A NOTE ON SPLITTING FIELDS OF REPRESENTATIONS OF FINITE

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1 A NOTE ON SPLITTING FIELDS OF REPRESENTATIONS OF FINITE GROUPS BY be a finite group, and let x be the character of an absolutely irreducible representation An algebraic number field K is defined to be a splitring field of x if can be written in the field K(x), where K(x) is the field generated by K and the values of x The existence of splitting fields with suitable properties has been of fundamental importance in the study of group characters In [1, Lemma 1], Brauer proved the following result (A) be a finite group of order g pare, where p is a rational prime and (p, m) 1 Then there exists an algebraic number field K with the following properties: (i) K contains the m th roots of unity era (ii) The prime p does not ramify in K (iii) K is a splitting field of every irreducible character x Fields K which satisfy the conditions of (A) are called splitting fields of least possible ramification A recent result by Solomon [9, Corollary, p 163] is related to (A) (B) be a finite group of order g pare, where p is a rational prime and (p, m) 1 Let Q be the rational field, and let K Q(e,) if p is odd, K Q(em, /-1) if p 2 If x is an irreducible character then the Schur index ink(x) of x with respect to K divides p 1 In this paper an improvement of Solomon s result will be given Namely, the conclusion of (B) can be strengthened to the following: The Schur index ink(x) 1 In particular, for p odd, the field Q(e) is a splitting field of least possible ramification A modification of the proof will show that for p 2, the field Q(e, /[) is a splitting field of least possible ramification The proofs are based on a theorem of Solomon together with some results from modular representation theory The use of the Hasse Theorem of class field theory, which up to now has been necessary in the proof of (A), can thus be avoided will always be a fixed finite group of order g pare, where p is a rational prime and (p, m) 1 A is elementary if, is a direct 9 X where N is a cyclic group, and where is a q-group for some prime q not dividing the order (N: 1) of N A subgroup is a p -subgroup if p does not divide (@: 1) Absolutely irreducible char- Received April 9,

2 516 PAUL FONG acters or representations will simply be referred to as the irreducible characters or representations 1 Results from modular representation theory In this section, the prime p will be fixed All reference to modular representation theory will then be with respect to this prime p The following lemma is due to Brauer [2, Theorem 5] The proof below is a modified version of the original proof LEMMA 1 Let qp be a modular principal indecomposable character of the finite Then qp is a linear combination with integer coeicients of characters of (R) induced by characters of elementary subgroups of p -order Proof By a theorem of Brauer [4, Theorem A], the 1-character can be expressed as 1 a*, where a are rational integers, and where is the character induced by an irreducible character of an elementary * subgroup g) In particular, The character g) vanishes on p-singular elements of g), and hence must be a linear combination of the principal indecomposable characters of g)i The coefficients in such a linear combination are moreover integers by [3, Theorem 17] But since g) is an elementary subgroup, g) can be factored into a direct product g) 3 X 9i, where is a p-group and 9 is a p -group Each principal indecomposable character of g) is thus the product of the character of the regular representation of 3 with a suitable irreducible character of 9 The principal indecomposable characters of g) are therefore induced by the irreducible characters of 9 This proves the lemma IEMMA 2 be a finite group of order g pam, where p is a rational prime and (p, m) 1 Let qp be a principal indecomposable character If llo is a representation with character qo then 1I can be written in the Proof By Lemma 1 (1) where the a are rational integers, and where is an irreducible character of a Rearrange the sum (1) so that p -subgroup g) where the integers b and cj are positive Each is the character of a representation which can be written in Q(em), since the order of g$ divides m

