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Books of special value and gift books, when the giver wishes it, are not allowed to circulate. 8817 ON PRIMITIVE GROUPS OF ODD ORDER A THESIS PRESENTED TO THE UNIVERSITY FACULTY OF CORNELL UNIVERSITY IN CANDIDACY FOR THE DEGREE OF DOCTOR OF PHILOSOPHY HENRY LEWIS RIETZ BALTIMORE Cge £orb QgSaftimore (fittee THE FRIEDENWALD COMPANY I904 ^ Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924032189692 ON PRIMITIVE GROUPS OF ODD ORDER A THESIS PRESENTED TO THE UNIVERSITY FACULTY OF CORNELL UNIVERSITY IN CANDIDACY FOR THE DEGREE OF DOCTOR OF PHILOSOPHY BY HENRY LEWIS RIETZ BALTIMORE Zfyt Boti> (gaUimott (P«ae THE FRIEDENWALD COMPANY I904 On Primitive Groups of Odd Order. By Henkt Lewis Rietz. Introduction. In his "Theory of Groups of Finite Order " (1897), p. 379, Burnside has called attention to the fact that no simple group of odd composite order is known to exist. Several articles* have recently appeared bearing on this question, in which, among other things, it was proved that no such group can be represented as a substitution group whose degree does not exceed 100. This result was obtained by showing that there is no simple primitive group of odd composite order whose degree falls within the given limits. Burnside determined all the primitive groups of odd order of degree less than lOO.f Since any primitive group of odd order is simply transitiye, a study of simply transitive primitive groups may throw light on the question of simple groups of odd order. Some important properties of simply transitive primitive groups have been given by Jordan, Miller, and Burnside. J The main objects of the present paper are ; first, to make a further study of primitive groups with special reference to those of odd order ; secondly, to extend the determination of the primitive groups of odd order to all degrees less than 243. It results that all groups arrived at in this determination are solvable. From this result it is evident that no simple group of odd composite order can occur * Miller, Proc. Lond. Math. Soc, Vol. 33, pp. 6-10. Burnside, Proc. Lond. Math. Soc, Vol. 33, pp. 162-185 ; 257-268. Frobenius, Berliner Sitzungsberichte (1901), pp. 849-858 ; 1216-1230. f At the time of the publication of this work, I had also made this determination with the same results. % Jordan, "Traite des Substitutions," pp. 281-284. Miller, Proc. Lond. Math. Soc, Vol. 28, pp. 633- 542. Burnside, loc. cit., pp. 162-185. 1 2 Rietz: On Primitive Groups of Odd Order. as a substitution group of degree less than 243, since if a simple group is repre- sented as a substitution group on the minimum number of letters, it is primitive. Part I contains a number of theorems, most of which apply to primitive groups whether the order is even or odd, but some use can be made of nearly all of them in determining all the primitive groups of odd order of a given degree. Part II contains the determination of the primitive groups of odd order whose degrees lie between 100 and 243. I desire to acknowledge my indebtedness to Professor G. A. Miller for help- ful suggestions and criticisms during the preparation of this paper. Part I. §1. — On the Number of Substitutions of Degree less than n contained in any Transitive Group of Degree n. Let G be any primitive group of composite order g on the elements a x , a z , , a n , and G, the subgroup leaving a given letter a s fixed. If n — A a is the degree of any substitution of G s , and fi a the number of substitutions of G e of this degree, then the total number of substitutions of degree less than n con- tained in G is m + mi + 7 VL+ .... +m? or n (j± + bl + j& + . . . . + j±\ , Aj Ag A3 A p \ Aj Ag A3 Ap / where p is the number of different degrees occurring among the substitutions of G s . Since ^ + [i z + [i 3 + .... + fi p = ^- , the above summation in the paren- n 1 theses may be considered as the sum of just -^- terms of the form — . "We may n A then rewrite the above expression for the number of substitutions of degree less than n in the form a a = 2- Let x denote the number of systems of intransitivity of G s , and look upon Cr 8 as having just one system of intransitivity when it is transitive. Then x + 1 * Jordan, Liouville's Journal, Vol. 17 (1872), p. 352. Rietz : On Primitive Groups of Odd Order. 3 is the average value of K since the average number of letters in the substitutions of an intransitive group is equal to the excess of the degree over the number of systems of intransitivity.* Hence we have 1 = l 2> n The Xs in this summation cannot all be equal, since identity is included among the substitutions of O, . Since the arithmetic mean of any number of positive quantities which are not all equal is greater than their geometric mean, it follows that a=£ n 2>» -l_>-C/a 1 X 8 ....^ (3) g n and « = £ Jl*Lt/JLJ_.. * " »/ 1 1 1 " > V^-^-----v (4) n From (3) and (4) it follows that 9_ n l= £ From (2) and (5) it follows that a = •Jordan, Comptes Rendus, Vol. 74 (1872), p. 977. Frobenius, Crelle's Journal, Vol. 101 (1887), p. 288. i = l , = £. (6) 4 Rietz: On Primitive Groups of Odd Order. From (1) and (6) we obtain Theorem 1. — In any primitive group G of degree n of composite order g there are more than — ^ — substitutions of degree less than n, where x is the number of x + 1 systems of intransitivity of the subgroup which leaves a given letter fixed. In particular, for a multiply transitive group, x = 1. Hence, Cor. 1. In a multiply transitive group of degree n more than one-half of the substitutions are of degree less than n . Cor. 2. If G is of degree kp (p a prime) and of order mp (m prime to p and p — 1), the subgroup G s has at least p + 1 transitive constituents. For a group of this order contains exactly m operators whose orders divide m.