ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS

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1 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS DANIEL SMERTNIG Abstract. For a fnte abelan group G and a splttng feld K of G, let d(g, K) denote the largest nteger l N for whch there s a sequence S = g 1... g l over G such that (X g 1 a 1 )... (X g l a l ) 0 K[G] for all a 1,..., a l K. If D(G) denotes the Davenport constant of G, then there s the straghtforward nequalty D(G) 1 d(g, K). Equalty holds for a varety of groups, and a standng conjecture of W. Gao et.al. states that equalty holds for all groups. We offer further groups for whch equalty holds, but we also gve the frst examples of groups G for whch D(G) 1 < d(g, K) holds. Thus we dsprove the conjecture. 1. Introducton and Man Result Let G be an addtve fnte abelan group. For a (multplcatvely wrtten) sequence S = g 1... g l over G, S = l s called the length of S, and S s sad to be zero-sum free f I g 0 for every nonempty subset I [1, l]. Let d(g) denote the maxmal length of a zero-sum free sequence over G. Then d(g) + 1 s the Davenport constant of G, a classcal constant from Combnatoral Number Theory (for surveys and hstorcal comments, the reader s referred to [3], [8, Chapter 5], [7]). In general, the precse value of d(g) (n terms of the group nvarants of G) and the structure of the extremal sequences s unknown, see [12, 1, 13, 10, 11, 4, 14, 15, 9] for recent progress. Group algebras R[G] - over sutable commutatve rngs R - have turned out to be powerful tools for a great varety of questons from combnatorcs and number theory, among them the Davenport constant. We recall the defnton of an nvarant (nvolvng group algebras) whch was used for the nvestgaton of the Davenport constant snce the 1960s. For a commutatve rng R, let d(g, R) N { } denote the supremum of all l N havng the followng property: There s some sequence S = g 1... g l of length l over G such that (X g1 a 1 )... (X g l a l ) 0 R[G] for all a 1,..., a l R \ {0}. If S s zero-sum free, R s an ntegral doman, a 1,..., a l R \ {0} and then c 0 0. Hence f 0, and t follows that f = (X g1 a 1 )... (X g l a l ) = g G c g X g, d(g) d(g, R). The followng Theorem A was proved by P. van Emde Boas, D. Kruyswjk and J.E. Olson n the 1960s (n fact, they dd not explctly defne the nvarants d(g, K) but got these results mplctly). Hstorcal remarks and proofs n the present termnology may be found n [7, Secton 2.2] and [8, Theorem 5.5.9]; see also [5] Mathematcs Subject Classfcaton. 11P70, 11B50, 20K01, 11B30. Key words and phrases. Davenport constant, zero-sum sequence, group algebras. Ths work was supported by the Austran Scence Fund FWF, Doctoral Program C75-N13 Dscrete Mathematcs. 1

2 2 DANIEL SMERTNIG Theorem A. Let G be a fnte abelan group wth exp(g) = n Let K be a splttng feld of G wth char(k) exp(g). Then 2. If G s a p-group, then d(g) = d(g, Z/pZ). d(g, K) (n 1) + n log G n. Note that for a cyclc group G of order n, the above upper bound mples that d(g) = d(g, K) = n 1, snce d(c n ) n 1 can easly be seen. Only recently, W. Gao and Y. L showed that d(c 2 C 2n ) = d(c 2 C 2n, K) ([6, Theorem 3.3]). We extend ther result, but we also show that Conjecture 3.4 n [6], statng that d(g) = d(g, K) for all groups G, does not hold. Here s the man result of the present paper. Theorem 1.1. Let G = C p C pn wth p P, n N and let K be a splttng feld of G. 1. If p 3, then d(g) = d(g, K). 2. If p 5 and n 2, then d(g) < d(g, K). 