Iteratoal Joural of Algebra, Vol 6, 0, o 3, 0-07 Factorzato of Fte Abela Grous Khald Am Uversty of Bahra Deartmet of Mathematcs PO Box 3038 Sakhr, Bahra kamee@uobedubh Abstract If G s a fte abela grou ad > s a teger, we say that G has the Hajós-roerty f from ay factorzato G = A A K A of G to a drect roduct of subsets such that each of the these K A s cotas the detty elemet of G, t follows that at least oe of A s erodc ad we say that G has the Réde--roerty f from ay factorzato G = A K A K A of G, t follows that at least oe of the A s s such that A G We shall vestgate some abela grous wth resect to these two roertes ad wll also study the relato betwee them NOTATIONS AND DEFINITIONS Throughout ths aer, by G s meat a fte abela grou, Elemets of G are deoted by the letters a, b, ck I artcular, e stads for the detty elemet ofg Subsets of G are deoted by A, B, C, K ad subgrous by H, K, JK The umber of elemets of a subset A s deoted by A So, artcular, G deotes the order of G ad a deotes the order the elemet a The varats of a grou are gve the usual way, whch we descrbe by a, q β s the drect roduct of cyclc grou of orders examle: a grou of the tye ( ), q β, where ad q are rmes If K A are subsets of a groug, we wrte G = AK A K A to mea that every elemet g G ca be uquely rereseted as
0 Khald Am g = aa Ka, where a A Moreover, ths case, we call G = AK A a factorzato of G I artcular, f H ad K are subgrous, G = HK meas that G s the drect roduct of H ad K, Namely, G = H K If G s a grou, A G ad g G, the ga deotes the set of all roducts ga, where a A Smlarly AK A deotes the set of all roducts a a Ka, where a A However, we must emhasze that the otato AA A K A K wll be used oly f every elemet AK A ca be exressed a uque way Thus, G = A K A K A A subset A of G s called erodc f there s a elemet Such a elemet g s called a erod of A g e A such that ga = A A grou G s sad to have the Hajós--roerty f from ay factorzato G = A K A K A of G, t follows that at least oe of the subsets A s erodc We say that G has the Réde--roerty f from ay factorzato G = A A K A, t follows that at least oe of the subsets geerated by A K A s such that A G, where A deotes the subgrou INTRODUCTION A famous cojecture of H Mkowsk [4] was frst roved by G Hajós [] He solved the cojecture after trasformg the roblem to a questo about fte abela grous Below, we state two versos of ths cojecture: Verso : Let G be a fte abela grou If a, a, Ka are elemets of G ad r, r, K, r are ostve tegers such that each elemet of G s uquely exressble the form x x x a a a K, where 0 x r the a r = e for some, A subset A of G s called cyclc f { r = e, a, a,, a } r = a, A = a, the subgrou geerated by a Verso : Let G be a fte abela grou ad let A K, where r a Observe that, whe A, K, A,, A subsets of G, each cotag the K detty elemet e of G IfG = A K A s factorzato of G ad each A s cyclc, the at least oe of the subsets A s a subgrou of G
Factorzato of fte Abela grous 03 L Réde [5] roved the followg theorem, whch ca be cosdered as a geeralzato of Verso of Hajos s theorem Let G be a fte abela grou ad A, K, A,, A subsets of G each cotag the K K A detty elemet e of G, such that G = A s a factorzato of G ad each A has a rme umber of elemets, the at least oe of the subsets A s a subgrou of G It turs out that t s ot always true that from ay such factorzatos, t follows that oe of the factors s a subgrou For examle, f we let G =< g >, the cyclc grou of order 8 wth geerator g ad choose A = { e, g, g 4, g 5 } ad A = { e, g } The t s straghtforward to check that G = AB s a factorzato of G It s also evdet that ether A or B s a subgrou of G Now, observe that f H s a subgrou of G, the for ay h H,hH=H Ths observato gave rse to the oto of erodc subsets of G Namely, a subset A of G s called a erodc subset f there s a elemet a e A such that aa = A HAJƠS GROUPS A Hajós grou s a grou G for whch from ay factorzato of the form G = AB of G, t follows that ether A or B s erodc The classfcato of Hajós fte Abela grous was acheved by Sads ad De Bruj These grous are lsted below: ( 3 (,, ),(, q),(, q ),( (, q,, ),(,3,3),(3,3),( (, ),(, ), q, r),(, q, r, s),,,),(,,),(,,,,),) where, q, r ad s are rmes ad s a teger Ay cyclc grous of order s a Hajós grou f s of the form where, q, r ad s are rmes ad s a tegers, q, qr, qr s, I the lterature, Hajós grous are also called good grous ad o- Hajós grous are called bad grous Ths oto s geeralzed as follows:
