Characterization of almost maximally almost-periodic groups

Let $G$ be an abelian group. We prove that a group $G$ admits a Hausdorff group topology $\tau$ such that the von Neumann radical $\mathbf{n}(G, \tau)$ of $(G, \tau)$ is non-trivial and finite iff $G$ has a non-trivial finite subgroup. If $G$ is a to…

Authors: S.S. Gabriyelyan

Characterization of almost maximally almost -p erio dic groups S.S. Gabriyely an ∗ Abstract Let G b e an ab elian group. W e pro v e that a group G admits a H ausdorff group top ology τ suc h that the vo n Neumann radical n ( G, τ ) of ( G, τ ) is n on-trivial and fin ite iff G has a non-trivial fi nite s ubgroup. If G is a top ological group, then n ( n ( G )) 6 = n ( G ) if and o nly if n ( G ) is not dually em b ed d ed. In particular, n ( n ( Z , τ )) = n ( Z , τ ) f or any Hausdorff group top ology τ on Z . W e shall write our ab elian groups additiv ely . F or a topo lo gical group X , b X denotes the group of all con tin uous c haracters on X . W e denote its dual group by X ∧ , i.e. the group b X endo w ed w ith the compact-op en topolo g y . Denote by n ( X ) = ∩ χ ∈ b X k er χ the v on N eumann radical of X . If H is a subgroup, w e denote by H ⊥ its annihilator. If A b e a subset of a group X , h A i denotes the subgroup g enerated by A . Let X be a top ological group and u = { u n } a sequence of elemen ts o f b X . W e denote by s u ( X ) the set of all x ∈ X such that ( u n , x ) → 1. Let G b e a subgroup of X . If G = s u ( X ) w e sa y that u char acterizes G and that G is char acterize d (b y u ). F ollow ing E.G.Zelen yuk and I.V.Prota sov [10], [11], w e sa y that a sequence u = { u n } in a group G is a T - se quenc e if there is a Hausdorff group top ology on G fo r whic h u n con v erges to zero. The group G equipp ed with the finest gro up top ology with this pro p ert y is denoted b y ( G, u ). Group top ologies on Z ( p ∞ ) with n ( Z ( p ∞ )) = Z ( p ) w ere considered in corollary 4.9 [3]. Although no explicit construction of suc h top ology was given there, it w as conj ectured b y D. D ikranjan that such top ology can b e f o und b y means of a n appropr ia te T -seq uence on Z ( p ∞ ). This conjecture w as successfully prov ed by G . Luk´ acs [7] for ev ery prime p 6 = 2. He called a Hausdorff top olog ical group G almost m aximal ly almost-p e rio dic if n ( G ) is non-tr ivial and finite and raised the problem of their description. He prov ed that infinite direct sums and the Pr ¨ ufer group Z ( p ∞ ) , for ev ery prime p 6 = 2, are almost maximally almost-p erio dic. A.P .Nguy en [8] generalized these results and prov ed that an y Pr ¨ ufer groups Z ( p ∞ ) a nd a wide class of torsion groups a dmit a (Hausdorff ) almost maximally a lmo st- p erio dic group top ology . Using theorem 4 [5], w e give a general c haracterization of almost m aximally almo st- p erio dic groups. Theorem 1. L et G b e an infinite gr o up. Then the fol lowin g s tateme n ts ar e e quivalent. 