* But all the substitutions of degree less than kp would be of orders prime to p. Hence from the above theorem we have x >■ J_ or x ~^>p — 1 . Since mp must clearly be an odd number, x must be even. Hence, x>p+l. While it is not our object to treat imprimitive groups, the above theorem can at once be extended to any non-regular transitive group. The only change in the argument is the substitution of x + m in expression (l) for x + 1, where m represents the number of letters of the transitive group left fixed by the sub- group which leaves a given letter fixed. Hence, Theorem 2. — In any non-regular transitive group of degree n of order g there are more than —£- — substitutions of degree less than n, where x and m are defined as above. When applied to known groups, I find that in many eases this simple formula gives very nearly the actual number of substitutions of degree less than the degree of the group. * Frobenius, Berliner Sitzungsberichte (1895), p. 1035. Rietz : On Primitive Groups of Odd Order. 5 §2. — Restrictions on the Order of G, when G s has a Transitive Constituent of Degree p, p a , pm or pq (p and q primes and m 0) ; this must be the case if the theorem is not true. It will be shown that this hypothesis leads to a con- tradiction. In H s all the substitutions whose orders are powers of p would gene- rate a group H' s of order 7Jp m invariant in G s . In the conjugate G r of G s , leav- ing fixed an element of T lt there occurs just 1/t of the substitutions of G„. Hence the subgroup Hi would be one of a set of t conjugates transformed by G r according to one of its transitive constituents T of degree t. In the invariant subgroup HI of G r corresponding to identity in T, all the substitutions are com- mon to G r and G s , since they transform H' s into itself. Now E' r would be of order %"p m ( a," prime to p) . Since in T x all the substitutions whose orders are not prime top are of degree t, all the substitutions whose orders are powers of p common to G r and G s are contained in H, . If H' r contained all the substitutions whose orders are powers of p which occur in H s , the subgroup H[ would be invariant in G s and G r . But this is impossible, since these subgroups are maximal. If Hi. contains only part of these substitutions, let P be such a substitution not contained in Hi. The order of * Miller, loo. cit., pp. 534, 535. t Jordan, loc. cifc, p. 284. 6 Rietz : On Primitive Groups of Odd Order. \H' r , P\ would then be divisible byp m + 1 and there would be common to G s and G r subgroups of order p m + 1 , which is impossible, since, by hypothesis, the order of H s is not divisible by p m + \ Hence the theorem. Cor. 1. If G s has a transitive constituent of prime degree p, the order of G s is not divisible by p 2 . Cor. 2. If any number of the transitive constituents of H s are of a given prime degree p, the constituent group formed of all these transitive constituents is formed by establishing a simple isomorphism between them. Cor. 3. If in G 8 all the transitive constituents of a given degree p a are of class p a — 1 , the order of G s is not divisible by p a + \ Cor. 4. If in G s all the transitive constituents of degree mp (p >■ m) have p sys- tems of imprimitivity, the order of G s is not divisible by p 2 . Lemma. When p and q are distinct primes each of the form 2 m + 1, there is no imprimitive group of degree pq of odd order whose order is divisible by both p 2 and q* ; and there is no primitive group of degree pq involving in its order only the primes p and q . The part of this lemma which relates to the imprimitive groups follows at once from the fact, that the only transitive groups of degrees p and q whose orders are odd are the cyclical groups of orders p and q. Suppose there is a primitive group of degree^ of order p a 'q°*. The maximal subgroup G u leaving a given letter fixed, is then of degree pq — 1 and of order p ai ~ 1 q ai ~ 1 . Take p^>q, then, since p 2 ^>pq — 1 , no transitive constituent can be of degree^> v (y ]>1). The transitive constituents cannot all be of degree p, since p is not a divisor of pq — 1. Since pq — 1 is not divisible by q, we may assume that some of the transitive constituents are of degree p while others are of degrees equal to a power of q. But the order of a transitive constituent of degree pis p, and would therefore not contain q as a factor, but every prime which divides the order one transitive constituent of G 1 divides the order of each of its transitive constituent. Theorem 4. — If p and q are distinct primes of the form 2" 1 + 1 , and if G s is of odd order, and has as a transitive constituent an imprimitive group of degree pq ; then, according as T has p or q systems of imprimitivity, the order of G s is not divisible by p 2 or q z . To make the conditions definite, suppose that T has q systems of imprimi- tivity. These systems are then permuted according to the cyclical group of Rietz : On Primitive Groups of Odd Order. 7: order q, and all the substitutions in the tail of T are of degree pq. Corre- sponding to identity in T, there is in G a an invariant subgroup H s of degree n — a (a <^ pq -\- I) . If we can show that the order of H 8 is not divisible by q % our theorem is proved. Let G T be the conjugate of G s which leaves fixed an element of T. Also let R s be the invariant subgroup of G s corresponding to the head of T. In G r the subgroup H s is one of a set of pq conjugates transformed by G r according to a transitive constituent T x of order p a 'q a *. According to the lemma, T x is imprimitive and its order is not divisible by both p 2 and q 2 . The subgroup T lt leaving a given letter fixed, would leave more than one letter fixed. Hence in G r the subgroup H, is transformed into itself by some of its conjugates. Let H Si be one of these conjugates such that H'~ 1 n,H ai = E 8 . H ti then occurs in both G s and G r . Hence, it occurs in B s . If R s contains operators of order q, they clearly occur in H s . Hence, H, and H ti have the same substitutions of order q. But H g is invariant in G s and H Si in G Si . The substitutions of order q in H s would then generate a group invariant in both G s and 6r Si . But this is impossible, since G g is maximal. Hence the theorem. • §3. — On Certain Subgroups Contained in G. Let p a be the highest power of a prime p which divides the order of G . and suppose that the number p is prime to n, the degree of G. Let P be any subgroup of order p a . It must be contained in some of the subgroups (•?!, G 2 , .... , G n , leaving a given letter