2. Prelmnares Let N denote the set of postve ntegers, P N the set of prme numbers, and let N 0 = N {0}. For real numbers a, b R, we set [a, b] = {x Z a x b}. For n N and p P, let C n denote a cyclc group wth n elements, v p (n) N 0 the p-adc valuaton of n wth v p (p) = 1 and F p = Z/pZ the fnte feld wth p elements. Let G be an addtve fnte abelan group. Suppose that G = C n1... C nr wth 1 < n 1... n r. Then r = r(g) s the rank of G, n r = exp(g) s the exponent of G, and we defne d (G) = r (n 1). If G = 1, then the exponent exp(g) = 1, the rank r(g) = 0, and we set d (G) = 0. If A, B G are nonempty subsets, then A + B = {a + b a A, b B} s ther sumset. We wll make use of a Theorem of Cauchy-Davenport whch runs as follows (for a proof see [8, Cor ]). Lemma 2.1. Let G be a cyclc group of order p P and let A, B G be nonempty subsets. Then A + B mn{ A + B 1, p}. Sequences over groups. Let F(G) be the (multplcatvely wrtten) free abelan monod wth bass G. The elements of F(G) are called sequences over G. We wrte sequences S F(G) n the form S = g G g vg(s), wth v g (S) N 0 for all g G. We call v g (S) the multplcty of g n S, and we say that S contans g f v g (S) > 0. A sequence S 1 s called a subsequence of S f S 1 S n F(G) (equvalently, v g (S 1 ) v g (S) for all g G). If a sequence S F(G) s wrtten n the form S = g 1... g l, we tactly assume that l N 0 and g 1,..., g l G. For a sequence S = g 1... g l = g G g vg(s) F(G), we call S = l = g G v g (S) N 0 the length of S and σ(s) = l g = v g (S)g G the sum of S. g G The sequence S s called a zero-sum sequence f σ(s) = 0, and t s called zero-sum free f I g 0 for all I [1, l] (equvalently, f there s no nontrval zero-sum subsequence). We denote by

3 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS 3 D(G) the smallest nteger l N such that every sequence S over G of length S l has a nontrval zero-sum subsequence; d(g) the maxmal length of a zero-sum free sequence over G. Then D(G) s called the Davenport constant of G, and we have trvally that d (G) d(g) = D(G) 1. We wll use wthout further menton that equalty holds for p-groups and for groups of rank r(g) 2 ([8, Theorems and 5.8.3]) (equalty holds for further groups, but not n general [7, Corollary ]). Group algebras and characters. Let R be a commutatve rng (throughout, we assume that R has a unt element 1 0) and G a fnte abelan group. The group algebra R[G] of G over R s a free R-module wth bass {X g g G} (bult wth a symbol X), where multplcaton s defned by ( a g X g)( ( a h b g h )X g. g G g G b g X g) = g G We vew R as a subset of R[G] by means of a = ax 0 for all a R. An element of R s a zero-dvsor [ a unt ] of R[G] f and only f t s a zero-dvsor [ a unt ] of R. Let K be a feld, G a fnte abelan group wth exp(g) = n N and µ n (K) = {ζ K ζ n = 1} the group of n-th roots of unty n K. An n-th root of unty ζ s called prmtve f ζ m 1 for all m [1, n 1], and we denote by µ n(k) µ n (K) the subset of all prmtve n-th roots of unty. We denote by Hom(G, K ) = Hom(G, µ n (K)) the character group of G wth values n K (whose operaton s gven by pontwse multplcaton wth the constant 1 functon as dentty), and we brefly set Ĝ = Hom(G, K ) f there s no danger of confuson. Every character χ Ĝ has a unque extenson to a K-algebra homomorphsm χ: K[G] K (agan denoted by χ) actng by means of ( χ a g χ(g). g G a g X g) = g G We call K a splttng feld of G f µ n (K) = n. Let K be a splttng feld of G and Ĝ = Hom(G, K ). We gather the propertes needed for the sequel (for detals see [8, Secton 5.5] and [2, 17]). We have char(k) exp(g), G = G 1 K K, G = Hom(G, K ), and the map h G Hom(G, K ) G K, defned by (χ, g) χ(g), s a non-degenerate parng (that s, f χ(g) = 1 for all χ Ĝ, then g = 0, and f χ(g) = 1 for all g G, then clearly χ = 1, the constant 1 functon). Furthermore, the Orthogonalty Relatons hold ([8, Proposton 5.5.2]), and for every f K[G], we have (see [8, Proposton 5.5.2]) f = 0 K[G] f and only f χ(f) = 0 for every χ Hom(G, K ). Moreover, f χ(f) 0 for all χ Hom(G, K ), then f K[G] ; explctly, a smple calculaton usng the Orthogonalty Relatons shows that f 1 = 1 ( χ( g) ) X g. G χ(f) For a subgroup H G, we set g G χ Hom(G,K ) H = {χ Ĝ χ(h) = 1 for all h H}. We clearly have a natural somorphsm H = Ĝ/H.

