2 10 8 6 4 14 12 9 1 8 13 5 12 4 11 3 10 2 15 7 14 6 9 l ll 3 13 5 15 7 8 10 2 12 4 14 6 2 10 8 6 14 4 12 3 11 1 9 7 15 5 13 2 10 3 11 6 14 17 15 8 1 9 4 12 5 13 8 10 2 12 14 4 6 9 11 1 3 13 15 5 7 8 10 9 11 12 14 13 15 2 l 3 4 6 5 7 11 10 3 2 15 14 7 6 98 1 13 12 5 4 11 10 9 8 15 14 13 12 3 2 1 7 6 5 4 °11 t0 V °21 t0 °22' °12 to °1V °13 t0 °H' °22 t0 °21' °14 t0 °12' °24 t0 °23' °24 t0 °23" °ll to °14' °2l t0 °24'^ °i2 t0 °iy °i3 to °i2> °22 to °23' °i4 t0 °ir o to o . o , to o , . 23 22' 24 21 4 5 6 7 1 2 3 12 13 14 15 8 9 10 11 4 5 12 13 1 8 9 6 7 14 15 2 3 10 11 7 5 6 4 3 1 2 15 13 14 12 11 9 10 8 7 5 15 13 3 1 11 9 6 4 14 12 2 10 8 13 5 12 4 9 18 15 7 14 6 11 3 10 2 13 5 15 7 9 l ll 3 12 4 14 6 8 10 2 6 14 4 12 2 10 8 7 15 5 13 3 11 1 9 6 14 7 15 2 10 3 11 4 12 5 13 8 1 9 12 14 4 6 8 10 2 13 15 5 7 9 11 1 3 12 14 13 15 8 10 9 ll 4 6 5 7 2 1 3 15 14 13 12 11 10 98 7 6 5 4 3 2 1 15 14 7 6 ll 10 3 2 13 12 5 4 9 8 1 51 All of the strings satisfy the conditions of Theorem 12, and there- fore constitute N„. To complete the algorithm we need only to convert rl the strings to cyclic notation. Having computed K. by the algorithm in Theorem 15, we can use the n following Lemma k and Theorem 16 to compute the normalizer of H in G , N~ (H). Lemma If is a restatement of a result from Scott [39; P- 12]. G n Lemma k - If t and cp are elements of S , then a cyclic decomposition of 2 cp\|rcp " is formed from such a decomposition of ty by replacing each minterm of "vjr by the minterm standing below it in a two-row form of cp. Proof: In a cyclic decomposition of "v|r _, the minterm m is followed (cyclically) by the minterm \|/ . The minterms standing under m and t in a two-row form of cp are cp and cp , respectively. Since m (cp\|Kp~ )cp = (cpi|/) = cp , cp is the minterm following cp in a cyclic m m decomposition of cpfcp" . I Theorem 16. For a subgroup H of G , L (H) c N, where N_ (H) denotes the —————— n G — n G n n normalizer of H in G . n Proof: Let cp € KL (H). We shall show that cp[F„] = F„ which means that — — — \j n ri n cp e N„ and that N^ (H) c N TT . n G _ H n Let f e F-.. Then f is the function or a disjunction of orbit n functions for H. Case 1 : If f = 0, then Ocp = and fcp e F u . 52 Case 2: If f = o n V o„ V . . . V o. where o. e 0„ for i = 1, 2, . . . , k, 12 k l H ' k 5 1, then fcp = o.cp V OpCp V ... V o cp. Now for each \|r in H, o.\|r = o. for i = 1, 2, . .., k, and for each \f in H there is a \|r' in H such that \JfCp = cpt 1 . So, for each \|r in H, t( ={(l), (0231)0*675), (03) (12) (kl) (-56), (0132)^567), (0536)(172^), (07)(16)(25)(34), (04) 0-5) (26) ( 37 X] of G y Since His a transitive subgroup of G,, that is, H has only one orbit, 5 Mfj = N H = G . However, taking (ll*)(36) in N R and (023l)(^675) in H, we 53 find that (lk) (36) (0231) (1^675) (1^) (36) = (026*0(1375) i H. Therefore, (llj.) (36) / N (H). Thus, H is not normal in N R , and N (H) S N r . 3 3 The containment H_ (H) c N does mean that if H i. J Step k- : Convert all remaining cp to cyclic notation. The transformations obtained constitute N~ (H). G n Proof : Every element 7 of H is a finite product of the generators, say 7=7. 7. ... 7. , 1 S i. Ik, lg j I i, Thus, for any cp e KL, l l x 2 ^ J H cp 7 cp" = cp 7 7 ... 7 cp' = cp 7 cp" cp 7 cp" cp. . . cp" 1 cp 7 cp" 1 . 