A talk given at Suzhou Univ. (June 19) and Nankai Univ. (June 25, 2008) Zhi-Wei Sun

A talk given at Suzhou Univ. (June 19) and Nankai Univ. (June 25, 2008) AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS Zhi-Wei Sun Department of Mathematics Nanjing University Nanjing 210093, P. R. China zwsun@nju.edu.cn http://math.nju.edu.cn/ zwsun Abstract. An interesting extremal problem asks for the smallest positive integer k such that an abelian group G can be irredundantly covered by k cosets of subgroups one of which has index n. We will talk about the solution to this problem provided by G. Lettl and the speaker via characters of abelian groups. Some related results and conjectures will be also mentioned in the talk. 1. Basic results on covers of groups by left cosets Let H be a subgroup of a group G with [G : H] = k <. Then we can partition G into k left cosets g 1 H,..., g k H, and {g i H} k forms a disjoint cover of G by left cosets. Let {Ha i } k be a right coset decomposition of G. Then {a i G i } k is a disjoint cover of G where G i = a 1 i Ha i. Observe that k G i = k h H a 1 i h 1 Hha i = g 1 Hg g G is the normal core H G of H in G (H G denotes the largest normal subgroup of G contained in H). In group theory, it is known that G/H G can be 1

2 ZHI-WEI SUN embedded into the symmetric group S [G:H] = S k and thus [ G : k ] G i = G/H G k!. An Example of M. J. Tomkinson. Let k > 1 be a positive integer, and let G be the symmetric group S k and H be the stabilizer of 1. Then G i = (1i) 1 H(1i) is the stabilizer of i for each i = 1,..., k. Clearly, {G 1, (12)G 2,..., (1k)G k } = {H, H(12),..., H(1k)} forms a disjoint cover of G with k G i = H G = {e}. Note that [G : k G i] = G = k!. A Basic Theorem on Covers of Groups. Let A = {a i G i } k be a finite system of left cosets in a group G where G 1,..., G k are subgroups of G. Suppose that A forms a minimal cover of G (i.e. A covers all the elements of G but none of its proper systems does). (i) (B. H. Neumann, 1954) There is a constant c k depending only on k such that [G : G i ] c k for all i = 1,..., k. (ii) (M. J. Tomkinson, 1987) We have [G : k G i] k!, where the upper bound k! is best possible. Proof (Tomkinson). We prove the inequality in (ii) by induction. We want to show that [ G i : i I k ] G i (k I )! ( I ) for all I {1,..., k}, where i G i is regarded as G.

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 3 Clearly ( I ) holds for I = {1,..., k}. Now let I {1,..., k} and assume ( J ) for all J {1,..., k} with J > I. Since {a i G i } i I is not a cover of G, there is an a G not covered by {a i G i } i I. Clearly a( i I G i) is disjoint from the union i I a ig i and hence contained in j I a jg j. Thus ( ) a G i = i I and hence [ ] G i : H i I j I j I a j G j a( i I G i) ( ( a j G j a i I G i )) [ G j ] G i : H ( I + 1))! = (k I )! i I j I(k where H = k G i. This concludes the induction proof. Definition of m-covers. Let m be a positive integer, and let A = {a i G i } k be a finite system of left cosets in a group G. If each element of G is covered by A at least (resp., exactly) m times, then we call A an m-cover (resp., exact m-cover) of G. If A is an m-cover of G but none of its proper subsystems does, then A is said to be a minimal m-cover of G. The Neumann-Tomkinson theorem can be extended to minimal m- covers of groups (cf. Corollary 1 of Z. W. Sun [Fund. Math. 134(1990)]); it also has applications in Galois theory, groups rings, Banach spaces, projective geometry and Riemann surfaces as pointed out by T. Soundararajan and K. Venkatachaliengar [Acta Math. Vietnam 19(1994)]. 2. Extremal problems for exact m-covers Let A = {a i G i } k be an exact m-cover of a group G with k G i = H.