1. G admits a T -se quenc e u such that ( G, u ) is almost maxi m al ly alm ost-p erio dic. ∗ The a uthor was partially suppor ted by Israel Ministry of Immigrant Abso r ption Key wor ds and phr ases . Character ized gr oup, T -s equence, v on Neumann radica l, dually embedded, a lmost maximally a lmost-p erio dic. 1 2. G has a non-trivial finite sub gr oup. Eviden tly that n ( n ( G )) 6 = n ( G ) if n ( G ) is no n-trivial and finite. This observ ation leaded G. Luk´ acs [7] to the problem of description o f ab elian groups G whic h admit a T -sequenc e u suc h that n ( n ( G, u )) is strictly contained in n ( G, u ). W e can generalize this question in the follo wing w a y . Problem. Whic h ab elian groups G admit a Hausdorff group top ology τ suc h that n ( n ( G, τ )) 6 = n ( G, τ )? By theorem 1, if G is a not torsion free group, the answ er is p o sitiv e. But for torsion free groups an a nsw er is negativ e in general. W e giv e a n answ er to this problem in the next t heorem. Theorem 2. L et G b e a top olo gic al gr oup. Then n ( n ( G )) 6 = n ( G ) i f and only if n ( G ) is not dual ly em b e dde d. In p articular, n ( n ( Z , τ )) = n ( Z , τ ) for a ny Hausdorff gr o up top olo gy τ on Z . The p ro ofs The following lemma pla ys an imp orta n t role for the pro ofs of theorems 1 and 2. Lemma 1. [5] L e t H b e a dual ly close d and dual ly emb e dde d sub gr oup of a top olo gic al gr oup G . Then n ( H ) = n ( G ) . Pro of of theorem 1 . It is clear that 1 ⇒ 2. Let us prov e that 2 ⇒ 1. Denote by H an y countable subgroup whic h con t a ins an elemen t of finite order. If H admits an almost maximally almost-p erio dic g r o up top ology , then w e can consider the top ology on G in whic h H is op en. Since any o p en subgroup is dually closed and dually embedded (lemma 3.3 [9]), b y lemma 1, G admits an almost maximally almost- p erio dic group top ology to o. Th us w e can and will pass t o coun table subgroups. In particular, we will supp ose that G is countable. There exist three p ossibilities. 1) G con tains an elemen t g o f infinite or der and an elemen t e of o rder p for some prime p . Th us G con tains t he s ubgroup h e, g i which is isomorphic to Z ( p ) ⊕ Z . By lemma 1, w e can assume that G = Z ( p ) ⊕ Z . Then G ∧ = Z ( p ) ⊕ T . No w w e define the follo wing sequence d 2 n − 1 = e + p n 3 − n 2 + · · · + p n 3 − 2 n + p n 3 − n + p n 3 , d 2 n = p n , n ≥ 1 . Let us prov e that d = { d n } is a T -seq uence. F or our con v enience, w e put f n = p n 3 − n 2 + · · · + p n 3 − 2 n + p n 3 − n + p n 3 , then f n < 2 p n 3 ≤ p n 3 +1 . F or an y 0 < r 1 < r 2 < · · · < r v and in tegers l 1 , l 2 , . . . , l v suc h that P v i =1 | l i | ≤ k + 1, w e ha v e | l 1 f r 1 + l 2 f r 2 + · · · + l v f r v | < ( k + 1) f r v ≤ ( k + 1) p r 3 v +1 , (1) Let g = ae + b 6 = 0 , 