K-Admissibility of S, S, S, S

JOURNAL OF ALGEBRA 185, 577 1996 ARTICLE NO. 00 K-Admissibility of S, S, S, S 1 1 14 15 Zara Girnius Department of Mathematics, Uniersity of California at Los Angeles, Los Angeles, California 9004-1555 and Steven Liedahl Department of Mathematics, California State Uniersity at Fresno, Fresno, California 9740-0108 Communicated by Walter Feit Received August 8, 1995 1. INTRODUCTION If k is a field and G is a finite group, then G is said to be k-admissible if there exists a Galois extension Lk such that GalŽ Lk. G and L is a maximal subfield of a finite dimensional k-central division ring. Equivalently, G is k-admissible if there is a k-division ring which is a crossed product for G. The notion of admissibility was introduced by Schacher in 1. A review of some basic results of that paper provides some context for the present work. It is familiar from Galois theory that for a given finite group G, there exists an algebraic number field k and a Galois extension Lk such that GalŽ Lk. G. This is seen by first embedding G as a subgroup of a symmetric group S n. By Hilbert s irreducibility theorem, there is a Galois extension LQ with GalŽ LQ. S n. Taking k to be the fixed field of G, one has GalŽ Lk. G. Current address: Department of Mathematics, University of Cincinnati, Cincinnati, OH 451. Current address: Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556. 57 001-86996 $18.00 Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.

58 GIRNIUS AND LIEDAHL By analogy, it is shown in 1 that a given finite group G is k-admissible for some number field k. Using a localglobal criterion for k-admissibility Ž Lemma 1 below., Schacher proved the existence of a number field K and an Sn-extension LK for which L is a maximal subfield of a K-division ring. Taking k to be the fixed field of G, one has GalŽ Lk. G and L is a maximal subfield of a k-division ring. If a number field k is specified in advance, then there are severe restrictions on which symmetric groups are k-admissible. Indeed, by 1, the only symmetric groups Sn which are Q-admissible are those for which n5. In fact, the only symmetric groups which are QŽ. i -admissible are S and S 8. More generally, Schacher s analysis of -subgroups is used in 1 to show that for a given k, there are only finitely many k-admissible symmetric groups. In that connection, the first author has answered a question raised in 1 by showing that the k-admissibility of SN need not imply the k-admissibility of S for n N Žsee 1 and Corollary. n. It is therefore desirable to determine, for given n, the number fields k such that S is k-admissible. By 1 n, the groups S and S are k-admissi- ble for all number fields k; the same is true for each group having only cyclic Sylow subgroups. In 9 it is shown that S 4, equivalently S 5, is k-admissible if and only if k contains two divisors of or ' 1 k. For 6 n 11, and n 16, 17, the number fields for which Sn is k-admissible are determined in. The main results of the present paper are the following. They characterize the number fields k for which the symmetric groups S 1, S 1, S 14, and S15 are k-admissible. Here denotes a prime of a number field k, and k denotes the -adic completion of k. We will use to denote any divisor of in a field extension of k. THEOREM. The symmetric group S, equialently S, is k-admissible if 1 1 and only if Ž. i the prime has at least two diisors in k, and Ž ii. the prime has at least two diisors n k such that k i Q Ž '., i1,. THEOREM. The symmetric group S is k-admissible if and only if 14 Ž. i the prime has at least two diisors in k, and Ž ii. the prime has at least two diisors n k such that k Q, i Q Ž '., i1,. The symmetric group S15 is k-admissible if and only if, in addition to conditions Ž. i and Ž ii., the prime 5 has at least two diisors in k.