04 Khald Am If G s a fte abela grou ad from each factorzato G = A K A of G to subsets each cotag the detty elemet e of G t follows that at least oe of the subsets A s erodc, we say that G s -good or has the Hajós--Proerty I [], t s show that the cyclc grou G of order, where s a rme has the Hajós-roerty I ths aer, we shall study some o-cyclc -grous wth resect to Hajós-roerty We shall study the Hajós--Proerty for a geeral grou G ad show that there are may grous whch do ot osses the Hajós--roerty We shall also study these grous wth resect to the Réde--Proerty by whch we mea If G s a fte obela grou ad from each factorzato subsets t follows that at least oe of the subsets the Réde--Proerty A s such that G = A A K A of G to K A G, we say that G has We start wth the followg Statemet LEMM Suose G has a subgrou H such that G = H for some rme If H does ot have the Hajós--Proerty The G tself does ot have the Hajós--Proerty Proof Let H = A A A A be a factorzato of H whch oe of the A s erodc, ad let G = b H + bh + + b H = BH, where { b, b,, b } s o-erodc TheG has the factorzato G = ( b + b + + b ) A A Now, oe of A,, A s erodc Also ( b + b + + b ) s ot erodc, for f t were erodc, the there must exst a elemet g = b ( ) a e, b j( ) B, a ad j () a ermutato of,,, such that g( b A ( b A j + b A + + b + b A + + b A A ) = b j ( ) a ( b + b + + b ( b + b + + b ) It follows that ) = That s b j( ) a = b Thus b j ( ) a = b = =
Factorzato of fte Abela grous 05 Hece b j ( ) = b Therefore b b j( ) = erod g = b b j( ) e ad ths s a cotradcto form whch we get that s erodc wth A easy ducto argumet, gves the followg THEOREM If G has a subgrou H whch does ot have the Hajós--Proerty The G tself does ot have the Hajós--Proerty We shall eed some kowledge about characters of abela grous Acharacter of G s a homomarhsm χ : G C, where C: the feld of comlex umbers If χ ( g) = for each g G, the χ s called the rcal character If χ s a character of a grou G ad A be a subset of G,the x(a ) deotes χ(a) ad the a A ( A) = χ : χ( a) = 0 s called the Ahlator of A G set defed by A { } We have the followg lemma LEMMA If A, B are subsets of G, G = A B ad each o-rcal character of G s A(A), the G = AB s a factorzato of G Next, we show that for a rme, grous of tye (, ) do ot have the Hajós--Proerty Note that the ossble otrval values for are = ad = 3 Clearly G has the Hajós-3- Proerty by Rede s theorem However we shall see that G does ot have the Hajós-- Proerty Let G = x x y, where x = ad y = We costruct a factorzato G = AB of G whch ether A = B = x U x Aor y U x B s erodc as follows: y U U x ad 3 ( ) ( ) { e, x y, ( x y),,( x y), x y, x y } y Now, we use Lemma, makg case by case secto to show factorzato ofg G = AB s deed a
06 Khald Am Observe that the ossbltes for order of x(x) are, ad ad the ossbltes for order of x(y) are ad so total, there are sx dfferet cases to cosder Case Order of x (x) Order of x (y) x (A) x (B) 0 3 4 0 0 5 0 6 0 It oly remas to show that ether exst a e A such that aa = A But the a A = A A or B s erodc If A were erodc the there would for all Now, a = or Clearly a = s mossble sce that would force A to be a subgrou of G,whch s ot the case, whle f a =, the a = ( x ) for some, O the other o elemet of the form ( x ) ca be a erod for A, smly because for examle x + ( ) ( xy) = x y A, for ay If B were erodc, the there would exst b e B such that bb = B But the β b B = B for all β Now, b = or Clearly b = s mossble sce the we would have tow may elemets If we assume b =, the B must be subgrou of G whch s ot the case sce for examle x y Bbut ( x y) B CONSEQUENCES 3 A alcato of our revous result gve us that grous of tye (,, ) do ot have the Hajós--Proerty for all, where < + + + s Fally, we would coclude from the ext result that these grous also do ot have the Réde--Proerty Theorem If G has the Hajós--Proerty ad of the A s such that A G G = A A A s a factorzato of G,the at least oe
Factorzato of fte Abela grous 07 Proof Let G have the Hajós--Proerty ad cosder a factorzato G = A A A of G We roceed by ducto o the umber of dstct rme factors k of G If k =, ths s obvous sce G = A A mles that oe of A, say (sce ths s oly a matter of reorderg of A ) has rme order ad the rest have order So { e}, A { } A = = for all Thus, A G for all G, A = e Now, suose k sce G has the Hajós--Proerty, some A s erodc It follows that we get the factorzato A = HC, where H s a roer subgrou of G From the factorzato G = A A A, we get the factorzato of the quotet grou G / H Namely, we ow have G H = A H / H A H / H A H H Now observe that G / H < G / H / / also has the Hajós--Proerty So by ducto assumto, we get that some A j ( ) H / H G / H where j() a ermutato of {,,,} Cosequetly, A j ( ) G Refereces K Am The Hajós Factorzato of Elemetary 3-Grous, Joural of Algebra 4, 4-47(000) G Hajós, Sur la factorzato des grous abeles, Casos Pest, MatFys74 (949), 57-6 3 OHKeller, Uber de luckelose Efull des Raumes Wurfel,/Ree AgewMaath63(930), 3-48 4 HMkowsjk, Dhatsche Aroxmatoem (Lezg, 907) 5 LRede, De eue Theore der edlch abelsch Grue ud Berallgemerug des Hautsatz vo Hajos, Acta mathacadhuar,6(965), 39-69 6 Sads, factorzaro of fte Abela Grous, Acta mathacadhug-(96)53-69 7 SSzabo, Grous wth the Rede Proerty, Le Matematche, VolLII(997)- FascII,357-364 Receved: August, 0