fixed, since its degree is prime to n. If P is of degree n — a, (a, > 1), it is proved by Burnside ("Theory of Groups," p. 202), that the subgroup of G, which contains all the substitutions of G which transform P into itself, permutes the /I elements not occurring in P transitively. It is our object to consider the case % = 1 . Let P' be a subgroup of order p* common to any two of the subgroups G it G z , G 3 , .... , G n such that there is no subgroup of order p y (y >-/3) common to any two of these subgroups. We shall first assume ft > . P' must be contained in subgroups P x and P 2 of order p a in those subgroups which leave a given letter fixed in which it occurs. Since, in a subgroup of order p a , any subgroup P' is transformed into itself by operators of the group not contained in P', it follows that P' is invariant in a subgroup P" of P x which is of degree n — 1 . Likewise in P 2 the subgroup P' is invariant in a subgroup P'" of degree n — 1. Hence, the subgroup P' is invariant in \P", P"'\ 8 Rietz : On Primitive Groups of Odd Order. of degree n. Since n = 1 raod^j, the number of elements of G not occurring in P is congruent to unity mod p . Also, since P" and P" are each of degree n — 1, it follows that \P", P'"\ has a transitive constituent of degree 1 + kp (&>0), formed of elements not occurring in P, and whose order is multiple of p. When (3=0, the subgroup P is clearly formed by establishing a simple isomorphism between regular groups. Hence, Theorem 5. — If p a is the highest power of a prime p which divides the order of G, and if a subgroup P of order p* is of degree n — 1 , then, unless P is a regular group or is formed by establishing a simple isomorphism between regular groups of order p*, G contains an intransitive subgroup of degree n having a transitive con- stituent of degree 1 -f- Jcp (Jc > 0) and of order Ip. Cor- lfp a is the highest power of a prime p which divides the order of G s , and if the degree of each transitive constituent of G s is divisible by p 1 *, but at least one of them is not divisible by p p + 1 , then either a = (3 or the group G contains a subgroup of degree n having a transitive constituent of degree 1 -f- lep {Jc >• 0) and of order equal to a multiple of p. It may be observed that the theorem and corollary just stated apply to any transitive group in which the subgroup which leaves a given letter fixed leaves only one letter fixed, as well as to a primitive group. §4. — On the Transitive Constituents of G s . Theorem 6. — If G s has an invariant subgroup H s of degree n — a (a > 1) , then G s has at least one transitive constituent whose degree exceeds the degree of any transitive constituent of M s . Suppose, if possible, that H s has a transitive constituent T such that its degree is equal to the degree of the transitive constituents of G s of largest degree. Consider a conjugate G r of G s , leaving fixed an element of G e not occurring in B„. Since H 8 occurs in both G 8 and G r , these two groups have at least one transitive constituent in the same elements, i. e., in the elements of T. The group \G r , G,\ would then be intransitive. But \G r , G,\ must be identical Bietz : On Primitive Groups of Odd Order. 9 with G, since G s is maximal. Hence the hypothesis that H s has the transitive constituent T leads to an absurdity. Cor. If all the transitive constituents of G 8 are primitive groups of the same degree t, then G, is formed by establishing a simple isomorphism between these tran- sitive constituents. This follows readily from the theorem if we remember that every invariant subgroup of a primitive group is transitive. Theorem 7. — If G s has as a transitive constituent a regular group T of degree t , and if the order of G s exceeds t , then G, has another transitive constituent of degree t which has the property that its subgroup which leaves a given letter fixed permutes all the remaining letters. Consider the invariant subgroup H s of G, corresponding to identity in T. In a conjugate H r of G s , leaving fixed an element of T, there occur just 1/t of the substitutions of G s and the subgroup H s is one of t conjugates transformed according to a transitive constituent T x . If, in the group T lt the subgroup which leaves a given letter fixed, leaves more than one letter fixed, H s is transformed into itself by some of its t conjugates under G r . But the substitutions of E s are the only substitutions common to G r and G,. Hence, the transitive constituent T has the property mentioned in the theorem. Theorem 8. — If G s has X systems of intransitivity, and if H s is the invariant subgroup of G s corresponding to identity in any transitive constituent T, while G s transforms H Si , H H , .... , H s _ (denned as in §2) according to a constituent group having u systems of intransitivity, then H s has more than — systems of intransi- tivity, excepting when /x = 1 , and then it has at least X .* If H, has as few as — systems of intransitivity, the [i conjugate sets under G s into which H s , H $ , , H s are divided could, at most, contain elements from f l)(i + l = a, — [i + l of the /I systems of intransitivity of G s , since each of the subgroups H Si , H H , ^ a -i must contain at least a cycle from T. * Of. Miller, loc. cit., p. 535 10 Kietz : On Primitive Groups of Odd Order. These subgroups could not then generate a group of degree n — 1 unless (i = 1 . Hence, by means of 1, §2, the theorem follows. Cor. If all the transitive constituents of G„ are primitive groups, H Si , H H , . . . . , H a cannot be a single conjugate set under G g . §5. — On the Transitive Constituents of G s when the Order of G is Restricted to be an Odd Number. Burnside recently proved the interesting theorem* that, if G is of odd order, G s has its transitive constituents in pairs of the same degree. Let a Si , a H , a H , , a st be the elements of any transitive constituent of degree t. The above theorem was proved by considering the quadratic function 8 = n /= 2 a * K + a n + •■■■ + «J, s = l which is transformed into itself by all the substitutions of G. In this summa- tion, a s occurs in the parentheses exactly t times. Hence the function /may also be