1 2 £ X l X 2 X £ From this it is clear that if {cp7.cp~ | H i I k) c H, then cpycp" eH and this conclusion is independent of the particular factorization used for 7. 54 The theorem follows from this fact and Theorem 16. I Example . We now continue the example given immediately after Theorem 15. Recall that H = <(0,9,3)(2,8,ll)(^,l3,7)(6,l2,15)> = {(i), (0,9, 3) (2, 8, 11) (4, 13, 7) (6, 12, 15), (0, 3, 9) (2, 11, 8) (4, 7, 13) (6, 15, 12)) is a subgroup of G. . In this case H is cyclic and has one generator. Let 7 = (0,9,3) (2,8,11) (4,13.7) (6,12,15). The computations are presented in tabular form. (Because of space considerations, only the first half of all the strings representing elements in N„ are given. ) rl cp g N„ (half strings ) n 01234567 18 9 4 5 12 12 31207564 3 1 11 9 7 5 15 13 9 1 8 13 5 12 4 9 l 11 3 13 5 15 7 2 10 8 6 14 4 12 2 10 3 11 6 14 7 15 8 10 2 12 14 4 6 8 10 9 11 12 14 13 15 11 10 3 2 15 14 7 6 11 10 9 8 15 14 13 12 45670123 4 5 12 13 1 8 9 75643120 7 5 15 13 3 1 11 9 cp 7l qf [6,12,15 cp 7l cp _1 e H? (0,90) ;2.8.ll)(4,l3,7) ) Yes (0,3.9) ;8,2,ll)(4,7,i3) : 12, 6, 15 ) Yes (3,9,o)( '2,11,8)(7,13,4) [6,15,12] 1 Yes (3,0,9) ;il,2,8)(7,4,13) [15,6,12] ) Yes (9,3,0) ;8,11,2)(13,7,4) [12,15,6 ) Yes (9,0,3) ;il,8,2)(i3,4,7) [15,12,6; > Yes (2,11,8 )(0,3,9)(6,15,12 )(4,7,13: ) Yes (2,8,11; )(3,0,9)(6,12,15. >(7,4,13: ) Yes (8,11,2; )(0,9,3)(12,15,6; )(4,13,7: ) Yes (8,2,11: )(9,0,3)(12,6,15: >(13,4,7: I Yes (11,8,2; )(3,9,0)(15,12,6; )(7,13,4; ) Yes (11,2,8; )(9,3,0)(15,6,12; )(13,7,4: 1 Yes (4,13,7; )(6,12,15)(0,9,3: )(2,8,li; 1 Yes (4,7,13: )(12,6,15)(0,3,9: 1(8, 2, li; 1 Yes (7.13,4: )(6,l5,12)(3,9,o; 1(2,11,8; ! Yes (7,4,13: )(15,6,12)(3,0,9: 1(11,2,8; I Yes 55 13 5 12 ^ 9 1 8 13 5 15 7 9 1 11 3 6 lk If 12 2 10 8 6 lk 7 15 2 10 3 11 12 lk k 6 8 10 2 12 lif 13 15 8 10 9 11 15 lk 13 12 11 10 9 8 15 lk 7 6 11 10 3 2 (13,7,0(12,15,6 (13,4,7) (15,12,6 (6,15,12) (k,l A3 (6,12,15) (7AA3 (12,15,6) (ij.,13,7 (12,6,15) (13 A,7 (15,6,12) (13,7,4 (15,12,6)(7,13^ )(9,3,o)(8,n,2) Yes ) (9,0,3) (11,8,2) Yes )(2,11,8)(0,3,9) Yes ) (2,8,11) (3,0,9) Yes )(8,11,2)(0,9,3) Yes )(8,2,ll)(9,0,3) Yes )(H,2,8)(9,3,0) Yes )<11,8,2)(3,9,0) Yes In this example, IT (H) = K,. G n H We conclude this chapter "by proving a relationship between NL and N . Theorem 18 . If H is any subgroup of G , then ML is a normal subgroup of N . Proof: Let cp e N . We want to show that cpNL = NLcp, or equivalently, that -1 for each \|r e NL, cp f cp e NL. This is also equivalent to showing that for each t £ NL and for each f eF , cp" tcp(f) = f. (For clarity we have once more changed the notation. ) Since cp" ^cp(O) =0, it is sufficient, by the definition of F , to show that for each i|r e NL and for each oeO , cp" i(rcp(o) = o. Since cpelL,, cp(o) = o* for some o' e TT . And n n cp" \|rtp(o) = cp" i|r(o') = cp" (o 1 ) = o. ■ Corollary 18.1 . If H O G then NL <1 G . Proof: It follows from Theorem 16 that if H O G then N TT - G . n H n 56 CHAPTER III ASSOCIATED LATTICE STRUCTURES In this chapter we shall prove results about the lattice structures associated with the group of symmetries 1VL. For subgroups H and K of G we shall use [HUK] to denote the subgroup of G generated by H U K. Definition llx. Let L E2 K. By the definition of [HU K], [HUK] = L. Therefore, F rTTMTr n is the inf of F^ and F^. The completeness follows from the fact |_n U K. J n is. that ^ is finite. ■ Definition 15 . Let #/ = {ML [ H is a subgroup of G } . Define the relation C on T^thus : For subgroups H and K of G , ML E \ i f and only if H c L Lemma 6 . The set 77^ with relation 5 is a poset. Proof : Again, this lemma follows from the fact that the subset relation on sets is a partial ordering. ■ Theorem 20 . Let ML U ML = Mr ■.