Interntonl Journl of Algebr, Vol 5, 011, no 30, 1481-1488 he Number of Rows whch Equl Certn Row Ahmd Hbl Deprtment of mthemtcs Fcult of Scences Dmscus unverst Dmscus, Sr hblhmd1@gmlcom Abstrct Let be X fnte BCI - lgebr, we shll prove tht the number of rows whch equl certn row, n the Cle tble of the operton, s the gret common dvsor of the orders of ll rows contned n tht row Mthemtcl Subject Clssfcton: 035, 06F35 Kewords: fnte BCI lgebr; gret common dvsor Introducton Suppose tht X s fnte BCI - lgebr, we wll prove tht the number of rows of X, whch equl certn row of X, s the gret common dvsor of the orders of rows contned wthn ths row, b row we men row n the Cle tble of the bnr operton, ths s justfed b Proposton () of [1] Exmple: here s fnte BCI - lgebr X, where X = {0,1,,3,4,5} nd the operton s gven b the followng tble: * 0 1 3 4 5 0 0 0 0 3 3 3 1 1 0 0 4 3 3 1 0 5 4 3 3 3 3 3 0 0 0 4 4 3 3 1 0 0 5 5 4 3 1 0
148 A Hbl So we cll the subset {0,3} the row of 0, nd of 3, the subset {0,1,,3,4,5}the row of nd 5, the subset {0,1,3, 4} the row of 1, nd 4 And we cn cll n of those subsets row of X Here re some prelmnres [3] Defnton (1) non-empt set X wth bnr operton nd dstngushed element 0 clled BCI -lgebr f the followng xoms re stsfed for ever x,, z X : I) (( x ) ( x z)) ( z ) = 0 II) ( x ( x )) = 0 III) x x = 0 IV) x = 0, x = 0 x = A prtl orderng relton cn be defned on BCI -lgebr X b: x If nd onl f x = 0 In n BCI -lgebr X, the followng Propretes hold for ever x,, z X : 1) x 0 = x ) x ( x ( x )) = x 3)( x ) z = ( x z) 4) x x z z, z z x Here we use the sme defntons nd nottons n [1], so we cn Wrght b corollr (1) of [1] tht = { x X : x = } whch men tht s the subset of ll elements of X for whch the rows equl, the row of Proposton (1): let be X BCI -lgebr, nd X n element of X then the subset hs the followng propretes: 1) 0 = 0 ) x 1, x then x 1 x s n tom, [4] 1) b propostons () nd corollr (1) of [1] S0 = X, 0 = S0 0 = X 0 = 0 ) x 1, x x = 1 x nd b proposton (1) of [1] we hve x 1 X = x X so we cn Wrght : ( x 1X ) x = ( x X ) x ( x 1x ) X = ( x x ) X = 0X nd fnll: x1 x = 0 So x 1x 0 = 0 whch b lemm () nd proposton (1) of [1] men tht x 1 x s n tom heorem(): let be X BCI -lgebr, nd X n element of X, then the subset wth the bnr operton x = x ( ), x, nd wth s the unt element, form n beln group Clerl snce Let x, we wnt to show tht: x
he number of rows whch equl certn row 1483 But = S, so we should prove tht x = x ( ) S nd x = x ( ) Now b corollr (1) of [1] x x = so we hve just to prove tht x *( * ) = or b proposton (1) of [1] X = ( x ( )) X But ths s cler snce: ( x ( )) X = ( x X ) ( ) = ( X ) ( ) = ( ( )) X = X = X So x whch men tht the defned bnr operton s ntern on he element s the unt element becuse f x then: x = x ( ) = x 0 = x Also, we hve b proposton (1) of [1] : x x = ( x) = x So, we cn Wrght x = x = x Now ever element x hs n nverse whch s ( x ) Of course ths s n element of, whch cn be shown s we hve dong bove to prove tht s closed under the bnr operton On the other hnd: x ( ( x )) = x ( ( ( x ))) = x ( x ) = And: ( ( x )) x = ( ( x )) ( x) = ( ( x)) ( x ) = x ( x ) = Now we prove tht the bnr operton s commuttve: x = x( ) = ( ( x)) ( ) = ( ( )) ( x) = ( x) = x Fnll the bnr operton s ssoctve: Let x,, z, then we hve:
1484 A Hbl x( z) = x ( ( z)) = x ( ( ( z))) B commuttve propret nd snce the element ( z) belong to : = ( ( z)) ( x) = ( ( x)) ( z) = ( x ( )) ( z) = ( x ) ( z) = ( x ) z So we hve tht (,, ) s n beln group, wth s the unt element, nd wth the defned bnr operton Notton: for n element X, the subset wll be clled the group of heorem (3): Let be X BCI -lgebr, for n elements b, X we hve tht the group s somorphc to sub-group of the group b * We shll show tht the functon ϕ : (, ) ( b, b) defned b ϕ ( x ) = x b for n x s well defned, njectve homomorphsm o do wth, we hve to prove tht: ϕ( x ) = x b b, whenever x In fct b proposton (1), nd corollr (1) of [1]: x x = x X = X ( x X ) b = ( X ) b ( x b) X = ( b) X x b= b x b b ϕ( x) b So, for n x, we hve tht x b, b b nd: ϕ( x) ϕ ( ) = ( x b)( b) = ( x b) (( b) ( b)) But ( b) ( b) nd s n tom b () of proposton (1), so we hve ( b) ( b) = nd we cn Wrght : ϕ( x ) ϕ( ) = ( x b) ( ) = ( x ( )) b = ( x ) b = ϕ( x ) So, the functon ϕ s homomorphsm of groups Fnll ths functon s njectve becuse for n x 1, x : ϕ( x 1) = ϕ( x ) x1 b = x b now snce x 1, x we hve b proposton (1) tht the elements x 1x, x x1re toms