written S = 71 /=2 (°« + a ** + ' • • • + °o a *> s=l and it is shown in the proof of the above theorem that the elements a si , a ai , a K , .... , a H are elements of a transitive constituent T of G s distinct from T. The constituents T and T' will be spoken of as a " pair of transitive constituents." Use will be made of the two ways in which / is written to prove some theorems in reference to the transformation by G, of its subgroups E Si , H 8i , . . . . , E s _ (defined as in §2) when H s corresponds to identity in T. It is known (p. 5) that these a — 1 subgroups are transformed by G 8 according to one of its constituent groups. But it is not known whether this constituent group ever contains elements occurring in E s . Form the conjugate G Sa of G s , leaving fixed an element of T. From the two ways of writing /, it is seen that in G Sa the element a s occurs in the transform of T' ; i. e., in B~ 1 T'B, where B is such that B~ l G s B = G Sa . But H s is transformed by G in the same manner as a 8 is replaced. Hence, * Loc. cit., p. 163. Eietz : On Primitive Groups of Odd Order. 11 Theorem 9. — Some of the subgroups E H , E H , .... , E s _ are transformed according to T 1 when H s corresponds to identity in T. Cor. If the subgroups E Si , E ti , .... , E t _ are a single conjugate set under G e , they are transformed according to T 1 when H, corresponds to identity in T. Suppose, next, that G s has only two transitive constituents T and T. If, cor- responding to identity in one of these constituents, say T, there is in G t an inva- riant subgroup E s , the subgroups E 8i , E H , .... ,E Snl (Cor., Theor. 9) are trans- formed by G s according to the elements of E s . Then, for any two of the n subgroups Si, E % , .... , E n , which are conjugate under G, one of two relations Hf^H f -H. or H7 l E^=H, (1) holds, but both cannot hold for any two of the subgroups. Let x be the number of elements common to E a and E fi ; then x is clearly the number of elements common to any two of the H's. Also, let x + y be the degree of H s . Let a lt a 2 , . . . . , a y , b 1: b 2 , . . . ■ , b x be the elements of H s , and b 1: b it . . . . , b x , c i> c 2> ••'•'• > c y * ne elements of H tl , one of the subgroups E h , E H , . . . . , H, n _ 1 con- tained in G s . Since E Si must be transformed according to an element of E e not contained in E h , it must be transformed according to one of the a's. There must be substitutions in E„ which do not transform E !t into itself. If iS is such a substitution, then S~ 1 E S] S contains all the a's, since E Si and /S~ 1 E h S have just x elements in common. But since a Sl , according to which E Si is permuted, occurs in S~ 1 E Sl S, this latter subgroup cannot transform E Sl into itself. By exactly the same reasoning E h cannot transform S~ 1 E H S into itself. But this is contrary to relations (1). Hence, Theorem 1 0. — If, in a primitive group G of odd order, the subgroup G s has only two transitive constituents, G„ is formed by establishing a simple isomorphism between them. Theorem 11. — If, in a primitive group G of odd order, G s has not more than four transitive constituents, and if these are all primitive groups, then it is formed by establishing a simple isomorphism between them. Since G s has an even number of transitive constituents, we need consider only the cases where it has two or four transitive constituents. Since any inva- riant subgroup of a primitive group is transitive, and since a simply transitive 12 Rietz : On Primitive Groups of Odd Order. primitive group of degree n cannot have a transitive subgroup of degree less than n, the theorem follows at once when G, has only two transitive constituents. If G s has four transitive constituents,- and is not formed according to the theorem, there corresponds to identity in some transitive constituent T of degree t an intransitive subgroup H s invariant in G s . It has two or three systems of intransitivity. Suppose, first, that H, of degree n — a has three systems. Then a — 1 = t. In the conjugate of G s , leaving fixed a letter of T, the subgroup H is one of t conjugates. But these a — 1 subgroups cannot be conjugate (Cor., Theor. 8). It remains to consider the case where H s has two systems of intran- sitivity ; then n — a (the degree of H s ) is an even number. Hence a — 1 is an even number and the a — 1 subgroups E x , ff Si , , H s _ could only be trans- formed according to a group T having two transitive constituents. But by Theor. 8 this is impossible. Hence the theorem. §6. — Certain Primitive Groups of Odd Order contained in the Hohmorjph of the Abelian Group P of Order p m (p an odd prime) of Type (1, 1, , l). Represent P as a regular group. Suppose that the order of its group of isomorphisms L is divisible by q n (q an odd prime). To any subgroup of order q n in L there corresponds in the holomorph of P a transitive group of degree p m and of order p m q u . The subgroup of this transitive group, which leaves a given letter fixed, is of order q n , and is clearly maximal, if m is the index to which p belongs mod q. Hence, Theorem 12.— If p m = 1 mod q n (n <£ l), m being the index to which p belongs mod q, there is a primitive group G of order p m q n contained in the holomorph of the abelian group of order p m of type (1, 1, , 1). Cor. 1. If q n is the highest power of q which divides p m — 1, there is only one group G satisfying the above conditions. Cor. 2. If p — 1 (p zp, 3) is not divisible by 3, there exists a primitive group G of degree p 2 and of order Zp % . Furthermore, if p — 1 is divisible by 3 , there is no primitive group of degree p* whose order is 3p z . The first part of this corollary is merely a special case of the general theorem. The second part may be proved as follows : By Sylow's theorem, a group of this order contains a single subgroup P of Rietz : On Primitive Groups of Odd Order. 