0 ≤ a < p, b ∈ Z , and k ≥ 0. Set t = | b | + p ( k + 1) and m = 10 t . W e shall pro v e that g 6∈ A ( k , m ). a) Let σ ∈ A ( k , m ) hav e t he follo wing form σ = l 1 d 2 r 1 + l 2 d 2 r 2 + · · · + l s d 2 r s = l 1 p r 1 + · · · + l s p r s = p r 1 · σ ′ , where m ≤ 2 r 1 < 2 r 2 < · · · < 2 r s and σ ′ ∈ Z . If σ ′ = 0, then σ 6 = g . If σ ′ 6 = 0, then | σ | ≥ p r 1 ≥ p 5 | b | > | b | , a nd σ 6 = g . b) Let σ ∈ A ( k , m ) hav e the follow ing for m σ = l 1 d 2 r 1 − 1 + l 2 d 2 r 2 − 1 + · · · + l s d 2 r s − 1 , 2 where m < 2 r 1 − 1 < 2 r 2 − 1 < · · · < 2 r s − 1 and in tegers l 1 , l 2 , . . . , l s b e suc h that l s 6 = 0 and P s i =1 | l i | ≤ k + 1 . Since n 3 < ( n + 1) 3 − ( n + 1) 2 and r s > 5 p ( k + 1), by (1), w e hav e | σ − ( l 1 + · · · + l s ) e | = | l 1 f r 1 + · · · + l s − 1 f r s − 1 + l s f r s | > f r s − ( k + 1) p r 3 s − 1 +1 = p r 3 s + · · · +  p r 3 s − r s ( r s − 1) + p r 3 s − r 2 s − ( k + 1) p r 3 s − 1 +1  >  since ( k + 1) p < p r s − 1  > p r 3 s > | b | and σ 6 = g . c) Let σ ∈ A ( k , m ) hav e the follow ing form σ = l 1 d 2 r 1 − 1 + l 2 d 2 r 2 − 1 + · · · + l s d 2 r s − 1 + l s +1 d 2 r s +1 + l s +2 d 2 r s +2 + · · · + l h d 2 r h , where 0 < s < h and m < 2 r 1 − 1 < 2 r 2 − 1 < · · · < 2 r s − 1 , m ≤ 2 r s +1 < 2 r s +2 < · · · < 2 r h , l i ∈ Z \ { 0 } , P h i =1 | l i | ≤ k + 1 . Since the num b er of summands in d 2 r s − 1 is r s + 2 > 5 p ( k + 1) + 2 and h − s < k + 1, there exists r s − 2 > i 0 > 2 such that for ev ery 1 ≤ w ≤ h − s w e ha v e either r s + w < r 3 s − ( i 0 + 2) r s or r s + w > r 3 s − ( i 0 − 1) r s . The set of all w suc h that r s + w < r 3 s − ( i 0 + 2) r s w e denote b y B (it may b e empt y or has t he form { 1 , . . . , δ } for some 1 ≤ δ ≤ h − s ). Set D = { 1 , . . . , h − s } \ B . Th us σ = ( l 1 + · · · + l s ) e + l 1 f r 1 + · · · + l s − 1 f r s − 1 + P w ∈ B l s + w d 2 r s + w + l s p r 3 s − r 2 s + · · · + l s p r 3 s − ( i 0 +2) r s + l s p r 3 s − ( i 0 +1) r s + l s p r 3 s − i 0 r s + l s p r 3 s − ( i 0 − 1) r s + · · · + l s p r 3 s + P w ∈ D l s + w d 2 r s + w . W e can estimate the expression in row 2, whic h we denote b y A 2 , as follows | A 2 | < X w ∈ B | l s + w | p r 3 s − ( i 0 +2) r s + | l s | 2 p r 3 s − ( i 0 +2) r s < 3( k + 1) p r 3 s − ( i 0 +2) r s < p r 3 s − ( i 0 +1) r s , (2) since 3( k + 1) < r s < p r s − 1. F o r the express ion in row 4, whic h w e denote by A 4 , w e ha v e A 4 = l s p r 3 s − ( i 0 − 1) r s + · · · + l s p r 3 s + X w ∈ D l s + w d 2 r s + w = p r 3 s − ( i 0 − 1) r s · σ ′′ , (3) where σ ′′ ∈ Z . By (1)- (3) , w e ha v e: if σ ′′ 6 = 0, then | σ − ( l 1 + · · · + l h ) e | = | l 1 f r 1 + · · · + l s − 1 f r s − 1 + A 2 + l s p r 3 s − ( i 0 +1) r s + l s p r 3 s − i 0 r s + A 4 | > p r 3 s − ( i 0 − 1) r s − ( k + 1) p r 3 s − 1 +1 − 2( k + 1 ) p r 3 s − i 0 r s > p r 3 s − ( i 0 − 1) r s − 3( k + 1) p r 3 s − i 