FINITE MULTIPLICATIVE SUBGROUPS IN DIVISION RINGS

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1 FINITE MULTIPLICATIVE SUBGROUPS IN DIVISION RINGS I. N. HERSTEIN 1. Itroductio. If G is a fiite subgroup of the multiplicative group of ozero elemets of a commutative field, the it is kow that G must be cyclic. This result clearly is ot true for geeral divisio rigs, as is show, for istace, by the example of the group G = (± 1, ± i, ±j, ± k) i the divisio algebra of the quaterios over the real field. Oe might ask, however, if there exist extesios or aalogues of the theorem for commutative fields to geeral divisio rigs. The results of this paper are i the directio of possible such extesios. Oe result we obtai is, i fact, that if K is of characteristic p, p / 0, the ay fiite multiplicative subgroup of the group of ozero elemets is ideed cyclic. For arbitrary K, the groups of odd order are at least metacyclic, ad i the special case of the quaterios the subgroups of odd order are all cyclic. Let K be a divisio rig with ceter Z, ad let K be the multiplicative group of ozero elemets of K. Wheever we write G C K, we mea that G is a subgroup of K uder the multiplicatio defied i K. 2. Results o fiite subgroups of R* for geeral K. Suppose that G C R* is both fiite ad Abelia. Let the elemets of G be g^9 g 2,, g R. Cosider T = Z(g t,, g ), the divisio rig obtaied by adjoiig g^,, g to Z. Sice G is Abelia, T must be commutative, so 7 is a commutative field; moreover G C Γ*. Thus, beig a fiite subgroup of the multiplicative group of a field, G is cyclic. So we have: LEMMA 1. If G C K* is a Abelia group of fiite order, the G is cyclic. Let p be a prime umber. If G is of order p or p 2, it is Abelia. So by Lemma 1 we obtai: LEMMA 2. If G C K* is of order p or p 2, the G is cyclic. Received April 5, Pacific J. Math. 3 (1953),

2 122 I. N. IIERSTEIN LEMMA 3. If G C K is of order p r, p a odd prime, the G is cyclic. Proof. Our proof is by iductio over r: (1) If r = 1, or r = 2, this is merely Lemma 2. (2) Suppose that all G C K of order p v, v < r, are cyclic. Let G C K be of order p r, r > 2. The by the iductio hypothesis all the proper subgroups of G are cyclic. Sice p is a odd prime ad r > 2, by Satz 88 [2, p. 72], G is cyclic. Thus Lemma 3 is established. THEOREM 4. If G C K is of odd order, the G is metacyclic. Proof. All the Sylow subgroups of G belog to odd primes, so they are cyclic by Lemma 3. Hece G is metacyclic [3, p. 145, Theorem 11]. 11]: As a cosequece of the metacyclicity of G we obtai [3, p. 145, Theorem THEOREM 5. If G C K* is of odd order, the there exist a, b L G which geerate all of G, ad which satisfy (1) a = b m = 1 (, m odd) (2) bab" 1 = a r. 3. The case where K is of characteristic p, p ^ 0. I this sectio we assume K is of characteristic p, p ^ 0. Let P C Z, Z the ceter of K, P the prime field of characteristic p. THEOREM 6. If G C K is of fiite order, the G is cyclic. Proof. Let Clearly ί/ is a group uder multiplicatio ad additio. Moreover, U is fiite, sice P is fiite. Sice ί/ is cotaied i K 9 it ca have o divisors of zero. Thus U is a fiite divisio rig. By Wedderbur's theorem, U must the be commutative; sice G C U*, we the have the result that G is cyclic. So for divisio rigs of ozero characteristic, the result for commutative fields carries over i its etirety. Oe might well ask how much of the result carries over i the case of divisio rigs of characteristic zero. We have ot

3 FINITE MULTIPLICATIVE SUBGROUPS IN DIVISION RINGS 123 solved this completely as yet; but i the special case that A' is the quaterio algebra over the real field, v\ 7 e obtai a fairly satisfactory aswer. >. K the real quaterios. I this sectio, A will deote the divisio algebra of the quaterios over the real field Z(Z is the, of course, also the ceter of A). The pricipal result we obtai is: THEOREM 7. If G C A* is of odd order, the G is cyclic. We first establish several prelimiary lemmas. Suppose that x C A'. The ormalizer of x 9 U(x), is defied by ϊ[(x) = {a K 1 ax = xa\. Trivially, Z C U(x); ad Y\(x) is a divisio algebra over the reals. Not beig the reals or the quaterios, U(%) must be isomorphic with the complex umbers. Thus there must be a t H(Λ ), with t 2 = - 1, so that every a C Yl{x) ca be writte as a = α 0 + (X ι t, where (Xj C Z. Let C \y CK I y = y Q + γ χ i, γ. CZ\. The C is also isomorphic to the complex umbers. There exists a isomorphism of ϊl{x) oto C which leaves the elemets of Z fixed. The ext two lemmas are cocered with establishig the ature of this isomorphism. Usig results about divisio subalgebras of divisio algebras 11, p. 42, Satz 3], we could obtai the results immediately; but, for the sake of self-cotaimet, we establish these results here. LEMMA 8. // t C K is such that t 2 = - 1, the there exists a S C K so that StS" ι = i. Proof. Suppose that t = τ 0 + τ^ + τ 2 / + τ 3 k (α m 's i Z). Sice t 2 = - 1, it follows that T o = 0 ad Ί 2 + T^ + T^ = 1. If 1 X = 1, the t = i ad there is othig to prove. So we suppose that 1 X Φ l A simple computatio the shows that StS" ι i, where ad where S = S o + S t i + S 2 j + S 3 k,