13 order p*. Since G is primitive, an invariant subgroup P must be transitive. The subgroup is, therefore, regular, and it must be the non- cyclical group of order p 2 . P would contain p + 1 subgroups of order p, these would have to occur in conjugate sets of three, since G cannot contain an invariant intransitive subgroup. But p + 1 is not divisible by 3 when p — 1 is divisible by 3 . §7. — On the Class of Primitive Groups G of Odd Order. By the class of a substitution group is meant the smallest number of ele- ments in any one of its substitutions besides identity.* Let n — [i represent the class of G. For all odd values of fi less than 7 there exist groups G of odd order of class n — p. Thus : For fi = 1 , in any non-cyclic invariant subgroup of a metacyclic group. For fi = 3, in the primitive group of degree 27 of order 27.39.f For (i = 5, in the primitive group of degree 125 of order 125.93.| It will now be shown that there is no primitive group of odd order in which (i is even and less than 6. G s has an even number of transitive constituents in pairs of the same degree (p. 10), and is clearly formed by establishing a simple isomorphism between its transitive constituents. If p = 2 , at least two of the transitive constituents must be non-regular, since they are in pairs of the same degree. Suppose that t x is the degree of one of these non-regular transitive con- stituents. It must clearly be of class t x — 1 . In the constituent of degree 1t y formed by combining these two, every substitution of degree t x — 1 would corre- spond to a substitution of degree t x , or G would contain substitutions of degree n — 3. But this is clearly impossible, since a transitive group of degree t t and of class t y — 1 , the order of the substitutions of degree t x is prime to the order of those of degree t t — 1 . It remains to consider the case when (i = 4. Here again G s must be formed by establishing a simple isomorphism between transitive constituents, not all of which can be regular. If t x is the degree of any non-regular transitive constitu- ent, this constituent must either be of class t t — 3 or t x — 1 . Suppose that all * Jordan, Liouville, Vol. 16 (1871), p. 408. t Burnside, loc. cit., p. 180. % See p. 30 of this paper. 14 Rietz: On Primitive Groups of Odd Order. the non-regular transitive constituents are of class one less than their degrees. Since there must be an even number of such transitive constituents, it readily follows that G 8 cannot have more than two such transitive constituents or G would contain substitutions of degree less than n — 4. But in this case G would have no substitutions of degree less than n — 3 . Hence G s must have at least one transitive constituent T of some degree t s of class t z —3. Now, G„ must have at least one more transitive constituent T' of degree t t . This constituent must be of class t z — 1 or t 3 — 1 . In combining T and T' into a constituent of degree 2t z , all the substitutions of degree t 2 — 3 in one must correspond to sub- stitutions of degree t z in the other or G would contain substitutions of degree less than n — 4. From this it is easily seen that substitutions of degree t z — 3 and those of degree t z to which they correspond must be regular. Hence all the substitutions of degree t % — 3 would be of order 3 . Consider the subgroup P s of G 8 corresponding to a subgroup of To? degree t 2 — 3 of order 3° such that there is no subgroup of T of order 3 a + 1 which is of degree t z — 3 . P 3 would then be invariant in a subgroup of G of degree n , and the 4 letters of G not occurring in P s would be transitively connected so that the order of G must be an even num- ber. Hence there is no primitive group of odd order of class n — 4. Part II. §8. — On the Primitive Groups of Odd Order of Degree less than 243. It is known* that all transitive groups of odd order of prime degree are invariant subgroups of the metacyclic group. Inasmuch as these groups are well known, we shall consider only those primitive groups whose degrees are not primes. As already stated in this paper (p. 1), the primitive groups of odd order have been determined for degrees not exceeding 100. It is the object of this part to extend this determination to all degrees less than 243. "We are then concerned with the groups whose degrees lie between 100 and 243. It is stated without proof by Burnside (loc. cit., note, p. 185) that any transitive group of odd order of degree 3/> (p a prime) is imprimitive. We have examined all the com- * Burnside, loc. cit., p. 177. Rietz : On Primitive Groups of Odd Order. 15 posite numbers within the given limits for the degrees of primitive groups of odd order, and the results agree with this statement. We shall, therefore, for the sake of brevity, omit these degrees. Represent by G n a primitive group of odd composite order of degree n . As in Part I, let G, denote the subgroup of G n containing all the substitutions which leave a given letter a a fixed. The method is, briefly, as follows : For each odd degree n (n not a prime nor 3 times a prime) it is assumed that a group G n exists. The degree of any solvable primitive group is a power of a prime.* Hence, in order to prove that no group G n exists, for a given value of n which is not the power of a prime, it is sufficient to prove that there is no simple group G", provided there is no simple group of odd composite order of degree less than n . As the latter condition is satisfied, it is further assumed that G n is simple when it is not a power of a prime. We write down for examination all those, and only those, systems of intran- sitivity of G s which are not excluded by the conditions, 1°, that every prime which divides the order of one transitive constituent divides the order of every transitive constituent ;f 2°, that the transitive constituents occur in pairs of the same degree ;J 3°, that when G n is simple, there can be no transitive constituent whose degree is a prime of the form 2 m + 1 ;§ 4°, that if the degree of one tran- sitive constituent is a prime of the form 2™ + 1 , all the transitive constituents are of this degree. This method of excluding transitive constituents of G s depends, of course, on a knowledge of the primes which occur in the orders of transitive groups of odd order of degree less than w/2. The following table shows the primes which may occur in the orders of transitive groups of odd order of degree less than 120. The primes written under the degree are the primes which occur in the orders of some transitive groups of odd order of the given degree in addition to those primes contained in the degree itself. * Lettre de Galois, a M. Auguste Chevalier, Liouville's Journ. (1846), p. 41. t Jordan, loc. cit., p. 284. t Burnside, loc. cit., p. 165. § Miller, loc. cit., p. 6. 