„■, and ML n ML = ML n „. Then the poset is a complete lattice with MUM denoting the sup of ML and ML. and ML n ML denoting the inf of ML and ML. Proof : By arguments similar to the proof of Theorem 19, the claims here can be verified. The completeness follows from the finiteness of 1tt . ■ Theorem 21 . Let PL, H„, . . . , K be subgroups of G so chosen that F , 1 ^ i ^ k, are distinct, ML , 1 ^ i ^ k, are distinct, * = {F H , F H , .... F ), and ^ = {M H , ML. ..., ML }. There is a one-to-one correspondence between £ and 7h defined by ?: ^ -*■ #7 where Y(F ) = L , 1 S m, such that i i 1. F ^ F if and only if ML 3 M . 2. Y(F H A F R ) = JL U Ky , and i J i J 3. M^ n Mg = Y(F H V F H ). 58 Proof: Clearly V is a well-defined operation and from its definition it is easily seen to be surjective. Since the sets are finite, ¥ is a one-to-one correspondence between fh and ^ . 1 F *F is equivalent to H. 2 H, which in turn is equivalent J- *u - h. i J l J to *v 3 Mg- 1 J 2. ^(F H A F H _) =¥(F [H _ UH id ! o = M [H.UH.] i 1 J 3 . *(f h v f h _) =Y(F HnH .) 1.1 i ri = M H.n H . ExamB le . ' We now give an example to illustrate this lattice anti-isomorphis ffi . Let n = 2; G- - «D, «a)<85), (06(13). (03) (12); (12), (0251), (O*), (0?)) "2 G ? has ten subgroups: E ± = {(1)} H 2 ={(1), (01) (23)} H_ = {(1), (02) (13)} H. = {(1). (03) (12)} H_ = {(1), (12)} 5 H 6 ={(1), (023D, (03) (12), (0132)} H ? = {(1), (03)} 59 Hq = {(1), (03)(12), (03), (12)} H 9 = {(1), (03) (12), (01)(23), (02)(13)} H 10 = G 2 The lattice i 1 contains the following sets of functions of two variables. We shall use F. to denote P and M. to denote M„ , for i i i = 1, 2, . .., 10. F l = B 2 F 2 = {0, 2 V 3, V 1, 1} F 3 = CO, 1 V 3, V 2, 1} F^(0,1V2,0V 3, 1} F = {0, 3, 1 V 2, 1 V 2 V 3, 0, V 3, V 1 V 2, L} F 6 - {0, L) P = [0, 2, 1 4 1V2, 0V3j 0V2V3, 0V IV 3, L] Fq = {0, 1 V 2, V 3, 1) F 9 = CO, 1} F 10 = t°> V The lattice 1f\ contains the following subgroups of G : \ = C(i)} IL, = {(1), (01)(23)} Mj = ((l), (02)(13)) \ = {(1), (03) (12), (12), (03)} ^ = [(I), (12)} 6o \ aG 2 Hj = {(i), (03)) % = i(l), (03)(12), (03), (12)} ^0 = G 2' Th e lattice diagrams for ^ and 7ft are the following: Jfe ■ M 9 ■ "10 6 9 10 These diagrams illustrate clearly the three parts of Theorem 21. Theorem 22 . Let H and K "be conjugate subgroups in G . Then -li 1. IVL and JVL. are conjugate subgroups in G , and if K = cp" x Hcp for cp e G , then JVL = cp~ Mcp. 2. N and N- are conjugate subgroups in G , and if K = cp" Hep h. n n for cp e G , then K_ = cp" N TT cp. n K T H Y Proof : 1. Suppose K = cp" Hep for cp e G . Let \jf e M„. Then the set of minterms of each cycle of f is a subset of an orbit of H. This means that 61 the set of minterms of each cycle of cp~ fcp is a subset of an orbit of cp~ Hep = K. Therefore, by Theorem 9, cp~ i|rcp e 1C_, and cp" iVLcp c M . If t eNL then the set of minterms of each cycle of t is a subset of an orbit of K. So, the set of minterms of each cycle of cp\|rcp" is a subset of an orbit of H, and cp\|rcp~ e rVL. Therefore, \(r e cp" IVLcp and 1VL. c: cp" IVLcp. 