4 124 I. N. HERSTEIN Now τ 3 - τ 2 τ 3 + τ 2 0, i zmzrr 2 3 2VI- τ ι 2Vl~T t 2 U(αc) = ί σ Λ' I o = α 0 + a ι t, di's i Z, t 2 = - 1 1, C =! g C K I g = y Q + y χ i, γ. 's i Z }. By Lemma 8, there exists a S so that StS" = i, whece S ΓI(Λ;)5" 1 have show: = C. So we LEMMA 9. //# ^ Z, the there exists a S G K so that Sϊl(x)S" 1 = C. LEMMA 10. If a, b C X, α 71 = b m = 1, α/iί/ 6α6" 1 = α Γ, ί/ie/z eiίaer α6 = ba or a" 1 b ~ ba. Proof. If a ζl Z, the α& = Z>α, ad the result is correct. Suppose the that a y- Z. Sice a ^1 ϊl(a) ^ K 9 by Lemma 9 there exists a S G A' so that SaS" 1 C. Thus (if we assume rc is the least positive iteger so that a = 1), 27TA 2τrλ _. λ = SαS" 1 = cos + i si [(λ, ) = 1] is a primitive th root of uity i C. Let = β 0 + β ί i + β 2 j + β 3 k, β i εz. From όαό" 1 = α r, we obtai δ^β" 1 = A r. If β = /3 =0, the AB = δ^ sice i 2 3 that case both A ad β would be i C So we suppose oe of them, say β 2, is ot zero. The BAB' 1 = A r yields (1) (j8 0 + β χ i + β 2 j+ β 3 k) cos + t si 1 / 2?7λr 2πλr \ = cos + i si (β + β i + β j + βλ). υ ι ι ό \ I Computig the coefficiets of / ad k i (1), we obtai, o x I 2πλ 2πλr\ I 2πλ 2πλr\ {2a) js I cos cos + β [ si + si = 0, 2 3 \ I \ I

5 FINITE MULTIPLICATIVE SUBGROUPS IN DIVISION RINGS 125,, / 2πλ 2πλr\ I 2πλ 2πλr\ λ (2b ) -p si + si I + β [ cos cos 1 = \ I \ I Sice β φ- 0, we have (3) cos 2πλ cos 2πλr 2πλ 2πλr m si -f si I 2πλ m 2πλr\ 2πλ I si + si I cos \ ' Expadig this determiat, we have cos 2πλr = 0. 2τ7λ(r+l) 2πλ 2πλr 2πλ. 2πλr (4) cos = cos cos si si = 1. Sice (λ, ) = 1, from (4) we obtai \ (r + 1.) Thus BA = A r B =s A" ι B 9 sice A 1; this gives correspodigly for a ad b the result that ba = a" ι b 9 which is the lemma. ab. COROLLARY. // a = b m = 1, ad m both odd, ad bab" 1 = a Γ, the ba = For if bab" 1 = a" 1 the b 2 a=ab 2 9 ad sice m is odd this gives ba = ab. Proof of Theorem 7. Sice G C X* is of odd order, by Theorem 5 there exist a, b G G which geerate G ad which satisfy (1) a = b m = l {, m odd), (2) bab~ ι =a r. Thus by the corollary to Lemma 10, ab = ba. So G must be a Abelia group; sice G is i X*, a applicatio of Lemma 1 yields the result that G is cyclic. This is Theorem 7. Oe might hope that a more geeral result would hold. Such a result might be that if K is a divisio rig, the ay fiite subgroup G, G C / *, which is of odd order, is cyclic. It would be eough to prove this for divisio algebras of fiite order over the ratioale. REFERENCES 1. M. Deurig, Algebre, Chelsea, New York, 1948.

6 L26 I. N. HERSTEIN 2. A. Speiser, Die Theorie der Gruppe vo edlicher Ordug, Third Editio, Dover, New York, H. Zassehaus, Theory of groups, Eglish Traslatio, Chelsea, New York, COWLES COMMISSION FOR RESEARCH IN ECONOMICS AND THE UNIVERSITY OF CHICAGO