4 4 DANIEL SMERTNIG 3. Proof of the Theorem We fx our notaton, whch wll reman vald throughout ths secton. Let G = C m C mn wth m N 2, n N and let e 1, e 2 G be such that G = e 1 e 2, ord(e 1 ) = m and ord(e 2 ) = mn. Furthermore, let K be a splttng feld of G, ζ µ mn(k), and let ψ, ϕ Ĝ be defned by ψ(e 1) = ζ n, ψ(e 2 ) = 1 and ϕ(e 1 ) = 1, ϕ(e 2 ) = ζ. Then ord(ψ) = m, ord(ϕ) = mn and Ĝ = ψ ϕ. Note that, n the case m = p P, θ : { F p ψ, ϕ n ψ, ϕ n (k + pz, χ) χ k, s an F p -vector space structure on ( ψ, ϕ n, ). Whenever ψ, ϕ n s consdered as F p -vector space t s done so wth respect to θ. The followng Lemmas 3.1 and 3.2 wll allow us to restrct ourselves to sequences consstng of certan specal elements n the proof of Theorem Lemma 3.2 s a generalzaton of a statement used by W. Gao and Y. L n ther proof of the case m = 2 [6]. Lemma 3.1. Let R be a commutatve rng, g 1... g l F(G) a sequence over G, and let a 1,..., a l R\{0} be such that (X g1 a 1 )... (X g l a l ) = 0 R[G] Then, for any k 1,..., k l N, also (X k1g1 a k1 1 )... (X k lg l a k l l ) = 0 R[G]. Proof. For all [1, l], from whch the lemma mmedately follows. k 1 X kg a k = (X g a ) X jg (a ) k 1 j, j=0 Lemma 3.2. Let R be a commutatve rng and G 0 = {e 1 } {ke 1 + p P, p m p up e 2 k [0, m 1], u p N 0 }. Let M N be such that, for every sequence S = g 1... g M+1 F(G 0 ), there exst a 1,..., a M+1 R\{0} such that f = (X g1 a 1 )... (X g M+1 a M+1 ) = 0 R[G]. Then d(g, R) M. Proof. By Lemma 3.1 and the defnton of d(g, R), t s suffcent to show that every element g G s a multple of an element n G 0. Let g = ke 1 +le 2 wth k [0, m 1] and l [0, mn 1]. If l = 0, g s obvously a multple of e 1. Consder the case l 0. Then l = p P,p m pvp(l) q wth q [1, mn 1] and gcd(q, m) = 1. Therefore there exsts an a [1, m 1] wth qa 1 mod m. From ord(e 1 ) = m, t follows that g = q(ake 1 + p P,p m pvp(l) e 2 ). Choosng k [0, m 1] such that k ak mod m, we obtan g = q(k e 1 + p P,p m pvp(l) e 2 ), whch s a multple of an element n G 0. Lemma 3.3. Let g G and χ, χ Ĝ. Then χ (g) = χ(g) f and only f χ χ g. Also 1. ke 1 + e 2 = ψϕ nk for k [0, m 1]; 2. ϕ n ke 1 + mle 2 for k [0, m 1] and l [0, n 1].

5 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS 5 Proof. Clearly χ (g) = χ(g) f and only f χ 1 χ (g) = 1,.e., χ χ g. 1. From ψ 1 (ke 1 + e 2 ) = ζ nk = ϕ nk (ke 1 + e 2 ), t follows that ψϕ nk ke 1 + e 2. Then ord(ke 1 + e 2 ) = mn and ke 1 + e 2 = G/ ke1 + e 2 mply ke 1 + e 2 = m, from whch ke 1 + e 2 = ψϕ nk follows. 2. Observe that ϕ n (ke 1 + mle 2 ) = ζ nml = (ζ nm ) l = 1 mples ϕ n ke 1 + mle 2. Lemma 3.4. Let H Ĝ and S = g 1... g l F(G). Then the followng statements are equvalent : (a) There exst a 1,..., a l K such that χ ( l (Xg a ) ) = 0 for all χ H. (b) There exst s [0, l] and χ 1,..., χ s H such that H s χ g. (c) H = or there exst χ 1,..., χ l H such that H l χ g. Proof. For H = all statements are trvally true. Let H. (a) (b) The extenson of χ Ĝ to K[G] s a K-algebra homomorphsm, and thus χ ( l (X g a ) ) = 0 f and only f there s an [1, l] wth χ(x g a ) = 0,.e., χ(g ) = a. Let s = { [1, l] there exsts a χ H such that χ(g ) = a } [0, l]. Wthout restrcton let g 1,..., g s and a 1,..., a s be such that there exst χ H wth χ (g ) = a for [1, s]. Let χ H. Then, by assumpton, χ(g ) = a for some [1, s]. Therefore χ 1 χ(g ) = 1,.e. χ χ g. (b) (a) Let a = χ (g ) for [1, s] and let a s+1 =... = a l = 1. Let χ H. Then, by assumpton, there exsts an [1, s] such that χ χ g,.e., χ(g ) = χ (g ) = a. Hence χ(x g a ) = 0. (b) (c) Obvous. Note that, n partcular, d(g, K) s the supremum of all l N 0 such that there exsts a sequence S = g 1... g l F(G) wth l χ g Ĝ for any choce of χ 1,..., χ l Ĝ. Or, equvalently, d(g, K) + 1 s the mnmum of all l N 0 such that, for any sequence S = g 1... g l F(G), there exst χ 1,..., χ l Ĝ such that Ĝ can be covered as above: Ĝ = l χ g. Consder m = p P. Our strategy for fndng an upper bound on d(g, K) wll be to subdvde Ĝ nto cosets modulo ψ, ϕ n and cover each of these cosets ndvdually. Lemma 3.2 allows us to restrct ourselves to certan specal elements g G n dong so, and from Lemma 3.3, we see that for these elements g contan (or are) 1-dmensonal subspaces,.e., lnes of the 2-dmensonal F p -vector space ψ, ϕ n. Then, for χ ψ, ϕ n, χ g s an affne lne n ψ, ϕ n contanng the pont χ, and our task essentally bols down to coverng n copes of ψ, ϕ n by such lnes (where the slopes are fxed by S). Before we do so, we study some smple confguratons n Lemma 3.5 and Lemma 3.6. The man part of the proof for the cases m {2, 3} then follows n Lemma 3.7. It s based on the proof by Gao and L of the case m = 2, but s stated n terms of group characters nstead of workng wth the group algebra drectly.