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 59. PRELIMINARIES The following lemma, due to Schacher, gives an arithmetic criterion for admissiblity over global fields. It is the basic tool in the subject. LEMMA 1 1, Prop. 4.6. Let k be a global field and let G be a finite group. Then G is k-admissible if and only if there exists a G-Galois extension Lk such that for each prime p diiding the order of G, there are two primes, of k such that the decomposition groups GalŽ L k. 1 contain a Sylow p-subgroup of G, i 1,. The proof of Lemma 1 uses the description of central simple algebras over global fields by Hasse invariants, as found in 10. In verifying the hypotheses of Lemma 1 for symmetric groups, we rely on the following results of Saltman, which allow us to focus solely on local considerations. For background and terminology pertaining to generic Galois extensions we refer the reader to 11. LEMMA 11. Ž. i If k is an infinite field, then each symmetric group has a generic Galois extension oer k. Ž ii. Let k be an algebraic number field and let G be a finite group. For distinct primes 1,...,N of k, assume Lk are Galois extensions for which there is an embedding GalŽ L k. G. If the group G has a generic Galois extension oer k, then there exists a G-Galois extension Lk haing the L i as completions. Remark. The conclusion of Lemma Ž ii. is proved in 1 in the case GS and the groups GalŽ L k. n account for each Sylow subgroup of S n. This is shown by combining Krasner s lemma with a careful analysis of Sylow subgroups of S as permutation groups. n According to Lemmas 1 and, in showing k-admissibility of Sn our task is to realize each Sylow p-subgroup, or larger, of Sn as a Galois group over two completions k, where the i may depend on p. If p,, or 5, the i Sylow p-subgroups of S 1, S 1, etc., need not be metacyclic. According to ramification theory, the primes i of k must then be chosen to be divisors of p. In many instances the following structural results on profinite Galois groups can be used fairly directly. The most basic is due to Shafarevich: LEMMA 1. Let k be a finite extension of the p-adic field Q p which contains no primitie pth root of unity. If k : Q p d then the Galois group of the maximal p-extension of k is isomorphic to the free pro-p group on d 1 generators.

60 GIRNIUS AND LIEDAHL If k is an extension of Q p and k contains a nontrivial group of p-power roots of unity, we let q denote the order of this group. The following analogues of Lemma, due to Demushkin, Labute, and Serre, treat the 1 1 case q 1. In Lemma 4 the commutator a b ab is denoted a, b. LEMMA 4 7. Assume k : Q p d and q 1. Then the Galois group of the maximal p-extension of k is a pro-p group defined by d generators x, x,..., x and one of the following relations, 1 d Ž. i q: x q 1x 1, xx, x4 x d1, x d 1 Ž ii. q, d odd: x 4 1xx, xx 4, x5 x d1, x d 1 Ž iii. q, deen: ½ f x1 x 1, xx, x4 x d1, x d 1, or f x x, x x x, x x, x 1, where f. 1 1 4 d1 d Remark. The relation in Ž iii. and the value of f are chosen according to 7, p. 1. In our applications we will allow for either relation. We conclude this section with a special case of a theorem of Jannsen and Wingberg. LEMMA 5 6. The absolute Galois group G of Q is isomorphic to a profinite group generated by elements,, x, y, where x, y generate a normal pro- subgroup, and these elements satisfy the defining relations 1, 4 4 4 4 4 4 4 x x x y y, y y, y,,. ˆ ˆ Here is the element of Z such that Z Z,, and p p p. b 1 1 1 Also, a bab, a, b aba b, and a, b4 Ža 4 b a Žb. b. for a suitably defined function. Ž i. i Remark. The function : G Z satisfies and Ž. 1. Refer to 6, pp. 7476, noting typographical errors in the description of x in 6, p. 76, Ž. a. 0. SYLOW SUBGROUPS OF S 1 We begin with the problem of realizing the Sylow -subgroup, or larger, of S as a Galois group over each finite extension of Q. By, a Sylow 1

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 61 -subgroup of S is isomorphic to Ž C C. 1 C, where C C denotes the semidirect product of C acting on the product C by cyclic permutation. THEOREM 6. Let k be a finite extension of Q and let H be a Sylow -subgroup of S 1. Then H occurs as a Galois group oer k if and only if k :Q. In the case k Q, there is a subgroup of S of order H 1 which occurs as a Galois group oer k. Proof. Let d k : Q and assume H occurs as a Galois group over k. As H is three-generated, Lemma implies d. For the converse, assume d. If k contains no primitive third root of unity then H occurs as a Galois group over k by Lemma. Therefore we may assume q. In this case, the defining relation q 1x x, x x, x x, x 1 1 4 d1 d of Lemma 4Ž. s easily seen to be satisfied by taking x1 as a generator of the C direct factor of H, and by taking x and x to be generators of the C terms in C C. In addition we set xi 1 for i. For the second statement, let G be the subgroup of S1 generated by the permutations Ž 10, 11, 1. x Ž 1,,. y Ž 1,4,7.Ž,5,8.Ž,6,9. Ž 1,.Ž 4,5.Ž 7,8.. The subgroup generated by, x, and y is a Sylow subgroup of S index in G. In the notation of Lemma 5, we claim 1 and has 1, and 4 4 4 4 4 4 4 x x x y y, y y, y,,. The first relation is immediate. For the second, observe that x x. Then 4 Ž x x. y Ž x. x