2. Suppose K = cp" Hep for cp £ G . Let t e 1\L. Then n H (cp~ \|rcp)~ Kcp" i|rcp = cp" \]T cpKcp" ijrcp = Cp \|r Hijrcp = cp~ Hep (since i|r eL) n = K. So, by Theorem 16, cp" ijrcp e 1NL. and cp" l\Lcp C 1\L.. iv n~~.iv Let ■>]/ e N . If 7 = ep\Jrcp~ then K 7~ H7 = (cp\|rcp~ )" Hcp\|rcp~ qa|r~ cp" Hcpnjrcp" ,-1 tr, -1 cp\|r Ktycp cp Kcp" (since \|r 6 1L.) IV = H. -1 -1 -1 1 So, 7c-N and since cp 79 = cp" (cp\|rcp~ )cp = \|/, 1\L. c cp" N TT cp. ■H. iv H 62 CHAPTER IV CONCLUSION The major results of this thesis are the determination of NL for an arbitrary subgroup H of G (Theorem 9 and its corollaries), necessary and sufficient conditions for an element of S to belong to G (Theorem 12), the algorithm to compute the subgroup M„ (Theorem 13) , the algorithm to compute the subgroup K, (Theorem 15 )> and the algorithm to compute the n normalizer of H in G , N (H), (Theorem 17). The determination of NL n completes the generalization question posed by Forbes [8]. It is quite conceivable that the algorithm presented here could be simplified, or modified to make the preparation of computer programs more feasible. Some of these algorithms raise interesting questions. For example, in the computation of ML,, what is the nature of the tree of choices of pairs? As we have seen in Chapter II, N (H) may be a proper subgroup of JL-. G n H It remains to be seen under what conditions these two subgroups are equal. It is also not known at this time exactly how N (H) fits into the G n relationships He JLcLcG . Examples exist for both H c M c L (H) n and H c N (h) c m . It would be of interest to determine simple conditions n under which one or the other relation holds. 63 If Y[ = {N„ | H < G ), it remains to be seen if 77 has any useful or interesting structure. The study of the structures of all the groups involved is still open to investigation. The relationship of such structures to the structure of B may prove very useful in obtaining further results concerning G and B . It seems likely that this work can be extended and used to prove new results in coding theory, circuit design, and other related subjects. 61* LIST OF REFERENCES 1. Arnold, R. F. , and M. A. Harrison, "Algebraic Properties of Symmetric and Partially Symmetric Boolean Functions," IEEE Trans. Computers , vol. EC-12, no. 3, pp. 21^-251, 1963. 2. 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Bernstein, "Optimal Binary Coding of Ordered Numbers," S1AM J. , vol. 13, pp. Iflfl-lflf3, 1965. If5. Stone, Harold S. , and Charles L. Jackson, "Structures of the Affine Families of Switching Functions," IEEE Trans. Computers , vol. EC-18, no. 3, pp. 251-257, 1969. 1+6. Todd, J. A., "The Groups of Symmetries of the Regular Polytopes," Proc. Cambridge Philos. Soc. , vol. 27, pp. 212-231, 1931. Ij. 7. Washburn, S. H. , "Relay 'Trees' and Symmetric Circuits," AIEE Trans . , vol. 68, Part I, pp. 582-586, I9I4.9. . If 8. Wielandt, Helmut, Finite Permutation Groups , Academic Press, New York, 196lf. if9« Young, A., "On Quantitative Substitutional Analysis," Proc. London Math. Soc. , vol. 31, no. 2, pp. 273-288, 1930. 