6 6 DANIEL SMERTNIG Lemma 3.5. Let s [0, m] and let S = g 1... g s+(m s)m F(G) such that ether g 1 =... = g s = ke 1 +e 2 wth k [0, m 1] or g 1,..., g s {ke 1 +mle 2 k [0, m 1], l N 0 }. Then there exst χ 1,..., χ s+(m s)m such that ψ, ϕ n s+(m s)m χ g. Proof. Let L = ψϕ nk n the case g 1 =... = g s = ke 1 + e 2, and let L = ϕ n otherwse. Snce L s a subgroup of ψ, ϕ n and has cardnalty L = m, there exst τ 1,..., τ m ψ, ϕ n such that ψ, ϕ n = m τ L. By Lemma 3.3, L g for [1, s]. Then s m ψ, ϕ n τ g τ L. =s+1 For j [s + 1, s + (m s)m], let χ j g j, and let L = {λ 1,..., λ m }. Then, for [s + 1, m], τ L = {τ λ j j [1, m]} m j=1 τ λ j χ 1 s+( (s+1))m+j g s+( (s+1))m+j. Lemma 3.6. Let m = p P, g {ke 1 + ple 2 k [0, p 1], l N 0 } and S = p 1 =0 (e 1 + e 2 )g. Then ψ, ϕ n p 1 =0 e 1 + e 2 g. Proof. By Lemma 3.3, p 1 ψϕ n ϕ n =0 p 1 e 1 + e 2 g. Let ψ k ϕ nl ψ, ϕ n wth k, l [0, p 1]. In the case k = 0, clearly ϕ nl ϕ n. Otherwse, there exsts an [0, p 1] such that k l mod p. Hence ψ k ϕ nl = (ψϕ n ) k ψϕ n. =0 Lemma 3.7. Let m = p P, G 1 = {e 1 } {ke 1 + p u e 2 k [0, p 1], u N}, and G 0 = {e 1 } {ke 1 + p u e 2 k [0, p 1], u N 0 } = {ke 1 + e 2 k [0, p 1]} G 1. If, for all sequences T = h 1... h rp 1 F(G 0 ) wth r [2, mn {p 1, n + 1}] and v g (T ) < p for all g G 0 as well as g G 1 v g (T ) < p, there exst χ 1,..., χ rp 1 Ĝ such that r 2 =0 ϕ ψ, ϕ n rp 1 χ h, then d(g, K) = d (G). Proof. Snce d (G) d(g) d(g, K) always holds, t s suffcent to show that d(g, K) d (G) = (pn 1) + (p 1) = (n + 1)p 2. By Lemma 3.2, t s suffcent to show that, for any sequence S = g 1... g (n+1)p 1 F(G 0 ), there exst a 1,..., a (n+1)p 1 K such that f = (n+1)p 1 (X g a ) = 0 K[G]. To see ths, we use Lemma 3.4 and show that there exst χ 1,..., χ (n+1)p 1 such that Ĝ = n 1 =0 ϕ ψ, ϕ n (n+1)p 1 χ g. We group the elements of S nto as many p-tuples of the forms (e 2,..., e 2 ), (e 1 + e 2,..., e 1 + e 2 ),..., ((p 1)e 1 + e 2,... (p 1)e 1 + e 2 ) and (g 1,..., g p) G p 1 as possble to obtan l [0, n] such tuples. Wthout restrcton, let these p-tuples be (g 1,..., g p ),..., (g (l 1)p+1,..., g lp ).