16 Rietz : On Primitive Groups of Odd Order. Degree : 3 5 7 9 11 13 15 17 19 21 23 Primes : 3 5 3 3 11 Degree : 25 27 29 31 33 35 37 39 41 43 45 Primes : 3 13 7 3, 5 5 3 3 5 3, 7 Degree : 47 49 51 53 55 57 59 61 63 65 67 Primes : 23 3 13 29 3, 5 3 3, 11 Degree : 69 71 73 75 77 79 81 83 85 87 89 Primes : 11 5, 7 3 3, 5 3, 13 5, 13 41 7 11 Degree : 91 93 95 97 99 101 103 105 107 109 111 Primes : 3 5 3 3 5 5 3, 17 53 3 Degree : 113 115 117 119 Primes : 7 11 3. We shall use &, S 2 , . • • • , S t to represent the sets of systems of intransi- tivity of G s which are not excluded by the conditions above stated. The tran- sitive constituents thus obtained are either shown to lead to impossibilities or the orders and number of the groups are determined. \ 2). But G m would then contain in its order less than 7 prime factors, and could not be a simple group.* If G, has systems S it the order of G would be 5 . 13 . 7 . 3" (Cor. 1, Theor. 3). Let p = 5 in Cor. 2, Theor. 1, and it follows that G m does not exist. G m . /8i= 19, 19, , 19. # 2 = 57, 57. G m would be of order 3 a . 5.19'. 23. Let p = 23 in Cor. 2, Theor. 1, and it follows that G 11 * does not exist. * Burnside, loc. cit., pp. 265-268. Rietz : On Primitive Groups of Odd Order. 17 G m . S x = 29, 29, 29, 29. The order of G 8 would be, by Cor. 2, Theor. 3, 29 . T (a > 1). • But #" 7 would then contain in its order less than 7 primes, and could therefore not be a simple group. G m - S 1 = 59, 59. G m would be of order 17.7. 59*. 29*. Letp = 17 in Cor. 2, §1, and it fol- lows that G n9 does not exist. G m . ^=3,3, . .,3. #3=15,15 15. S z = 5, 5, ,5. S i =15, 15, 45, 45. If G s has systems Si the order of G m would be 121. 3.* By Cor. 2, §6, there is a group of order 3 . 121, and it is easily seen that there is only one such group. With S x , the order of G 121 would be 121. 5. Any group of this order contains at least one invariant subgroup of order 11. As this subgroup would be intransitive, it cannot occur in a primitive group. If G s has systems S 3 or S it the order of G m would be of the form ll 2 . 3°. 5^ If any transitive constituent of degree 15 has 5 systems of imprimitivity the order of G s is not divisible by 5 2 (Theor. 4). G m would then be of order ll 2 . 3 a . 5. The subgroup of order 3 a would then be of degree 120. If a >■ 1, from §3, G m would contain an intransitive subgroup having a transitive constituent of degree 1 -f- 3k (k > 1), and whose order is divisible by 3. The number of the form 1 + 3k could only be 55 and be a divisor of the order. But no transitive group of degree 55 of odd order has its order divisible by 3. The transitive constitu- ents of degree 15 must then all have 3 systems of imprimitivity, and 6r m would be of order 1 21 . 3 . 5 3 . If /3 >-l , it follows from §3 that a subgroup P of greatest order common to two subgroups of G m , leaving given letters fixed, is invariant in a subgroup of order ll 2 . 5", 11 . 3 . 5 ? or 11 . 5 3 . The conjugate set to which * Miller, loc. cit., p. 536. 18 Rietz: On Primitive Groups of Odd Order. P belongs under G m would then be transformed by G m , according to a transi- tive group of degree 3, 11 or 33. Now, it is easily seen that /?> 2. We shall show that one group G m exists when (3 = 1, and none when (3 =■ 2. A group of order 1 l 2 . 5 3 . 3 ((3 > 2) must, by Sy low's theorem, contain an invariant subgroup P m of order ll 2 - This subgroup in G m must be of type (1, 1), since the cyclical group would contain a single subgroup of order 11 which would be an invariant intransitive subgroup of G 1Z1 . The group G m must then occur in the holomorph of P m . The group of isomorphisms L of P m is of order 120 . 110 = 3 . 5 2 . 2 4 . 11. Now, L contains just 55 subgroups* of order 3, each being invariant in a subgroup of order 3 . 5 . 2 4 - Hence L contains subgroups of order 15 and they are all conjugate. It results, therefore, that the transitive groups of order 121 . 15 in the holomorph of G m are conjugate. The subgroup of order 15 which leaves a giving letter fixed, is clearly maximal. Hence the group is primitive. Similarly, examining L, it is found to contain no subgroups of order 3.5 a . Hence there is no group G m of order ll 2 . 3.5* contained in the holomorph of P m . G m . /Si = 31, 31, 31, 31. The order of G m would be 5 a . 31 . 3" (a = 3 or 4, (3 = or 1, Cor. 2, §2). By Sylow's theorem, G m would contain 1 or 31 subgroups of order 5\ If it contained only one, a — 3; for, if a== 4, G s and its conjugates would contain more than 5 4 substitutions of order 5. If it contained 31 subgroups of order 5°, they would be transformed by G m according to a non-cyclic transitive group of degree 3 1 isomorphic with G m . The subgroup of G m corresponding to identity on this quotient group would contain a single subgroup of order 5 3 . Any group 6r 125 then contains an invariant subgroup P m of order 125. The group P m must be the abelian group of type (1, 1, 1), since all other groups of order 125 con- tain characteristic subgroups, and a characteristic subgroup of P m would be an invariant intransitive subgroup of G m . The group G m is then contained in the holomorph of P m . The group of isomorphisms L of P m is of order 2 7 . 3 . 5 3 . 31. The group L contains, by Sylow's theorem, 1, 2 5 , 5 3 or 2 5 . 5 3 subgroups of order * This is shown from the composition series of L . The factor groups of L are the cyclic group of order 10, the simple group of order 660 representable on 11 letters, and the group of order 2. Rietz : On Primitive Groups of Odd Order. 1 9 31. That it could not contain 1 or 2 5 such subgroups is easily seen from tne fact that L occurs as a transitive group of degree 124. In I a subgroup of order 31 is then invariant in a subgroup of order 31 . 2 3 . 3 or 31 . 2 7 . 3. Hence L contains subgroups of order 31 and 31 . 