4 ZHI-WEI SUN By the Neumann-Tomkinson theorem, [G : H] k!. How to provide a sharp lower bound of k in terms of G and H? An Example of Š. Znám. Let n > 1 be an integer with the factorization r t=1 pα t t, where p 1,..., p r are distinct primes and α 1,..., α r Z + = {1, 2, 3,... }. Then 0(mod n) and the following f(n) = r s=1 α s(p s 1) residue classes jp α 1 1 pα s 1 s 1 pα 1 s (mod p α 1 1 pα s 1 s 1 pα s ) (α = 1,..., α s ; j = 1,..., p s 1; s = 1,..., r) form a disjoint cover of Z whose moduli have the least common multiple n. As a convention we define f(1) = 0. The function f is called the Mycielski function. An Example of Z. W. Sun. Let H be a subnormal subgroup of a group G with finite index. Let H 0 = H H 1 H n = G be a composition series from H to G. For each i = 0,..., n 1, write H i+1 \ H i = [H i+1 :H i ] 1 j=1 b (i) j H i. Then the following d(g, H) = n 1 i=0 ([H i+1 : H i ] 1) left cosets b (i) j H i (0 i < n; 1 j < [H i+1 : H i ]), together with H and m 1 copies of G, form an exact m-cover of G by m + d(g, H) left cosets of subgroups whose intersection is H. (In the case H = G we define d(g, H) = 0.)

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 5 Relation between the Mycielski Function f and d(g, H) (Z. W. Sun, Fund. Math. 1990; European J. Combin. 2001). Let H be any subnormal subgroup of G with finite index. Then d(g, H) f([g : H]) log 2 [G : H]. Also, d(g, H) = f([g : H]) if and only if G/H G is solvable. Mycielski s Conjecture. (J. Mycielski, 1966) If {a i G i } k is a disjoint cover of an abelian group G, then k 1 + f([g : G i ]) for all i = 1,..., k. Related Results on Exact m-covers. Let A = {a i G i } k m-cover of a group G with k G i = H. be an exact (i) (I. Korec [Fund. Math., 1974]) If m = 1 and G 1,..., G k are normal in G, then k 1 + f([g : H]). (ii) (Z. W. Sun [European J. Combin., 2001]) If G 1,..., G k are subnormal in G, then k m + d(g, H), with the lower bound best possible. Note that Korec s result is stronger than Mycielski s conjecture, and also Sun s result has the following consequence. A Corollary (Sun [Fund. Math., 1990]). Let H be a subnormal subgroup of a group G with [G : H] <. Then [G : H] 1 + d(g, H G ) 1 + f([g : H G ]) and hence G/H G 2 [G:H] 1.

6 ZHI-WEI SUN Proof. Let {Ha i } k be a right coset decomposition of G where k = [G : H]. Then {a i G i } k is a disjoint cover of G where all the G i = a 1 i Ha i are subnormal in G and k G i = H G. So the desired result follows. 3. An extremal problem for minimal m-covers of abelian groups Korec s and Sun s results on exact m-covers can be extended to minimal m-covers of Z, see R. J. Simpson [Acta Arith., 1985] for the case m = 1 and Z. W. Sun [Internat. J. Math. 17(2006)] for general m 1. However, they cannot be extended to minimal m-covers of abelian groups as illustrated by the following example. An Example of G. Lettl and Z. W. Sun. Let G be the abelian group C p C p where p is a prime and C p is the cyclic group of order p. Then any element a e of G has order p. Let G 1,..., G k be all the distinct subgroups of G with order p. If 1 i < j k, then G i G j = {e}. Thus {G s } k s=1 forms a minimal cover of G with k s=1 G s = {e}. Since 1 + k(p 1) = k s=1 G s = G = p 2, we have k = p + 1 1 + f([g : G s ]) = 1 + f(p) = p. However, ( k = p + 1 2p 1 = 1 + f([g : {e}]) = 1 + d G, and the last inequality becomes strict when p > 2. k G s ), s=1

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 7 An Extremal Problem on m-covers of Abelian Groups. Let m and n be positive integers. Is m + f(n) the smallest positive integer k such that for any abelian group having a subgroup of index n there is a minimal m-cover of G by k cosets of subgroups one of which has index n? G. Lettl and Z. W. Sun has provided an affirmative answer to this problem. A Theorem of Lettl and Sun [Acta Arith. 131(2008)]. Let A = {a s G s } k s=1 be an m-cover of an abelian group G by left cosets. Assume that a G is covered by A exactly m times. Then k m + f(n a ), where N a = [ G : s I a G s ] & I a = {1 s k : a a s G s }. In particular, if {a s G s } s t fails to be an m-cover of G, then k m + f([g : G t ]). This theorem implies the following conjecture of W. D. Gao and A. Geroldinger [European J. Combin. 2003] who proved it for elementary abelian p-groups. Gao-Geroldinger Conjecture (W. D. Gao and A. Geroldinger). Let G be a finite abelian group with identity e. If G \ {e} is a union of k cosets a 1 G 1,..., a k G k, then we have k f( G ). In fact, if we set a 0 = e and G 0 = {e} then {a s G s } k s=0 forms a cover of G with a 0 G 0 irredundant and hence k + 1 1 + f([g : G 0 ]) = 1 + f( G ).