6 GIRNIUS AND LIEDAHL as x has order. Next we note that and 1. Therefore 4 4 4 4 4 4 y, y y, y,, y, y y,14 y,1 4,4 y, yy 1, so that the second relation is satisfied. It follows from Lemma 5 that the group G occurs as a Galois group over Q. Next we consider the occurrence of Sylow -subgroups, or larger, of S 1 as Galois groups over -adic fields. A Sylow -subgroup of S1 is isomorphic to ŽŽ C C. C. D by 8. THEOREM 7. Let k be a finite extension of Q and let H be a Sylow -subgroup of S 1. Then H occurs as a Galois group oer k if and only if k :Q. In the case k : Q, there is a subgroup of S1 of order H which occurs as a Galois group oer k if and only if k Q Ž '.. Proof. We begin with the first statement. Let d k : Q and assume H occurs as a Galois group over k. As His a five-generator group, Lemma 4 implies d. Conversely, assume d. Let x, x4 be involutions which generate D 8, let x 1, x, x5 be generators of each C term in Ž C C. C, and let x 1 for i 5. Then each relation in Lemma 4 is i easily seen to be satisfied. Our proof of the second statement is involved, and requires a series of lemmas. We assume k is a finite extension of Q such that d and Ž '. 4 4 kq. Let f X X 4X and g X X X, and let K be the splitting field of the set fž X., gž X.4 over k. The desired extension of k will be obtained by embedding K in a field L such that Lk is Galois with GalŽ Lk. S and L : k H 1. LEMMA 8. The splitting fields of fž X. and gž X. oer Q are distinct, and hae Galois group S 4. Proof. As an Eisenstein polynomial over Q, fž X. is irreducible. The resolvent cubic r X X 8X16 5, p. 61 f has no root in Z,so GalŽfŽ X. Q. A or S. The discriminant d d equals 8 Ž 5. 4 4 f r, which is not a square in Q. Therefore GalŽfŽ X. Q. S 4. Similarly, the polynomial gž X. is Eisenstein over Q. It has resolvent cubic rg X X 8X4, which is irreducible. The discriminant dg d equals 4 Ž 101. r, which is not a square in Q. As before, we conclude that GalŽgŽ X. Q. S 4. Let and denote roots of fž X. and gž X., respectively. As fž X. and g X are Eisenstein, we have by 14, -- that the sets 1,,, 4 and 1,,, 4 form integral bases of the respective rings of integers of

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 6 Q Ž. and Q Ž.. The discriminants of these bases, given above, fail to differ by the square of a unit in Q ; therefore the fields Q Ž. and Q Ž. are not conjugate. It follows that the splitting fields of fž X. and gž X. are distinct. Note. We express our thanks to Walter Feit for suggesting the polynomials fž X. and gž X. and their properties. LEMMA 9. Assume k is a finite extension of Q. There exists an S-exten- sion of k if and only if k contains no primitie third root of unity. If an S-extension of k exists then it is unique. Proof. Assume k. If is a prime of k then the polynomial X has Galois group S over k. Conversely, assume Lk is an S -extension. Then L contains a field K such that K : k and Kk is totally and tamely ramified. By 14, -4- there is a prime of k such that K is obtained by adjoining a root of X to k. Then k according to the fact that Kk is not Galois. In addition, K is unique up to conjugacy as k k is cyclic of order with as a generator. Now we are prepared to describe the Galois group of KQ, where K is 4 the splitting field of f X, g X. We let K4 denote the direct product of two copies of the Klein 4-group. There is a natural action of S on K 4, and diagonally on K4. We use K4 S to denote the resulting semidirect product. LEMMA 10. The Galois group of KQ is isomorphic to K S. 4 Proof. Let FQ denote the S-extension of Lemma 9. From S4K4 S, the splitting fields of fž X. and gž X. each contain F. But their intersection cannot have degree 1 over Q as this would imply S4 contains a normal subgroup of order. By Lemma 8 we conclude that 5 K :Q. We now have Gal KF K, and S GalŽ FQ. 4 acts on K in the manner described above, and the group extension 4 1 K4 GalŽ KQ. S 1 is clearly a split extension. ŽBy restricting to Sylow subgroups, it is easy to Ž see that H S, K. is in fact trivial.. This proves the lemma. 4 For the next step we construct a wreath product of K4 by S as follows. If we compose permutations according to Ž 1,,.Ž 1,. Ž,., then there is a right action of S on K given by Ž x, y, z. Ž1,,. Ž z, x, y. 4 and Ž x, y, z. Ž1,. Ž y, x, z.. The resulting semidirect product is referred to as the wreath product, and is denoted K4 S. Each element of K4 S is uniquely expressible as a pair Ž,., where C 6 Ž K. 4 and S.In Ž K S we have the multiplication rule u,, u,.. 4

