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 “TNE ANALYTIC REPRESENTA LION OF SUB. 
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 NUMBER OF LETTERS WITH A. DIS 
 CUSSION OF THE LINEAR GROUP. 
 
 
 
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 A DISSERTATION PRESENTED TO THE FACULTY OF ARTS, 
 LITERATURE, AND ~CIENCE, OF THE UNIVERSITY OF 
 CHICAGO, IN CANDIDACY FOR THE 
 PEGIREE.-+OF. “ROC TOR )F 
 PHILOSOPHY. 
 
 By 
 LEONARD EUGENE DICKSON. 
 
 Reprinted srom the Annals of Mathematics, 1897. 
 
 
 
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 ve ae 
 
 
 
THE ANALYTIC REPRESENTATION OF SUBSTITUTIONS ON A 
 POWER OF A PRIME NUMBER OF LETTERS WITH A DIS- 
 CUSSION OF THE LINEAR GROUP.* 
 
 By Dr. Leonarp Eveene Dickson, Chicago, Ill. 
 
 _ TABLE OF CONTENTS. 
 
 Preface with literature. 
 
 PART I.—AwnatytTic REPRESENTATION OF SUBSTITUTIONS. 
 
 Section I.— General Theory. 
 
 Arts. 1- 2. Definitions. 
 3. Uniqueness of representation. 
 
 4— 5. Lagrange’s interpolation formula, 
 
 6— 7. De Polignac’s result anticipated. 
 8. Matrix property of substitution quantic. 
 
 9-11. Generalization of Hermite’s Theorem. 
 
 12. Reciprocal of a substitution quantic. 
 
 13-15. Residue of a multinomial coefficient (mod p). 
 16. Reduced form of substitution quantic. 
 17. Degree not a divisor of p” — 1. 
 18. &* a substitution quantic if % is prime to p" — 1. 
 
 Szotion Il.—Quanties of degree prime to p. 
 
 20-45. Complete determination of reduced quantics of degree = 6 suitable on p” letters. 
 46-50. Preliminary study of septics. 
 51-56. Quantics with an infinite range of suitability. 
 
 Section ITI.— Quantics of degree a power of p. 
 
 57-59. Quantics with all exponents powers of p. 
 
 60-68. General theorems on quantics of degree p”. 
 
 69-74. Determination of all reduced SQ[p"; p"], for p” = 7 and partially for p” = 11. 
 75-77, General types of SQ[ p”; p”]. 
 
 Section 1V.—Degree a multiple, not a power, of p. 
 78-82. Determination of all reduced SQ [6; 3%]. 
 82-86. Special results on SQ[6; 2”]. 
 Miscellany. 
 
 87. Table of all reduced SQ[%; p”"], k= 6. 
 88-89. Analytic generators of substitutions on 7 and 5 letters. 
 90. Enumerative proof of Wilson’s theorem. 
 
 
 
 * Dissertation accepted by the University of Chicago in partial fulfilment of the requirement 
 for the degree Doctor of Philosophy. 
 
 149084 
 
66 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 PART II.—Linzar Grovp. 
 
 Section I.—Linear homogeneous group. 
 
 1. Definition of a linear homogeneous substitution. 
 2. Literal and analytic composition ; group. 
 3. Restriction to prime modulus. 
 4, Order of linear homogeneous group. 
 5. Transformation of indices; determinant invariant. 
 — 7. Decomposition of a linear homogeneous substitution. 
 2. Factors of composition of linear homogeneous group. 
 3. A triply infinite system of simple groups. 
 Srotion IIl.— Linear fractional group. 
 14. Definition, exhibition, and order. 
 15. Group of linear fractional substitutions with determinant unity is simple. 
 16-17. Remarks on systems of simple groups. 
 Section III.— The Betti-Mathieu group. 
 
 18-19. Identification with the linear group. 
 20. Order of the group. 
 21-24. Mathieu’s special type of substitutions. 
 
 PREFACE. 
 
 This paper is an application of the Galois Field theory, a conception of 
 fundamental importance in Higher Algebra. This theory is here presupposed 
 and will be used in the abstract form given it by E. H. Moore.’ Reference 
 may also be made to Galois,’ Serret’, Jordan*, Borel et Drach’, in a note by 
 the latter the Galois Field being developed in its abstract form. 
 
 The aim in Part I is two-fold: (1) the complete determination of all 
 quantics up to as high a degree as practicable which are suitable to represent 
 substitutions on p” letters, p being a prime, 2 an integer; (2) the determi- 
 nation of special quantics suitable on p” letters, where for each quantic the 
 combination ( p, 2) takes infinitely many values. 
 
 It is a remarkable fact that, on the one hand, the conditions necessary 
 and sufficient to determine a substitution quantic are found by multinomial 
 expansions,—on the other hand, one of the observed types of substitution 
 quantics having an infinite range of suitability is a multinomial expansion 
 
 ‘Moore, Proceedings of the Congress of Mathematics of 1893, at Chicago. 
 * Galois, Bulletin des Sciences mathématiques de M. Férussac, vol. 13, p. 428, 1830; reprinted 
 in Liouville’s Journal de Mathématiques, vol. 11, pp. 398-407, 1846. 
 *Serret, Algebre supérieure, fifth edition, vol. 2, pp. 122-189. 
 ‘Jordan, 7’raité des substitutions, pp. 14-18. 
 °Borel et Drach, 7'héorie des Nombres e¢ Algébre supériewre, 1895, pp. 42-50, 58-62; Note, pp. 
 343-350. f 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 67 
 
 (multiplied by a power of the variable) and the other type a reverse-binomial 
 expansion (see $§ 53-54). 
 
 The results of this investigation warrant the conjecture that there exist a 
 small number of types of substitution quantics of such wide ranges that together 
 they represent all the py”! substitutions on p” letters. Examples where appar- 
 ently isolated quantics fall under a general type (when it is not reduced) are 
 given in § 77. 
 
 While the aim in Part II* is primarily to generalize the work of Jordan 
 on the linear homogeneous group, the treatment has been considerably modi- 
 fied to render the subject more accessible. The desire being chiefly non-¢yclic¢ 
 simple groups, the modulus is supposed prime, which affords much simplifica- 
 tion. On the other hand, many amplifications occur and also the correction 
 of several errors (indicated by foot-notes). 
 
 In the same investigation there may occur marks of the Galois Fields of 
 orders p', p”, amd p”™, n > 1, m > 1, when (as a useful notation) we use 
 small Roman, small Greek, and capital Roman letters respectively. 
 
 Literature on the analytic representation of substitutions : 
 
 M. Hermite, Sur les fonctions de sept lettres, Comptes Rendus des Seances 
 de L’ Académie des Sciences, vol. 57, pp. 750-757, 1863. 
 
 Hermite’s results (in whole or part) are given by: 
 
 Serret, Cours D’ Algébre supérieure, vol. 2, pp. 883-389 and 405-412 ; 
 
 Netto, Swbstitutionentheorie, ch. 8 ; 
 
 Jordan, 7raité des Substitutions, pp. 88-91 ; 
 
 Borel et Drach, Zhéorte des Nombres et Algebre supérieure, pp. 305-807, 
 1895. 
 
 Enrico Betti, Sopra la risolubilita per radicali delle equazioni algebriche 
 irriduttibili di grado primo, Annali di Scienze Matematiche e Fisiche, vol. 2, 
 pp. 5-19, 1851; Sudla risoluzione delle Equazioni algebriche, ibid, vol, 3, pp. 
 49-115, 1852 ; Sopra la teorica delle sostituzioni, ibid, vol. 6, pp. 5-34, 1855. 
 
 Emile Mathieu, J/émoire sur le nombre de valeurs que peut acquérir une 
 Jonction quand on y permute ses variables de toutes les maniéres possibles, 
 Journal de Mathématiques pure et appliquées, second series, vol. 5, pp. 9-42, 
 1860 ; Mémoire sur Vetude des fonctions de plusieurs quantités, sur la maniére 
 de les former et sur les substitutions qui les laissent invariables, ibid, vol. 6, 
 pp. 241-323, 1861. 
 
 F. Brioschi. Des substitutions de la forme 
 
 n—3 
 
 I iy he ul en 
 
 * For the suggestion of this generalization as leading to a triply infinite system of simple 
 groups, as also for much yaluable aid throughout the investigation, I am indebted to Professor 
 Moore. 
 
 
 
68 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 pour un nombre n premier de lettres, Mathematische Annalen, vol. 2, pp. 467- 
 470, 1870. 
 
 De Polignac, Sur la représentation analytique des substitutions, Bulletin 
 de la Societé Mathématique de France, vol. 9, pp. 59-67, 1881. 
 
 A. Grandi, Un teorema sulla rappresentazione analitica delle sostituziont 
 sopra un numero primo di elementi, Giornale Matematico del Prof. G. Batta- 
 glini, vol. 19, pp. 288-245. The conditions are found under which 
 
 5 gies 
 
 eP-*§ 4 an 2 -+ ba 
 
 
 
 shall represent a substitution on p letters. A generalization by the same writer 
 is given in’ Reale Istituto Lombardo di scienze e lettere, Leendiconti, Milano, 
 vol. 16, pp. 101-111. For abstracts of each see Fortschritte der Mathematik, 
 vol. 13, p. 118, 1881, and vol. 15, p. 116, 1883. 
 
 J. L. Rogers, On the Analytical [representation of Heptagrams, London 
 Mathematical Society, vol. 22, pp. 87-52, 1890. 
 
 L. E. Dickson, Analytic Functions Suitable to Represent Substitutions, 
 American Journal of Mathematics, vol. 18, pp. 210-218, 1896. 
 
 
 
 PART I.—AnatytTic REPRESENTATION OF SUBSTITUTIONS ON A POWER OF A PRIME 
 NuMBER OF LETTERS. 
 
 Section I.—General Theory. 
 
 1. Let ¢ be any mark of the Galois Field of order p”, p being a prime and 
 n a positive integer. Also let ~(&) be an integral function of ¢ having as 
 coefficients certain marks of the @#'[p"]. The replacing of the letter lz by 
 the letter dy. defines a substitution on p” letters, in notation, 
 
 e == O45). 
 
 2. In order that ¢ (¢) shall indeed be suitable to represent a substitution 
 on the marks 
 
 Vi ie OM ee po) 
 it is necessary and sufficient that v (™), ¢ (M4), ---, Y (Mp1), be identical with 
 Moy Py +++ Upn—1 except as to order. If an integral function of degree / belong- 
 
 ing to the GF’ | p”"| satisfies these conditions, it will be called a substitution 
 quantic of degree & on p" letters and denoted thus 
 
 SQ[z; p"). 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 69 
 
 The degree /# will be supposed < p” owing to the equation 
 
 eee, 
 satisfied by every mark of the field. 
 3. Theorem. Two different quantics ¢(&) and $(§) belonging to the 
 GH'[ p"| cannot represent the same literal substitution. 
 For if the substitution replace the index yp; by py, fori = 0,1,..., p? —1, 
 then must 
 Gh Wee tip i hh ( 14,) (tee Oe ps 1); 
 
 CO ie) ae U 
 
 is of degree p" — 1 at most but has ~” distinct roots y; It is thus an identity 
 in &. 
 
 Thus 
 
 4. The most evident substitution quantic is the integral function ¢ (¢) 
 which gives Lagrange’s interpolation formula: 
 n—]l es 
 
 egg D vl) 
 
 ¢ (é) ; = (= ms ti) Fm) (1) 
 
 where @), 4, ..., @pn_; is any permutation of 0,1,..., py’ —1, and where 
 
 pr—1 
 
 UES) = af (¢ — 
 
 and /” denotes its derivative. Thus 
 
 represents the substitution 
 
 Me. Pee rise gees | 
 
 Cae Pays «+> ig ee J 
 
 5. Following Hermite’s method* for the case n = 1, we may give (1) a 
 simpler form. ; 
 
 In the G/F'[ p"), 
 Le vie Pa or (eet Le 
 
 
 
 pl 5, pt may é) 
 is : VW fa\> 
 Mats oan 
 a i=0 Ser vi 
 Taking 4, = 0, so that 
 pe = 1 = 0 (Hee ey 2 1) 
 
 - * Cf Serret, 1. c. 2, pp. 384-5. 
 
 
 
 
 
 
 
70 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 and performing the division by § — yy, 
 
 pr—] 
 
 L > © = 
 g (E) = — pa, (€?" — 1) — 2 Halen te fee ee ees 
 ‘i= 
 
 Arranging according to powers of ¢ and noting that* 
 
 prt 
 fa eas 
 = 
 we reach the desired quantic, 
 pr—2 
 es eee 
 ¢ (§) = os" (2) 
 j= 
 where 
 pr— 
 = v a ee a 
 a; ne ed ai . be ij (7 : 0, ix a fie <> 2). (3) 
 
 i=0 
 
 It follows from § 3 that every substitution quantic on p” letters which 
 
 belongs to the G/’[ p”] is contained in the form (2). 
 6. The conditions on a, that an arbitrary quantic 
 
 n—] 
 g(§)= = af! 
 
 j=0 
 belonging to the G/’[ p"] shall represent a substitution on its p" marks yp, are 
 
 pr—l1 
 v pees es : 
 2, Gps SOX) (ee eee) 
 j=0 
 where o(4,) — fq, form a permutation of y, Applying the lemma proven in 
 § 10, we have on adding the p” equations (4), 
 pr—l 
 aS ss 
 ee Opn a D (fi) =.0, 
 i=0 
 Taking a,,_, zero and dropping the first one of the equations (4), we have the 
 system of conditions 
 
 fy Gp = 04) ie Lee , pn —1). (4) 
 j=0 
 The determinantt ; 
 | | = Th, — 8 
 
 where’?,. 8 S12)... pa ere 
 We may thus express a; linearly in terms of 4, By equation (2) of § 5, 
 this is done by formula (3). 
 
 
 
 * pr—1 pr—1 
 s% sy 
 “ a= + My = 0 by § 10. 
 
 i=) : i=0 
 
 + Baltzer, Theorie und Anwendung der Determinanten, p. 85. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. (io 
 
 7. The gist of De Polignae’s paper is to actually solve equations (4) for 
 the case 2 = 1. Using @ and m for the integers corresponding to the marks 
 a and yp, his result is given in the form 
 
 Lay teen be RE oe 
 a= 2 {(—1)’ —2 } m 
 
 eed | 
 
 aid 
 
 where previously m,, has been taken to be zero. Subtracting 
 
 pot 
 (—1). 2 m,, = 0 (mod p), 
 i=l 
 re Pens 
 = (= 1 Cae ra = aes We ae — es ers THU ake foe) 
 — = 
 
 8. We may independently verify the inverse character of the linear rela- 
 tions (3) and (4) between the coefticients a; of any substitution quantic ¢ (¢) 
 and the marks ¢(4;) = ,, By (4’) the matrix 
 
 4 
 
 2 r—? ~ 
 (hee ST Ae ee ee ee 
 ine je ee ? : VF 
 2 pr—2 
 el b] pr» UL pr—1) eine e) <9 (Tres 
 expresses 9(1;), 9 (M2), ---, G (Mp1) linearly in terms of @, %, ..., Gn» 
 
 Inversely, by (3) the cine 
 
 So ee ea DS ae en ee ee | 
 me pr—2 pr—2 eos, pr—2 | 
 ty és me 925) 3 3 Tae Upn-1 
 * " « © 
 at pr-3 = p"--3 nae 3 pr—3 ee 
 | eae P LE ha ea ean 
 ‘9 —— 'y ; == 2 ) Sino! Sty L Ayyn—1 J 
 
 expresses the latter linearly in terms of the former. To prove that the product 
 of the two matrices is the identity, let c’ denote the term in the product derived 
 from the 7th row of the first and the jth column of the second matrix. Then 
 
 pr—2 te Ae pr— 3 a. pr—2 
 ij pe M5 aes fi Kj ‘ 
 
 Cy =—_—- i L 
 Thus 
 
 C= (pl) 4 1 (tg et nee 1 77 ere hs) 
 
72 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 For 4,7 =.) 2). De oe 
 
 ee pn 
 i Ls 
 
 i) als pr—l pr—2 pr—2 pr—1 
 es a =—- IF shat SUR Sih peti ew UTE + By és 
 ae 2 J 
 
 i Nee 
 
 Hence ¢, = 0.if ¢ 9: 
 
 Fromrthe matrix expressing ¢ (y,) in terms of @, G,..., Gn—2, We derive 
 by reflection on its main diagonal and change of sign of all its elements a 
 matrix which expresses 4, Gpn_o, Zyn_3, ---, @, In terms of 
 
 gM) (i 1) Oe ep eed 
 
 9. Lemma. , 4), +--+ Mp1, being the marks of the G/’'[ p"] and ¢a posi- 
 tive integer, 
 n—] ; 
 —1l1fort?=p"—1 
 
 hier 
 i=0 
 
 For let ¢, denote the sum of the ¢th powers of the roots of the equation 
 belonging to the G/'[ p”] 
 Bye" =0 (yw =1). (5) 
 
 i=0 
 Applying Newton’s identity (as is allowable) 
 Opt Yi Fe + Ye Opn Fe HAG + hy, HO. (hl) 
 If for (5) we take the equation whose roots are y,, 
 ae eo eae 
 
 the proof of the lemma follows. | 
 10. Lemma. If, in the notation of § 9, 
 
 ss (teal oe, oon 
 Toad ae Girone) |z + 0, (mod p) Ube 
 
 and if all the roots of equation (5) be marks of the G/’{ p"], then will (5) 
 take the form 
 
 prs yee dae 0s 
 Applying Newton’s identity cited above, we find 
 
 Wa Sat, (ae Pye 
 Ves (2 + 0 (mod p) ey 
 
 CP ge ips 8h aide ee iN 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 73 
 
 To determine ¥,, Yop, - ++» Ypn—p» apply the identity 
 
 Dre Oe tke eo Vg be = O (AS p"); 
 
 which reduces to 
 
 Oy, oa YnFk—p > YopFr—2p si\3 pases oe @ pn—1F pnt af Y pr F —pn — 0 . 
 
 < 
 
 By our assumption on the roots, 
 
 IF, = Ferpn-i+ 
 
 Hence fork = p, + p—1, 
 
 YpF pp = 0. 
 Generally, for k = p" + Ilp—1, le p™ —1, 
 yl, Fyn == 0 = 
 
 Since o,,_, $ 0 by hypothesis, ¥,, = 0. 
 
 11. Generalization of Hermite’s Theorem. 
 
 The necessary and sufficient conditions that ¢(&) shall be suitable to 
 represent a substitution on p” letters are : 
 
 (1) Every tth power of ¢(&), for i << p" —1 and prime to p, shall reduce 
 to a degree = p"™ — 2 when we lower the exponents of € below p” by means of 
 
 the equation 
 Epn — = = O . 
 > = > 
 
 (2) There shall be but one distinct root of 
 
 3 p»—1 
 
 7 =; 
 
 g (§) ome 
 jem D 
 
 Proof: Suppose 
 
 After reducing exponents below p” let 
 
 rn 
 
 x p =f as 
 [y is S= > ames! ; 
 
 1=0 
 
 Give to € the values of the p” marks yw; of the field and add the resulting 
 
 equalities. 
 
 pr—l1 pr—1 
 
 re [e(y)]™ —— pra” — a,” 2 yy af kp =f. ag, ay ip 
 = [= 
 
 Hence by § 9, form < p" —1, 
 
 yn—] 
 (m) 
 
 ~ [eo (4) = — ay . 
 j=—O 
 
74 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 But if g (€) represents a substitution, 
 
 pr—l pr—1 
 2 [¢y (4) |" = py ign ee 0 
 
 7=0 7=0 
 ifm < p"—1. Hence a necessary condition is 
 
 Co Oe (Wie lion a Oe ce) 
 
 wr —-1 
 
 Also there must be one and only one of the marks ¢ (y;) equal to zero, i. e. 
 but one distinct root of g(¢) = 0. ‘ 
 Inversely, suppose (1) and (2) are satisfied, so that 
 
 pel te Ne ae sa) 
 p ee 4) ( te : 4 
 
 = l¢ (4) | = Apr = ( ¢ $ O (mod Pp) j 
 pr— 
 
 I 
 [eo (4) ]?"7* = —140. 
 