8 ZHI-WEI SUN The proof of the Lettl-Sun result was obtained via characters of abelian groups and algebraic number theory; below is a key lemma used for the proof. A Lemma of Lettl and Sun ([Acta Arith. 131(2008)]). Let n > 1 be an integer. Then f(n) is the smallest positive integer k such that there are roots of unity ζ 1,..., ζ k different from 1 for which k s=1 (1 ζ s) 0 (mod n) in the ring of algebraic integers. For a finite abelian group G, let Ĝ denote the group of all complexvalued characters of G. One has Ĝ = G. For any subgroup H of G let H denote the group of those characters χ Ĝ with ker(χ) = {x G : χ(x) = 1} containing H. Then there is a canonical isomorphism H = Ĝ/H by putting χ(ah) = χ(a) for any a G and any χ H. Furthermore, for each a G \ H there exists some χ H with χ(a) 1. Proof of the Lettl-Sun Result. Choose a minimal I {1,..., k} such that the system {a s G s } s I forms an m-cover of G. As I a = {1 s k : a a s G s } has cardinality m, I a is contained in I. So we can simply assume that A is a minimal m-cover of G (i.e., I = {1,..., k}) and thus H = k s=1 G s is of finite index in G. Instead of the minimal m-cover A = {a s G s } k s=1 of G, we may consider the minimal m-cover Ā = {ā sḡs} k s=1 of the finite abelian group Ḡ = G/H, where ā s = a s H and Ḡs = G s /H (hence [Ḡ : Ḡs] = [G : G s ]). Therefore, without any loss of generality, we can assume that G is finite. Put H a = s I a G s ; then H a = [G : H a ] = N a.

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 9 Note that J = {1 j k : a a j G j } has cardinality k m. For each j J we may choose a χ j G j with ζ j := χ j (a 1 a j ) 1. For any x G \ H a we have ax s I a ag s = s I a a s G s. Since A is an m-cover of G, there exists some j J with ax a j G j, and therefore χ j (x) = ζ j by the choice of χ j and the definition of ζ j. For x G we define Ψ(x) = j J (χ j (x) ζ j ). If χ H a and χ(x) 1, then x H a and hence Ψ(x) = 0 by the above. Thus Ψχ = Ψ for all χ H a. Observe that Ψ(x) = I J ( j I ) χ j (x) j J\I ( ζ j ) = ψ Ĝ c(ψ)ψ(x), where c(ψ) = ( ζ j ) Z. I J j J\I j I χ j=ψ Let C be the complex field. As the set Ĝ is a basis of the C-vector space C G = {g : g is a function from G to C}, for any χ Ha we have c(ψχ) = c(ψ) for all ψ Ĝ because Ψχ 1 = Ψ. Clearly (1 ζ j ) = Ψ(e) = c(ψ)ψ(e) = c(ψ). j J ψ Ĝ ψ Ĝ

10 ZHI-WEI SUN Let ψ 1 Ha ψ l Ha be a coset decomposition of Ĝ where l = [Ĝ : H a ]. Then l l l c(ψ) = c(ψ r χ) = Ha c(ψ r ) = N a c(ψ r ). ψ Ĝ r=1 χ Ha r=1 r=1 Therefore N a divides j J (1 ζ j) in the ring Z of all algebraic integers, and the lemma of Lettl and Sun gives k m = J f(n a ). If {a s G s } s t is not an m-cover of G, then for some a a t G t we have I a = m, hence k m f(n a ) f([g : G t ]). A Conjecture of Z. W. Sun (2004). If A = {a i G i } k forms a minimal m-cover of an abelian group G by left cosets or an exact m-cover of a solvable group G by left cosets, then we have k m + f(n), where N is the least common multiple of the indices [G : G 1 ],..., [G : G k ]. When {a i G i } k forms an exact m-cover of a solvable group G, the inequality k m + f([g : G t ]) was shown by Berger, Felzenbaum and Fraenkel [Colloq. Math. 1988] in the case m = 1 and proved by the speaker [European J. Combin. 2003] for general m. Concerning covers of abelian groups by subgroups, Song Guo and the speaker have made the following conjecture. A Conjecture of S. Guo and Z. W. Sun (2004). If {G i } k forms a minimal m-cover of an abelian group G with [G : k G i] = r t=1 pα t t, where p 1,..., p r are distinct primes and α 1,..., α r are positive integers. Then we have r k > m + (α t 1)(p t 1). t=1