64 GIRNIUS AND LIEDAHL Ž. LEMMA 11. There is an isomorphism K S K S K. 4 4 4 Proof. Let K be the subgroup of K S generated by the elements 4 Ž Ž 1,0,1,0,1,0., e. Ž Ž 0,1,0,1,0,1., e., where e is the identity in S. Let H be the subgroup of K S generated 4 by the elements a b c Ž 0, Ž 1,,.. Ž 0, Ž 1,.. Ž Ž 0,0,1,0,1,0., e. Ž Ž 1,0,0,0,1,0., e. Ž Ž 0,0,0,1,0,1., e. d Ž Ž 0,1,0,0,0,1., e., where 0 is the identity of C 6. A routine verification shows that K K 4, HK S, and K S H K. 4 4 LEMMA 1. The commutator quotient of K S is cyclic of order. 4 Proof. In the group H K S of Lemma 11, we have K H 4 4 according to Ž. Ž. 1 1 0, 1,, b 0, 1,, b a Ž. Ž. 1 1 0, 1, a 0, 1, a b Ž. Ž. 1 1 0, 1,, d 0, 1,, d c Ž. Ž. 1 0, 1, c 0, 1, c1d. Also, Ž 1,.Ž 1,.Ž 1,.Ž 1,. Ž 1,,., which shows that H : H. But Ž0, Ž 1,.. H, as otherwise the multiplication rule given for K4 S would imply Ž 1,. S, which is false. Therefore H : H. We return to the extension KQ, where K is the splitting field of 4 f X, g X. LEMMA 1. There is an embedding of K in an extension LQ such that LQ is Galois and GalŽ LQ. K S. 4 Proof. It follows from Lemmas 10 and 1 that K contains a unique quadratic subfield. Denoting this field by K, the proofs of Lemmas 9 and

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 65 10 show that K is generated by a primitive third root of unity; therefore K Q Ž '.. Let L KŽ ' 1, '.. According to Q Ž ' 1, '. K Ž Q, we have Gal LQ K S. 4 K 4, which, by Lemma 11, completes the proof. The following lemma is well known, but we sketch a proof for the reader s convenience. LEMMA 14. Assume charž F., and assume KF is a Galois extension with GalŽ KF. K 4. Then K may be embedded in a D8-extension EF if ' ' and only if K may be expressed as FŽ a, b., where a, b F and ab is a norm from FŽ ' a. to F. Proof. Assume K FŽ ' a, ' b. with NŽ xy' a. ab, x, y F. Let ' ' ' ' xy a and x y a. Then F : F 4, and the conjugates of over F are,. Let E FŽ,.. Then E is the splitting 4 field over F of T xt ab, so GalŽ EF. S. But FŽ. 4 and FŽ.,soE:F8, which implies GalŽ EF. D 8. For the converse, we regard D8 as the group generated by elements, 4 1 1 subject to 1 and.if FŽ ' a. is the fixed field of ², : and FŽ b. is the fixed field of, then FŽ a, b.: F4 and ab is a norm from F a to F. We omit the details. ' ' ' Ž '. The field L is a composite of three K4-extensions of F, where F is the unique S-extension of Q. Denote these extensions by K 1, K, K. In addition F contains three subfields F, F, and F for which F : Q 1 i. In particular, let K1 be the fixed field of the subgroup of K4 S generated by the elements Ž Ž 0,0,1,0,0,0., e. Ž Ž 0,0,0,1,0,0., e. Ž Ž 0,0,0,0,1,0., e. Ž Ž 0,0,0,0,0,1., e.. Ž Let F be the fixed field of the subgroups of K S generated by K, e. 1 4 4 and Ž0, Ž,... For Ž0, Ž 1,.. and Ž0, Ž 1,.. we let K K, F F 1 1 K K, F F. 1 1 The next lemma shows that the extensions K if are definable over F i.