 LY 
 vO 
 
 Then by § 10 the equation 
 
 pr—1 
 ff [7 — ¢ (w)l= 9 
 j= 
 takes the form 
 ea a Yor 7] + Yon == 0 
 
 7] (1 “= Won 4) ae Yoon == 0 * 
 
 or 
 
 If y,,_, + — 1, this linear equation is satisfied by the p” marks ¢(y,). These 
 must therefore be equal, and since their sum is zero, each must be zero. But 
 in this case the equation belonging to our field, 
 
 pr—2 
 
 gE) ==. 50,5 = 0 
 
 1=0 
 
 would have p” distinct roots ¢ = y; (i = 0, 1, ..., p” —1) which is impossible. 
 Hence y,,_, = — 1 and then, since one of the ¢ (,)’s are zero, y,, = 0. Thus 
 
 the marks  (#,) are identical apart from order to the roots y; of 
 7" re y) —s @) , 
 12. Reciprocal* of a substitution quantic. Given a substitution quantic 
 
 pr—2 , 
 z = OAS) eee eer 
 
 
 
 * Rogers, l. ec. for p» = 7. His proof is objectionable, since he makes £° — 1 (mod 7) while 
 ~ is to be taken = 0. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 75 
 
 Suppose its reciprocal is 
 pr—2 
 Ee VS Q.pr—1-—t 
 = 2 pyr. 
 11 
 Then 
 yn—] 
 Sy = S [gy (E) = 2 aor", 
 
 I 
 
 after reduction of exponents asin $11. On the other hand, 
 
 pr—2 
 
 ft SV Q.-,p"—1—i+2z Q -t—1 9 --t—2 QD ag pr—I i2 agtt1 
 eee oe Sapna et pa se Pp 
 
 i=! 
 
 epr—1 
 
 Since 7 = ¢(€) is a SQ[ ; p”], then 7’ contains no term ¢’"” as long as 
 j <p” —1. Hence the term a,” &"" must be derived from #7". But 
 
 y= Oniand only if¢ == 0,so-that 7° = ¢. Hence 
 
 1) page ate <b 
 é Pe a” 
 or f, is the coefticient of €”" in [¢ (§)]*. 
 Residue of a multinomial coefficient modulo p, p being prime, $$ 13-15, 
 13. A number m can be written in one and but one way in the form 
 n 
 
 Dees ODS, 
 
 i=) 
 
 
 
 where a{” is zero or a positive integer << p. Then,* if 7’,, denotes the highest 
 
 power of p which divides m!, we have 
 
 P=(m— 3 a)/(p —1). 
 
 iz} 
 
 14. Theorem. The multinomial coefficient 
 
 m | 
 Ti Wn i ete 
 where m, + m, +... + m= m, is prime to p if and only if, when each mm, 
 is written in the form 
 nr ' 
 Ta a ea, 
 sO) 
 we have as numerical equalities 
 t 5, 
 2 Oe ee a) gece W), Ly cixiees It). 
 
 k=1 
 
 
 
 * Bachmann, Zahlentheorie, I, p. 33. 
 
76 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 Proof.—The necessary and sufficient condition is 
 
 Se ee FO os jest, 
 
 or 
 i=0 i=0 0) i) 
 
 It follows that if m (written in the above form) be partitioned into m, + m, 
 +... -—+ m, the partitioning must take place in the coefficients a{™ of the 
 powers of p each independently of the others. 
 
 Example. 42 = 2? +- 2° + 2; hence the binomial coefficient c;? is odd 
 only if 4 = 2,272 + 2 224 2), 2? 4-2) 2 ee ee D. 
 
 15. Theorem. If the multinomial coefticient be prime to p, it is congruent 
 modulo p to 
 
 as! 
 
 6 ae! ain! ree 
 where s! = 1,ifs = 0. * 
 
 In proof let 
 
 M=PA+T NV =P +, etc., 
 
 where 7,, 7, ... are positive integers < py. ‘Then 
 
 
 
 my= (1.2.38 1..p =) pl pe lt pe ap eee 
 UP(MP+1...9p +7) 
 
 
 
 == (pS ger mods) 
 =(— pp). gir,s: 
 Similarly, 
 gi! =(— p)” - gq! 7! (mod p), ete. 
 ahd (oe ny Vine IESE bs 
 
 Hence 
 
 But if /'(s) denotes the greatest integer in s, 
 5G, =a Limp) =P, 
 Thus 
 m! = (— p)? Hai! 
 
 Proceeding likewise for m,!, m,!,..., m,! and applying § 14, we have the 
 proof of the theorem. 
 
 16. Reduced form of substitution quantics. Let g (€)= 4 €* + a,€* 1+... 
 bea SQ[A; p"]. If on the right and left of the substitution 
 
 S.== oe) 
 
 * Bachmann, Zahlentheorie, p. 32. 
 
 
 
s 
 “I 
 
 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 we apply linear substitutions of the form 
 
 iS — Bé + oe (3, t 0) 
 
 we obtain a substitution 
 aes fie (B,'E aid 8’) + Bo = ¢,(S) . 
 
 Take 3,, = 1 and dispose of the indeterminate /, so that the coefficient of =’ 
 in ¢,() shall be unity. If a, + 0, the coefficient £8, + f,a, of €*' in o(€) can, 
 by choice of 7,, be made zero, if and only if & be prime to p. Finally 3, is 
 chosen to make the constant term zero. Then 9,(¢), in which the coefficient 
 of ¢* is unity, the constant term zero, and when £ is prime to p the coefficient 
 of ¢*~ zero, will be called the reduced form* of o(€) for the GL'[ p"). 
 
 lie ieorenees (-td.c mae. 2. eg sf 4s) not.a substitution 
 quantic on p” marks if p” be of the form mk + 1,k > 1. For on raising it 
 to the power ( 7” — 1)/k the coefticient of ¢”""' is 1 + 0. 
 
 18. Theorem. €*is a SQ [k; p"] if and only if k be relatively prime 
 to p” — 1. 
 
 For, / being any integer < p” — land prime to p, // must not be a mul- 
 tiple of p” — 1 (by § 11). 
 
 Corollary.—The extraction of /th roots in the GP’ py is always possible 
 (and then uniquely) if and only if / be prime to p” — 1. 
 
 Secrion Il.— Degree k prime to p. 
 
 19. Using the same method as in my papert giving a complete list of 
 SQ [(k; p'] for & < 7 and p any prime, I shall first determine all SQ[k; p”] 
 for k < 7, p any prime not a divisor of #, and n any integer. After a pre- 
 liminary study of septics, I shail conclude Section II with the derivation of a 
 remarkable class of substitution quantics of arbitrary odd degree /. 
 
 Complete determination of reduced quantics of degree k < 7 suitable to 
 represent substitutions on p" letters, p being prime not a divisor of k, $$ 20-45. 
 
 20. S is suitable for every p”. 
 
 21. is the reduced quadratic and is rejected by $17 since p” is odd. 
 
 22. zs Ae On. 
 
 (a) The case p” of the form 3m + 1 is rejected by § 17. 
 
 (b) The case p” = 38m + 2. 
 
 Then (& + a€)"t! gives (m + 1) aas the coefficient of ¢”"t'. Hence, by 
 
 
 
 * By making use of the indeter aay By ® further reduction may often be. made in ¢,(&). 
 The simplest form thus obtainable is called ultimately reduced. Thus 4 + 35 reduce to £4 + 32, 
 + American Journal of Mathematics, Vol. 18, pp. 210-218, 1896. 
 
78 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 §$11,a = 0; forif m + 1 were divisible by p, then would also 3m + 3 or 
 p" +1. The remaining form & is suitable by § 18. 
 
 23. ¢¢ + a& + Bé. 
 
 The only case to consider here is p" = 4m + 3. 
 
 The m + Ist power requires (m + 1)a=0 ora = 0. 
 
 (& + &)"+? requires Mee ah ie soda) f* = 0, provided p" > 7. Hence 
 8 = 0; for if m + 2 be divisible by p then is also 4m + 8 or p” + 5,1. €. 
 p = 5, while p must be of the form 4/ + 3. 
 But & is rejected by § 18, viz, by the power / = 2m be 
 J J > P WF 
 
 For the case above excluded, p = 7,” = 1, we have Hermite’s result, 
 
 the suitable quartics € + 36. 
 
 Reduced quintic & + a& + B& + 7&, S$ 24-44. 
 24. The case p” = 5m + 2. Hence v is odd. 
 The power m + I requires 
 
 (m+ 1)7 + Me. genie a= ();, 
 
 But m + 1 is divisible by p only ifp = 3. Thusif p ¢ 3, 
 By = (1) 
 The power m + 2 requires if p” > 7 
 
 (m + 2)(m 
 Toa 
 
 1)” (6apy + f+ (im — 1) 088} = 0. 
 Hence if p + 2 and p” > 7, 
 5 (6af7 + #) —Ta fs = 0. (2) 
 From (1) and (2), if p + 2 and ¢ 3 and if p” ¢ 7, 
 Ye am Ti he (3) 
 
 The power 7 + 3 requires if p” > 7 
 
 
 
 hes doy? + 66%)? + = F (100%? + 30a%8% + 5465) 
 
 (m —1)(m — 2 ne 2 m —1)(m — 2)(m — 3), 
 4 zat *) (Gaby + bata") + | Aire Ue 
 
 Now m + 3 is divisible by p only if» = 13. Hence if p is neither 2 nor 13 
 we may divide off the binomial factor ¢,"**. Multiplying the resulting equa- 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 19 
 
 tion by 5*, replacing 5 (m — 1) by — 7, etc., we have for p” not 2”, 7', 13” the 
 condition 
 
 20a (5y)* + 1508? (57)? — 700? (57)? — 1050026? (57) — 875u/f' + 8405 (57) 
 
 + 10500%f? — 3407 — 0. (4) 
 
 Applying (1) to (4): 
 15006? — 8750p = 0, 
 
 from which by (3) we find # = 0. 
 For the proof that the resulting form 
 
 Be + Bas? + aff 
 
 does represent a substitution on p” = 5m + 2 letters, a being arbitrary, see 
 Secbe. 
 
 2h. ne case p” == 13" == Ben + 2. 
 
 The powers m + 3, m + 4, m + 5 give identities. The power m + 6 
 
 requires 
 Co else nr =.0 or a B=. 0: 
 
 For, 5m + 2 — 0 (mod 13)1. e. m = 10 (mod 13). Thus c¢/"* or ¢/%** is + 0 
 Oily 0K cal 2 o.10, 14, 15,°16,26,...,, by §'14. 
 
 It follows from this equation and (3) that 8 = 0. 
 
 26, Che. case p —- 1, n= 1. 
 
 Hermite gave the complete list of suitable forms : 
 
 SVy 
 
 Snes 
 & + af + & + 8s, a = quadratic non-residue of 7. 
 & + af + 3a, a = arbitrary. 
 The last may be written 5& + 5a& + a6. 
 27. The case p” = 3” = 5m + 2. 
 The powers m + 2 and m + 3 require by (2) and (4) 
 
 2h = ap (2’) 
 a — oP + afft—a = 0. (4') 
 
 Since n S 3, 5m + 2 = 0 (mod 27) or m = 5 (mod 27). Hence e"** or of"*" 
 IocteOnl yl Awaeeeleeneo, LOLI or. 27. 
 The power m + 6 thus requires, if 2 > 3, 
 
 Gye Up eens 3 OK (5’) 
 
80 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 Thus § = 0 and (4’) becomes 
 a(iyt@)(7+y7e? —a)=—0. 
 Ifa+0,7 = — @; for if the last factor vanishes, 
 (Com ene ee ; 
 
 while — 1 is a not-square in the G/’[3'] and hence in the GF [8"], n being 
 odd. 
 If a = 0, the power m + 9 of & + 7€ requires 
 
 MOA alae ca eee 
 ete ge ty pt a) 
 
 The possible form is thus 5 + 506 + o&& when n > 3. 
 28. The case p” = 3°. 
 
 The powers 11 and 13 require 
 
 B+ af =0 (5”) 
 et deo wT aes ae eo (6") 
 Suppose § +0. Then by (2’), # = — a’, so that (5’) is satisfied and (4’) 
 
 becomes 
 Og i ee 
 
 But if either 7 = 0 or 7 = @, (6”) requires that 8 =0. Since f = 0 we have 
 y = — @ as in the case n > 3. 
 
 29. The case p” = 2" = 5m + 2. 
 
 Since 7 S 5, m = 6 (mod 32). The power m + 5 requires 
 
 Cite ee mt bot ae (oregon 
 Applying (1), 7 = a’, this becomes 
 e+) =O. © (7) 
 The power m + 7 requires Ligtre tO, 
 
 af+a¢h + a*=—0. 
 
 Applying (1) this becomes a’f* = 0. Hence # = 0. 
 
 For n = 5, the 13th power requires a condition which on applying (1) 
 becomes exactly the fourth power of (7). 
 
 For n = 5, the 15th power requires 
 
 P+ BP + all + BP + ah + byt + BP + by + RP + By + aft = 0. 
 
 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 81 
 Applying (1) this becomes 
 aft (a + aG* + BY) —0. (8) 
 
 If 6 + 0, #° = @ by (7), whence (8) becomes a’ = 0. Hence must 8 = 0. 
 30. Summary of §$§ 24-29. The only possible reduced substitution quintic 
 on p” == 5m + 2 letters is 
 
 beverebas 4 aes) « arbitrary, 
 except for p = 7, = 1 when we have two additional forms : 
 
 £4 of 
 
 t 
 
 P4 ah + 4 80%, 
 
 where @ is a quadratic non-residue of 7. 
 
 31. The case p” = 5m + 3. 
 
 The power m + 1 requires (m+ 1)f8=0. Hence ifp+2,8=0. The 
 power m + 2 of & + a& + 7€ requires 
 
 Cae - ae Ae : Say ae ee a = 0) , (1) 
 If p + 7 we may divide out ¢,’"**, giving readily 
 257? — l5dey + 2a = 0. 
 
 Hence, if p + 2, + 7, either 57 = @ or 57 = 2a”. 
 The power m + 4 requires for p” > 13, 
 
 Oy ero eee LUC a la 4 Cot, Sale Gt a == 0. (2) 
 If p is not 2,3, 7 or 17, we may divide out (m + 4) (m + 3) (m + 2) 
 (m + 1)m, multiply by 5*.7! and replace 5(m — 1) by — 8, ete., giving 
 2104 (57)* — 8 .140a' (57)? + 8. 13. 21a (57)? — 8.13 . 18a’ (57) + 23. 260° — 0. 
 If 5; = @’, this becomes 
 
 a’ (210 — 8.140 + 8.13.21 —8.13.18 + 23.26) =—0, 
 
 in which the coefficient of @ is identically zero. If 5; = 2a’, the coefficient 
 of a’ reduces to — 10. Hence in this case 4 = 7 = 0. 
 
 Thus for p” not 2”, 3", 7, 17" or 18, the only possible quintic on py” = 
 5m + 3 letters is reducible to 
 
 SIS Ya ST aie 
 
82 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 32. For p = 13, n = 1, I have elsewhere* shown that the only suitable 
 quintics are 
 56° + 5u& + a&, a@ = arbitrary. 
 5& + 5a& + 2a, a = quadratic non-residue of 13. 
 33. The case p” = 17" = 5m + 3. 
 The value 57 = 2a’ is rejected since the equation 
 be? - halt = Aas = 0 
 has a solution + 0 for every a when p = 17. Thus 
 5 (€* + a& — 3a’) = 0 
 has the solutions ¢? = 44 and — 5a. Now — 5isa not-square in the G/’[17'] 
 and hence also in the G/’[17"], x being odd. Thus whether a be a square or 
 a not-square, we have a solution ¢ + 0 belonging to the field. 
 
 34. The case yp" = 7” = 5m + 3d. 
 The condition (2) of § 31 becomes 
 
 5.8. 9a, — 230 = 0. 
 
 Thus either == 0) ory ion, 
 
 But the power m + 6 of © + 7€ requires 7’ = 
 
 The only possible form is thus 5& + 5aé* + a6. 
 
 35. The case p” = 3” = 5m-+8. 
 
 The condition (2) of § 31 becomes since m — 21 (mod 27), 
 
 Oe ; 200°? - oh = a + ii — O . 
 
 Hence 57 = a’. 
 
 36. The case p” = 2” = 5m + 3, supposing x > 3. Thus 5m — — 3 
 (mod 2’) or m — 25 (mod 128). 
 
 The powers m + 2 and m + 4 of & + a& + B& + 7€ give 
 
 7 ay + af? = 0 
 at +a=0. 
 
 Heneaif.a == 0,7 =.0 Sila 2100 eee 
 But the power m + 22 of & + #¢ requires 
 
 Cue ‘ jes — p® — O : 
 
 The only possible form is again 5& + 5a& + a. 
 
 
 
 * American Journal of Mathematics, vol. 18, p. 213. 
 
DICKSON. 
 
 ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 83 
 
 37. The case p” = 2°. 
 The third and fifth powers require respectively 
 
 Yta@ayrit+ af = 0 
 TY ay' + a8 ==1()7 
 
 of which the former is the square of the latter. Hence either 4 = 7 = 0 or 
 
 Die pate 
 
 But & + f¢* vanishes for ¢ = f®, every mark in the GH [2°] being a 
 
 cube by § 18. 
 
 By replacing = by a'*¢ and 7 by a’, the quintic 
 
 becomes 
 
 = 
 - 
 
 Y (é) == ra aE ae -- (a*y ae aty*) &2 at ve 
 
 OUP +O + +E + rh 
 
 Hence the quintic ¢(€) will vanish only for ¢ = 0 if the resultant of 
 
 
 
 P+&E + (747) 4 7é and & = 1 
 
 is +0. This resultant may be written as a cyclic determinant whose first 
 
 row is 
 
 Ore Oe ee 7 eer Oe 
 
 On expansion it becomes 
 
 
 
 rete tae fae he DG aH)» 
 
 
 
 which is + 0 only when 7 = 1. 
 The only suitable quintic on 8 letters is thus 
 
 e 4 06° + ws. 
 
 38. Summary of $$ 51-37. The only possible substitution quantic on 
 p” = 5m + 3 letters is reducible to the form 
 
 Bot bak oe, a arbitrary , 
 
 except for p” = 13 when we have the additional form 
 
 Dee tOUs 20, 
 
 a being a quadratic non-residue of 13. 
 
 39. Theorem. 
 
 5 + 5as3 + oS, where «a is an arbitrary mark of the 
 
 GF'{ p], is suitable to represent a substitution on its p® marks, if p is a prime: 
 number of the form 5m + 2 and n is odd. 
 