0 r s > p r 3 s − i 0 r s > p r 2 s > p | b | > | b | and σ 6 = g ; if σ ′′ = 0 , then, b y (1) and (2), | σ − ( l 1 + · · · + l h ) e | = | l 1 f r 1 + · · · + l s − 1 f r s − 1 + A 2 + l s p r 3 s − ( i 0 +1) r s + l s p r 3 s − i 0 r s | > p r 3 s − i 0 r s − ( k + 1) p r 3 s − 1 +1 − ( k + 2) p r 3 s − ( i 0 +1) r s > p r 3 s − i 0 r s − 3( k + 1 ) p r 3 s − ( i 0 +1) r s > p r 3 s − ( i 0 +1) r s > p r 2 s > p | b | > | b | 3 and σ 6 = g to o. Th us d is a T -sequence. Let us prov e t hat s d ( Z ( p ) ⊕ T ) = 0 ⊕ Z ( p ∞ ). Let 0 ≤ ω < p, x ∈ T and ( d n , ω + x ) → 1. T hen ( d 2 n , ω + x ) = ( p n , x ) → 1. Hence x ∈ Z ( p ∞ ) ( see [1] or remark 3 .8 [2]). Let x = ρ p τ , ρ ∈ Z , τ > 0, then for n > 2 τ we hav e ( d 2 n − 1 , ω + x ) = ( e, ω ) → 1 only if ω = 0. Hence Cl( s d ( G ∧ )) = T . By theorem 4 [5], n ( G, d ) = Z ( p ) is finite. 2) G is a to rsion group a nd it contains a subgroup of the form Z ( p ∞ ). Then, b y lemma 1, we can assume that G = Z ( p ∞ ). Let u ∈ Z ( p ∞ ) and β its order. G .Luk´ acs [7] defined the follo wing sequence d 2 n − 1 = u + 1 p n 3 − n 2 + · · · + 1 p n 3 − 2 n + 1 p n 3 − n + 1 p n 3 , d 2 n = 1 p n , n ≥ 1 . He prov ed that if p > 2 then d = { d n } is a T -se quence. Now w e give a simple pro of that d is a T -seq uence for ev ery p . Let g = b p z 6 = 0 ∈ Z ( p ∞ ) , b ∈ Z , b e a n ir r educible fraction and k ≥ 0. Then, for some q ≥ z , 1 p q generates the subgroup con taining u and g . Set t = p ( k + 1) + q and m = 10 t . W e shall pro v e that g 6∈ A ( k , m ). a) Let σ ∈ A ( k , m ) hav e t he follo wing form σ = l 1 d 2 r 1 + l 2 d 2 r 2 + · · · + l s d 2 r s = l 1 p r 1 + · · · + l s p r s = p r 1 · σ ′ , where m ≤ 2 r 1 < 2 r 2 < · · · < 2 r s . Then | σ | = | l 1 d 2 r 1 + l 2 d 2 r 2 + · · · + l h d 2 r h | ≤ h X i =1 | l i | p r i ≤ k + 1 p r 1 < k + 1 p k +1+ q < 1 p q and σ 6 = g (mo d1). b) Let σ ∈ A ( k , m ) hav e the follow ing for m σ = l 1 d 2 r 1 − 1 + l 2 d 2 r 2 − 1 + · · · + l s d 2 r s − 1 , where m < 2 r 1 − 1 < 2 r 2 − 1 < · · · < 2 r s − 1 and in tegers l 1 , l 2 , . . . , l s b e suc h that l s 6 = 0 and P s i =1 | l i | ≤ k + 1 . Since n 3 < ( n + 1) 3 − ( n + 1) 2 and r s > 5 p ( k + 1), we ha v e σ = z ′ p r 3 s − r s + l s p r 3 s , where z ′ ∈ Z . Since | l s | < k + 1 < r s p < p r s − 1 , w e ha v e the following: if σ = z ′′ p α , z ′′ ∈ Z , is an irreducible fraction, then α ≥ r 3 s − r s + 1 > 5 q and σ 6 = g (mo d1). c) Let σ ∈ A ( k , m ) hav e the follow ing form σ = l 1 d 2 r 1 − 1 + l 2 d 2 r 2 − 1 + · · · + l s d 2 r s − 1 + l s +1 d 2 r s +1 + l s +2 d 2 r s +2 + · · · + l h d 2 r h , where 0 < s < h and m < 2 r 1 − 1 < 2 r 2 − 1 < · · · < 2 r s − 1 , m ≤ 2 r s +1 < 2 r s +2 < · · · < 2 r h , l i ∈ Z \ { 0 } , P h i =1 | l i | ≤ k + 1 . Since t he num b er of summands in d 2 r s − 1 is r s + 2 > 5 p ( k + 1 ) + q + 2 and h − s < k + 1, there exists r s − 2 > i 0 > 2 such that for ev ery 1 ≤ w ≤ h − s w e ha v e either r s + w < r 3 s − ( i 0 + 2) r s or r s + w > r 3 s − ( i 0 − 1) r s . 