66 GIRNIUS AND LIEDAHL LEMMA 15. For i 1,,, we hae K FŽ a, b., where a, b F. ' ' Proof. By conjugating, it is enough to prove K FŽ a, b. 1, with a, b F 1. This is equivalent to the existence of a normal subgroup N of Gal LF such that Gal LF N K and N K GalŽ LK. 1 1 4 4 1. The subgroup N of GalŽ LF. generated by GalŽ LK. and Ž0, Ž,.. 1 1 has these properties. As with our definition of the group K S GalŽ LQ. 4, the right action of S on the Cartesian product D8 defines a wreath product D8S. We observe that D8 S contains a Sylow -subgroup H of S1 and D S H. 8 Proof of Theorem 7. It remains to prove the second statement in the theorem. First, let k Q and consider the extension LQ constructed ' ' above. We have K FŽ a, b. 1, with a, b F 1, by Lemma 15. We claim that ab is a norm from FŽ ' a. to F; equivalently, that the quaternion algebra Ž a, ab. is split by F. AsF:F F 1, this is a consequence of 1 local class field theory. We conclude from Lemma 14 that there is an embedding of K1F in a D8-extension L1F. Then L1 has two other conjugate fields over Q of the form L K and L K. Let L be the composite of the L i. We prove the lemma by showing that L : F 8. First, we observe that L1 L F. This is clear as L1 L is normal over F, which implies L1 L K1 K F. Therefore LL 1 :F 8. Similarly LLL 1 is normal over F, which implies LLLK 1 1K K F, as desired. Next we assume k : Q and k Q Ž '.. The field K splitting field of fž X., gž X.4 over Q has Q Ž '. as its unique quadratic subfield; therefore Kkk is Galois and Gal Kkk K4 S. The square class group k Žk. has order 16; therefore we may choose a K4-exten- sion Ek such that K E k. Then GalŽ EKk. K4 S by Lemma 11. Exactly as before, EK can be embedded in a D8 S-extension of k. Finally, we claim there is no subgroup of S1 of order H which occurs as a Galois group over k Q Ž '.. Let G be any subgroup of S! of order H. As the Sylow -subgroups of S1 are self-normalizing, the Sylow theorems imply G contains precisely three Sylow -subgroups. The transitive action of G on these subgroups by conjugation induces a surjective map from G to S. But there is no S-extension of k by Lemma 9. ' '

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 67 4. SYLOW SUBGROUPS OF S AND S 14 15 In this section we realize Sylow p-subgroups, or larger, of S14 and S15 as local Galois groups, where p, or 5. Let H denote a Sylow -subgroup of S 1. Therefore a Sylow -subgroup of S14 is isomorphic to H C. THEOREM 16. Let k be a finite extension of Q and let H C be a Sylow -subgroup of S 14. Then H C occurs as a Galois group oer k if and only if k : Q 4. In the case k : Q, there is a subgroup of S14 of order H C which occurs as a Galois group oer k if and only if kq,q Ž '.. Proof. If k is a finite extension of Q and H C occurs as a Galois group over k, then k : Q 4 by Lemma 4. Conversely, if k : Q 4 then H occurs as a Galois group over k by Theorem 7. The commutator 5 quotient of H is isomorphic to C. Therefore if GalŽ Lk. H then L contains 5 1 quadratic extensions of k. But k has at least 6 1 quadratic extensions. We may choose one which is independent from L; therefore H C occurs as a Galois group over k. Next assume H C G and G H C. Then G does not occur as a Galois group over Q as otherwise the fixed field of H C has degree over Q, which contradicts Lemma 4. Suppose k : Q and k Q Ž '.. According to the proof of Theorem 7, let Lk be a Galois extension with Galois group D8 S. Then Lk contains 7 quadratic subextensions. There are 15 extensions of k. Therefore we may choose one which is disjoint quadratic from L over k, so that Ž D S. 8 C occurs as a Galois group over k. On the other hand, if G satisfies H C G S and G H C 14, then G cannot occur as a Galois group over Q Ž '.. Any such group G has S as a quotient, and there is no S -extension of Q Ž '. by Lemma 9. Assume k : Q and kq is unramified. From Theorem 7 we have GalŽ LQ. D S and Lˆ k Q. Therefore GalŽ Lkk ˆ. 8 D8 S. As in the quadratic case, Ž D S. 8 C occurs as a Galois group over k. For the case k : Q with kq ramified, we modify the method used to realize D8 S as a Galois group over Q. We have seen kq Ž., where is a prime of k which satisfies the polynomial 4 4 X. Let fž X. X X and gž X. X X. LEMMA 17. The splitting fields of f X and g X oer k are distinct, and hae Galois group S. 4