84 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 For if not we must have for two different values of €, say 7 and g, the 
 following equation : 
 
 57? aN Sax? SE an — dy? a 5ag® =" ao 
 
 or 
 (9 — 9) {50 eet IG Ora Ne Para erate earn 
 Hence since 7 ¢ ¢, 
 Sift Ee +e +ae+yg+¢)p+e=0. (1) 
 
 (a) Suppose p > 2. 
 Making the substitution 
 
 
 
 i oe i ee 
 5 {5M + LOR? + pt + 8a”? 4+ av?} + & = 0. 
 Multiply by 16 and substitute 
 202 = 40—a, 4 = 46 —a4. 
 We thus reach the simple form , 
 
 16 (eo? + 1006 + 50°) = 0, 
 or 
 
 (0 -|- 5a) = 2007 = 5 (20) F 
 
 But* + 5 is a quadratic residue of no odd number of the form 5m + 2 
 or 5m + 8. Hence 5 is a not-square in the G/'[ p"], n being odd and 
 Dp = OM. seas 
 
 (b) Suppose p = 2. 
 
 Making in (1) the substitution 
 
 Pa Para a reals 
 M+ ah + a = (Ph + pt a). (2) 
 Put ? = v and multiply through by » + pw: 
 Ytar+av=w + a/+ en. (3) 
 
 But, by § 18, = is suitable to represent a substitution on 2” letters, 2 odd, and 
 hence is also 
 
 §+a~pt+@[8 4 a 4+ @. 
 
 Hence (3) has no solution in the G/’[2”] except » = yp, when by (2) we find 
 
 y= pL == @; 
 
 
 
 
 
 
 
 * Gauss, Disquisitiones Arithmetical, Art. 121. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 85 
 
 Hence if (1) is satisfied, we have 
 
 2 
 
 Ths a Ue 
 But this set of values for 7 and ¢ is impossible since 
 
 wo + afw + a0 
 
 has no root in the GF'[2"], n odd, if a+0, i. e. if 7 +g. For writing 
 w = @4, it becomes 
 
 eee ee OF 
 
 We thus need only prove that # — 1 = 0 has no root other than 6 = 1 in 
 the G/F’ [2"], n odd. But if there exists a root + 1 of 
 
 Ce 1) 
 (and thus having the exponent 3) which is also a root of 
 
 Goan Lea 
 then* must 2 be even. 
 A general theorem comprising the one here proven is given in § 54. 
 40. The case p” = 5m + 4. 
 The power m + 1 of & + a& + BE + yé gives a = 0. 
 The power m + 2 of & + fs + 7€ requires 
 
 Geen Oe, 
 Hence if p + 2, + 3, we have fy = 0. 
 The power m + 3 requires, if p” > 9, 
 (Pah gap aa reed ott UF 
 Now m + 3 is divisible by p only if p = 11, when p” is not of the form 
 
 5m + 4. If p = 2 (and thus n S 6, we have m — 12 (mod 64). Hence for 
 
 every p” > 9, 
 SF pice =e oe 
 
 Thus for p + 2, + 3, 8 = y = 0 and © is the only suitable form. 
 41. The case p” = 3" = 5m + 4,2 > 2. 
 The power m + 4 requires, since m — 1 (mod 9), 
 
 oh4 1087 = B= 0. 
 
 Hence © is the only suitable quintic. 
 
 
 
 *Moore, 1. c. § 46. 
 
86 
 
 tively, 
 
 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 42. The case p” = 37. 
 
 The fourth, fifth, and seventh powers of ¢ + f€ + 7€ require respec- 
 
 rdit+r)+F#=0 
 Pau+/7)=0 
 aad +7) =0. 
 
 The solutions of these equations are 
 
 B30 Sy s=0 Gand. OFS 0 eal eee 
 The suitable forms are thus 
 
 SUC CNY ete igh a 
 
 since if the latter vanished for ¢ + 0, €* = 2. Indeed # + 23° represents the 
 literal substitution, 
 
 (0) (1,1 =e O19) (= 0 ee eee 
 
 43. The case p” = 2" = 5m + 4. 
 The power m + 11 requires, since m = 64/ 4+ 12, 
 
 Cee F has == Bon ee 0) : 
 
 Hence § = 7 = 0 and © is the only suitable form. 
 44. Summary of $$ 40-43. © is the only reduced substitution quintic on 
 
 nr 
 
 Pp 
 
 = 5m + 4 letters, except for p” = 3’ when we have also ¢ + 2c. 
 
 45. & + af + BE + 7s? + O86. 
 p” = 6m + 5, since here p is prime to 6 and since p” + 6m + 1. 
 
 
 
 The power m + 1 requires 4 = 0. 
 The power m + 2 of & + fe" + 7& + OF requires 
 
 230 + 7? = 0. 
 The power m + 3 requires, if p” > 11, 
 6y0" + m (20 + 3/7”) = 0. 
 The power m + 4 requires, if p” > 17, 
 
 o +2 (20pa* + 208/49 + 7°) + M4) emo + 15044) = 0. 
 
 (1) 
 
 (2) 
 
 (3) 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 87 
 
 Substituting 7° from (1) into (2) and (3), 
 
 670? — 4mf?o = 0 (4) 
 A OWbie oss m\(m — 1) a5 
 eH ae Bd? — 4m il 1) os = 0. (5) 
 I 5 : 
 
 From (4) and (5) 
 M4 4m (2m == 1) 
 Oa 
 5 
 
 od = 0. (6) 
 
 Multiplying (1) by m° and subtracting from (2), 
 670? + 2m??? — 0. 
 
 Ii p = 5, m = 0 (mod 5), so that yo? = 0 and then by (6)0 = 0. Ifp: 5, 
 suppose; +0. Then 
 mPr = — 30. 
 Substituting this in (5), 
 230* — 4m(m — 1) f'0 = 0. 
 Combining with (6), 
 
 (41m — 18) ot = 0 or 3130 = 0. 
 
 Since 313 is a prime not of the form 6m + 5,¢0 = 0. Hence must 7 = 0, so 
 that fo = 0 and by (6) d= 0. But the power 3m + 2 of & + f& gives 
 get? and no other term with exponent divisible by 6m + 4. Hence there is 
 no suitable sextic when p” > 17. I have shown* that p = 17, = 1 leads to 
 no suitable septic ; while p = 11,7 = 1 leads to the following substitution 
 quantics : 
 
 A preliminary study of septics, $$ 46-50. 
 ay £5 Q&4 We Ne2 oe 
 Set Ose ee oun 11 oe rr Okeer tet, 
 
 46. The case p” = Tm + 6. 
 The power m + 1 requires a = 0. 
 The power m + 2 requires, if p + 2, 
 
 Tpe + Tyo — B= 0. (1) 
 
 
 
 ~ American Journal of Mathematics, Vol. 18, pp. 216-217. 
 
 
 
88 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 The power m + 3 requires, if p > 5 and p” + 13, 
 
 9 9 a2 la¥. Dw) 20 gt ia la 
 (xe? + Oe) — 7 (6Pre + 38/0" + 68770 4 9 7) + 9 Cer (2) 
 
 
 
 Rejecting the special values, 2, 5, 13, etc., we may prove that if any one 
 of the four coefficients /, 7, 0, ¢ be zero, then all are zero. To handle (1), (2) 
 and the very lengthy conditions given by the powers m + 4 and m + 5; 
 when f, 7, 0, ¢ are all + 0, is perhaps impracticable. 
 
 Suppose, for example, 7 = 0. Then (1) and (2) become 
 
 Thus =-0 4 foriiynot 
 
 Te 32 and thusggy—= of - 
 
 The power m + 5 of & + fé + e€ requires 
 
 Teo — 7! 15 Ret + 73. 6522 — 7, 180 fhe? + a 27 g, _ 13.51 
 
 - Pe — 14 Baie) 
 If # + 0, # = Te and the last equation becomes — f° = 0. Thus if 7 = 0, 
 then 8 = 0 = ¢ = 0, certain values of p” being excepted. 
 
 47. The case p” = 7m + 5. 
 
 The power m + 1 requires 6 = 0. If either a, 7, or ¢ be zero, we can 
 prove that all are zero. If 0 = 0, the conditions given by the powers m -++ 2 
 and m -++ 4 (of 5 and 19 terms respectively) are satisfied by 
 
 Cte) 2G" ite aes. 
 48. The case p” = 7m + 4. 
 The power m + 1 requires, if p + 3, 
 7 = 20". 
 
 
 
 We may prove that if a ='0, then B= jy =6 =e = 0; thatil p — 0 then 
 0 =) andre = 0% 
 
 49. The case p” = Tm + 3. 
 
 The power m + 1 requires, if p ¢ 2, 70 = 3af. 
 
 We may prove that if a = 0, then B = 7 = 0 =e = 0; if 8 = 0 then 
 0 = 0 and the conditions given by the powers m + 2 and m + 4 (containing 
 seven and twenty-one terms respectively) are seen to be satisfied by 
 
 \ Yemen Lee ees 
 50. The case p® = Tm + 2. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 89 
 
 The conditions are quite unwieldly even when a = 0. If 3 = 0, then 
 GeO ma AOS eS Oe 
 
 Quantics with an infinite range of suitability, $$ 51-56. 
 
 51. We have found that the quintic 
 
 b=” 4 bak? aE , 
 
 a. being an arbitrary mark of the G/’'[ p"], is suitable to represent a substitu- 
 tion on its p” marks, if and only if p" be of the form 5m + 2. Also in our 
 preliminary survey of septics, the quantic 
 
 CE ae 2. Take? 4 ae , 
 
 a. being an arbitrary mark of the G/'[ p"], stood out in a prominent way as 
 probably suitable on its py” marks if and only if p” be of the form 7m + 2 or 
 7m + 3. Note further that there is no suitable cubic other than €. Thus is 
 suggested the possible existence of a quantic of odd prime degree & which is 
 suitable to represent a substitution on the marks of every G/'[ p"], except 
 when p” is of the form Am + 1. 
 
 52. Suppose the reduced quantic belonging to the G/’[ p"], 
 
 Si SAC a ks es Co eae SiR ad a (1) 
 
 whose degree / is an odd prime number ¢ 7, is suitable to represent a substi- 
 tution on the p” marks of the field for every p” of the form Am + 2, km + 38, 
 km + 4, ..., or, km + (k — 3). We do not at first assume it suitable on 
 p" = km + (k — 2) letters, in which case the power m + 1 requires 
 
 (m4 Ly a, == a, == 0, 1f pt 2. 
 For p” = km + (k — 3), the power m + 1 requires 
 
 (m +1)a,+ (m = 1)m fit ee A), 
 
 or, if p + 3, 
 k= 33 
 2 
 
 9 
 hie Ce 
 
 For p" = km + (k — 4), the power m + 1 requires if p ¢ 2, 
 ka, = (k — 4) aa, . 
 For p" = km + (k — 5), the power (m + 1) requires, if p + 5, 
 
 _ k(k—5) 
 
 : ger as iy 
 Ka, 9 (45° + 20,44) + ( Z ase ) Gees 
 
90 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 or 
 2 A) : kik —5 
 ; kG, = ( kts E ») Os ae ( 7 5) Oy 
 For:p” = km + (& — 6), the power m + 1 requires, if p + 2, + 3, 
 ka, — k (k —B6) (020 + O04) + (ie) ee ay eer Ae 
 or 
 
 Ka. = ( 
 
 aes 
 
 k — 5)(k—6) _, 
 9 a. 
 
 Similarly, for p” = km + (k& — 7), the power m + 1 requires, if p ¢ 7, 
 
 (ees 31.2) @—T) 1 4 b= 8) UB a 
 
 Ka, = 
 
 for p" = km + (k — 8), p ¢ 2, the power m + 1 requires 
 
 io, — 4 —9) © cil (k= 8) p54 4 Bk =e ws BS) 
 
 for p" = km + (k — 9), p + 8, the power m + 1 requires 
 
 5 he 
 ee ee ( Cate 
 
 ae (he — 6) (eB) ee) 4 k— T){k— 8) (k—9) 
 ee 2.3.4.5 4 
 It would be impracticable to attempt to calculate the general coefficient in this 
 way. Itis to be noted that if p > & every coefficient is expressed uniquely 
 in terms of a, and 4. 
 53. The sum s, of the /th powers of the roots of the cubic 
 e Os 
 
 is given by Waring’s formula thus : 
 
 DY m(A or A, sic A, — 1) r ne 
 Bae tin p ye  - 285} 3 
 * acre (Apes ea SPOS SES (3) 
 
 where z (¢) = ¢! with the convention that z(0) = 1, and where the summation 
 extends over /,, 4,, 4, such that 
 
 peas Say yee Bee 7 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 91 
 
 Arranging according to descending powers of €, we have 
 
 > feo) i ‘ e383 
 Pia Se dace se “Egy 
 
 apa , K—A = 
 azgk—t 1 © gg Eb-s 
 
 ke 
 
 
 
 
 
 
 
 Ne es ah eek Os deal See Ie 2y,Ek—7 
 ba oth 0) (el k —6)(k—T7 2) ses 
 iE DIO, er WED caste 
 k — 6) (k — 7) (k—8) k — 7) (k — ay 
 ESSE DE) 3, PDR 8) 0} on 
 
 (k — 6) (k —1) (k —8)(k—9) 5 , (E—T)(k—8)(k—9 
 = BI A ho 4h 
 
 Thus the first ten coefticients in s, are exactly the corresponding coefii- 
 cients of the quantic (1) as calculated in § 52. A complete identification of 
 s, with (1) will be carried out for the most interest case, viz, when a, = 0, 
 which happens when the range of suitability of (1) excludes only the com- 
 binations p” = km + 1. For then by § 54 the quantic s,(€) will satisfy the 
 conditions derived by the method of § 52 which have an unique solution in 
 terms of 4. 
 
 For a, = 0, (3) reduces to 
 
 Pe (k —1—1) oe. 
 
 AG Zest \ ; Ek-2l 
 
 i eh esa i 
 
 Thus 
 
 Cee ead (i ba) i(k cae eat) Ek a 
 
 k ay 0 ae lh > 
 ein ee TA c= a Dea a 
 
 (5 
 where 7, and 7, are the roots of the quadratic 
 
 7 — iy —7 = 0. (6) 
 
 a 
 
 54. Theorem. Zhe quantic 
 
 (k—-1)/2 i af : 
 AG ajae+ oP EID 20+ 1) aera 
 
 where k is any odd integer® not divisible by p, and a any mark except} zero of 
 the GF'[ p"), is suitable to represent a substitution on its p” marks, if and only 
 ego: be be relatively prime to k. 
 
 * kis ast necessarily prime. ‘The proof in § 52 that 0, .(&, @) is the on ly quantic w Ay ‘the range 
 
 of suitability p" = km-+ 2, +3,...-+ (k — 2) requires that & be a prime number. 
 +See § 18. 
 
 ) Oia Sager it Ae 
 
 (4) 
 
92 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 We are to prove that under the named restrictions 
 0.8, a) = 8 (7) 
 
 has a solution € in the G/’[ p”], § being an arbitrary mark of that field. 
 Now by the transformation 
 
 Coal : Hone Aee a. / 
 Sa hart (6) 
 the quantic is given the form 
 ee la ; 
 Coie a (5’) 
 Thus equation (7) becomes 
 77* — Byk — at = 0. (7) 
 
 Substituting Y = 7, this becomes 
 Y?— BY +4 (—af¥=0, 
 
 which belongs to the G/'[ p"] and is for every § resolvable in the G/’'[ p*”"] but 
 not in the GF'[ p"). Call its roots Y and Y. 
 _Now the equation 
 pn Ve 
 is solvable in the G/F'[ pp], Y being an arbitrary mark of that field, if and 
 only if p’” — 1 be relatively prime to 4. 
 Then if 7 be a mark of the G/’'[ p*"| satisfying (7), we must prove that 
 
 falls into the lower field G/’'[ p"]. Since Y and ¥ are conjugate marks with 
 respect to the G/'[ p"] whose product is (— a)*, we have 
 
 = — 4. 
 Hence 
 
 f= 9 —a/gaygt+a=F 
 so that indeed* € belongs to the G/'[ p"]. 
 55. The algebraic roots of (7) are 
 
 pene ge” (iON Lea ey 
 
 where 
 ots h ila/2 el Gaara 
 
 : * Moore, 1. ChsipLs 7 
 
 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 93 
 
 and ¢ is a primitive “th root of unity. This is a straight generalization of 
 Cardan’s formula for the roots of the cubic and also of Vallés’ solution of the 
 quintic* 
 
 a , 
 at Pa). 
 
 m we Oa? = 5 
 
 It is evident that the algebraic solution of (7) for a and arbitrary is 
 equivalent to the extraction of the Ath root of an arbitrary complex quantity 
 and hence equivalent to the partitioning of an arbitrary angle into /# equal 
 parts. 
 
 56. The study of the quantic ¢, (€, a, a,) derived as in § 52, or conjec- 
 turally (when a, + 0) as in § 53, is made here only for the case a4, = 0. It is 
 to be expected that, for a, + 0, it is a substitution quantic at least for certain 
 special values of &, p, n, a4, and a, Thus if & = 5, p = 7, we find 
 
 oo oe = Gas “ts BOy Ss 
 which by § 30 includes the three types of quintics suitable on 7 letters, as well 
 as the one suitable on 7” letters, » being odd. 
 Section III.— Degree a Power of p. 
 
 Quantics with all exponents powers of p, $$ 57-59. 
 57. Theorem.t The reduced quantic 
 m & : 
 ¥ (x ) = D: A,X a Re 
 it 
 belonging to the G/’[ p’"], will represent a substitution on its p”” marks if 
 and only if 
 
 7(X) = 0 EL} 
 
 has no root in the GF'[ p””] other than XY = 0. 
 For it will be a substitution quantic if and only if it be impossible to find 
 two different marks A, and X, of the G/'[ p””] such that 
 
 1(X,) = 1 (44) 
 Veet 0. 
 
 or 
 
 Corollary. A?" — AA” represents a substitution on p”” letters if and 
 only if A = O or A isa not (p — p") power in the Cie. 
 
 
 
 *M. F. Valles, Formes imaginaires en Algebre, Vol. i pp. 90-92, 1869, 
 
 + I reached this result independently. Cf. Mathieu, 1. c. Vol. 6, 1861, p. 275 ; also Betti, 1. ¢. 
 Vol. 3, p. 74, 1852. For the connection with linear substitutions see Part II, Section III. 
 
94 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 58. Since every mark except Y = 0 of the G/'[ p”"] satisfies the equa- 
 tion 
 
 Xr eee (2) 
 
 it follows from § 57 that y(X) will represent a substitution on p”” letters if 
 and only if the resultant of (2) and (1) is different from zero. This resultant 
 is 
 
 A 1 ? A 2 ’ ae A m 
 ¥ 
 AD See oA 
 UP os A ae ee eee (3) 
 nim—)) nim—I1) n(m—1) 
 a in ’ A ‘ ’ ae) A ain —1) 
 
 The proof is analogous to Sylvester’s dialytic method. I set up m equa- 
 tions linear and homogeneous in the m quantities 
 
 >) n(m—i) ° 
 ae —1, (Zens A tae?) 
 
 such that the system of 7 equations is equivalent to the system (1) and (2). 
 Thus, raising 7 (X ) to the p™ power, 
 
 si A ~ DC kant! = 0 . 
 
 i=l 
 
 Applying (2) multiplied by X to the first / terms, 
 
 Ft ica pa > yrk—i) \" pak Yrpnntk—i) 
 _ A i a + ad A 2 oA — 0 . 
 ii] : i=kh+1 4 
 
 Introducing in each a new summation index, 
 at 
 
 m—k E 
 \' pk i 7 ym—J) v prk } x pnim—)) 
 J ~ Pek ste om Xx +- a OL jtk va —— 0 ’ 
 j=m—k+1 j=1 
 
 Combining and replacing the index 7 by 7, 
 v k 7 mWn—1) 
 = Oe Rails Gey Ged Ue (4) 
 = 
 
 where, if 7 + 4 > m, we are to understand 
 
 , A isp — Agia, : 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 95 
 
 Dividing out. X from (4) we obtain for & = 0,1, 2,..., m — 1a system 
 of m equations equivalent to the system (1) and (2), since from the former we 
 can pass back to the latter. 
 