he number of rows whch equl certn row 1485 Also ( x 1 b) ( x b) x1 x so, ( x 1b) ( x b) = x1 x nd x1 x = 0 In the sme w we hve x x1 = 0, so we got x 1 = x So we cn s tht the beln group (, ) s somorphc to subgroup of the beln group ( b, b) he mentoned subgroup s, ϕ ( ) = b Corollr (1): For n element X we hve tht: dvdes B proposton () of [1] mples nd snce the subsets lke re equls or dsjonts, so cn be wrthen s dsjont unon of such subsets, where the elements re of the form x b the proposton () of [1] So, = n = U whch mples = 1 x = n = now b the bove theorem nd the = 1 x so-clled lgrnge theorem n group theor we hve dvdes * x for = 1,,, n nd clerl: dvdes Corollr (): let X be n fnte BCI -lgebr For n elements b, X, we hve tht: dvdes b In others words: he crdnl of group of n element X dvdes the crdnl of n row contned n the row of B the precedent Corollr we hve b * dvdes b *, but dvdes b * b theorem (3) nd Lgrnge theorem So dvdes b * B the precedent Corollr we hve clerl: Corollr (3): Let X be n fnte BCI -lgebr For n X the order of the group dvdes the gret common dvsor of the orders of the rows of X contned n the row of Notton: the gret common dvsor of the orders of the rows of X contned wthn the row of, wll be denoted b CDR heorem (4): Let be X fnte BCI -lgebr, X n element of X, then the number of elements of X whch ts rows re equl to the row of, s CDR As we noted before = { x X : = } so we hve just to prove tht the order of equl CDR Frst, b Corollr (3) prove the converse x dvdescdr So for prove the theorem we hve just to
1486 A Hbl he converse wll be proved b nducton on the order of, the row of Frst t s cler tht when = 0 then = 0, snce b propostons () of [1] 0 In ths cse, when row x s contned n then x = 0 Also b proposton (), (3) nd corollr (1) of [1] 0 = 0 Now suppose tht for n row b of X whose order strctl less thn CDRb dvdes b Now, s we hve seen n Corollr (1) we cn Wrght = n = U, where = 1 we hve tht ; = 1,,, nre elements of, nd the groups ; = 1,,, n dsjonts Here of course, the rows,,, re ll the rows contned n, whch s cler b the eqult = n = U = 1 1 n Clerl we cn consder tht 1 =, n = 0 So, the eqult: = n = U whch gve us = = n = = n = + nd = (1) = And here t must be ;,3,, n = so, snce 1 < = If there s {,3,, n}, we hve clerl = =, whch s mpossble snce 1 such tht: =, but ths men tht So, b the sttement of nducton we hve tht for ech {,3,, n} dvdes Now snce the rows contned wthn CDR re clerl contned wthn, we got tht CDR dvdes CDR, so CDR dvdes, nd for ths reson the number CDR must dvdes the rght hnd sde of the eqult (1), consequentl t dvdes So, = CDR, nd the theorem s proved Exmples: From [5] we hve the followng BCI -lgebr, number 61: 0 1 3 4 5 0 0 0 3 3 1 1 0 5 4 3 0 3 0 3 3 3 3 0 0 4 4 1 5 0 3 5 5 3 4 1 0
he number of rows whch equl certn row 1487 he order of the row of 5 s 6, nd the orders of rows of elements 0, 1,, 3, 4 re respectvel 3, 6, 3, 3, 6 so the gret common dvsors of ll rows contned n the row of 5 s 3 whch equl the number of rows who re equl the row of 5 Also from [5], for BCI -lgebr number 43: 0 1 3 4 5 0 0 0 0 3 3 3 1 1 0 0 4 3 3 1 0 5 4 3 3 3 3 3 0 0 0 4 4 3 3 1 0 0 5 5 4 3 1 0 he order of the row of 1 s 4, nd the orders of the rows of 0, 4, 3 re respectvel, 4, so the gret common dvsors of the orders of rows contned n the row of 1 s, nd the rows whch equl the row of 1 s clerl tself nd the row of 4 Remrk As fnl result we hve new proof to the theorem (4) of [] 4[] heorem: he gret common dvsor of the orders of rows n Cle tble CDR, whch represent the bnr operton, dvdes the gret common dvsor of the orders of columns n ths tble CDC B theorem (4) the number CDR, defned n the bove text, dvdes the order of n group of X And snce ever column of X s the unon of dsjont groups, so the number CDR dvdes the order of tht column, nd lso dvdes the numbercdc References [1] A Hbl, A fnte BCI -lgebr of KL -product, Dmscus Unverst Journl for the bsc scences, 4(008), No 1, 11-16 [] A Hbl, Propert of fnte BCI lgebrs, JP Journl of Algebr, Number heor& Applctons, 19(010), No1, 11-17 [3] K Isek, On BCI -lgebrs, Mth Sem Notes 8(1980), 15-130 [4] J Meng, nd X L Xn, Chrcterztons of toms n BCI lgebrs, Mth Jponc, 37(199) 359-361
1488 A Hbl [5] J Meng, Y B Jun nd E H Roh, BCI -lgebrs of order 6, Mth Jponc, 47(1998), No 1, 33-43 Receved: June, 011