4 The set of all w suc h that r s + w < r 3 s − ( i 0 + 2) r s w e denote b y K (it ma y b e empty or has the form { 1 , . . . , a } for some 1 ≤ a ≤ h − s ) . Set L = { 1 , . . . , h − s } \ B . Thus σ = l 1 d 2 r 1 − 1 + · · · + l s − 1 d 2 r s − 1 − 1 + P w ∈ K l s + w d 2 r s + w + l s p r 3 s − r 2 s + . . . l s p r 3 s − ( i 0 +2) r s + l s p r 3 s − ( i 0 +1) r s + l s p r 3 s − i 0 r s + l s p r 3 s − ( i 0 − 1) r s + · · · + l s p r 3 s + P w ∈ L l s + w d 2 r s + w . Since n 3 < ( n + 1) 3 − ( n + 1 ) 2 , t hen the expression in ro w 1 can b e represen ted in the fo r m c p r 3 s − ( i 0 +2) r s , for some c ∈ Z . Since r s > 5 p ( k + 1), then 1 1 − 1 /p r s < 32 31 and 2 k < p 2 k < p r s . Th us       l s p r 3 s − ( i 0 − 1) r s + · · · + l s p r 3 s  + X w ∈ L l s + w d 2 r s + w      < | l s | p r 3 s − ( i 0 − 1) r s · 1 1 − 1 p r s + k p p r 3 s − ( i 0 − 1) r s +1 < 1 p r 3 s − ( i 0 − 1) r s  k 32 31 + k 1 p  < 2 k p r 3 s − ( i 0 − 1) r s < 1 p r 3 s − i 0 r s , and w e hav e the following: if σ = c ′′ p α , c ′′ ∈ Z , is an irreducible fraction, t hen α ≥ r 3 s − ( i 0 + 1) r s > 5 q and σ 6 = g . Th us { d n } is a T -sequenc e. No w we can rep eat the pro of of theorem 4.4(b) [7]. If x ∈ s u ( X ) and ( x, d 2 n ) → 1, then ( x, χ ) = exp(2 π imχ ) , ∀ χ ∈ Z ( p ∞ ), for some m ∈ Z (example 2.6.3 [11]). Since 0 ≤ 1 p n 3 − n 2 + · · · + 1 p n 3 − 2 n + 1 p n 3 − n + 1 p n 3 ≤ n + 1 p n 3 − n 2 → 0 one has ( x, d 2 n − 1 ) → ( x, u ). Th us ( x, u ) = 1. On t he other hand, if x 0 = (0 , . . . , 0 , 1 , 0 , . . . ), where 1 o ccupies the p osition β + 1, then ( x 0 , d n ) → 1. Since the closure of h x 0 i is h u i ⊥ (remark 10.6 [6]), w e hav e n ( b X , u ) = h u i . 3) G is a torsion g roup and it con tains no a subgroup of the form Z ( p ∞ ). Then, b y the Kulik ov theorem (corollary 24.3 [4]), there exist some prime p 1 and 1 ≤ k 1 < ∞ such that X ∧ = Z ( p k 1 1 ) ⊕ H 1 . Since H 1 is infinite, ag a in b y the Kuliko v theorem, there exist some prime p 2 and 1 ≤ k 2 < ∞ suc h that H 1 = Z ( p k 2 2 ) ⊕ H 2 . Con tin uing this pro cess w e c an s ee that G mus t con tain a subgroup of the form L ∞ i =1 Z ( p i ) with prime p i . Th us w e can assume tha t G has this form. By theorem 3.1 [7], G admits a n almost maximally-almost p erio dic group top ology .  Remark 1. I.Protaso v and E.Zelen yuk [10] prov ed that there exists a T -sequence d suc h that ( Z , d ) has only the trivial c haracter. No w w e g iv e a simple c onc r ete suc h T -sequence . Let G = Z and γ and q > 1 b e any p o sitiv e in tegers. Set d 2 n − 1 = γ + q n 3 − n 2 + · · · + q n 3 − 2 n + q n 3 − n + q n 3 , d 2 n = q n , n ≥ 1 . Then d = { d n } is a T - sequence . If γ = q , then n ( Z , d ) = q Z . If γ = 1, then n ( Z , d ) = Z . (Indeed, exactly t he same pro of whic h is in item 1) of the pro of of theorem 1 (putting a = 0 and t = | b | + ( k + 1)( m + γ )) sho ws that d is a T - sequence . 