 59. Owing to the objection that may be made against the usual proof of 
 Sylvester’s dialytic method of elimination, the following proof that (3) is the 
 resultant of (1) and (2) is given. 
 
 Lemma.* The necessary and sufficient condition that m marks 2, %,..., 
 2, of the GF[p””"] shall be linearly independent with respect to the 
 GF'| p"| is that the determinant 
 
 O O ( 
 
 oF es ae ees, 
 
 QQ pn QQ p” Qp 
 
 AoE peed, ad eegige rae ee 
 
 QO p» QO p> Op —!| Qpr | 7 5 
 Coy > Ih | ) Ct Gs of ts ea | ee bly \9 () —— 0, hi LE) Le —— 1) (5) 
 O pran—1) O prm—1) O prm—1) 
 
 — 0 9 -_— 1 eT el BAe _— m—tl 
 
 shall + 0. 
 It is sufficient; for if 2, 2, ..., 2,_, be linearly dependent in the 
 GF’ [ p"),i. e. if a relation : 
 
 peepee (6) 
 
 holds where 7, 7, ---; 7m—1 are marks of the GF'[ p"] not all zero, then will 
 the determinant (5) vanish. 
 It is necessary ; for if (5) be zero, write 
 
 m—1 
 = 9 > j ee 
 Repairs Syd eR! Cea OF ee i) 
 
 iO 
 
 where /? is a primitive root of the G/’[ p””"] and the y,;s are marks of the 
 
 ie 
 Gio [pela Len 
 Co enon. ble is (ig == 0, 1. me — 1) 
 
 
 
 *A more general theorem is given by E. H. Moore, A two-fold generalization of Fermat's 
 theorem, Bulletin of the American Mathematical Society, second series, Vol. 2, April, 1896. 
 
96 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 where, writing in full the determinant in /?, we have* 
 
 ager ul Re fer 
 
 Sie ; fr : ¥ pe 
 
 PR? Re pwn) = || CPs — Wrap) 
 b] | oe eee ‘ 
 
 Sot aU, le ee eS 
 
 eS t 
 es face aks ne Roeper) 
 which ¢ 0, # being a primitive root in the G/’'[ p™”"]. Hence | »,,| = 0, so 
 
 that a linear relation (6) exists, in which not every 7; is zero, or 2, 
 2,,, are linearly dependent in the GF'[ p"]. 
 The condition on A,, Az, i254, that 
 
 ta Ip pe) 
 
 a ee A" > ymin—i) 
 aN Mies i A,A (8) 
 shall represent a substitution on p”” letters is the same as the condition under 
 which 7 Ay,..., 4,’ shall be linearly independent with respect to the 
 Gi[p")|, when it is given that 4), X,,..., X,, are similarly independent. 
 By our lemma, the latter condition is 
 
 | AG S25 (597 == 0, 18 oe 5 Me 1) 
 Applying (8) this determinant becomes 
 
 >) nim— 1) > n(m—?2) r > ,n(m—1) r 
 Aix? ys A,X? +- eee + ASA eae JA iA? -+- aie at + BAe 
 
 ny n pprum—i n Vpn an a ? 
 AY Xx, ee A PN OS ee SA A es ee ee eee 
 
 Db m7 
 
 
 
 Ae Yom = a omy, 1 etd YPN Aap Nae eee 
 
 m m ML 
 
 which equals the product of determinant (3) of § 58 by 
 
 yar % 
 
 7 »ynm—)) Me prm—l) 
 er 2 > ee ey m 
 
 1 3 
 
 ” 
 “pr ym—2) 
 Ad 
 
 Aue _9 
 yee 2) y pas 2) 
 41) ’ “12 ’ PL m 
 
 
 
 CEE TP GIT PG ats ee 
 
 * Baltzer, Determinanten, p. 85. 
 
 
 
 
 
 
 
 
 
 
 
DICKSON. ANALYTIC REPRESENTATION 
 
 U7 
 But by (5) this ¢ 0, Ay, ..., 4, being supposed linearly independent in the 
 Cel pg" |: 
 
 OF SUBSTITUTIONS. 
 
 General theorems on SQ [ p”; p"], $$ 60-68. 
 60. In studying the quantic belonging to the G/’[ p”], 
 
 jie aa Epr—k iL (e 
 = = Oy? re Bigs? =F og ee = i DyyAS 5) 
 
 in which «, + 0, and & < p” — 1 it is found desirable to compute the power 
 ite = hire +. re = Any. =I Di) =| hs 
 where ¢ is the least positive residue of x modulo 7 and 4 < p’*’. 
 Let the general term of the expansion be 
 
 (EP"\%o . (4,6? *)% : (ayer en ; 
 
 Cpe) leat 
 where we thus have 
 
 © PO Ga +. + Oy y= pp’ + kp s+... th, 
 
 (1) 
 9= pa, + (p'— hk) a + (p" — & — 1) Gey +... + Ayr = p" 
 
 - —1. (2) 
 I shall use the abbreviations 
 & = pay + (p" — &) ty + Ops + Apyg +--+ Opa 
 S5= pty + (p" — Bae + (PP — 1) (ten + deg + + ys) 
 61. By equation (2) 
 
 dy = ( p” es 1)/p" = poe ; 
 Hence by an application of § 14 to (1), 
 
 Be gD ily sea ee el dae eee ea Np 
 Suppose at first 
 
 Ny = k : yale = if ; ee ! : 
 where 
 
 
 
 tee) town ott 
 
 ire ee ts ee Ee oh hy. 
 Then by (1) 
 Gy = Opa pe se A Ope = pO 
 If a, = p”" + y, where 0 < y < 2, we readily find 
 
 8, =p” + kp + kp +... + Ap’ —(p"— let (p’—k—ly 
 
 Sp t+ kph? +...+ hp" —pe>p —1. 
 Hence s 5 
 
 > 8 > p” — 1, contrary to (2). 
 
98 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 lf a, = y, then 
 
 
 
 8 = (p—YDp rth. pr t+kp’4+...+hp—(k+1)e+y. 
 
 Hence s < s, << p" — 1. 
 Applying § 14, we conclude that a < Ap" *’ and thus 
 
 a =< (kK—1)p*" + kp? * +... +k. prt +h. 
 62. We may prove by induction the theorem 
 a Z(k—V (wee ee 
 Thus, to make the general step, suppose 
 by =k 1) Ce ae ee | 
 so ea ay Onegai acim y Meth BI) <2 
 
 where 
 0 2S a! = ke ( Cae. + yp cs 2 a eae — pido ae A ‘ 
 Then ; 
 Ly, + nay + ieee a Gard — oad a8 Daan + “aie —- y Lee § a. a ‘ 
 Ife, =p? tp 4p” sy, 0 a ee then 
 By OP cee pe a DO a MT ee 
 
 Hence s Ss, > p"— 1. 
 If a, —. (is | yma ms et a Daan | Ys 
 
 oe = 7 al + ki Cp ees + Pe ee + yaaa + hp” ee (hk + 1) a’ ae y ‘ 
 
 Hence s = s, < p”® — 1. 
 
 If a, have a value less than that last supposed, much more will s be < 
 p" —1. Hence 
 
 a, <(k al} (gr Eg pe ) Eee 
 Hence finally, 
 ay (k—1) (po 4 2 pe he eee eee 
 63. Let g denote a positive integer and write 
 dy (bh L) O™ y aiit) ea (3) 
 
 Here g is not a multiple of p, since then would also @, + a4; +... + G1 
 and by § 14 s would likewise be a multiple of p. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 99 
 From (1), (2) and (3) we readily find 
 Op Oya oe or ApS Pe pn eet pag (4) 
 (p —kh)a, + (p" —k Vga t+... + a4 =p" —1 — pra, 
 ee eee aa Pt ee (ep + gh) 1 « * (5) 
 Multiplying (4) by p” — & and subtracting (5), 
 Ops + 2dgre + BQy43 +... (p" —k— 1)a,_, =p" (A — kp) — kg +1. (6) 
 Thus 
 
 
 
 g <i p (h — kp!) +1} /k 
 chi Unless fa ha 1 hes 0: (7) 
 
 The calculation of the condition given by the power 
 Tae =f ie (poe a goes = ete = pare an h 
 consists in taking for g in turn each of the values* 
 7 eee fs t 
 fossa 2 ee Ee 
 L J 
 
 and determining by (6) every possible set of values for p41, @p4o) 26+ © Gyr—t 
 which form a partition of 
 
 jos i =e ees = bor { peg g 
 
 of the kind required by § 14. Then (4) gives the a, and finally (8) the a, 
 corresponding to each set. 
 64. The value g = 1 may be discarded if 
 
 he see (ke Hey 1) gf. 
 
 provided £ > 1 when the equality sign holds. 
 For by (8), 
 Ay, Say # pani “f- youre’ a ay und: 
 and hence by (4), 
 ay ey a8 gosta? ae x See ae go - i} : 
 Then 
 
 3; = pe oad 1) Cpe + | et + ts a atl ot hp” — kh =e ee Co Ss k — 1) ; 
 
 Hence 
 8 = fe <p" ais hp” a (k a 1) prt peat fs <p 
 if 
 Bee (eel peor hak 1) pk > 1. 
 
 * H(«) denotes the greatest integer in @. 
 
 
 
100 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 65. Theorem. Jf ¢ = 0,i.e. n be a multiple of r, the coefficient a, = 
 
 For p" = 2, € + 4 is suitable on 2” letters (by § 57) if and only if it 
 vanishes only when ¢ = 0i.e. if a4, = 0. 
 For p’ > 2, consider the power 
 
 aD ae =e pi a me ae p ae ik ; 
 
 Making 4 = £ = 1, ¢t = 0, we have g = 1 and hence 
 
 
 
 @= 0) aS Pp" 7 |] ea DO Le Og 
 
 The condition is thus 
 Par see fp ti ==) } 
 
 66. Theorem. Jf the quantic belonging to the GF [ p"\, p > 2, 
 
 7 . — “ " = . © . is 
 opr || £( pr—1)/2 &( pr—3)/2 je 
 g ft Dayr4 $ -+ Qe pr+3)Q -- Shane -j- Oyr3S 
 
 be suitable on p” letters, then Gran = 9. 
 Consider the power 
 
 
 
 ; ret ne a . 
 jie + 1Y eh Gy cease + Vb ea ar | Dae DP) f 1 , 
 Thus 
 ko(p st D2, p=2 eS pogo 
 Hence by (6) and (4) 
 Qisy = Oyyy Se Ope ee Og a ee 
 
 The condition is thus a,“ = 0. 
 67. Less frequently will be studied the power 
 
 g ube Al kh es au aan =} Ly: — pe a h : 
 
 where g and £ are < p" andi < p"*’. 
 Similarly as in $$ 61 and 62 we may prove that 
 
 Og <9 TO Oe at ia 
 Taking s = ¢ (p” — 1), we find as in § 63, 
 p= (g TD) BOL) ea hat ae) ty ee 
 Oks, + 2p, + 8Gny3 +... + (p" —k—1)a,_, =p" (h— kp’) —ko + 9, 
 
 where g > 0, g + 0 (mod p). 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 101 
 The parts of the condition given by 
 s=n(p"—l), n<g 
 
 can not be satisfactorily obtained by a similar analysis. See the direct method 
 used in § 73 (f). 
 68. Exactly as in $$ 61-63, we may prove that for the 
 
 saree ae ip ( Dhaene te Da Be peas =e yang ce (k ae Liges Pale h 
 
 power of the quantic (in which the term @,,¢”*"' is absent) 
 
 EP" + gpePr—k 4 gy Pe? + A dy 6, 
 ta =(k—1)(p™*4+...¢+ yp") + kp’ +h—g,951; 
 Or Ora Cee at Spa Pp ep + 
 
 
 
 2nry + Bang +... + (p” —k— 1) ap, = prt + hy” — kp’ — kg +1. 
 
 _ Determination of all reduced SQ (| p"; y"\ for p' = T and partially for 
 p’ = 11, §§ 69-74. 
 
 G2 p z= 2. 
 By § 65 we need only consider <*, which is suitable on 2” letters by § 18. 
 pee. 2. 
 
 By § 65 the form is & + «4,€, which is suitable on 3” letters by § 57, corol- 
 lary, if and only if a4, = 0, or — a, is a not-square in the G/'[3"]. 
 
 1. p=? & 4 af 4+ a6? + a.€ on 2” letters. 
 
 If n be even, a, = 0 by § 65. 
 
 If m be odd, then also a4, = 0. For if not we may remove the term < by 
 a linear transformation. Consider then the power 2”? -+ 2"™* 4+... 2°43 
 of 
 
 e+ ae + as. 
 
 Since a, S 0, it follows from (3) of § 63 that g = 3; also thatg +2. By § 64, 
 g > 1. The remaining value g = » gives in equation (6) of § 63, 
 AG ees es ced OL? gee LY 
 Then by (4) 
 a = 2? 4+ Qr4*4...4+3 42. 
 
 
 
 The condition is thus 
 o9n—2 93 
 Oe +5242 42 gn a), 
 
 Hence would 4, = 0. But ¢ + a,é vanishes for =a,. Hence a, ¢ 0 leads 
 to no substitution quantic. 
 
102 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 Our quantic thus becomes 
 >: “f- nS = CRS 
 
 which is suitable on 2” letters if and only if a determinant of the form (3) in 
 § 58 is + 0. 
 
 dene at oe 
 Cases depending on which is the first coefficient + 0. 
 (ayan: 
 
 Suitable on 5” letters by § 18. 
 
 (b) & + af, a, + 0. 
 
 Suitable by corollary to § 57 if — a, is a not fourth power in the GF'[5"}. 
 
 (c) & + a€* + a,§, a, + 0. 
 
 Rejected by § 66. 
 
 (d) & 4+ 46° + af, a, ¢ 0. 
 
 The coefficient of ¢? has been removed by a linear transformation. The 
 lowest power giving a condition is 
 
 fet 2 bt 2 een 
 
 
 
 Thus g = 2or8=# | 5-), since g > 1 by § 64. 
 J 
 
 For 9, = 2,0, 1p0,.= 5°70" = eer 
 for. g== 3,00 eas 
 The condition is thus (using § 15) 
 
 Qn-2 31 ae +e F541 pee» Qn—2 ge te F548 3%) 
 or 
 - 
 4, > 40,7 - 
 
 The quintic is thus ¢ (€’ — 2a,)*, which is indeed suitable on the 5” marks of 
 the G/’[5"], if 2a, be a not-square in the field. For if, when 7 ¢ g, 
 
 7 (7 — 2a,)’ = ¢ (¢? — 2a,)" 
 on dividing out 7 — 9g, 
 
 (7—¢) + a(7 + 4¢ + @) + 4a2 = 0. 
 
 
 
 Substituting 
 ZA Sf, CSL — Bs 
 
 (w — 2a,)? = 24,7? . 
 
 (e) The case a, + 0 is rejected by § 65. 
 yb Py se 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 103 
 
 (a) € is suitable on 7” letters. 
 (b) 67 + a,€. 
 Suitable if — a, is a not-sixth power in the G/’[7"]. 
 (oye ae a £0. 
 Consider the power 
 My -- ae — 2, : 71 + A(T*? + 7n—3 ap set 28 7) = 5 ; 
 If 7a, + 2a, = 2(7" — 1), then we find 
 
 Fo aa ey (aoe Ry fare 0 a 
 
 contrary to the method of partition required by § 14. 
 
 If 7a, + 2a, = 7” — 1, we find 
 
 ee ONT ee TOs a ee OOP OTP 87 4 8. 
 
 Hence a, = 0. 
 
 or 
 
 or 
 
 (d) & + 4,8 + 4,8 + a,€. 
 
 Rejected by § 66. 
 
 (e) & +. 4,€* + 4,67 + af, a, + 0. 
 
 For the power 7" + 3 (777? + 77% 4+ ...4+ 7) + 4,9 = 2 giving 
 
 “n—l by 
 ae aCe ha 2 a; = 0, or a; = 0 A 
 
 For the power 7"? + 3(7"? + ...+ 7) + 5,g =3, 4, or 5, giving 
 
 
 
 ee er (mA ae TES (fae ae AE 
 (31)"2. 5! Ae ae oie 6 Oe Coes ? = 0, 
 
 ? Qn on—2 3) ae Qn-2 5 5! ) 
 
 
 
 (lie oo ace (4, —- 2a") = 0 ; 
 The only possible form is thus 
 & (&° — 3a,)’. 
 
 () 4 af 4 of + af + af, #0. 
 The power 777! + 2.7%? + 2.7"% + ...+ 2.7 + 3 requires 
 
 9 - —1 719 -n—I = ‘ ~n—1 - 
 2r 3tf{4 eee arta lai pe mee af Ser eee tls Meaney == () 
 Oga, + a2 + 2a,0, = 0. (1) 
 The power 7°71? 4 2.7"? 4+ 2.7" % + ...4 2.7 + 4 requires 
 
 ‘ 7ni—l - ¢ ¢ ¢ * 5 
 Qn Alat t- Masta? + aa,+4 aa? +440} = 0. 
 
 , 
 
104 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 (Disgaea re [s : fo : ; so that by the partition requirements we have 
 J 
 
 L 
 from (4) § 63 that g = / or 4.) 
 Applying (1), either a, = 0 or else 
 
 O20; = — Ont, (802 + 5a, (2) 
 
 vo 
 
 and thus — a,4, would be a square in the G/'[7"]. 
 Consider the power 2.7" + 2.7"*% + ...42.7+4 2. 
 For s = 2(7" — 1), we apply $67 for ¢ = 4 =k 2 oe ence 
 
 9g 1, 4, Sa, 6, =", 0) Soa ee eee 
 
 
 
 > - fe . wN—1 ‘ 
 The coefficient of 2 (7 — 1) is this 2%," > 
 
 For s = 7" — 1, we have 
 4t+ata+a+a=2(M1+...4+7+41) 
 Ta + 5a, + 3a,+ 2a, +4a,=6(714+...4+741))=7"—-1. 
 
 Hence 
 4a, + 2a, — a, — 2a,= 0. 
 