7 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS 7 For each [1, l], the tuple (g ( 1)p+1,..., g p ) fulflls the condtons of Lemma 3.5 wth s = p. Therefore, there exst χ ( 1)p+1,..., χ p such that ϕ n ψ, ϕ n p j=( 1)p+1 χ j g j. It remans to be shown that χ lp+1,..., χ (n+1)p 1 can be chosen such that n l 1 =0 ϕ ψ, ϕ n (n+1)p 1 j=lp+1 χ j g j. In the case l n, ths s trvally so, and therefore t s suffcent to consder l n 1. By T = g lp+1... g (n+1)p 1 we denote the subsequence of S consstng of the remanng elements. We have T = S lp = (n+1 l)p 1. In the process of creatng p-tuples, we parttoned the elements of G 0 nto p + 1 dfferent types. If there were at least p elements of one type, we could create another tuple, n contradcton to the maxmal choce of l. Thus we must have v g (T ) < p for all g G 0, g G 1 v g (T ) < p, and T (p + 1)(p 1) = p 2 1, whch mples n + 1 l p. Altogether, we have n + 1 l [2, p]. In the case n + 1 l p 1, we set r = n + 1 l [2, mn {p 1, n + 1}]. Then, by assumpton, χ lp+1,..., χ (n+1)p 1 can be chosen such that r 2 =0 ϕ ψ, ϕ n (n+1)p 1 j=lp+1 χ j g j. Snce r 2 = n l 1, ths already means Ĝ (n+1)p 1 χ g. In the case n + 1 l = p, we have T = p 2 1 = (p + 1)(p 1). Ths can only happen f each of the p + 1 dfferent types of elements occurs exactly p 1 tmes. Therefore p 1 p 2 p 2 T = (je 1 + e 2 ) p 1 h j = j=0 =0 =0 ( p 1 ) (je 1 + e 2 ) h wth h 0,..., h p 2 G 1. Wthout restrcton, for [0, p 2], let g lp+(p+1)+1... g lp+(p+1)+(p+1) = p 1 j=0 (je 1 + e 2 ) h. For every [0, p 2], we set χ lp+(p+1)+1 =... = χ lp+(p+1)+(p+1) = ϕ. Then, from Lemma 3.6, t follows that ϕ ψ, ϕ n (p+1)+(p+1) j=(p+1)+1 χ lp+j g lp+j. Due to n l 1 = p 2, ths agan mples Ĝ (n+1)p 1 χ g. j=0 Proof of Theorem For p = 2,.e. G = C 2 C 2n, ths follows trvally from Lemma 3.7, snce there are no admssble sequences. Consder p = 3,.e., G = C 3 C 3n. Let G 1 = {e 1 } {ke u e 2 k [0, 2], u N} and G 0 = {e 2, e 1 +e 2, 2e 1 +e 2 } G 1. Then, by Lemma 3.7, t s suffcent to show that, for T = h 1... h 5 F(G 0 ), we can choose χ 1,..., χ 5 Ĝ such that ψ, ϕn χ 1 h 1... χ 5 h 5. We dvde the elements nto four types: e 2, e 1 +e 2, 2e 1 +e 2 and elements from G 1. Snce T = 5, one of these types must occur at least twce. Wthout restrcton, let h 1 and h 2 be of the same type. Thus we have ether h 1 = h 2 = ke 1 + e 2 for some k [0, 2] or h 1, h 2 G 1. Then T fulflls the condtons of Lemma 3.5 wth s = 2, and t follows that χ 1,..., χ 5 can be chosen such that ψ, ϕ n 5 χ h. The followng Lemma 3.8 recaptulates a few smple facts, whch are well known n the context of affne lnes, and wll be used extensvely n the constructon of a counterexample n the case p 5 and n 2. Lemma 3.8. Let m = p P, g 1 = k 1 e 1 + e 2, g 2 = k 2 e 1 + e 2 wth k 1, k 2 [0, p 1], χ Ĝ and χ 1, χ 2 χ ψ, ϕ n.