3, but it does not contain any subgroup of order 31 • 5 or 31 . 15. Corresponding to these subgroups, there are in the holomorph of P m transitive groups of orders 5 3 . 31 and 5 3 .31.3. These groups are evi- dently primitive groups. That there is only one group G m of each of these orders follows from the fact that in L the subgroups of each of the orders 31 and 31.3 form a single conjugate set. G m . $ = 11, 11, ,11. #4 = 27, 27, 39, 39. S z = 11, 11, 55, 55. # B = 27, 27, 13, 13, , 13. S t = 33, 33, 33, 33. If G s has systems S t or S z , the order of G m would be of the form 19 . 7 . 11". 5". Let p — 19 in Cor. 2, §1, and it follows at once that G 133 does no exist in the case under consideration. If G s has systems # 3 , the order of (r 133 would be 19 . 7 . 11". 5 s . 3 Y . From §3 it follows that a = 1 . Now, it is easily seen that an imprimitive group of odd order of degree 33 whose order is not divisible by ll 2 , does not have its order divisible by 5 8 . Hence, the order of any transitive constituent T would be 11 .3" l .5 Pl - (/?! }> 1). Corresponding to identity in T, there could be no substi- tutions of order 5 or there would also be substitution of order 11, and the order of G, would be divisible by ll a . Hence, (3 >1. If (3 = 1, G m contains an inva- riant subgroup of index 5.* If (3 = 0, G m contains an invariant subgroup of index 11. Hence, G m does not exist if G s has systems S 3 . If G s has systems S A or S & , the order of G 133 is of the form 7 . 19.3". 13". The transitive constituents of degree 27 must be primitive groups. Let T x and T 2 represent the transitive constituents of degree 27. Consider the invariant subgroup H s of G B corresponding to identity in T x . If any elements of T 2 occur in H s , the latter must have a transitive constituent of degree 27, which is an invariant subgroup of T % . Now, it is easily seen that an imprimitive group of degree 39 of odd order cannot contain a transitive subgroup of degree 27. Hence a conjugate G r of G s in which M s occurs would contain a transitive constituent of * Burnside, loc. cit., pp- 261- 20 Rietz : On Primitive Groups of Odd Order. degree 27 having the same elements as T % . \ G r , G s \ would then be an intran- sitive group. But \G r , G s \ must coincide with G m , since G s is maximal. Hence H s has no transitive constituent of degree 27. T x and T 2 must then be combined by establishing a simple isomorphism between them, and H s could only be of degree < 39 . 2. From the fact that 2 . 27 of the conjugates of H s under G m contained in G e would be transformed accord- ing to the constituent of degree 54 in G s , this assumption as to the degree of E' readily leads to impossibilities. Hence, the order of G s is equal to the order of its transitive constituent of degree 27. The order of G m would then be 133 . 3 a . 13 (a }> 4), but a group of this order could not, by Sylow's theorem, con- tain more than 39 subgroups of order 19. Hence 6r 133 does not exist. G m . S 1 = 67, 67. S z = 27, 27, 27, 27, 13, 13. If G s has systems S l , the order of G m would be 67 . 5 . 3°. II 3 (a >4, £ = or 1, Cor. 2, Theor. 3). Let p = 5 in Cor. 2, Theor. 1, and it follows that G m does not exist. If G s has systems S 2 , the order of (r 135 is 5 . 3 a . 13. By Sylow's theorem, a group of this order contains not more than 13 subgroups of order 3 a , Hence G m does not exist. G li \ /Si = 71, 71. G ua would be of order 11 . 13. 71 . 5 a . 7" (a< 1, /?< 1) Cor. 2, Theor. 3. As this order would be the product of distinct primes, G li3 cannot be a simple group* and therefore does not exist. 145 G S 1 = 9,9, ,9. S„ = 9, 9, , 9, 27, 27. £ 3 = 9, 9, ,9, 27, 27, 27, 27. The group G m would be of order 29 . 5 . 3 a . Let p =29 in Cor. 2, Theor. 1 and it follows that G m does not exist. * Frobenius, Berliner Sitzungsberichte (1893), p. 337. Kietz: On Primitive Groups of Odd Order. 21 G u \ /Si = 73, 73. By Cor. 2, Theor. 3, the order of G m would be 7 2 . 73. 3* (a > 3). But a group of this order would, by Sylow's theorem, contain a single subgroup of order 7 2 - As this subgroup would be au invariant intransitive subgroup, no group G w exists. G m . S 1 = 19, 19, , 19. S 2 = 19, 19, 57, 57. The group G m would be of order 17 . 3\ 19 3 . If in Cor. 2, §1, we make p ■=■ 17, it follows that G m does not exist. G m . 2) , but a group of this order cannot be a simple group as this number contains less than 7 prime factors. Hence 6r 155 does not exist. 22 Rietz : On Primitive Groups of Odd Order. Si = 15, 15, 15, 15, 25, 25, .... , 25. # = 27, 27, 27, 27, 13, . . . . , 13. The order of G m would be of the form 7.23.5*.5M3*. Let p = 23 in Cor. 2, Theor. 1, and it follows that G m does not exist. G m . S x = 41, 41, 41, 41. By Cor. 2, Theor. 3, the order of G m would be 11.5".3.41 (a > 2), and Gr 165 would, by Sylow's theorem, contain a single subgroup of order 11. Hence G m does not exist. G m . Si = 3, 3, .... ,3. S 6 = 21, 21, .... , 21. £ 2 = 7, 7, ,7. /Si = 21, 21, 63, 63. # 3 = 7, 7, . . . • , 7, 21, 21. S 8 =7,7, .... , 7, 49, 49. Si = 7, 7, ..... 7, 21, 21, 21, 21. £ 9 = 7, 7, 7, 7, 21, 21, 49, 49. S 5 = 7, 7, .... ,7, 21, 21, 21, 21, 21, 21. S 10 = 7, 7, . . . . , 7, 63, 63. If (? g has systems £ x , the order of G? 169 is 13 2 .3, but by Cor. 2, §6, there is no group G m of this order. If G s has some transitive constituent of degree 7, the order of 6r 169 is of the form 13*. 7 . 3 a (Cor. 2, Theor. 3). G m would contain 13, 91 or 169 subgroups of order 3", if a > 0. For, since the number of such subgroup in G„ must be 7, 7 169 the total number in G m is - 1 - — ■ , where 'X is the number of letters of G m left fixed by a subgroup of order 3% and a > 1 . If the number is 13 or 91, by considering the isomorphic group of degree 13 or 91, according to which the conjugate set is transformed, it is easily seen that a > 3. If the number is 169, a subgroup of order 3 a is invariant in a sub- group -ST of order 3 a .7 and of degree 169, since in any subgroup leaving a given letter fixed, a subgroup of order 3" is one of 7 conjugates. Since a substitution of order 7 in G m is of degree 168, every transitive constituent of K has its order divisible by 7. Now there is no transitive group of odd order of degree 3 P when 3 3 < 169, which contains in its order the factor 7. Hence the degree of every transitive constituent of K is a multiple of 7. But 7 is not a divisor of 169. Rietz : On Primitive Groups of Odd Order. 