68 GIRNIUS AND LIEDAHL Proof. The polynomial fž X. is Eisenstein over k, hence irreducible. Its 7 4 resolvent cubic rf X equals X X, which has no root in k. Therefore GalŽfŽ X. k. A4 or S 4. The discriminants d f dr equal 8 Ž 19. 19 7. Let 7, which is a unit in k. Every unit in k is uniquely expressible in the form 1a1a a, ai 0,1 4. 6 Accordingly we have 1 Ž higher terms.. Setting 1a1a a and comparing coefficients, we see that and, therefore, d f are non- squares in k. This implies GalŽfŽ X. k. S 4. The polynomial gž X. is handled similarly. It has resolvent cubic r Ž X. g 7 4 Ž. X X and discriminant dg 7. Again 7 6 1 Ž higher terms., which is not a square in k. We conclude that GalŽgŽ X. k. S 4. Finally, comparing d with d shows that the splitting fields of fž X. f g and gž X. over k are distinct. Proof of Theorem 16. Assume kq is ramified of degree. Let K be the splitting field of fž X., gž X.4 over k. Then K contains the unique S-extension of k by Lemma 9. Now Gal Kk K4 S and K embeds in a Galois extension Lk with GalŽ Lk. D8 S. We choose a quadratic extension of k which is disjoint from L and the theorem is proved. Let H be a Sylow -subgroup of S. Then H C C C C. 15 THEOREM 18. Let k be a finite extension of Q and let H be a Sylow -subgroup of S 15. Then H occurs as a Galois group oer k if and only if k :Q or k Q Ž '.. The group S15 contains subgroups of orders 4H and H which occur as Galois groups oer Q and oer the quadratic extensions of Q, respectiely. Proof. If k : Q and k, then H occurs as a Galois group over k by Lemma. If kq is any finite extension such that k then, if one lets x, x 4, x 1, x denote generators of the C terms above, the relation in Lemma 4Ž. s satisfied, so H occurs as a Galois group over k. Assume H occurs as a Galois group over k. If kthen k : Q by Lemma. If k then Q Ž '. k.

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 69 For the second assertion, let k Q. Let G generated by the permutations Ž 1,4.Ž,5.Ž,6. Ž 1,.Ž 4,5. y Ž 1,,., be the subgroup of S 1 15 1 and note that y 4, 5, 6 and y y. Let G be the subgroup of S generated by Ž 7, 8, 9. and the product Ž 7, 10, 1.Ž 8, 11, 14.Ž 9, 1, 15.. 15 Then G C C and G G 4H 1. In the notation of Lemma 5 1 Ž 4. we have. Then, taking x 1, we have x x y 1 and 4 4 4 4 4 4 4 x x x y y, y y, y,, y, y y,14 y,1 4, 4 4 1 y, yy y, 1 4 Ž. Ž 1.Ž. y, yy y y. ² : The second term in this commutator lies in the abelian group y, y ; hence the commutator equals 1. By Lemma 5 there is a Galois extension K Q with Galois group G. Let K Q be an extension with Galois 1 1 group G, according to Lemma. Then K1 K Q. In fact, G1 has no quotients of order or 9, as a Sylow -subgroup Ž e.g., ², :. is not normal, and the commutator group G equals ² y, y : 1. This proves GalŽ K K Q. 1 G1G, and the isomorphism G1G1 K4 proves the assertion for the three quadratic extensions of Q. Now let H denote a Sylow 5-subgroup of S 15. Thus H C5. THEOREM 19. Let k be a finite extension of Q5 and let H be a Sylow 5-subgroup of S 15. Then H occurs as a Galois group oer k if and only if k :Q. There is a subgroup of S of order H 5 15 which occurs as a Galois group oer Q 5. Proof. If H occurs as a Galois group over k then k : Q 5 by Lemma. Assume k : Q 5. If k contains no primitive fifth root of unity then H occurs as a Galois group over k by Lemma. If k contains