8 8 DANIEL SMERTNIG 1. χ 1 χ g = ϕ ns g wth s [0, p 1] for {1, 2}. 2. χ 1 χ g = {ψ u ϕ nv u, v [0, p 1] wth k u + v s mod p} for {1, 2}. 3. (a) χ 1 g 1 χ 2 g 2 = 1 f and only f g 1 g 2. (b) χ 1 g 1 χ 2 g 2 = 0 f and only f g 1 = g 2 and s 1 s 2. (c) χ 1 g 1 χ 2 g 2 = p f and only f g 1 = g 2 and s 1 = s 2. Proof. 1. Let {1, 2} and χ 1 χ = ψ u ϕ nv wth u, v [0, p 1]. By Lemma 3.3.1, g = ψϕ nk. Therefore ϕ n(ku+v) χ 1 χ = ψ u ϕ nku g, and hence χ 1 χ g = ϕ ns g wth s [0, p 1] chosen such that s k u + v mod p. 2. In vew of Lemma 3.3.1, we have, for u, v [0, p 1], ψ u ϕ nv χ 1 χ g = ϕ ns ψϕ nk f and only f ψ u ϕ nv = ψ w ϕ n(s kw) for some w [0, p 1]. Ths s the case f and only f u w mod p and v s k w mod p,.e., f and only f u w mod p and k u + v s mod p (recall by Lemma that g ψ, ϕ n ). 3. By 2, we have χ 1 χ 1 g 1 χ 1 χ 2 g 2 = {ψ u ϕ nv u, v [0, p 1] wth k 1 u + v s 1 mod p and k 2 u + v s 2 mod p}. Snce χ 1 χ 1 g 1 χ 1 χ 2 g 2 = χ 1 g 1 χ 2 g 2, t s suffcent to consder the number of solutons of the lnear system k 1 u + v s 1 mod p and k 2 u + v s 2 mod p for u, v [0, p 1] over F p. In the case g 1 g 2,.e., k 1 k 2, t possesses a unque soluton. In the case g 1 = g 2, t possesses no soluton for s 1 s 2. For s 1 = s 2, the two equatons concde, and we obtan p solutons. In the constructon of the counterexamples, we use the same characterzaton of d(g, K), derved from Lemma 3.4, as n the proof of Theorem except now we show that t s not possble to cover Ĝ wth such subsets. To do so, we frst consder a specal type of sequence n Lemma 3.9, whch wll turn out to be the only one whch cannot be dscarded wth smpler combnatoral arguments, as wll be gven n the proof of Theorem that follows the lemma. Lemma 3.9. Let m = p P, p 5 and k 1, k 2, k 3 [0, p 1] be dstnct. Let l [2, p 1], T = (k 1 e 1 + e 2 ) l (k 2 e 1 + e 2 ) l (k 3 e 1 + e 2 ) l F(G), and χ Ĝ. For [1, 3] and j [1, l], let χ,j Ĝ. Then ( 3 j=1 l χ,j k e 1 + e 2 ) χ ψ, ϕ n < l(3p 2l). Proof. We set g = k e 1 + e 2 for [1, 3]. Let [1, 3] and j [1, l]. We can assume χ,j χ ψ, ϕ n snce otherwse χ,j g χ ψ, ϕ n = (due to g = ψϕ nk ψ, ϕ n ). Usng Lemma 3.8.1, we can furthermore assume χ 1 χ,j = ϕ ns,j wth s,j [0, p 1]. And we can then also assume, wthout restrcton, s,j s,j for j [1, l] \ {j}, snce otherwse χ,j g = χ,j g. For [1, 3], let E = l j=1 χ,j g. Then and ( 3 j=1 E 1 E 2 E 3 = l χ,j g ) χ ψ, ϕ n = E 1 E 2 E 3 3 E 1 < 3 E E + E 1 E 2 E 3.