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 11 4. Two Conjectures on Disjoint Cosets First we mention a challenging conjecture arising from the speaker s study of Huhn-Megyesi problems and covers of groups. A Conjecture on Disjoint Cosets (Z. W. Sun, [Internat. J. Math., 2006]). Let G be a group, and a 1 G 1,..., a k G k (k > 1) be pairwise disjoint left cosets of G with all the indices [G : G i ] finite. Then, for some 1 i < j k we have gcd([g : G i ], [G : G j ]) k. Z. W. Sun [Internat. J. Math. 2006] noted that this conjecture holds for p-groups as well as the special case k = 2. Recently, W.-J. Zhu [Int. J. Mod. Math. 3(2008), no. 2] proved the conjecture for k = 3, 4 via several sophisticated lemmas. K. O Bryant [Integers 2007] confirmed the conjecture for G = Z in the case k 20. Soon after his invention of covers of Z, Erdős made the following conjecture: If A = {a s (mod n s )} k s=1 (k > 1) is a system of residue classes with the moduli n 1,..., n k distinct, then it cannot be a disjoint cover of Z. A Result of H. Davenport, L. Mirsky, D. Newman and R. Rado. If A = {a s (mod n s )} k s=1 is a disjoint cover of Z with 1 < n 1 n 2 n k 1 n k, then we must have n k 1 = n k. Proof. Without loss of generality we assume 0 a s < n s (1 s k). For z < 1 we have k s=1 z a s k 1 z n = s s=1 q=0 z a s+qn s = n=0 z n = 1 1 z.

12 ZHI-WEI SUN If n k 1 < n k, then = lim z e 2πi/n k z <1 z a k 1 z n k = lim z e 2πi/n k z <1 ( ) k 1 1 1 z z a s 1 z n <, s s=1 which leads a contradiction! The following conjecture extends the above conjecture of P. Erdős on covers of Z. Herzog-Schönheim Conjecture (1974). Let {a i G i } k (k > 1) be a partition of a group G into left cosets of subgroups G 1,..., G k. Then the indices n 1 = [G : G 1 ],..., n k = [G : G k ] cannot be distinct. It is known that any finite nilpotent group is the direct product of its Sylow subgroups. Using this fact and lattice parallelotopes, Berger, Felzenbaum and Fraenkel [Canad. Bull. Math. 1986] confirmed the above conjecture for finite nilpotent groups. A Result of Z. W. Sun [J. Algebra 273(2004)]. Let G be a group, and A = {a i G i } k (k > 1) be a system of left cosets of subnormal subgroups. Suppose that A covers each x G the same number of times, and n 1 = [G : G 1 ] n k = [G : G k ]. Then the indices n 1,..., n k cannot be distinct. Moreover, if each index occurs in n 1,..., n k at most M times, then log n 1 eγ log 2 M log2 M + O(M log M log log M) where γ = 0.577... is the Euler constant and the O-constant is absolute.

AN EXTREMAL PROBLEM ON COVERS OF ABELIAN GROUPS 13 The above theorem was established by a combined use of tools from combinatorics, group theory and number theory. One of the key lemmas is the following one which is the main reason why covers involving subnormal subgroups are better behaved than general covers. A Lemma on Indices of Subnormal Subgroups (Z. W. Sun). Let G be a group, and let P (n) denote the set of prime divisors of a positive integer n. (i) [European J. Combin. 2001] If G 1,..., G k are subnormal subgroups of G with finite index, then [ G : k ] G i k [G : G i ] and hence P ([ G : k ]) G i = k P ([G : G i ]). (ii) [J. Algebra, 2004] Let H be a subnormal subgroup of G with finite index. Then P ( G/H G ) = P ([G : H]). Here is another useful lemma of combinatorial nature. A Lemma on Unions of Cosets (Z. W. Sun [J. Algebra, 2004]). Let G be a group and H its subgroup with finite index N. Let a 1,..., a k G, and let G 1,..., G k be subnormal subgroups of G containing H. Then k a ig i contains at least 0(mod n i) {0, 1,..., N 1} left cosets of H, where n i = [G : G i ]. This lemma implies the following result of Z. W. Sun [Internat. J. Math.

14 ZHI-WEI SUN 2006]: If G 1,..., G k are normal Hall subgroups of a finite group G, then k a i G i k G i. (A subgroup H of a finite group G is called a Hall subgroup of G if H is relatively prime to [G : H].) We also need the following deep theorems in analytic number theory. The Prime Number Theorem with Error Terms. For x 2 we have π(x) = x ( ) x log x + O log 2, x where π(x) = p x 1 is the number of primes not exceeding x. Mertens Theorem. For x 2 we have p x ( 1 1 ) ( ) = e γ 1 p log x + O log 2. x