70 GIRNIUS AND LIEDAHL primitive fifth roots of unity then k : Q 5 4. Let x 1, x, x be generators of each C5 term in C5 and all other xi 1. Then the relation of Lemma 4Ž. s clearly satisfied. This proves the first statement. For the second statement, we claim that the subgroup of S15 generated by a Ž 1,,,4,5. b Ž 6,7,8,9,10. c Ž 11, 1, 1, 14, 15. d Ž 6, 11.Ž 7, 1.Ž 8, 1.Ž 9, 14.Ž 10, 15. occurs as a Galois group over Q. We note that H ² a, b, c: 5 has index and dbd 1 c. Let TQ be the unramified quadratic extension, with GalŽ TQ. 5 5 ² :. Then T Q Ž. 5, where is a root of unity of order. Let L1T be unramified of degree 5. Our claim is proved if we can exhibit a cyclic 5-extension L of T which is not Galois over Q 5. For then L has only one other conjugate field L over Q5 and the composite of the Li has the desired Galois group over Q 5. By local class field theory, the existence of L is equivalent to the existence of a subgroup of the multiplicative group T such that T C5 and is not invariant under the action of Ž1. Gal TQ 5. Let UT denote the group of units of T which are congruent to 1 modulo 5. Reducing modulo 5, the classes of and form a basis of Ž1. the residue extension TF 5. It follows from 4, p. 8 that UT is a free multiplicative Z5-module with basis u1 1 5, u 1 5. Therefore we have a direct decomposition ²: ² Z 5 : ² Z 5: T 5 u u. 4 1 For we take the subgroup of T generated by 5,, u Z 5, and u 5Z 4 1 5. Then T C and u u, which proves the theorem. 5 1 5. MAIN THEOREMS We now let k denote an algebraic number field. For a given symmetric group we will be especially interested in Sylow subgroups which are not metacyclic. A group G is said to be metacyclic if it has a cyclic normal subgroup N such that GN is cyclic.

K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 71 THEOREM 0. The symmetric group S, equialently S, is k-admissible 1 1 if and only if Ž. i the prime has at least two diisors in k, and Ž ii. the prime has at least two diisors n k such that k i Q Ž '., i1,. Proof. Assume S1 is k-admissible. By Lemma 1 there is an S1-exten- sion Lk such that for two primes i of k the decomposition groups GalŽ L k. contain a Sylow -subgroup of S. We have Syl Ž S. Ž 1 1 C C. C, which is not metacyclic. By 14, -5-4, tamely ramified Galois extensions of local fields have metacyclic Galois groups. Therefore the primes are divisors of, which proves Ž. i i. Similarly, there are two primes of k such that the decomposition groups GalŽ L k. i contain a Sylow -subgroup of S. We have H Syl Ž S. ŽŽ C C. C. 1 1 D 8, which is not metacyclic. Therefore the primes i are divisors of. For these primes we have L : k H or H by a theorem of Feit i 1, Prop. 10.5. According to Theorem 7 we have k Q Ž '., i1,. i Conversely, assume k is a number field satisfying conditions Ž. i and Ž ii.. We claim that for each prime p dividing S 1 there are two primes i of k and Galois extensions L k such that Syl Ž S. GalŽ L k. S. p 1 1 For p, this follows from Theorems 6 and 7. A Sylow 5-subgroup of S1 is isomorphic to C5 C 5. Let p 1, p be distinct rational primes which split completely in the field kž. of fifth roots of unity over k, and let 5 i be divisors of pi in k. Then for i 1,, the composite of the unramified extension of degree 5 over k with the splitting field of x 5 p has Galois i group C5. Next let i be any two primes of k other than the above. If we choose the unramified extension L k of degree 7 11, then the claim i is proved. By Lemma there is an S1-extension Lk having the L i as completions. The k-admissibility of S1 follows from Lemma 1. If S is k-admissible, then the argument above shows that Ž. i and Ž ii. 1 must be satisfied. For the converse, it is enough to replace the two unramified extensions above with those of degree 7 11 1. Of course S1 is not Q-admissible. However, using basic results on the splitting behavior of primes in quadratic extensions, Theorem 0 gives the following: ' COROLLARY 1. Let k QŽ d., where d is a square-free integer. Then S, equialently S, is k-admissible if and only if d 1 Ž mod 4.. 1 1 In our final results we consider the groups S, S, and S. 14 15 16 i i