3 SPLITTING FIELDS OF FINITE GROUPS 517, We may assume that the are actually representations written in Q(c) If is the representation induced by then (2) implies that (3) where the sum of representations denotes the direct sum The representations b *, c- are written in Q(c) If these have common components over Q(), remove the component from both representations Repeat this until (3) reduces to where and are representations --- of go written in Q(), such that and![9 have no common components over Q() If is not the zero representation, then and would have a common, absolutely irreducible component, which is impossible Therefore lip and the lemma is proved LEMMA 3 Let be a finite group with a normal cyclic Sylow p-subgroup? If /? is abelian, then the decomposition numbers d,p are 0 or 1 Proof We proceed by induction on (: 1) Let be a maximal normal subgroup =< < The index (: ) is then a prime q p The irreducible modular representations contain in their kernel, and can therefore be identified with the ordinary irreducible representations of / In particular, if (@ ) r, has r irreducible modular characters, and these are all linear Let x, be an irreducible character and let be the decomposition of x into a sum of irreducible modular characters, Case If x, I is irreducible, then (x, ]) E d,,(, ]) is the modular decomposition of (x ) The numbers d,, are therefore summands of the decomposition numbers of (x, ); by induction the dvp can only be 0orl Case 2 If x,l is reducible, let, be an irreducible constituent of the,,, induction the are 0 or 1 Since /3 is abelian, each irreducible modular character 6, of has q distinct extensions to and indeed, the character of 3 induced by 6p is the sum of these q extensions For 6, # 6, the characters of induced by 6,, 6, respectively have no common modular constituent restriction Let,,, 6 be the modular decomposition of, by Now :, induces x,, and for p-regular elements, this is the character induced by, 6 Therefore the d,p are 0 or 1 This proves the lemma LEMMA 4 Let be a finite group with a normal cyclic Sylow p-subgroup 3 Then the decomposition numbers d, are 0 or 1 Proof If p 2, then by a theorem of Burnside [5, p 3 X, where 3 is a group of odd order The lemma then follows immediately We may therefore assume p is odd; the proof for this case is by induction on

4 518 PAZL (" 1) If () is the centralizer of then () is a p -subgroup In fact, must be the maximal normal p -subgroup of is isomorphic to a subgroup of the automorphism group must be cyclic Let x, be anirreducible character and let x, d,o 40 be the modular decomposition of x, If x,l involves two distinct irreducible characters of as constituents, then by [6, Theorem 2B], X, is induced by an irreducible character x, of a proper subgroup of Moreover, the decomposition numbers of x, and x, are the same, so that the d,o are 0 or 1 by induction We may therefore assume x,] is the multiple of one irreducible character of By [6, Theorem 2D], there is a group :), having a cyclic p -subgroup - in its center, such Moreover, there tt is an irreducible character x, of such that the decomposition numbers of t! x, and x, are the same Let be a Sylow p-complement of Then -< and :5/ 3/() :5 is then a cyclic extension of a central subgroup, and consequently :5 is abelian The decomposition numbers of x, are therefore 0 or 1 by Lemma 3, and thus the d,p are 0 or 1 then the field Q( e,) is a splitting field for every irreducible character of Proof Let x, be an irreducible character of By Lemma 4 there exists a principal indecomposable character p of which contains x as a constituent with multiplicity 1 Since the representation with character can be written in Q (e) and hence fortiori in Q(, x), it follows by the work of Schur [8], that Q(s) is a splitting field of LEMMA 5 Let be a finite group of order h pare, where p is a rational prime and (p, m) 1 If the Sylow p-subgroup of is cyclic and normal, 2 The main theorem --, can be identified with the integer i, i being uniquely determined Let be a primitive gth root of unity, and Q() the cyclotomic field Q(e)" The Galois group of Q(e)/Q is isomorphic to the multiplicative group of residue classes of integers modulo g If is an automorphism of Q()/Q and modulo g Let L be an algebraic number field; the subgroup in the Galois group of Q(s)/Q corresponding to Q() n L can therefore be identified with a certain multiplicative group (L) of integers modulo g An integer i is in (L) if and only if the automorphism of Q(s)/Q defined by s --+ si leaves Q (v) n L fixed It is clear 6(L) can also be regarded as the Galois group of L(s)/L be a finite group of order g If A is an element define the L-normalizer of A to be the subgroup of all elements X such that X-lAX, A for some integer i in t (L) A subgroup is L-elementary if is a product :5, where is a normal cyclic subgroup/a} of where is a q-group for some prime q not dividing (?I: 1), and where the L-normalizer of A in 55 is