 Using the notation 
 
 iat SES hae oN tl (S="0, 2, 45: 6) 
 we know by § 14 that each ¢” is 0, 1, or 2 such that 
 10 (2) 4 5 ee. reehe 
 Cp” se 6 bee el oh Cia (gree Oe ett |) 
 
 We have immediately 
 
 46 + 26° — 6, — 2¢,° = Tm 
 
 m being an integer. But m must be zero. By induction, 
 
 46 + 22 — of == Deh == 0 2 tg OF ae) 
 
 Then c® = 0 or2. Butifc® = 2,c = ¢® = ¢® = 0 and the last equa- 
 tion is not satisfied. Finally, ¢ = 0.. Hence @, = a4, =—0 and a, = q,. 
 Then 
 
 2a, + & = 2784 4-7. ot 
 
 from which it follows that a, takes exactly the 2” values 
 
 nm = 5 ; 
 Be C. f ”) . 7) 
 jJ=0 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 105 
 
 where c;? = 0 or 1. We thus obtain 2” terms, 
 
 (48°) (63) (a8) 
 or 
 (Ay)? ( oe) gs a ree ora 
 
 Hence, as far as their literal parts, these 2” terms are identical with those 
 ’ b] 
 given by the expansion 
 
 (24,0, + geet ge att é 
 
 In order that the numerical coefficients be congruent modulo 7, we must 
 have by § 15, 
 
 
 
 Dt on Sy, (4) | 
 Sate (6!) 
 / j=0 > 
 But if Z of the ¢,°’s are = 1, / of the c¢;*”s are zero and hence n —// ure = 2, 
 Hence 
 R=) : 
 Le AN SS 28 
 
 5D 
 
 
 
 n—1 Fe {2) n—1 2) 
 Dee OU ae TT (Dynes OF 
 j=0 x 
 
 vi 
 
 The identification is thus complete. 
 The complete condition given by the above power is thus 
 
 Raine oS + (24% + Ciiene = Ue 
 Applying (1) and remembering that «@, + 0, 
 
 = (— may oo" EE ee (3) 
 ence 
 (= Pie Ete (25\" i eet i 
 
 so that — a,a, is a not-square in the G/’[7"]. Hence by (2) a, = 0. 
 The power 777! + 2.7"? + 2.77% 4 ...4+ 2.7-+ 5 requires 
 
 m1 n—2 2 7 ; . 
 ae i al {24° + 4a,0,’4, + Ga;'a, + a,°a,' + 4a,'a,0,' + a,'a,'a, 
 + 6a,°a7a,’ + 20,4,’ + 2a,'0, + @,"a,’ + 3a,'°a, + 4a,°} = 0, 
 
 the last four.terms not occurring when 7 = 2. 
 Applying (1) to eliminate «,, 
 
 4a3 + 40,70,’ + ata + 2a,°a) + 2a,8a,! + 50°F + 2a,4a, + 4a,°=—0, (4) 
 
 the last two terms not occurring when n = 2. 
 
106 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 For n = 2, we may give (4) the form 
 
 4a; Clore 2a,”) (a, — 40,”) (a, — 5a”) [4,— (1p 3) ay"| [a, — te 3) dy f= Oy 
 
 But by (3) . 
 (0,0, ) == 2a 
 
 Hence «a, is a not-square in the G/’[7’] and 
 aS 2a? . 
 
 Thus (+ 2)? = 4; 44 = 2; (1 + /3)* = 5, each modulo 7. 
 
 For n > 2, (4) may be written 
 
 4 (a, + 2a,”) (a — 2ufa,? + 3a,a,* + 2a,°)=0, 
 
 the last factor being irreducible in the G/'[7']. 
 
 By using § 68, I find that the power 
 
 (et 200? fo 2a PB ST 
 
 (4’) 
 
 gives for zn > 2 a condition of six terms which on applying (1) to eliminate a, 
 
 reduces to an identity. But the power 7"? +4 2.7"?4...4+2.7+3.7+4 
 
 requires for n > 2 
 
 -n—1 2 - 7 2 - 7 ooo x 
 a +...47 {4a,a,°a,' =e 2a,/a, | 40,270,085 a 645) 06a, eis 5a,‘a,'a,! - 34,4, 
 
 + 24," + a.a,/a"2 + 2a,'a,a,4 + a,/a,'° + 20,"a,'a, + 24,94, 
 + 64,"*a,’ + 3a,"4a,%a,5 + 6a,."a,"} =0. 
 Applying (1) to eliminate a, (the 4th, 6th and 14th terms cancel), 
 a, {6a,° + aa, + 3a,"4a2 + 4a,'%a,' + 3a,%a, + 42,%) = 0, 
 4 a, (a4, + 64,”) (a, + 2a,”)* = 0. 
 Hence by (4’) the only possible values, when 2 > 2, are 
 iy i 
 The same holds also for n = 2, since the set of values 
 bi = i DGS 0 = 
 is excluded by the condition, given by the 18th power, 
 
 Ut, + Ba, a0,' + 5a,)a,' + 34,5474, + Lala’ + 4a) + 3a,4,'4,° 
 
 + 5a,a,' -+ 3a,%a8a,! + a,'aya.° + 2a,'a,'0,' + 4a,%a,2 = 0, 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 107 
 
 which reduces to 6a," = 0 by substituting the latter set of values, is an 
 identity for 
 t= 20, a4,=— aS: 
 
 The only possible form is thus ¢ (€’ — 2a,)*, which may be proved suitable on 
 7” letters, if 24, be a not-square, by the method used in § 39 or as below. 
 
 NAcep cen Lh. 
 
 (a) Et A aye S. 
 
 Suitable if — a, be a not-tenth power in the G/’[11"]. 
 
 (b) " + af, a +0. 
 
 The power 3 .11"* 4+ 7(11"? + ...+ 11) + 9 requires a, = 0. 
 
 (c) &" + a,§* + a, a, t 0. 
 
 For the power 2. 11" + 6(11"? + ... + 11) + 8, we must have 
 s = 11" — 1 and then a, = 6(11"7 4+ ...+11)+ 8. We may prove by 
 induction that 
 
 
 
 
 
 ape Ali 11? + 11) 8. 
 Thus suppose 
 a =4(11"? + 117% 4+...4 11"") + 6(11" *'+...+11) +8 —a, 
 Oe ee FICS Olt * 4 a 11) 4 8. 
 
 
 
 Then 
 Cait ee ree 2 Lt tot aD LL te, 
 
 lide elie. ed. 1" fy. where 0 = y = @, 
 ere Le OL all) 8 Ll 10e Dy S110. 
 
 Tide et Le oe BLL eH Le ty, 
 
 Pe Let ere et) 6 Ie + 112) 
 + 8.11 —10e + 24y< 11"—1. 
 Hence 
 ieee ere LL ee Ol Lt 
 and thus 
 pee ee te LIP ee OI et 1d) + 8. 
 It is now easily shown that 
 Gee eee Li 1) a 2 119 + ... 4  T1 8, a, = 1, 
 
 as a smaller value for a, would require s < 11" — 1. The condition is thus 
 
 Hiliah «2 Syne en = (Pe 
 fe ty = 0 or a = 0. 
 
108 DICKSON. 
 
 ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 The power 2.11" + 6(11"? +... + 11) + 10 then requires that a, = 0. 
 (a) & $ an8! + a? + a8, @ #0. 
 
 The powers 11" + 7(11"? + ...+ 11) 4+ 9, 1177? +7(11"? +...+11) 
 + 10, and 2.11"'+ 4(11"7 + ...+ 11) + 8, require in turn, 
 
 a—=0, a =9, 
 
 (e) €" + a€° + ...is rejected by § 66. 
 (f) €" + 4€ + af + a6 + a€ + af, a, + 0. 
 The power 11"7 + 5(11"7 +... + 11) + 6 requires a, = 0. 
 The powers 117 + 5(11"7 + ...+11) + Tand 11"1'+4 5(11"° 4+... 
 + 11’) + 6.11 + 4 require respectively 
 
 tee is 
 
 =i 
 Gee ee (4y° + 2a.) + 5a;"a,) = 0 
 
 n—1 © 
 ip) FIG. (g” + 2ag%) + 4a,'a,) = 0. 
 Hence a, = 0 and then a, = 0. 
 
 The power 11." + 5(11"° +... +11) + 9 of &" 4+ a6 + a,§ requires 
 
 n—1 
 qj 1+... 4141 (q,, — 3a,2)t = 0. 
 The only possible form is thus ¢ (© — 5a,). 
 
 (g) The cases a, + 0, u, + 0, a, #0 I have not attempted. 
 75. Theorem.* 
 
 a, =, 0 by.8-60. 
 Lf d be any divisor of p" — 1, the quantic 
 & (E4 — y)(Pr—lyd | 
 where vis a not dth power in the GF [| p"), represents a substitution on its p” 
 marks. 
 We are to show that 
 
 &(é4 _ y)\(pr-Did — 8 (1) 
 has a solution ¢ belonging to the G/’[ p"], § being an arbitrary mark of that 
 field. The statement being evident when / = 0, we will suppose that f/ + 0. 
 Writing ¢ = ¢* — » our quantic becomes 
 
 gpd : (¢ aH y)ua = (g?” -b yor) ; 
 It is then sufficient to prove that 
 
 g + vy? = ft = 6 
 * From the results of §§ 70, 72, 73, and 74, for ‘Di = 3, Osa and Nal respectively, I venture the 
 ple type 
 
 (2) 
 conjecture that all substitution quantics of degree p suitable on p” letters are reducible to the sim- 
 
 
 
 E (€4 — y)(p—-0e 
 where d is a divisor of p — 1. 
 
DICKSON, ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 109 
 
 has a solution ¢ belonging to the G/’[ p"], ¢ being any dth power and » any 
 not dth power of the field. For if there be such a solution ¢ (which + 0, since 
 3 + 0), then 
 
 é — (¢ + y)ua = B/gP ne 
 
 will belong to the G/’[ p”| and satisfy (1). 
 Writing in (2) 
 ¢ — 1/o 
 
 and multiplying by w”’, we obtain 
 
 Lee sur Dw”. 
 If we make 
 
 r 
 
 OO Lee baat ar 
 
 0 being thus a dth power and Y a not dth power in the field, the last equation 
 
 becomes 
 Go = Oa 
 
 But this always has a solution in the field ; for by § 57, corollary, the quantic 
 wo? — Yo 
 
 represents a substitution on p” letters, »’ being a not ¢th power and hence a 
 not (p” — 1)st power, d being a divisor of p” — 1. 
 
 For 7 = n, this theorem is a special case of the following theorem :* 
 
 16. Tf ris prime to and < p” —1,tfsisa divisor of p" — 1, and if 
 Ff (&) is a rational integral function of = belonging to the GF '| p"| which can 
 never vanish, then the quantic 
 
 er fe) 
 
 represents a substitution on p” letters. 
 
 For if the quantic be raised to the /th power, / being not divisible by s, 
 we have a set of terms whose exponents are of the form ms + dr and thus are 
 not divisible by s and hence not by p" — 1. But if 
 
 bes isi p= 1") 
 we get the term =”, since by the hypothesis on 7 (¢*) we have 
 Lee) = 1. 
 
 But /7 is not divisible by p" — 1. 
 
 
 
 * Proved for 2 — 1 by Rogers, |]. ¢. p. 41. I give a modified proof. 
 
110 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 77. The substitution quantics given by § 76 and, when x < 7, by § 75 are 
 
 not reduced. 
 Thus, for p > 2, we have the suitable quantic 
 
 yp 
 
 mE pip ta : = 5 
 er(& 2 — yp)? = — Df EF —1/2(v+1/) 
 
 v being a not (p" — 1)/2 power in the G/’[p"] and thus any mark except 
 +1,—1,0. For the p” — 3 values of v, we get for each value of 7, (p" — 3)/2 
 
 different substitution quantics on p” letters. For if 
 ytlf=v4+1/V 
 
 then either »y = Y ory =1/. Alsovt1/v. 
 Examples of the above quantic : 
 
 == 1,” Dp Ses0 ean i 
 
 &¢+ 3¢ and €, + 2c, Hermite’s forms. 
 
 at eg a 
 Poms Lp aL ee NG ne Aeecsseao 
 nes 2D =o oe ee ae cuLsere seas 
 
 For the values » = 1, p = 7, we have if » = — l, 
 
 
 
 é (62 — yp)? = — By ( — v6 +4 82%) 
 & (E? — yp) = 2 {EE Ove 4B ue] 
 
 which together give the known quantic on 7 letters, 
 
 G4 gE 4 By, a = arbitrary. 
 
 Srorion LV.— Degree a multiple, but not a power, of p. 
 
 78. Attempting no general investigation, I will confine myself to the deter- 
 mination of all sertics suitable on 8” letters, together with a few special results 
 
 on sextics suitable on 2” letters. 
 g(§)=—& + af + a f* + a6 + af + a,€ on 8” letters, §§ 79-82 
 79. Applying a linear transformation (in the G/’[3"]), 
 
 g(F + y) =F + a8 + (a, + 2a) &* + (a, + an + ay? + 27?) 
 “(ag sk &7P) FS = (Gg a 2a 4 ag? aa) ae 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 11] 
 
 Hence if a, ¢ 0, either the coefficient of €* or that of €* can be removed by 
 choice of 7. But if 4, = 0, no other coefficient can be removed in general by 
 a linear transformation. 
 
 80. The case a, = 0. 
 
 w+ 
 
 The second, fourth and fifth powers require respectively, 
 
 Ss a (1) 
 1l+ai+aéi=0 (2) 
 
 O° + 2a,a, + 4,5 + 2a3a, + 2aa,0, + a,°a, -+ 2a,a,4 + Zana; + 20,0; 
 
 + 2a,aa; + aga" = 0. (3) 
 
 The seventh power requires exactly the cube of (3). 
 From (1) and (2), a, is a square ¢ 0 in the G/'[3?]. 
 From (1) and (8), 
 
 (43 + 2ay%Gs0; + O°") + 2 (Gs + GHz) (4° + Gy4;") = O. 
 
 Multiplying by «,* and applying «,' = 1, 
 
 (a, + G0)? {a, + 2(a, + a,a,)?} = 0. 
 Hence either 
 Gx ht OL 0 Oa ee 
 The case 4, = 2a,4, is excluded below. The quantic 
 oo a ae a6 -- a6" + (2a,a, a 2,5") € (4) 
 becomes by writing ¢ = + a,'"7, a, = + a,Pa, 
 
 ie {7 ae 7 =a a7) ae 7 = (2a a 1) y) \ ; 
 
 The resultant of the equations 
 
 
 
 y+t7+ (Qa+1)y7=—0;7=1 
 is 
 
 oo t+ tol (e+ 1) (¢ —a— 1). 
 
 Hence « may have any value in the G/'[8"] except 2 and 2 + 2?” 
 
112 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 The ultimately reduced form of (4) 
 
 es bape ate ees hg 1é 
 
 
 
 represents the following substitutions on the marks of the G/’[3"] : 
 
 
 
 
 
 
 
 (0) (1) (— 1) (212, 21 — 1, 918 +1) (— 918, _ 912 _1 _ 918 4 1) fora = 0. 
 (0) (1) (= 1). (2, = 218 1) ol 1) ok oi ee) fore 
 
 
 
 
 
 (0) (1) (— 1) (— 91) (_ 942 4.1) (— 912 1) (22, 22 4.1, 212 — 1) for a = 21, 
 
 
 
 (0) (1) ‘eo 1) (Q42, ne ils 42 1, 91/2 il. 91/2°__ 1 2/2) fora =—1 + 217, 
 (b) n > 2. Write 3" = 6m + 3, m being thus of the form 9% + 4. 
 The powers m + 1, m + 3, and m + 4 require respectively 
 GO, =a, (1) 
 
 a, (20° + a,°) = 0 
 
 Os! + Ogt0;° + aga? + afah + 2a,°0, + a°a.” + 24,47 + a,'a,°a, 
 + a%d,'04-+ a,7d,° + 2a,%0, + 4''=0, (2) 
 the last two terms not occurring when 2 = 3. 
 
 Applying (1) to (2), we have whether 2 = 3, 
 
 Oy (Us! fF OglhgIs? + Oy°s'Us + Oy'Oy!) = Uy (U5 + Og)’ = 0. 
 
 
 
 
 
 Consider the power 3"! + 387 +...43+1 7 aq+a+4,+ 4+ 4. 
 For s — 6a, + 4a, + 3a, + 2a, + a, = 3 (8" — 1), we have 
 
 
 
 
 
 Gy Se BA Se 8 Se ae a a a re ee 
 
 For 6a, + 4a, + 3a, + 2a, + a, = 2 (8" — 1) = 3" 4+ 2.3"1 + 2,3" 
 +... + 2.3 + 1, it follows that a, = 0. For, in the notation of § 73 (f), if 
 
 of = 1, then of = 1, .. cf; == 1; while ah ote ewouldmives = 20.071), 
 Also a,.— 0; fornt cP? then 2 a 
 
 Thus 2a, = a, == 0, DIO, == 0, ee OF a ee 
 For s = 3" — 1, we have at once 
 
 
 
 
 
 
 
 Q, == 88) + BP ee BEAL, $a p= ee 
 
 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 113 
 The condition is thus 
 
 n—1 - an—1 
 jee i Fak OtT 4 ap” fins A OtE . () 
 
 Hence by (1), 4 is a square ¢ 0 in the G/’[3"]. 
 The only possible form is thus 
 
 & + gf + 46 + a2 + 2a, , 
 
 which is nof suitable on 3” letters since it vanishes for ¢ = a,!? + 0. 
 81. The case v4, + 0. Removing the term a‘, we consider 
 
 Sta?+a484 46+ 4€. 
 
 (a) n = 2. The second, fourth and fifth powers require respectively, 
 Gp 20 Ore (1) 
 1+ a%a, + ae, + a46=0. (2) 
 
 a, + 2a, -+- 2a,%a, + 2a,?4, + 4,707 + 2a,4,%0, + 2a,aa,' + af307—=0. (3) 
 
 The seventh power requires the cube of (3). 
 
 First, a2 = 1. For if a4, = 0, then a, = 0 by (1) and then «a, = 0 by (8), 
 which is contrary to (2). 
 
 By squaring (2), 
 
 ads’ + Qafas + a%a° = af + 2a, + 1 = 2(a, + 1) 
 = 0,0, 1-20," ; (2 ) 
 But a, +0; for if a, = 0, then by (3) 
 
 0° (03° 4- 2a,°0, + 2a,°a,* +- a,’a,') = 4, (a, — 4,°) (43° + 4°) = 0. 
 
 If either a, = a,° or a, = 2'a,°, then by (1) a,* = 1, contrary to (2) 
 for a.== (0! 
 
 Hence by (2’) 
 
 ieee Naas a a? ae Da® te Oa? 2.0), 
 
 Writing a, = ya,’ this becomes, aside from the factor @,’, 
 
 @+)@+)@—7-1) =0. 
 
114 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 If 7 = — 1, a,4 = 1 by (2), while (3) becomes 
 a, + 2af%a, + 2a,°a,° + aa,* = a,’ (a, — a°)' = 0. 
 The resulting quantic 
 £6 ae a2 che ape wy ose PA ape 
 
 is suitable on 3° letters. Thus for a, = 1, it represents the following substi- 
 tion on the marks of the G/’[3?] : 
 
 (0) (1) (— 1) (277, — 2%?) (24? + 1, — 24 — 1, — 23? + 1, 23" 1). 
 If 7? = — 1, a, = 21a’, while (2) and (3) become 
 i — a 
 (1 — 21) g,? — 2¥¢° + 2a3a, — 234,34, + a,?a,4 = 0. 
 