7 GIRNIUS AND LIEDAHL THEOREM. The symmetric group S is k-admissible if and only if 14 Ž. i the prime has at least two diisors in k, and Ž ii. the prime has at least two diisors n k such that k Q, i Q Ž '., i1,. The symmetric group S15 is k-admissible if and only if, in addition to conditions Ž. i and Ž ii., the prime 5 has at least two diisors in k. Proof. Assume S14 is k-admissible. A Sylow -subgroup of S14 is not metacyclic, so Lemma 1 and 14, -5-4 imply that there are at least two divisors of in k. In addition, there are two primes i of k such that Syl Ž S. GalŽ L k. S. Again 1, Prop. 10.6 implies L : k 14 14 HC or H C, where H Syl Ž S. 1. Theorem 16 gives k i Q,Q Ž '.. Conversely, assume k is a number field satisfying conditions Ž. i and Ž ii.. We claim that for each prime p dividing S 14 there are two primes i of k and Galois extensions L k such that Syl Ž S. GalŽ L k. S. p 14 14 For p,, this follows from Theorems 6 and 16. The primes 5, 7, 11, and 1 may be treated as for S1 above. The k-admissibility of S14 follows from Lemmas 1 and. If S is k-admissible then clearly conditions Ž. i and Ž ii. 15 hold. A Sylow 5-subgroup of S15 is isomorphic to C5, which is not metacyclic. Therefore the prime 5 has at least two divisors in k. Conversely, assume k satisfies Ž. i and Ž ii., and k contains at least two divisors of 5. We claim that for each prime p dividing S 15 there are two primes of k and Galois extensions L k such that Syl Ž S. i p 15 GalŽ L k. S 15. This is proved for the primes p, 7, 11, and 1 just as for S 14, while for the primes p and p 5 this follows from Theorems 18 and 19. Lemmas 1 and imply that S is k-admissible. Remark. If S is k-admissible then k : Q 4. If k : Q 14 4, and R denotes the ring of integers of k, then condition Ž ii. of Theorem is equivalent to the existence of two prime divisors P 1, P of in R such that RP1 P. The same holds for S 15. Let us consider the following question raised in 1. If k is a number field, then does the k-admissibility of SN imply that of Sn for all n N? This was first answered negatively in 1, where it is shown that for each integer m 1 there is a number field k such that S m is k-admissible 15

and S m 1 K-ADMISSIBILITY OF S, S, S, S 1 1 14 15 7 is not. Here we present the minimal such example: COROLLARY. Ž. i For n N 15, if SN is k-admissible then Sn is k-admissible. Ž ii. Let k QŽ ' d., where d is a square-free integer satisfying d 1 or 49 Ž mod 10.. Then S is k-admissible, but S and S are not. 16 15 14 Proof. Statement Ž. i follows by combining the results of, 9, 1 with Theorems 0 and. For Ž ii., it is shown in that S16 is k-admissible if and only if the primes,, 5 each have at least two divisors in k. But S 15 and S are not k-admissible by the preceding remark. 14 ACKNOWLEDGMENTS We thank Murray Schacher and Walter Feit for their helpful suggestions. The second author s research was supported in part by a grant from the School of Natural Sciences, California State University, Fresno. REFERENCES 1. Z. Girnius, K-Admissibility of Finite Groups over Quadratic and Cyclotomic Fields, Doctoral dissertation, UCLA, 1995.. Z. Girnius, k-admissibility of certain symmetric groups over algebraic number fields, J. Algebra 177 Ž 1995., 7796.. M. Hall, Theory of Groups, Macmillan Co., New York, 1959. 4. H. Hasse, Number Theory, Springer-Verlag, New York, 1980. 5. N. Jacobson, Basic Algebra I, nd ed., Freeman, New York, 1985. 6. U. Jannsen and K. Wingberg, Die Struktur der absoluten Galoisgruppe -adischer Zahlkorper, Inent. Math. 70 Ž 198., 7198. 7. J. Labute, Classification of Demushkin groups, Canad. J. Math. 19 Ž 1967., 1061. 8. S. Liedahl, QŽ. i -Division Rings and Admissibility, Doctoral dissertation, UCLA, 199. 9. S. Liedahl, Presentations of metacyclic p-groups with applications to K-admissibility questions, J. Algebra 169 Ž 1994., 96598. 10. I. Reiner, Maximal Orders, Academic Press, London, 1975. 11. D. Saltman, Generic Galois extensions and problems in field theory, Ad. in Math. 4 Ž 198., 508. 1. M. Schacher, Subfields of division rings, I, J. Algebra 9 Ž 1968., 451477. 1. I. Schafarevich, On p-extensions, Mat. Sb. 0 Ž 1947., 516; Amer. Math. Soc. Transl. Ž. 4Ž 1956., 597. 14. E. Weiss, Algebraic Number Theory, McGrawHill, New York, 196.