5 SPLITTING FIELDS OF FINITE GROUPS 519 THEOREM 1 be a finite group of order g pm, where p is a rational prime and (p, m) 1 Let Q be the rational field, and let K Q(v,,) if p is odd, let K Q(,, %/-1) or K Q(,,, /) if p 2 Then K is a splitting field of every irreducible character of go Proof Let x be an irreducible character Let L be the field K(x) Define a character of any subgroup to be a character in L if its values lie in L; in particular, x is a character in L By [9, Theorem 2], it follows that where the ai are rational integers, and where the are characters induced by characters i in L of L-elementary subgroups * Suppose we can show each is the character of a representation written in L Then by an argument exactly like that used in the proof of Lemma 2, it follows that the representation E with character x can be written in L, and this proves the, theorem We have thus reduced the theorem to the following case Let 91 be an L-elementary subgroup If is a character in L of then the representation with character can be written in L We may assume is an irreducible character in L, that is, is not the sum of two characters in L of This implies that where the are s distinct irreducible characters which form a complete set of algebraic conjugates over the field L We show each is the character of a representation i of which can be written in L(i) Consider the case p odd first If p does not divide (@:1), then L contains the (:1) th roots of unity, and can be written in L(i) If p divides (: 1), then L contains the (?I: 1)h roots of unity, and hence can be written in L() by [9, Theorem 5] If p divides (/:1), has a normal cyclic Sylow p-subgroup can therefore be written in L(i) by Lemma 5 Now the case p 2 If K Q(sm, %/--1) the argument used for the case p odd applies, since the result of Solomon can still be used If K Q(vm, /), this result cannot be used for the subcase where! is a 2-group But now we use the fact is an L-elementary subgroup (so far we have really used only the is a semidirect product of a cyclic group?i by a q-group :, q a prime not dividing the order of?i) is the L-normalizer of A and since L contains the (t:1) t roots of unity, it follows from the definitions that i is in the center Therefore l X It is then sufficient to show that K is a splitting field of the irreducible characters of Roquette has shown in [7] that Q is in fact a splitting field for: unless is a (generalized) quaternion group Moreover, for the irreducible characters of a (generalized) quaternion group which are not split by Q, the corresponding skew field is the ordinary quaternion algebra over Q [7, Bemerkung, p 249] The skew field in question contains Q(/)

6 520 PAUL FONG as a maximal subfield, and hence K is a splitting field of the characters of [10, Chapter VII, 9] To complete the proof, let 1 be written over L(I) If 1,,, are the s automorphisms of L()/L, we may take i to be [, where i is applied to the coefficients in each of the matrices of The representation i 0 is therefore equivalent to a representation in L, and its character is As a corollary we have TIEOnEM 2 be a finite group of order g pare, where p is a rational prime and (p, m) 1 Let K Q(e,) if p is odd, let K Q(e,, /[) if p 2 Then K is a splitting field of least possible ramification Proof K is a splitting field by Theorem 1; furthermore, the prime p does not ramify in K by the well-known behavior of the decomposition of primes in cyclotomic extensions [11, Chapter III, 12] REFERENCES 1 R BRAUER, Investigations on group characters, Ann of Math (2), vol 42 (1941), pp , Applications of induced characters, Amer J Math, col 69 (1947), pp , A characterization of the characters of groups of finite order, Ann of Math (2), col 57 (1953), pp R BRAUER AND J TATE, On the characters of finite groups, Ann of Math (2), vol 62 (1955), pp W BURNSIDE, Theory of groups of finite order, 2nd ed, Cambridge, University Press, P FONG, On the characters of p-solvable groups, Trans Amer Math Sou, col 98 (1961), pp P ROQUETTE, Realisierung yon Darstellungen endlicher nilpotenter Gruppen, Arch Math, vol 9 (1958), pp iber endliche Substitutionen, Sitzungsber Preuss Akad Wiss 9 L SOLOMON, The representation of finite groups in algebraic number fields, J Math Soc Japan, col 13 (1961), pp o M DEURING, Algebren, Berlin, J Springer, H WEYL, Algebraic theory of numbers, Princeton, Princeton University Press, 1940 I SCHUR, A rithmetische Untersuchungen Gruppen linearer (Berlin), 1906, pp UNIVERSITY OF CALIFORNIA BERKELEY, CALIFORNIA