 Multiplying by a,’, and placing a,'a,' = a, = — 1, « = ¢a,’, 
 
 
 
 ‘tae Q12g8 a (1 ran 21/2) ¢? 29 al + 21/2) = 0 ’ 
 
 whose roots are 
 
 Lie Ota eT ero ee 1 oe 
 
 But & + «4& + gas — gas? + 2%a,°€ vanishes for € = — 2'*a, when g = 
 — 1 — 2%", while for g = 1 — 2!” or gy = — 1 + 2%? it represents a substi- 
 tution on the marks of the G/’[3?]. Thus, taking ¢, = 1, 
 
 e446 1 (4 21/2) a abe Qu2) G2 4 gre 
 
 
 
 &6 as & (4 1 a 21) ra a (— il an 23) = ao yl aa 
 represent respectively the substitutions 
 
 (0) (— 1, 284 1) (1, 2 1, OR on oe ee 
 
 
 
 (0) (— 1) (1, 2%? — 1, 23”) (21? + 1, — 218, — 232 4 1, — 232— 1). 
 If 7? =7-+ 1, a, = (2'2 — 1) a’, and (2) and (3) become 
 ees 
 
 — 21g? + (1 — 2%*)a,° + 24a, + 2'a,°a,> + aa, = 0. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 115 
 Multiplying by a,’ and placing a,‘a,4 = a, = 1, 4, = ga; 
 21g Le D122 roles (212 a 1) aed) 
 
 Its roots are — 1, 1 + (— 21? — 1)!”, of which the last two are not marks of 
 
 the G/F’ [37] since (— 21? — 1) = — 1. 
 The quantic given by ¢ = — 1, 
 6 as ae Pe (gas = ae oh (212 ~s 1) ae 
 
 represents when 4, = 1 the substitution on the marks of the G/’[3"] : 
 (0) (222) (21? — 1)( — 2'? + 1)(1, 24 4 1, — 1, — 2!", — 212 _ 1). 
 dee eae 
 
 The fifth, seventh, eighth and thirteenth powers require respectively, 
 
 & + aoa, +a*=0. (1) 
 a,0° + af + aa? + a70,%0, = 0. (2) 
 Osby | 2apas + 0°05? 4- 4,0,°0, 4- 20,°a, + o,'a,0,° + 2a,°a,'a," 
 + G70. + 24,0,’ = 0. (3) 
 I + afajfa, + aa + aafa,7 + a2 = 0. (4) 
 
 The tenth power gives an identity ; the eleventh requires exactly the 9th | 
 power of (2). 
 Applying (1) to (2) 
 a, (a> + 24,843 + a,"o,) = 0 
 or 
 pe 0, dy a 2a. (5) 
 (b,) If a, = 0, then a, = 0, a4, = — a,‘ and the sextic obtained, & + a,&° 
 — 4,'§ is suitable on the mark of the G/’[38*]. Thus &° + & — & represents 
 the substitution 
 
 
 
 OM (-Y)GP-L—F+7-L—-P+I4+LI-LP+4—7 
 —f-Ujt+he—j,—P-J-1L—P-J4+N(-h-F +3, 
 
 
 
 2 2 | 
 
 ee a ae Pa tL Pe 7 LL, 
 ume geririels J == 1). 
 
 
 
 where 7? = j + 1 is the irreducible equation defining the G/’ [3°]. 
 
116 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 (b,) If a, + 0, we have on applying (1) and (5) to (3), 
 a,°a, (24,5 + 2a,? + a,°a,? + a,'a,") = 0. 
 
 Writing «, = 72,°, this becomes 
 
 
 
 vi 7 i 1 — O : 
 Multiply by 7’, replace 7" by 7" +” and 7” by 1: 
 je te pects Awe 
 
 Extracting the ninth root, 
 
 ae ape Fa ee 2 fant eae apd 4 — 
 i ee Mel Meme Remcttergl pam UE oe 98 Oi iy hha 
 Now 7 = 1 is excluded since € + a,¢° + a,°¢° + a,'€? vanishes for § = 
 Again 7” + — I, since then 7* = — 1. 
 If7v=7 +7 +1, then 7? = — 7 + 7,7" = 1. 
 
 From (1) and (5) 
 Oy == 90, 0, = — (4 + Ta a, ey ee 
 
 Substituting these values in (4), using 7° = 1, we have 
 
 12 
 
 7 7° 
 
 
 
 FP a a ats ae 
 which is readily seen to be inconsistent with 
 Siecle hor ce he 
 (c) 2 > 3. Write 3” = 6m + 3. 
 The powers m + 1 and m + 2 require respectively, 
 a, + a0, + a = 0 
 
 aa° + af + afa? + afa,'a; + a,"a, 24,7’ 0- 
 
 From these 
 a (5° + 2a°a,° + a") = 0 
 
 or 
 O, = 0,74, + 2a,°. 
 But 
 oS G6 a5 Oar) f° + (4,"a, + 2a,°)¢ 
 vanishes for € = — g, and is thus excluded. 
 
 — O,. 
 
 82. Summary of $$ 79-81. The only reduced SQ [6; 3”] are the last five 
 
 quantics in the table § 87. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 117 
 Introductory study of sextics on 2” letters, $§ 83. 
 83. Ifig(@)=& +46 4+ «€f + 4,6 + a? + a6, then 
 ¢g (¢ + 4) = Sa Cer e a9 + 4) Ss gs b ee sy + 44) oh 
 + (ayy + 457? + 45) & + (7) 
 Hence in general no term can be removed. If a, = 0 we can make the coeffi- 
 cient of €' zero; if 4, = 0, we can make that of € zero. 
 
 84. The case n even. Then 2" — 6m + 4. 
 (a) n = 4. Of the “ power conditions,” two are independent, 
 
 err, 3 
 tly ac 
 aos + af + a, + ata, + asa? + a,° + afafa, + afaa,? + a%a." + afa, 
 + 0370s! + G's) + O,°O,U, + asd; + aaa, + aao—0. 
 (b) n > 4. 
 The power 7 + 1 requires a, = «,°. Applying this to the condition given 
 by the power m + 5: 
 ads + OG, + afayas + aafa + 4,450; + 4,°o,'a,' + a°a'a, + 4,4, 
 + a Mata, + afa$ + afafa?2 + 4” + a@afa2 + aa? + a'a8 + aja =0. 
 
 By the method of proof used in (f) § 73, we find the power 
 2m +1= 274 Or44 ...4+RP 41 
 
 requires 
 
 ee ee eae UOT Ot. 0% 
 
 85. The case n odd. Then 2” = 6m + 2. 
 (a) zn = 3. The third power requires 
 
 a, + a7 + a%a, + aa? + a,’a, + aa7 =0. (13) 
 
 The fifth power requires exactly the 4th power of this. 
 (b) » > 3. The power m + 2 requires 
 
 ads + sae + aa, + a,'u, + a°a, + aa, + a,'a70, + a 3a =0. (15) 
 
 The power m + 6 requires for rn > 5 and n = 5 respectively 
 (Og AO) (G, Ga” id,” ag) (43° Pa) (Qa, O02 4 a,fo,)—0. (2) 
 a, + a” + aa, + af (a, + a3) + a,° (a,0, + aaf + a,’a,)—0. (2;) 
 
 The powers m + 8 and m + 18 lead to (1). 
 (b,) Suppose a, = 0, so that by § 83 we can take a, = 0. 
 If a, + 0 and n > 5, then by (1) and (2) we find a, = a, = 0. 
 
 
 
 
 
118 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 But & + a,€° vanishes for § = a,'%; while every mark in the G/’([2”"], n 
 odd, is a cube by § 18. 
 If a, + 0, and n = 5, (1) and (2,) give 
 
 7h tle . = 5, 4 
 Dai Tomes (0 Py lid 34 ren 0 a 
 Thusteither, a,’ Ss" a,%== Orca te Oe ee a 
 =6 £3 222 le} is S 1 as, ol, eee iS} 
 But & + a6 + a,”6 + a,"¢ vanishes for § = a,” = a, 
 
 If a, = 0, then the power 2m + 1 = 27 4+ 2744...42?4 241 
 requires 4, = 0. But & + a;€ vanishes for € = a;'", a mark of the G/'[2 ], 
 n odd. 
 
 Hence every suitable sextic on 2” letters, 2 odd and 738, in which the 
 coefficient of € is zero, is reducible to the form €°. , 
 
 (b,) Suppose a, = 0,4,+0. Then we may take a, =—0. Then a, = 0 
 by (1). Bor 7 > bare 0 by (2h: for enone, elves 
 
 a, + 4°a" = 0, 
 hence either a, — 0 ora, = a;. 
 
 But & + a€ vanishes for 2 = «, and is excluded. 
 
 The sextic & + a,¢° + a, repr aa a substitution on the marks of the 
 Galt [2 viz tor aie ls 
 
 (URGORG TORN ce hm pncwy =o biyhiae UR ae Rypiccs) cath 
 oct bat ie sok vase om OLGA cr cee hal sai tea eh 
 PEPADE ED GT Ey Le eye a 
 PEP PIS LI ITA LRT Bee ee ee 
 FI LG oF es Fo) ee 
 
 the G/’[2°] being defined by the equation 7? = 7? + 1. 
 (b,) Suppose a, = a, +0. Then by (1) 
 
 
 
 On'=. a> + afd, 10,6, 
 (2) and (2;) and (1;) are seen to be satisfied ; but 
 OP OS a Ge GR (Gy a Oy ea 
 vanishes for § = 4. 
 
 86. Summary of §§ 83-85. If a sextic represent a substitution on the 
 marks of the G/’[2”"], then, for n even, 
 
 Us =O, ob 0s 
 for n odd and > 3, 
 at Oat A) at ae 
 
 except for the suitable quantics 
 
 e° on 2” -létters eS" oe” oe Ont a letbers: 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 119 
 
 87. Table.* 
 
 Except for sextics on 2” letters, the following is a complete list of all 
 reduced quantics o (€) of degree = 6 which are suitable to represent substitu- 
 tions on a power of a prime number of letters. From them (by § 16) all 
 suitable quantics whatsoever of degree = 6 are obtained by the formula 
 
 ap(§ + f)+7. 
 
 Reduced quantic. Suitable for p” = 
 E any 
 = 2” 
 c ; 3”, 3m + 2 
  — g& (a = not-square) : ; 3” 
 fi BE. : ; : 7 
 cA on 
 ot + Og” + TS . ; z 4 F 
 
 (when it vanishes oul fOr Go BE Gy 
 
 ge é : : ; 5", 5m + 2, 5m + 4 
 & —a§ (a= not 4th power) . , 5" 
 
 gF Bite ; 3? 
 Cceeedcoo, ; ; 7 
 
 eo teas? eS? aE 3a7E anes a Late Seranral f ; : 7 
 
 567 + 50? + oS (a = arbitrary). ; 5m + 2 
 & + a& + 30°F (a = not-square) . : : 13 
 
 oS UN LAE Ei a eiies = not-square) . ; 5? 
 
 eo : ; 4 . ; ; ; 2”, n odd. 
 Ves 2 ; : ; 11 
 
 G+ w+ ag? + 5E (a = square) 11 
 
 + 40765 + af? + 45 (a = 0 or not-square) . “ Et 
 
 6° + a €* + ao? + a,7& + (2a a, + 0,5!) € : ; on 
 
 (a, = square +0; a, = 0, + 2¥%¢8?, + a3?) or + 2)? + 1) 2,37, the signs to 
 correspond to that of + a,°”). 
 
 & + a& — of (4 = arbitrary) : ; ; ; aha 
 Ge ae a oe eae (a =—arbitrary):. : 3° 
 4+ af + oa — pats? + 212a€ (4 = arbitrary) . 3” 
 
 (where g = + (1 — 2")). 
 f+ af — ofS? + aff + (212 — 1) a°F (a = arbitrary) 3° 
 In this table 21? occurs always as a symbol for ezther of the two marks of 
 the G/’ [37] satisfying the equation 2? — 2 = 0. 
 88. Theorem. Ad/ substitutions on T letters may be derived from the 
 two 
 C= ae 3b: we = 2. 
 [a | [a ) {2 An a ie 1 
 Le +5) Lat + BF) [Bota 4 208) = Lab — Bat + deat 4 Bbw) 
 ce Compare coe tcan Journal of Mathematics, vol. 18, p. 218, § 19. 
 
 
 
120 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 so that we reach the form 
 
 e+ ae+a?+ 3a°x, aw = quadratic non-residue of 7. 
 
 pa ped ete vy ) 
 La — b>} ba®J le He -- b8a:7 + 3d*z J | Bote | 3 J 
 te 
 
 i 
 = La? — bt at 3022 | a) 
 
 so that by writing 6 = c’ we reach the form 
 
 x — Can + 8cta. 
 
 [wc | [ w ae fee 2 | 3 
 Lae) Cae —b49% + i, Lave J La? + 2072? ++ 3 (20?)Pa J * ko] 
 
 Hence by (1) and (2) we reach the form 
 e + ae + 8e7x, a = arbitrary. 
 
 : 
 
 (x 1 fa Ta [ @ ) |; @) 
 
 | | w [x 
 La + 30°} La? +b) 1a? — be? + bo? + Sha) Lb’e. Ob? | =i + 46%x 
 
 ee) 
 
 so that we reach a* + 32. 
 
 [ # 7} 
 
 (21 fw er eee % 5 
 Aba is La® + Qe J? (5) 
 
 La?) Lat + 40%) 
 
 so that we reach a + 2z. 
 89. E. Betti proved (1. c. vol. 2, pp. 17-19, 1851) that all substitutions on 
 5 letters are derivable from 
 
 e = an + b> 7 = ws 
 
 90. Enumerative proof of Wilson’s theorem. 
 Of the p! literal substitutions on a prime number p of letters, p (p — 1) 
 have a linear representation 
 az+6, a0. 
 The remaining ones are represented by quantics of degree > 1 which fall into 
 sets of p’ (p — 1) each, viz, 
 
 ag(« + b)+e, a +0, 6 and ¢ arbitrary. 
 Hence p! — p( p —1)isa multiple of p’ 2 — 1), so that (p — 1)! + 1 is 
 divisible by p. 
 
 Enp oF Part I. 
 
THE ANALYTIC REPRESENTATION OF SUBSTITUTIONS ON A 
 POWER OF A PRIME NUMBER OF LETTERS WITH A DIS- 
 CUSSION OF THE LINEAR GROUP. 
 
 [ CONTINUED. | 
 
 By Dr. Leonarp Evucrne Dickson, Chicago, Ill. 
 
 PART IJ.—LrygEar Grovprs. 
 Section L.—Linear Homogeneous Group. 
 1. We may define p”” letters 
 
 le. £., EIN, Em 
 
 characterized by m indices, each being an arbitrary mark of the Galois field 
 of order p"”. The general linear homogeneous substitution A on these letters 
 replaces /; .. t, DY 4z,,..., ems Where 
 
 v 
 
 joa Sg 
 
 o¢ =. 
 J 
 
 Ay; OR ee af) a 
 
 Il has 
 ANY 
 
 — 
 
 where the a,’s are marks of the G/’[ p"]. But (1) will indeed be a substitu- 
 tion on the p”” letters if and only if the determinant 
 
 |} A {=|a,;| 40. Cerveaen ly 2, ert 7e) 
 
 For there must be one and only one system of m indices ¢, which (1) replaces 
 by a given system &’; and hence an unique set of values ¢; satisfying the equa- 
 tions 
 
 ke 
 
 POT em ee (hee ee 970) 
 
 ga) 
 Remark. If the substitution (1) be identical with 
 
 m 
 
 ote sy (i ses, nv) 
 il 
 then must 
 Op Ops Cee WOR Pobre eri §) 
 This follows if we take in turn, for 7 = 1, 2,...m, the particular set of values 
 
 é=1;6=0. ES RO eee 8 
 
122 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 2. Let & denote a second linear homogeneous substitution 
 
 = S iS i (7 se eae ner a) ad) 
 so that 
 |B) =} 8, £0- 
 
 The result of applying first the literal substitution 
 
 ey ien 
 Meee ey 
 and afterwards the literal substitution 
 A= (ls, | == (le, | 
 
 Pehla Wisk: 
 where by (1) 
 55 = & Aye, 
 
 is the same as applying the single substitution 
 
 bee 
 BA=| |, 
 &", 
 where 
 érr oe = 
 OS FS 
 V foot | 
 if 
 
 Tik = ~ OB - ” 
 =I] 
 
 We may think of this compounding of two substitutions analytically. 
 Laying aside the method of composition by matrices,* it will be found con- 
 venient here to think of the composition as follows. The result of applying 
 the analytic substitution A first and 6 afterwards is denoted by BA. A 
 replaces the index &; by 
 
 s Gish 4 » 
 
 j= 
 
 
 
 *Tn the matrix notation the substitution (1) is denoted by (a). The composition formula is 
 
 then 
 da , 
 (4°) = (4 ij) (4%) 
 
 ™m 
 
 where 
 vs te y , 
 a nS = 4x y 
 je ja 
 
 provided the matrix (4,,) operates first. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 123 
 
 which in turn B replaces by 
 
 eka et ee 
 < 4; [ = BaF | = 2 Vik? k » 
 j=) =i — 
 if 
 m 
 ss }. 
 lik = <= Bix 
 emt | 
 
 The compound /A is indeed a substitution since 
 | BA |=l\ral=lay!-| &el=lF St hea ea Vie 
 
 Moreover the 7,,'8 are marks of the G/’'[ p"}. ; 
 cae 
 The literal substitution product LA is thus the same as the analytic 
 Ppp! 
 substitution compound BA. The transformed of 7’ by S is always 
 
 CA! Foes iS pass 
 
 For literal substitutions, S~' operates first; while for analytic substitutions, 
 S operates first. The result is the same in both cases. 
 
 ~It follows from the composition that the totality of linear homogeneous 
 substitutions (1) forms a group, which is called the linear homogeneous group 
 of degree p™. It is a generalization of Jordan’s linear homogeneous group* 
 of degree p” in which the m indices and m? coefficients are integral marks, i. e. 
 marks of the G/’[ p']. It is however contained in the Jordan group on mn 
 indices. It is identicalt with the group investigated by E. Betti in 1852-1855 
 and in more detail by E. Mathieu in 1861. 
 
 3. Necessarily the group I study is of degree a power of a prime. But 
 all that is of essential interest in Jordan’s general linear homogeneous group 
 of degree g” centers in the case when g is a prime. Thus by Jordan, l. ¢. 
 Arts. 127-131, the factors of composition of the linear group on m indices 
 modulo g are simply the factors of composition of the groups on m indices 
 taken respectively modulo p/* where 
 
 qT De, 
 t=1 
 Di Px» +++, Pr being the different prime factors of g. While by Arts. 132-134, 
 the factors of composition of the group modulo p/‘ are those of the group 
 modulo p; together with factors all equal to p;. Hence for the study of non- 
 cyclic simple linear groups the case of a composite modulus offers nothing 
 beyond that of a prime modulus. 
 
 * Jordan, T’raité des Substitutions, pp. 91-110, 1870. 
 +See Section III; also literature given in Preface. 
 
 
 
124 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 4. Theorem.* The order 2(m, n, p) of the linear homogeneous group of 
 degree p"” on m indices is 
 
 Ouse Lae 1) Ges. PF. p”) Giese A oat » pr) 5 ir (ae ae. 2) 2 
 Let V be the number of linear substitutions 
 
 fi, = 1, Is fee CO) Ley, 
 
 which leave the index ¢, invariable, and let 7’ be a substitution which replaces 
 it by 
 eee 
 _ Dy 555 . 
 j=) 
 
 T TR, TR, ..., TRe, 
 
 The JY substitutions 
 
 will replace ¢, by this same linear function, and no other substitution has this 
 property ; for if (/ were such a one, 7’! UW would leave €, invariable and hence 
 be a certain /2,, so that 
 
 Ue Tie 
 
 The marks «,, may be taken arbitrary except that not all can be zero. 
 Hence the system a, is susceptible of p”" — 1 different sets of values. To 
 each set correspond JV substitutions. Hence 
 
 2im, 2,9) = (pw — 1). 
 