68 VITA Carol Shigeko Abe Edwards was born on October 8, 1938* in Hilo, Hawaii. After attending the University of Hawaii in Hilo, she trans- ferred to the University of California at Berkeley where she received a B.A. in mathematics, with honors, in i960. She was elected to Phi Beta Kappa in I96I. After a year as a research and teaching assistant at the University of Hawaii in Hilo, she was selected as a participant in the National Science Foundation Academic Year Institute in mathematics at the University of Illinois. She was elected to Pi Mu Epsilon during the year and received her M.A. in 1962. From 1962 through 1971 she was on the faculty at the University of Hawaii in Hilo, advancing from instructor to assistant professor in 1968. She returned to the University of Illinois from 19&5 ^o 1967 and from 1969 to 1973 to continue work toward the Ph.D. An Academic Year Institute grant and a National Science Foundation Science Faculty Fellowship supported two years of her study during these periods, and she was also a teaching assistant. She was elected to Phi Kappa Phi in 197 2 - IBLIOGRAPHIC DATA MEET Title and Subtitle 1. Report No. UIUCDCS-R-73-597 3. Recipient's Accession No. SUBGROUPS OF THE GROUP G 5. Report Date October, 1973 n 6. Author(s) Carol Shigeko Abe Edwards 8. Performing Organization Rept. No. Performing Organization Name and Address Department of Computer Science University of Illinois at Urbana-Champaign Urbana, Illinois 61801 10. Project/Task/Work Unit No. 11. Contract /Grant No. Sponsoring Organization Name and Address Department of Computer Science University of Illinois at Urbana-Champaign Urbana, Illinois 61801 13. Type of Report & Period Covered 14. I Supplementary Notes I Abstracts j n this thesis the group, G n , of permutations and complementations of n ridependent variables is embedded isomorphic ally in the symmetric group on the 2 n rnterms of n variables, S n . These minterms are represented in decimal notation. For a subgroup H of G n , the orbits of H play a key role in obtaining the r.jor results. Necessary and sufficient conditions for an element of S n to belong to are also essential. A subgroup H of G n is said to fix a set of n-variable Boolean functions if eery member of the subgroup leaves each function in the set invariant. For an eibtrary subgroup H of G n , the largest subgroup Mjj which fixes the set of functions fxed by H is determined, and an algorithm is given to compute Mjj. Algorithms are also given for the computation of %, the largest subgroup of 3 which leaves setwise invariant the set of functions fixed by H, and of N G (H), the 17Key Words and Document Analysis. 17a. Descriptors normalizer Of H in G n » Associated lattice structures are also studied 3Dup of permutations and complementations Iblean functions Chits 71 Identifiers/Open-Ended Terms 'eCOSATI Field/Group •• .'ailability Statement Unlimited. ,R NTIS-35 (10-70) 19. Security Class (This Report) UNCLASSIFIED 20. Security Class (This Page UNCLASSIFIED 21. No. of Pages 71* 22. Price USCOMM-DC 40329-P7 1 tf* \°^ mm Ss 'flmlBVmmnmniwn HBW o..w°«^' ,0,,, ....mil mill! ■MM HHH mUsH H sBHHl ^H H m hi ' . ■ ■■ n m &« VH ■ ■Mm "'7i H nBBl ■ H H hbhgk mil ,- *av HHH HB H H Ba wSE ffl i hem RB Hi HHHHVH HHH wBBm nan