5 Let x ∈ T and ( d n , x ) → 1. Then ( d 2 n , x ) = ( q n , x ) → 1. Hence x ∈ Z ( t ∞ 1 ) ⊕ · · · ⊕ Z ( t ∞ α ) , where t 1 , . . . , t α are the primes that divide q (remark 3 .8 [2]). Hence , for enough large n we ha v e ( d 2 n − 1 , x ) = ( γ , x ). Th us: if γ = q , then s d ( T ) = Z ( q ) a nd n ( Z , d ) = q Z ; if γ = 1, then n ( Z , d ) = Z .)  Pro of of theorem 2 . Let n ( G ) b e dually em b edded. Since n ( G ) is dually closed, then n ( n ( G )) = n ( G ) b y lemma 1. Con v ersely , if n ( n ( G )) 6 = n ( G ), then there exis ts a nonzero c ha racter χ ∈ n ( G ) ∧ . If χ is extended to e χ ∈ G ∧ , then, by the definition of n ( G ), w e m ust ha v e ( e χ, n ( G ) ) = ( χ, n ( G )) = 1 and χ is tr ivial. It is a contradiction. Let G = Z . Then any its non t rivial subgroup H has the form p Z for some p ositiv e in teger p . If H = n ( Z , τ ), then H is closed. Since G/H is finite and discrete, H is op en and, hence, dually em b edded [9]. The assertion follow s.  References [1] Arma co st D.L. , The structure of lo cally compact ab elian groups, Monographs and T ext- b o o ks in Pure and Applied Mathematics, 68, Marcel Dekk er, Inc., New Y ork, 198 1 . [2] Barbi er i G . , Dikran j an D., M ila n C., W eb er H., T op olo gical torsion related to some sequences o f in tegers. Math. Nachr. (7) 281 (2008) 930- 9 50. [3] Dikranj an D., Milan C ., T ono lo A., A c haracterization of the maximally a lmo st p erio dic ab elian groups, J. Pure Appl. Algebra, 197 (2 005) 23-4 1. [4] F uc hs L. , Ab elian groups, Budap est: Publishing House of the Hungarian Academ y of Sciences 1958, Pergamon Press , London, third edition, reprin ted 1967. [5] Gabri yely an S. S. , On T -sequences and c hara cterized subgroups, a rXiv:math.GN [6] Hewitt E., Ross K.A., Abstract harmonic analysis, v o l. 1, Springer-V erlag, Berlin and Heidelberg , 1964 . [7] Luk´ acs G. , Almost maximally almost-p erio dic group top ologies determined by T - sequence s, T op o logy Appl. 15 3 ( 2 006) 2922-29 32. [8] Nguyen A. P . , Which infinite ab elian torsion groups admit an almost maximally almost- p erio dic group top o logy? arXiv:math.GN/0812.041 0. [9] Noble N. , k - groups and duality , T rans. Amer. Math. So c. 151 (197 0 ) 551-561. [10] Zelenyuk E. G., Prot asov I.V. , T op olog ies on abelian groups, Math. USSR Izv . 37 (1991) 445- 460. Russian original: Izv. Ak ad. Nauk SSSR 54 (1990 ) 1090-1107. [11] Zelenyuk E. G., Prot asov I.V. , T op ologies on gr o ups determined by sequences, Mono- graph Series, Math. Studies VNTL, Lviv, 1 999. Department o f Mat hemati cs, B en -G u r ion Universit y of the Negev , Beer-Shev a, P .O. 653, Israel E-mail addr ess : saak@math.bgu.ac.il 6

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