 But the substitutions 7, are of the form 
 
 ™m 
 
 Sa Set le a (PeS2 ia. 5 72) 
 Yok 
 
 where the m — 1 coefficients a, are arbitrary and the coefticients a, a3, ..., 
 Aim are such that their determinant is ¢ 0, which can happen in 2 (m — 1, n, p) 
 ways. Hence 
 
 LV == PDO (ie ne) a 
 Thus 
 2(m, n, p) = pe (pm — 1) 2(m — 1, n, p) 
 
 — poo (ps ett 1) ; pe Ge: i 1) pick (p” — 1) 
 a Cee — 1) ie — p”) es Ga = oe) J 
 since evidently for a single index, 
 
 #2 (1, 2, p) =p" — 1. 
 * Stated Jordan, 1. c. Art. 169; proof for mn = 1, Art. 123. 
 
 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 125 
 
 5. Transformation of indices. Let A be the linear homogeneous substi- 
 tution on p”” letters : 
 
 ™m 
 
 Fee (re Lean a ie) (1) 
 ied 
 We define m new quantities 
 Aono ie ms (eaaatl eA arate oa aT 
 k=1 
 
 where the £;,’s are marks of the G/’[ p”] such that 
 
 
 
 Beata Vr0 | 
 We may thus solve (2), 
 a?) 4" ik A = 
 Se ee By ates (Ihe ek a ed 
 where £,, denotes the adjoint of 8, in || =| P|. 
 
 Instead of distinguishing the letters 7; |. <,, from each other by the vari- 
 ation of the indices ¢,, we can distinguish then by the variation of the 7,’s con- 
 sidered as independent indices. Thus A will replace one of these new indices 
 4, by 22 By, a455;; So that, applying (3), we may write the substitution A 
 
 kj 
 
 qi Se tak (2 eee heey m) (4) 
 where 
 
 la B,, 
 ta Py Pin Beg + Z 
 
 ky j |B) 
 
 
 
 The coefficients £,, may be chosen to simplify the expression of the substitu- 
 tion A. 
 
 Remark. The transformation of indices (2) leaves the determinant | a, | 
 of the substitution A invariant. 
 
 For, applying (2) to (4), 
 a ies 2 Ya Pie SK (aa ear petie | 
 k=1 lk 
 Applying (1) and changing a summation index on the right, 
 »' Bin Og & = ~* ra By § - CS ee 
 kj l,j 
 Since this holds for arbitrary €/s, we have [see remark § 1] 
 
 = Pie Gy = ae By 5 (7 == Le 2, as m) 
 
 k= 
 
ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 126 DICKSON. 
 Hence 
 
 Bi b= | pt 47 ball 
 or 
 
 | xs | == | Ta - 
 
 6. Denote by Bz, xs, the substitution 
 re i 
 
 mr ol = - = —_— 
 Sr — Spr =F AS 53 3 = 
 
 d 
 
 
 
 affecting only the index &,. Its determinant is 1 if 7 # s, butis 1 + Aifr=s. 
 Theorem. very linear homogencous substitution S can be placed under 
 
 the form* BL, where B is derived from the m(m — 1) ( p” —1) substitutions 
 Be. rte 5 (Fy os LD Rae) 
 
 1 being any mark +0 of the GF'[ p™], and where L is a substitution which 
 leaves unchanged every index except the last which it multiplies by the deter- 
 
 minant of S. 
 Let the given substitution S be 
 
 mm 
 cf = 4 0 F (2 = 1,2, , m) 
 i 
 
 Suppose 
 Dr¢ ¢ 0 5) O15 =0 2 
 
 If e > 1, the substitution 
 
 nm 
 ‘7. 7 
 T= i Bg, Gijt; 
 j=e 
 will replace ¢, by 
 m™m = x" Ps 
 y’ a2 5° = 
 rig be eet a 4 
 
 8 Y= 
 If ¢ = 41; the substitution 7 >== 
 Bz, af, ° Bz,, se, Bz, yée ott Ds ee 
 
 will accomplish the same result if 
 
 1+ f=, ~+a(l + fy) = 4%, 
 
 which may be satisfied by any ; + 0, since a, + 0. 
 We thus have S = 7'S,, where S, is a new substitution which leaves ¢, 
 
 unchanged and thus has the form 
 
 "= 6; €/ = 3B, 6,. (tien 23 eae eee 
 
 5) 
 
 
 
 * The variation from Jordan, Art. 121, is due to the correction of 1. 17, p. 94, where S, 7’ should 
 
 read 7'S,. 
 
ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 127 
 
 DICKSON. 
 
 To decompose S,, I build the substitution 
 if = II Be v8, viz. 
 Donna, 270) 
 
 | 
 
 fade fag 5 fag Ce ~ . 
 = 915 SH a HH (2 
 
 where the 7;,’s are chosen later. 
 Denoting by S, the substitution 
 
 Yo att at =e Ph 
 == Sis 
 
 (2: SS eer on 970) 
 
 dVyp 
 
 L 
 
 , m) 
 
 we will have S, = 7'S, provided 
 Pe Opie 
 
 m 
 c= By he = Ba : 
 
 I==2 
 
 oy 778)) 
 
 But the 7's can be determined so as to satisfy these equations since 
 (2, yy = 2, oe 
 
 By | =| 8,) =| S| #0. 
 
 Hence 
 Pevdome AS RASS 
 
 Proceeding with S, as we did with S, etc., 
 
 S = eds <6 Lees Tie Som—a ’ 
 
 where S,,,,_, denotes the substitution 
 Sete Joel At We Lee, oe A lh Oe 
 
 7. Theorem. fa linear homogeneous substitution S is commutative with 
 
 all the m (m — 1) substitutions 
 ote wiensenes heb) 
 
 Bz, mel» 
 where pis a fixed mark ¢ 0 of the GE'| p"), then S simply multiplies all the 
 
 indices by the same mark. 
 Let the substitution S be 
 Ey (¢ =1,2,...,m) 
 
 Tt a 
 iS Susy 
 
 j= 
 
 We then find that S. B:, ,¢, and Lz, , . S are respectively 
 
 mm , 
 j= 
 
 foal ’ ° 
 si= 3 yé; 5 Le oe. ee ot 2-t Jc) 
 
 m 
 ee at ol 
 o, = ae + pay) §5 5 
 j= 
 
 ™m 
 all 4 = \' »~ 
 G4 SS MGS, = By; - 
 j=l 
 
128 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 The conditions that the two expressions for &,’ shall be identical are, since 
 pt 0, 
 
 Gy = Gee Fy 
 
 =a C7 eae Ler Pe Ly 
 
 The conditions that the two expressions for €/ (i + £) shall be identical 
 are 
 Oem (ae LA eee eee) 
 Since the two sets of conditions are to hold for # and / = 1, 2,..., m, 
 k+l, S must be of the form 
 rie CRS SS re 
 where’@, == 0,)°== vas Cane 
 Factors of composition® of the generalized linear homogeneous group G of 
 degree p™, §§ 8-12. 
 8. Let p*»—_1l1—=p,.p,... py, where p,,...p, are equal or different 
 prime numbers. Let p be a primitive root of the equation 
 
 
 
 Ep" —1 —l1=0. 
 
 Denote by G,,, Gp», --. the sub-groups formed of those substitutions of 
 G whose determinant is a power of p”, p””,..., respectively ; finally, by 
 Gn = I that formed by the substitutions of G having determinant unity. 
 
 Let d be the greatest common divisor of mand p" —1; py, erat ie 
 the prime factors, equal or different, of d; p’ a primitive root of 
 
 Sti 02 
 
 Denote by //, //,,,, H,,,»,, ... the groups formed of those substitutions of 
 I which multiply all the indices by the same powert of p’, by the same power 
 of p?>,..., respectively. 
 
 9. Theorem. The factors of composition of G are 
 
 2 (m, n, p) raeitt: ; 
 Pir Por +++ Pres Ge Pris Pors++> Pi» 
 
 
 
 except when (m, n, p) = (2, 1, 2) or (2, I, 3). 
 
 
 
 *Some of the ideas used by Jordan in his decomposition of the linear group are to be found 
 in a different form in the earlier work of Betti, 1. c. vol. 3, pp 79-83, 1852, and vol. 6, p. 31, 1855. 
 The latter shows that his group of degree p” (see Section IIT) has as self-conjugate sub-group (of 
 index 2) the group Z of substitutions of determinant a square in the GF'[ p1]; that Z has as self- 
 conjugate sub-group of order p — 1 the group of those substitutions which multiply all the indices 
 by the same factor 2, a primitive root of p. He proves that in the case p” = 2° or 3? all the factors 
 of composition are prime [thus giving the exception in theorem § 9]. 
 
 + If a substitution of /’ multiplies all the m indices by the mark yp, then »”™ = 1, and hence 
 
 y* =1. Thus p is some power of p’. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 129 
 
 Tt is clear from § 6 that the groups 
 
 
 
 Fa Gas Gia ee! aA fe iat Line lees COE: 1 
 have respectively the orders 
 DGnenen). Q(m,n, Pp) 2(m, n, P) eae ee 1p) See @ ad ahs, 
 Pr PiP2 pa) Pi PiPr 
 
 and that each is a self-conjugate sub-group of the preceding. Further, the 
 numbers p; and p,; being prime, there can be interpolated between G,, and 
 G,.». for example, no new group which contains @,,,, and is contained in 
 G,,- It remains therefore to prove that, aside from the two exceptional cases, 
 His a maximal self-conjugate sub-group of /. To prove this we show that 
 if a group J be self-conjugate under /' and yet more general than //, then J 
 will necessarily contain all the substitutions of /° 
 
 10. Let 
 
 oi B OS; CSA a) 
 jee 
 be a substitution S of J but not in /, which therefore does not multiply all 
 the indices by the same factor. Then by § 7, 8 is not commutative with every 
 B:z,,x¢. 4 being a particular mark + 0,e.g.4 = 1. To fix the ideas, suppose 
 Sis not commutative with BL: ,<,. The substitution 
 
 i ie See es Ly 1, Ae 
 
 will thus not reduce to the identity. Further 7’ belongs to 7; for the trans- 
 formed of S by Bz, ,¢, belongs to /, since Lz, ¢, has the determinant unity and 
 thus is in /* 
 
 To express the substitution Z, we note that, for 7 > 1, Bz, yz, leaves §; 
 
 ay os . 
 unchanged ; that S replaces ¢, by + 4,¢,, which B;',, replaces by 
 i fad . 
 
 e 
 ta, 
 2 Oy5; — Ady, , 
 
 Sh 
 which in turn S~' replaces by ¢, — 4a,y, if g denotes the linear function by 
 which S~! replaces €,. Thus Z’ has the form 
 
 m 
 
 ay Spel te ae ee ’ eels 
 
 es = 2 Bess Gi, = 5; — Adin (ia eer oath) 
 j= 
 
 where it is not necessary to determine the /,,'s more closely. In this expres- 
 sion of 7’, which is the basis of the further developments, there is no longer 
 any trace of the ¢,-specialization in Bz, yz. 
 
130 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 11. Let us first suppose that m > 2. 
 
 (1) Suppose a, = 0 (2 = 2, 3,...,m). ‘Then #,, = 1, the determinant 
 of 7 being 1; while f,, (7 = 2,..., m) are not all zero, since 7 is not the 
 identity. Let us take /,, + 0, changing if necessary the previous order of the 
 independent indices &,, €, ..., &n- 
 
 To simplify Z’ we take a new system of independent indices 7,in place of 
 the former ones ¢,, viz 
 
 is 
 
 (2 ==; Lae ete POET EOD 
 
 ss 
 IL he 
 mee) 
 Syyp 
 os) 
 = 
 SVp 
 
 which is permissible since the determinant of transformation is f,, +0. Then 
 
 ne m 
 
 
 
 ta Sn ay, af iS. i= LYN Ops fa yy 
 =! S40 = Pie Busi + = Sys; = 7 + 4 
 Yj 92 
 ; = 
 ‘4 ’ ald f) i _ 
 43 = = PSs = = Puss = 4 - 
 j=2 4==2 
 AN Se akg pee ay a es ik 5 
 Hi ‘FZ Pe ee Cay Fae Big es AT 
 
 Hence 7’ becomes simply B,,,,. But by the Remark of § 5, [ contains 
 the substitution Z (of determinant unity) 
 
 PAE ae Pitter SUL eR tn 2 eC “ie 
 M =H 3 Gy = es Hi = 15 (2 = 3,..., m) 
 
 where / is an arbitrary mark + 0 of the G7 [p"]. Hence / contains Z7'7'Z, 
 i.e. B 
 
 nine Also / contains the substitution V7, of determinant unity, 
 
 am + Chad A 3d Spytita’ ps Agni ge ee “ aM ae coe (a= Sa eee et 
 De es Dh 8 Oh, ee a oe litk k>2 
 2} 
 
 Hence J contains /;;' B,,,,,U7;,, which gives for k = 3 the substitution 
 BG... xn, wad for > 3 the substitution B,,,,,. Likewise 7 contains the trans- 
 formed of these substitutions, etc., so that clearly 7 will contain all the substi- 
 tutions 
 
 hoe ae (0° = Fete ete) 
 
 But by § 6, these substitutions combine with each other to give every substi- 
 tution of /} i. e. of determinant unity. 
 
 (2) If not every a, (2 = 2,3, ..., m) is zero, suppose a, +0. Take 
 Tis Noy +++) Ym a8 independent indices where 
 pA eee ee en ain f eine ae | 
 Th Sie U2 as. haat, aoe eee (ire= ? sistema die) 
 “21 
 
 the determinant of transformation being unity. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 131 
 
 The m — 2 indices 7; (i > 2) are unchanged by 7’ (of § 10); thus 
 wi 5 GC; — 5 f= G; . 
 je (5; sa 20,0) SAS coe hy) =< Hi 
 ees Foy 
 
 Hence the substitution 7’ takes the form 
 
 a 
 
 m 
 tik eee SY Lee Aus abet, Ree SS fia 
 ieee ee acer Tyla i eI 2), 
 a= = 
 where (by Remark of § 5) the determinant of 7’is 
 tute — fen = 1. 
 
 In this form of 7’ the indices 7,, 7, play the same role. 
 
 (2a) Suppose 7, and 7,; (2 = 3, ..., m) are not all zero. Changing if 
 necessary the order of 7, 7, .--, 4m, We may suppose that 7,, and 7., are not 
 both zero. 
 
 The text in Jordan (I. c. p. 108, 1°) is here incorrect, but may be modified 
 as follows, if our linear substitutions be on four or more indices ¢ : 
 I will contain the substitution 
 
 Wheel Fe I ste. TB. 
 
 N3y M4 Ns, MNa ? 
 
 (~ being an arbitrary mark + 0 of the GF'[ p"]) viz 
 I = — fT Te =e — Uw MH =. (1 =3,..., m) 
 
 Suppose for definiteness that 7,, + 0, and take w,,..., w,, a8 independent 
 indices, where 
 
 Op Fi Tin 5 Og = To — Ta/ M13 Ot = Em “(1 = 8; 2) 
 Then for the substitution 7) we find 
 
 ~ , ) 
 pe Se at ed BYE ee 
 {= — =: 
 hs 113 
 
 wO, + LO, 
 
 , , , : om | 
 Le ede = I == (2 — LY 3hs) — & Ditrs eo ihe 
 
 On Ne WO, (2 een a) 
 
 Hence 7) takes the form B,,,.,, which combined with its transformed by 
 the substitutions of J’ will give as in (1) all the substitutions of /. 
 
152 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 For the case of three indices,* I contains the substitutions 
 
 Ti [AR em Ty viz 
 
 Nis PMNS Niy KN3) 
 
 ~) 
 
 I = GL) OS i Sh 
 {jit = Yip aw GT 3 eee iand3» Fa = 43- 
 
 Now Z, and 7, are both of the form U: 
 =I + O35 Yo = Yo + F235 Ys = 7 - 
 If both Z} and 7; reduce to the identity, 7’ becomes 
 = + 113933 Ye = M2 + Fo3%33 Is = Ns» 
 
 which is of the form l/, Hence in any case we have a substitution of the 
 form Y and not the identity. We then take ,, ,, w,, defined above as inde- 
 pendent indices and obtain for U/ the form B,,.... 
 
 (2b) Suppose 
 hi = 7x =O. (irs ov aye) 
 
 Since 7’ does not veduce to the identity, we can not have simultaneously 
 
 
 
 Yr = Ye = 1, Y= Yr =O. 
 Hence if we form 7, and 7’, above for m indices, e. g. 7 is 
 
 vie Seed epee 4 (gaia) 433 7 = Yo — PY¥n93 > Hi = 73 (a eta rye eC) 
 we see that one of the two must not be the identity and we proceed as in (2a). 
 
 12. The case m = 2. 
 
 Let S be a substitution of 7 but not in /7, which therefore does not mul- 
 tiply both indices by the same factor. Let 7, be a function of the indices §, 
 and € which S does not multiply by a constant factor, e. g. 7, = ¢, + &; let 
 7, be the new function by which S replaces 7,. Since 7, and 7, are linearly 
 independent, we may take them as new indices, when S becomes (having 
 determinant 1) 
 
 NM = hs Gy = —H + OH. ‘ 
 
 Now / contains the substitution 7’ 
 
 Wi 0s Ys a (a + 0) 
 
 
 
 *M. Jordan was kind enough to send mea correction of the error above mentioned. His proof 
 appears valid however only for three indices. It is given here with the usual modifications. 
 
» 
 
 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 13% 
 Hence / will contain UV = 7 “1S7S, viz 
 My = — 2 5 YY = — O(1 + a”) 4 — O,. 
 
 (1) Suppose od ¢ 0. 
 Then / will contain (/*, which for 2 = 1 becomes 
 i =| 3 My = 409, + Ap. 
 If p > 2, we can replace 7, by another index 
 @ = Adn,. 
 Then U/? becomes G,, 5. But /' contains the substitution V, 
 
 ete se oe 
 
 Thus / contains V~*0*V or Bb, _.,,. 
 If Z denote the substitution belonging to [ 
 
 
 
 CET AED p= AN, (A = any mark + 0) 
 
 f will contain the substitution Z“B,, 4 or By, a24: 
 By virtue of the property 
 
 Bie APp * ie Arh aaa ae (A? + Az) > 
 
 we must reach a substitution 2, ,4, where » is a not-square in the G/’[ p”), 
 p being > 2 by hypothesis. For, if not, the totality of the (p” — 1)/2 squares 
 in the G/'[ p"], together with the mark zero, would form not only a multipli- 
 cative but also an additive group and therefore* a Galois field of order say 
 p” included in the GF'[ p"]. But p’ = p, qua additive period of every mark 
 7 Ose Lbus 
 
 res . Deby seal De, 
 which is impossible. 
 
 Having one, we reach every not-square by the formula 
 —l i) 
 L No, vb L oe ae ae A2vp * 
 
 Thus for 0+0 and p > 2, Z will contain all the substitutions 5, ,4, and 
 By. ung -. being an arbitrary mark + 0 ef the GF [_p"], and hence all the sub- 
 stitutions of 1 
 
 (2) Suppose 0 = 0. 
 
 (2a) If p” = 2? or if p” > 5, we can find a mark a:0 of the GF'[7"] 
 such that a + 1 and then a mark / such that 
 
 B (at — 1) = pw = any mark : 0. 
 
 
 
 * Moore, l. c. § 40. 
 