9 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS 9 For, [1, 3] dstnct, we show E = lp, E E = l 2 and E 1 E 2 E 3 < l 2. Then E 1 E 2 E 3 < 3lp 3l 2 + l 2 = l(3p 2l). Let [1, 3]. By Lemma 3.8.3b, χ,j g χ,j g = for j, j [1, l] wth j j, and g = ψϕ nk = p (by Lemma 3.3.1). Therefore E = lp. Let, [1, 3] be dstnct. For j, j [1, l] dstnct, we have χ,j g χ,j g = and χ,j g χ,j g = (by Lemma 3.8.3b). Ths mples that, for E E = ( l χ,j g ) ( j=1 l j =1 χ,j g ) = l j=1 j =1 l (χ,j g χ,j g ), the unon s dsjont. By Lemma 3.8.3a χ,j g χ,j g = 1 for j, j [1, l], and therefore E E = l 2. Assume E 1 E 2 E 2 l 2. Then, snce E 1 E 2 = l 2, E 1 E 2 E 3 = l 2. For a Z, let a = a + pz F p. Let u, v [0, p 1]. By Lemma 3.8.2, χψ u ϕ nv E 1 E 2 E 3 f and only f there are b {s,1,..., s,l }, for [1, 3], such that k 1 u + v = b 1 k 2 u + v = b 2 k 3 u + v = b 3. Snce k 1, k 2 and k 3 are parwse dstnct, (k 1, 1), (k 2, 1) and (k 3, 1) are parwse F p -lnearly ndependent. For [1, 3], we defne Φ : χ ψ, ϕ n F p by Φ (χψ u ϕ nv ) = k u + v. Then the lnear ndependence of (k 1, 1) and (k 2, 1) mples that Φ = (Φ 1, Φ 2 ) : χ ψ, ϕ n F 2 p s bjectve. We have Φ(E 1 E 2 E 3 ) {s 1,1,..., s 1,l } {s 2,1,..., s 2,l }, and due to l 2 = E 1 E 2 E 3 {s 1,1,..., s 1,l } {s 2,1,..., s 2,l } = l 2, equalty holds. In partcular, Φ 1 (E 1 E 2 E 3 ) = {s 1,1,..., s 1,l } and Φ 2 (E 1 E 2 E 3 ) = {s 2,1,..., s 2,l }. Because (k 1, 1), (k 2, 1) and (k 3, 1) are parwse F p -lnearly ndependent, there exst x, y F p such that (k 3, 1) = x(k 1, 1) + y(k 2, 1). Hence Φ 3 = xφ 1 + yφ 2. Now xφ 1 (E 1 E 2 E 3 ) = yφ 2 (E 1 E 2 E 3 ) = l. Also, snce x, y 0, we have (smlar to Φ) that (xφ 1, yφ 2 ) : χ ψ, ϕ n F 2 p s a bjectve map. Thus, n vew of xφ 1 (E 1 E 2 E 3 ) = yφ 2 (E 1 E 2 E 3 ) = l and E 1 E 2 E 3 = l 2, we see that (xφ 1, yφ 2 )(E 1 E 2 E 3 ) = xφ 1 (E 1 E 2 E 3 ) yφ 2 (E 1 E 2 E 3 ). Therefore Φ 3 (E 1 E 2 E 3 ) = xφ 1 (E 1 E 2 E 3 ) + yφ 2 (E 1 E 2 E 3 ), where the ncluson s obvous and follows snce for any α, β E 1 E 2 E 3 we can fnd θ E 1 E 2 E 3 such that (xφ 1 (α), yφ 2 (β)) = (xφ 1 (θ), yφ 2 (θ)), and hence n partcular xφ 1 (α) + yφ 2 (β) = xφ 1 (θ) + yφ 2 (θ) = Φ 3 (θ). From the Cauchy-Davenport Theorem (Lemma 2.1), t then follows that Φ 3 (E 1 E 2 E 3 ) mn {2l 1, p} > l, a contradcton, snce Φ 3 (E 1 E 2 E 3 ) {s 3,1,..., s 3,l }. Proof of Theorem Consder m = p P 5 and n 2. Let k 1,..., k 4 [0, p 1] be parwse dstnct and set g = k e 1 +e 2 G for [1, 4]. Furthermore, set m 1 = (n 2)p+(p 1), m 2 = m 3 = p 1 and m 4 = 2. We consder the sequence S = 4 g m F(G) and, for any choce of χ,j Ĝ for [1, 4] and j [1, m ], show that 4 m j=1 χ,j g Ĝ.

10 10 DANIEL SMERTNIG Then, by Lemma 3.4 and the defnton of d(g, K), d(g, K) S = p + pn 1 > p + pn 2 = d (G). Let χ,j Ĝ for [1, 4] and j [1, m ] be arbtrary. Assume, to the contrary, 4 m j=1 χ,j g = Ĝ. For [1, 4] and j, j [1, m ] dstnct, we can wthout restrcton assume χ,j g χ,j g. For any permutaton σ S n (whch wll be fxed later), n Ĝ = ϕ σ(ν) ψ, ϕ n. ν=1 For gven [1, 4] and j [1, m ], we have by Lemma 3.3 that χ,j g ϕ σ(ν) ψ, ϕ n for a unquely determned ν [1, n]. For [1, 4] and ν [1, n], we can therefore defne B (ν) = { χ,j j [1, m ] wth χ,j g ϕ σ(ν) ψ, ϕ n }. We also defne n (ν) = max { B (ν) [1, 4]} as well as l (ν) = 4 Let ν [1, n]. By assumpton, 4 ϕ σ(ν) ψ, ϕ n = χ g. χ B (ν) B(ν), for ν [1, n]. Thus, snce ψ, ϕ n = p 2 and g = p for all [1, 4], we have l (ν) p. On the other hand, n (ν) p because otherwse there would exst [1, 4] and j, j [1, m ] dstnct such that χ,j g χ,j g, but ths would already mply χ,j g = χ,j g, contrary to assumpton. Fx σ S n so that there s a k N 0 such that n (1),..., n (k) < p and n (k+1) =... = n (n) = p. Snce m < p for 2, we see (for ν [1, n]) that n (ν) = p s only possble f B (ν) 1 = p. Due to m 1 = (n 2)p + (p 1), ths s possble for at most n 2 dfferent ν [1, n]. Thus k 2. We can also estmate 4 χ B (ν) χ g n a dfferent way: Assume for the purpose of showng (1) (the other cases are argued dentcally) that n (ν) = B (ν) 1 B(ν) 2 B(ν) 3 B(ν) 4. Each of the characters χ B (ν) 1 contrbutes χ g 1, and therefore exactly p characters, to the unon. Each of the characters χ B (ν) 2 contrbutes at most p B (ν) 1 characters, snce χ 1 g 1 χ g 2 = 1 for all χ 1 B (ν) 1. Smlarly, each of the characters χ B (ν) 3 contrbutes at most p max{ B (ν) 1, B(ν) 2 } = p B(ν) 1 characters, snce χ 1 g 1 χ g 3 = 1 for all χ 1 B (ν) 1 and χ 2 g 2 χ g 3 = 1 for all χ 2 B (ν) 2. Contnung ths thought for B (ν) 4, we obtan Therefore p 2 = 4 χ B (ν) χ g 4 (ν) p B 1 + (p B(ν) 1 )( =2 B (ν) ) = pn (ν) + (p n (ν) )(l (ν) n (ν) ). (1) (n (ν) (l (ν) p))(n (ν) p) = pn (ν) + (p n (ν) )(l (ν) n (ν) ) p 2 0. Thus ether n (ν) p (and therefore already n (ν) = p) or n (ν) l (ν) p. For ν [1, k], we obtan n (ν) l (ν) p. Due to B (ν) 4 m 4 = 2, we also have l (ν) = 4 B(ν) 3n (ν) + 2. Then 3l (ν) 3n (ν) + 3p = 3n (ν) p 2 l (ν) + 3p 2, and hence l (ν) 3 2 p 1 for all ν [1, k]. Because of n l(ν) = S = pn + (p 1) and l (ν) n (ν) = p for all ν [k + 1, n], we have l (1) l (k) pk + (p 1). For the remander of the argument, we consder ν [1, k].

11 ON THE DAVENPORT CONSTANT AND GROUP ALGEBRAS 11 Then, by the above, k, ν l(ν) (k 1)( 3 2p 1), and hence ( ) 3 (2) (k 1) 2 p 1 + l (ν) pk + (p 1), whch mples ( ) 3 l (ν) pk + (p 1) (k 1) 2 p 1 = pk + p kp + k p 1 = 3 2 p + (p 2) + k 1 2 pk = 3 2 p + (p 2) k (p 2) 2 Hence, snce k 2, t follows that l (ν) 3 2 p. 1 Together wth l (ν) 3 2 p 1, ths mples l(ν) = 3 2 p 1 2. Snce B (1) B(k) 4 m 4 = 2 and k 2, there exsts a ν [1, k] wth B (ν) 1. Then B (ν) 4 1 and 4 B(ν) B (ν) 1,..., B(ν) 3 n(ν) l (ν) p = 1 (p 1), 2 = l (ν) = 3 2 (p 1) + 1. Therefore we must have B (ν) 1 = B(ν) 2 = B(ν) 3 = n(ν) = 1 (p 1) 2 and B (ν) 4 = 1. Wth the help of Lemma 3.9, we show that ths leads to a contradcton. Consder T = g 1 2 (p 1) 1 g 1 2 (p 1) 2 g 1 2 (p 1) 3 F(G). Then, by Lemma 3.9 (wth l = 1 2 (p 1) and χ = ϕσ(ν) ), Thus, wth B (ν) 4 = {τ}, ( p 2 3 = χ B (ν) 3 χ B (ν) χ g 1 < (p 1)(2p + 1). 2 χ g ) τ g 4 3 < 1 2 (p 1)(2p + 1) (p + 1) = p2, χ B (ν) χ g + (p n (ν) ) 4 a contradcton. 4. Acknowledgements I am ndebted to Alfred Geroldnger for hs constant feedback and help durng the creaton of ths paper. I would also lke to thank Davd Grynkewcz and Günter Lettl for ther comments on prelmnary versons of ths paper. In partcular, G. Lettl suggested the use of the Cauchy-Davenport Theorem n Lemma 3.9, whch sgnfcantly shortened the proof, compared to an earler verson. 1 Alternatvely (2), together wth l (ν) 3 2 p 1, p 5 and k 2, already mples k = 2, whch yelds the same estmate for l (ν).

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