134 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 Then / contains the substitution 
 
 LB U B 
 
 | : 
 N1y BN2 U » VIZ 
 
 = 
 N15 Bn2 
 m= + BC —1)%3 He = ts 
 
 or B,. un,» Which combined with its transformed by the substitutions of J will 
 produce J. 
 (2b) If p" = 5, we may take as independent indices 
 
 wo, = 2m, + 3 Oy = — 2p + 
 Then S will become 
 
 WO; == 20,30 == AO 
 Thus / contains the substitution 
 
 27 BS yet 
 SB; S Das = Devas 
 
 19 2 
 
 and hence also its cube 5, .,. 
 
 (2c) For p" = 2, the linear homogeneous group on two indices has the 
 order (27 — 1) (2? — 2) = 6. It contains a subgroup of order 3 formed of 
 the powers of the substitution 
 
 These ea + &, 
 
 which represents the literal substitution (¢))) (4)4:/,)). The index of this sub- 
 group being 2, it is self-conjugate. The factors of composition are thus 2 
 and 3. 
 
 (2d) For p" = 3, we may prove that the (3? — 1) (3? — 3) = 48 substi- 
 tutions of the linear homogeneous group & of degree 9 are derived from 
 
 (A) Fo ae C3 Sc ee ee 
 (B) cies Pee sy ett 
 (C) Cee es eee 
 
 (D) Cf = Sa S25) Se Se ae 
 (E) Pee tes eS 
 
 Thus for the groups 
 {FEY {#, DEA, DOV AL, D, OC, By ED eC eeaa 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 135 
 we may verjly the following generating relations : 
 Ae We 
 TB IK IND cee WOE. 
 OMe PAO eee OS GD ah DC.. 
 Dim eb bee hee be DP. 
 Ae eee Ape) = GAN eA (1 DAs AB en Ob A. 
 
 From these relations it follows at once that the above groups have for orders 
 2, 4, 8, 24, 48, respectively, and that each is a self-conjugate subgroup of the 
 following. Hence the factors of composition of @ are 2, 3, 2, 2, 2. 
 
 13. Since the quotient-group* of any two consecutive groups in the series 
 of composition of any group is a simple group, we have the result that //H 
 is a simple group of order 
 
 2(m, n, Pp) Mi ies fen) 1) Co ws pe) a) you EEA Ea) 
 d(p"—1) d(p” — 1) 
 
 
 
 except in the cases (m, n, p) = (2, 1, 2) and (2, 1, 3). 
 Compare section IT. 
 
 Srcrion II.—Zinear Fractional Group.+ 
 
 14. In our generalized linear homogeneous group @ of degree p™” on m 
 indices, let us class into the same system the p” — 1 letters 
 
 l 
 
 Men, Bega = oy bEm 
 
 where the €/s are fixed but » runs through all the marks except zero of the 
 GH { p"], or, otherwise expressed, the p” — 1 letters corresponding to the 
 
 same values of the ratios ¢,/&,, (¢ = 1, 2,..., m — 1). 
 Each substitution of G will replace the letters of any one system by letters 
 all of some one system. We have thus (p”” — 1)/(p” — 1) systems, whose . 
 
 displacements (by the substitutions of G) form a group // of degree ( p”” — 1)/ 
 (p" —1). But there are exactly py" — 1 substitutions of G which do not dis- 
 place any system, viz, 
 . m Gs Lee eet) 
 : oe Lu = any mark ¢ 0} 
 
 wVy 
 ! 
 
 
 
 
 
 * Holder, Mathematische Annalen, vol. 34. 
 + Compare Jordan, 1. c. Arts. 315-317; Betti, 1. c. vol. 3, p. 74, 1852. 
 PM ° 
 
136 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 The substitutions of /7 may be exhibited thus: 
 
 m nm mm” 
 (ex, pps hone a fant 8s Ve mlat 7 ge 8M is 
 Sy We ele es. \° bie a ae = AyjS5 . =F y5; om (et Ohne =F yjSj . 
 j-l yt Fil 
 
 , 
 
 es 
 
 Cae 
 can 
 
 The order of //, being the total number 2(m, n, p) of the substitutions 
 in G divided by the number of those leaving the systems invariant, is thus 
 Q(m, n, p)/(p” — 1). 
 
 15. The linear fractional group // is clearly meriedrically isomorphie with 
 the linear homogeneous group G. To the subgroup G” formed of those sub- 
 stitutions of G whose determinant is unity there corresponds a subgroup /T7’ 
 formed of those substitutions of /7 whose determinant is unity, or, as is equiv- 
 alent, whose determinant is any th power in the G/’[ p"]. For the substi- 
 tutions of /7 remain unchanged by multiplying the m homogeneous indices by 
 the same mark yv, which multiplies the determinant by yp”. 
 
 Let d be the greatest common divisor of m and p" — 1. Then* the 
 proportion of marks (excluding zero) of the G/’[ p"|] which are mth powers 
 is 1:d. Hence the order of 7’ is 2(m, n, p)/d (p” — 1). 
 
 Now,t the maximal self-conjugate subgroup of G" is, for (sn, n, p) ¢ (2, 1, 2) 
 or (2, 1, 3), the group of substitutions which multiplies every index by the 
 same mark. The subgroup of 7’ isomorphic with it is evidently 1. Hence 
 the group //7’ of the linear fractional substitutions on mm — 1 indices having 
 determinant unity (or any mth power) is a stmple group of degree (p"" — 1)/ 
 (p" — 1) and order 2 (m, n, p)/d (p” — 1), with the two exceptions (m, n, p) 
 = (2, 1 2jiand. (2515-3): 
 
 16. The triply-infinite system of simple groups thus reached affords a 
 direct generalization of the doubly-infinite system obtained} differently, 
 
 BOD) ep a tty eon) 
 P(e" —1), p=%, (wnt, 1). 
 
 
 
 *Proof. A mark / + 0 is an mth power in the GF'[p"] if and only if 
 (p” —1)/m 
 
 p = ih, @)) 
 By raising (1) to the power m/d, we find a necessary condition, 
 p —1)/d 
 ui? ye 1a: (2) 
 
 But the condition (2) is sufficient. For, m/d being prime to p” — 1, the extraction in the G#H'[ p”] 
 of the root mid is (by Part I, § 18) always possible and indeed uniquely. Hence from (2) we 
 derive (1). Thus the ( p” —1)d marks satisfying (2) [cf. Moore, 1. c. § 32] give the mth powers 
 in the GF'[ p”]. 
 
 t Section I, §§ 8-12. 
 
 {E. H. Moore, A doubly-infinite system of simple groups, Bulletin of the New York Mathemat- 
 ical Society, vol. 3, pp. 73-78, 1894. 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. tor 
 
 
 
 Thus 
 2 (2, n, D) ra Gae ies 1) Gis tes Ve) rs p" Ge 1) 
 d(p" —1) d(p" — 1) d 
 where d is the greatest common divisor of 2 and p" — 1. 
 
 17. Our triply-infinite system of simple groups yields exactly 13 having 
 an order < 10,000, viz :* 
 
 60: = ((2, 1, 5)) = ((2, 2,2)) = 4.5! 
 TGS (els 1) ((8e162)) 
 360 ent), 22 3)\=d 6! 
 
 504 = ((2, 8, 2)). 
 
 Gale (2.1 11). 
 1092 = ((2, 1, 18)). 
 2448 =- ((2, 1, 17)). 
 3420 = ((2, 1, 19)). 
 4080 = ((2, 4, 2)). 
 5616 = ((3, 1, 3)). 
 6072 = ((2, 1, 23)). 
 7800 = ((2, 2, 5)). 
 9828 — ((2, 3, 8)). 
 
 The simple group of order 4 . 7! = 2520 is not found in this list. Of interest 
 is that of order 
 
 20160 = ((4, 1, 2)) = ((8, 2, 2)) =4.8! 
 
 By noting that the number 2 (m, n, p)/d(p" — 1) contains the factor p 
 exactly to the power nm (m — 1)/2, we may prove that 4. V! is not of the 
 form ((m, %, 7)) for 8 <n <= 16: 
 
 Dnusstoney == 901 O28 335 p20. <Evidently » +:5; + °7,-since. then 
 feat ae li p-— 3; then 2m (m — 1)/2 — 4, so thatim— 2;n — 4; 
 while ((2, 4, 3)) = 2*.3*.5.41. Lastly p = 2 leads to the possibilities 
 ((4, 1, 2)), ((8, 2, 2)) and ((2, 6, 2)) all of which are excluded. 
 
 Analogously, we may verify that the numbers ((3, 2, 3)) = 27. 3°.5.7. 138, 
 oero ee ean nse (ro, cow, ,o)))—ta eo. LS ( wi bt, ((4,-2, 2)) = 
 
 *((m, n, p)) is used as an abbreviation for Q (m, n, p)/d( pn — 1). 
 
 
 
138 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 23. OT OTT, and 4; 2, 3))'= 290 OS lo metd ear ce cach elven 
 but once in the form ((7, n, p)). 
 
 It follows that the system of simple groups of order ((m, , p) reached in 
 this paper includes simple groups not in Jordan’s system (x = 1) and not in 
 Moore’s system (m = 2). 
 
 Section Ill.—TZhe Betti-Mathieu Group. 
 Identification with Jordan’s Linear Group, $$ 18-20. 
 
 18. It was shown in Part I, §§ 58-59, that the quantic 
 rix)= Sacre, 
 pe 
 
 belonging to the G/ [ p"”], represents a substitution on its p”” marks if and 
 only if 
 3 fab rues I scan weed 
 
 ah Att vt 
 Pp vi vi 
 Ag?) 26 Age oh peered 
 
 +0. 
 
 
 
 
 
 n(m—1) n(im—l) 
 p p } 
 A 5 A, ) 
 
 ab 
 
 =A pum!) 
 Ht 1 
 
 m— 
 
 Let 7? be a primitive root of the G/'[ p””], thus satisfying an equation of 
 degree m belonging to and irreducible in the G/’[p"]. Then any mark X of 
 the GF [ p””] may be expressed in the form 
 
 a m—1 : 
 ae — Da€;R* 
 t= 
 where the €,;s are marks of the GF'[ p”]. 
 If we thus express A,, XY and A’ in the substitution 
 
 Lies dA eee (1) 
 
 Vil h 
 
 and reduce the powers of # to a degree = m — 1 by means of our irreducible 
 equation, we may equate coeflicients of like powers of /?. Since 
 
 Oy a es (= Op, eee eee) 
 j=0 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS 139 
 
 where the €,’s and the a,,’s belong to the G/’[ p"}. 
 
 19. Inversely, every such substitution (2) on the marks of the GF'[ p"] can 
 be represented in the form (1). 
 
 It is sufficient to prove that a set of marks A,, A,,...., A,, in the 
 GE'{ p"”| may be found such that (1) will transform any given set of m marks 
 of the G/’'[ p””] linearly independent in the GF'[ p”], as 
 
 Pepe Rte) Rm 
 
 into any other such set of m linearly independent marks, 
 
 m—1 . . 
 B, = 58,8 Ce] ae 
 gp 
 
 where thus we have | §;;| +0. For then on account of the property of /(Y), 
 
 germ eee 1, ao LX) (Ages ob AL), 
 
 m—1 
 
 the p”” marks 2'/,;/2' of the G/’'[ p’™] will be transformed into the marks 
 i) 
 
 m—1 
 
 +1,8,; all distinct. 
 
 i=0 
 
 Now the conditions on the A,’s for the above transformation are 
 
 m 
 
 SA(RH)\"™ — B,, (= 0el eT) 
 ei 
 
 which may be solved in the field for A, since the determinant in /? is + 0 [see 
 Part les hie 1) |: 
 
 This theorem may be shown indirectly thus. Generalizing the work both 
 of Betti* and Jordan,t we may prove that the maximum group commutative 
 with (1. e. the group of transformation into itself of) the group 
 
 Gr ae OSA. 6 + C} = 1$; = ye +. 0, ((=90,.. 5 dt —1} 
 
 p 
 
 is on the one hand 
 {A= 34 ga Bt 
 and on the other 
 
 m—l1 
 
 ee Onna y(t —— 0, Le Xml lp 
 
 j=—0 
 
 
 
 * Betti, 1. c. vol. 3, pp. 72-73, 1852, and vol. 6, p. 26, 1855. 
 + Jordan, 1. c. Arts, 118-119. 
 
140 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 20. Order* of the Betti-Mathieu group. If as in § 18 we express X of 
 the substitution (1) in terms of /?, we have 
 
 where 
 
 m é 
 ls = DA ,Ri F prum—1) \ 
 jot 
 
 But if (1) is indeed a substitution, then See § 57 of Part I, X’ is zero if and 
 only if X = 0,1... h&, == 6, ==. "6, = 0 en cet he ococeat yan 
 sufficient condition that (1) fall be a substitution on the marks of the 
 GF [ pe) is that £5, 1), ..., Ln be lenearly independent in the GF p”). 
 To give a more direct proof, let /5, 13,..., [j,_; be a particular set of m 
 marks. Then &,/) takes p” values ; ¢,/, + ¢,/, takes p” values if and only if 
 
 [hearin 
 
 for any mark » in the G/'[ p"). Finally, Say [‘=, takes the required number of 
 wa 
 
 values p™ if and only if (for & = 1, 2,..., m — 1) 
 
 Pet V0, 
 
 7] 
 
 for any set of marks 4, 4, ..., fp 1n the GF'[ p”). 
 
 We may now readily enumerate the substitutions (1). /, may be chosen 
 in p™” — 1 ways, the value /, = 0 being excluded since then X’ = 0 when 
 A=1. After a definite choice of /), we may take [| in p”™ — p” ways, etc.; 
 
 nn 
 
 finally /},, in p’™ — p™"-? ways. Hence the number of substitutions is 
 
 7 
 
 Q (m, 1, P) — (7 _ 1) ane — p") (rare —s Tee) miaet Gas — per) ; 
 
 Mathiews special type of substitution, §§ 21-24. 
 21. Mathieu states in general (1. ¢. vol. 6, p. 304) and proves for the case 
 n = 1 (pp. 294-300) the theorem that from the substitutions, belonging to the 
 GFT joi 
 m—1 5 
 Li = LAA SAL (2) 
 all the substitutions (1) of the Betti-Mathieu group may be derived. 
 22. The condition that (2) shall represent a substitution on the marks of 
 
 
 
 
 
 
 
 * Betti, 1. c. vol. 6, pp. 13 and 27, 1855; Mathieu, 1. c. vol. 6, pp. 282-5 and 302-3 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 141 
 
 the GF'[ p””] is, writing the columns of the determinant of § 18 in reverse 
 order, that 
 
 AH 1, AHP AHS cael re 
 APH ; Av Fe" =te c A ph Fy po coe sa Fe) 
 
 Ap TT Ane Fyre Av yr f Loe AB TT prim—1) 
 ? ? ’ 
 
 
 
 
 
 n(m—1) yn(vm—1) nv prm—1) 77 p2n snin—1) jn(m—1) 
 irae Ff ABT Oh A PD) i De A en Fr oD yy 
 
 shall not vanish. Multiplying the 7th column by H?"~?""~” and subtracting 
 from the 2 + Ist this becomes 
 
 Pliner ee et teas): 0 
 emis 1 0 
 Meio ert. [Oe 1 03 0 
 
 
 
 Ae 0 0 ect ge L ad. ee) 2) 
 
 
 
 BA eae: HT (anos 0 cemeR Os 1 
 
 Expanding with respect to the elements of the first column, the first minor 
 of A” /T7 is 
 (= Leh Pe ee : ie pe) “ae YE vise es (= Liar 
 
 given by the product of the terms in // just above the main diagonal as far as 
 the £th row and the terms 1 in the main diagonal beginning with the (& + 2)th 
 row. ‘The condition is thus 
 
 m—1 
 ole rer alee Oe 
 k=0 
 We may derive this condition directly from the decomposition 
 
 VERA se oe Z) iE {2 (uzLy" oe x, 
 
 j=0 4—0 
 
142 DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 where a, runs through all the marks of the G/[p"]. Thus any mark J of 
 the GH'[ p'™”] satisfies an equation of the form 
 
 m—1 : 
 DR OERA ee aris 
 7=0 
 where a, is a suitably chosen mark of the G/’[ p”]. 
 We seek the condition under which 
 m1 - 
 Z+A +(HZ)""=—0 
 i=0 
 has no solution except Z = 0 in the G/F'[ p™]. But if it has a solution 
 Z, + 0, then 
 m—1 : 
 Si HZ) eye ee 
 fe) 
 
 1 
 
 Hence 
 Ly + Adz, ai) ° 
 so that 
 m—l1 3 . 4 
 2 (Gy, HA)? San A ee 
 i=0 
 Hence 
 
 m—1 A 
 3 (HA ele 
 r=0 
 
 23 Za) = Suz)" has the properties in the GF'[ p””] 
 f(Z,+ 4) =F (Z) + F(Z). 
 [P(Z)]" = F(Z). 
 LLM TAL: ) ae OM ee 7 tes 
 Hence the result of applying first the substitution 
 Le Te Neal Le) (2) 
 and afterwards the substitution 
 Li = LAB HZ) 
 is the same as applying the single substitution 
 | Zi SD +p AR AeA BY neh (Zk (3) 
 Thus the square of the substitution (2) is 
 
 Z=Z+ASI4 F(A F(Z). 
 
DICKSON. ANALYTIC REPRESENTATION OF SUBSTITUTIONS. 
 
 Using (3) we have by induction that the /th power of (2) is 
 
 
 
 7 es Plea CA.) etre) y 
 Leno Fi} WesCZy 
 
 143 
 
 24. The substitution (2) leaves unchanged the p””” letters whose indices 
 
 satisfy the equation 
 
 "S (HZ eu axes 0 : 
 
 If #'(A) = 0, the *#th power of (2) reduces to 
 Lit Ase La), 
 
 when (2) is a regular substitution on p”” — p””™ letters composed of cycles 
 
 of p letters. 
 
 If (A) = a, + — 1, a, being a mark + 0 of the GF'[ p”") and if e is the 
 
 least integer such that 
 
 ieee aay cle 
 
 (2) represents a regular substitution on p”” — p’™™ letters composed of 
 
 cycles of ¢ letters. 
 
 The substitutions (2) constitute the totality of those substitutions (1) which 
 
 affect only p”™ — p”—” letters (Mathieu, 1. c. vol. 6, p. 288). 
 
 UNIVERSITY OF CuHiIcaGco, April 27, 1896. 
 
 
 
 ERRATA. 
 Page 68, line 8, read (py — s + 1)/2; line 9, read shall not. 
 Page 71, line 5 from bottom, read (;,. 
 Page 73, line 10, for y/, read y,,. 
 Pace 8 Peline L937 ioriC,"' read: C,"**: 
 Page 84, note, read Arzthmeticae. 
 Page 89, line 2, for 70 read Ty. 
 Page 90, line 12, for «a,’a,* read a,7a,?. 
 Page 91, bottom, for p”' read p™ — 1. 
 Page 94, middle, read Y?"” 
 Page 95, equation (5), for 7’ read J, @. 
 Page 100, bottom, for 4,,, read @,,,. 
 Page 102, line 19, for n + 1 read n — 1. 
 Page 103, bottom, for $ a,’a° read $ a,7a,. 
 Page 104, line 9, read 2°", 
 Page 110, line 12, read & + 26€?. 
 Page 112, line 12, for a,°a,’ read a,°a,°. 
 
 Page 117, line 1, read $$ 83-86; line 23, for (18) read (1,); line 26, for (1;) 
 
 read (1). 
 Page 118, line 8, read G/'[2"]; line 10, for 73 read > 3. 
 

 
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