23 Hence there cannot be 169 subgroups of order 3 a . We have then shown that a is not greater than 3 when G s has a transitive constituent of degree 7. In all the remaining cases G m would be of order 13 2 .7 a .3". Let p = 7 in Cor., §3, and it follows that a = 1. If T is a subgroup such that there is no sub- group of greater order common to two subgroups of order 3", then T is of order 30-1 From the reasoning of §3, it follows that T is invariant in a subgroup of degree 169 of one of the following orders : 1313", 13.7.3", 13.3" or 7.3". The subgroup T would then be one of 7, 13, 91, or 169 conjugates in G m and just as before it follows that /3 ^> 3 . It remains to consider the group G m of order 13 2 .7.3" (/3 ^> 3). By Sylow's theorem a group of this order contains 1 or 27 subgroups of order 13 2 . It could not contain 27 ; for, on account of the limitations on f3, they would be trans- formed according to a regular group of degree 27. Hence G m contains a single subgroup order 13 2 , and is contained in the holomorph of the abelian group P of order 13 2 of type (1, 1). The group of isomorphisms L of P is of order 2 5 .3 2 .7.13. It remains to examine L for subgroups of order 7.3" (/#3> 3). Such a subgroup contains a single subgroup of order 7. The group L contains just 78 subgroups of order 1* Bach of these is then invariant in a subgroup of order 7.3.2 4 . From thisf it follows that L contains subgroups of order 7 and 21 but none of order 7.3" (/? >■ 1). The subgroups of order 21 are conjugate in L. Hence there is in the holomorph of P just one subgroup of each of the orders 169.7 and 169.21. The subgroup leaving a given letter fixed is in each case maximal. Hence the groups are primitive. G m . S l = 15, 15, ..., , 15, 25, 25. tf 2 = 15, 15, 45, 45, 25, 25. $ = 86,85. If G s has systems S r or S z , by Theor. 4, the order of G s cannot be divisible by 3 2 , since the order must be divisible by 5 2 on account of the transitive constit- uent of degree 5 2 . The order of G m is then of the form 19.3 3 .5 a . Let p = 5 in Cor., §3, and it follows that a=l. As a group of 19.3 3 .5 contains a single subgroup of order 19, there is no simple group G in which G s has systems Si or S,. * Shown by considering the composition series of L- The factor groups of L are the cyclical group of order 13, the simple group of order 1093, and the group of order 3. t Ibid. 24 Rietz : On Primitive Groups of Odd Order. If G a has systems S„ by Theor. 4, the order of G m would be 19.3 8 .5.17° or 19.3117.5 s - But groups of these orders cannot be simple.* Hence G m does not exist. G m . /Si =29, 29, ,29. £ a = 87, 87. If G g has systems S lt by Cor. 2, Theor. 3, the order of G m would be 7 a .5 2 .29 (a> 2). But a group of this order cannot be a simple group, since it contains not more than 5 prime factors. If G s has systems S 2 , the order of G m is 7*.5 2 .29' s .3 1 '. hetp = 29 in Cor., §3, and it follows that /?= 1. Since G a has only two transitive constituents, its order is equal to the order of each of its con- stituents (Theor. 10). Making use of this fact, it is easily seen that 7 cannot occur to a higher power in the order of G s than the power to which 29 occurs. Hence a >/3 + 1 . If either transitive constituent of G s has 3 systems of imprim- itivity, the order of G™ is not divisible by 3 2 . But the order of G m would then contain not more than 6 prime factors. Hence each transitive constituent must have 29 systems of imprimitivity. G s would then contain an invariant sub- group of order 3 V . Now the subgroup of such an imprimitive group which leaves a given letter fixed would leave 3 letters fixed. The subgroup K, containing all the substitutions common to G s and any one of its conjugates G r , is of order 7a _i 3v _i an( j j g i nvar i an t; m subgroups of G s and G r whose orders are equal to 7 a-1 .3 Y and which contain a single subgroup of order 3 y . The subgroup of G m which contains all the substitutions which transform K into itself contains 1 + 3A;(A;>0) subgroups of order 3 7 . Its order could then only be 7 a .3 T or 7 a_1 5 3 .3 V (other assumptions lead to transitive representations of G m of degree less than 175). If subgroups of these orders occur in G m , it can be represented as a transitive group of degree 725 or 203. Since any divisor of these numbers is less than 175, these representations would be primitive. As a group of degree 725 the subgroup which leaves a given letter fixed would be of order 7".3 Y . As it is easily shown that there is no transitive group of odd order of degree 3 s (8 2). As this number contains not more than 5 prime factors, the group cannot be simple. Hence G 203 does not exist. Q.205 # = 51, 51, 51, 51. The order of G* m would be of the form 5.41. 3U7". Let p = 5 in Cor. 2, § 1 , and it follows that G m does not exist. G m . # = 103, 103. The order of G m would be of the form 23. 3M 03. 17 (Cor. 2, Theor. 3). Let p = 23 in Cor. 2, §1, and it follows that G m does not exist. # = 13,13, .-,13. #=13, 13, .... , 13,39, 39. # = 13, 13, 13, 13, 39, 39, 39, 39. If G s has systems # or #, it is easily seen that the order of G 209 is equal to or a divisor of 11.19.13.3 ; but a group of this order cannot be simple, as the order is the product of distinct primes. If G s has systems S 3 , the order of G' m is 11.19.13 a .3 3 . Letp = 11 in Cor. 2, §1, and it follows that G m does not exist. G m . #=107,107. #=13,13,13,13,27,27, 27. # = 13, 13, , 13, 81, 81. The order of G m would be of the form 5.43.107".53 3 .13 Y .3 5 . But a group of this order cannot be simple.* Hence G m does not exist. * Burnside, loc. cit., p. 170. Rietz : On Primitive Groups of Odd Order. 27 6 s ". Si = 9, 9, ,9. S 5 = 27, 27, 27, 27, 27, 27, 9, 9 , 9 S 2 = 9, 9, ,9, 27, 27. S 6 = 27, 27, , 27. &,= 9, 9 ,9, 81, 81. St= 27, 27, 81, 81. £, = ?, 9, 9, 27, 27, 27, 27. For all these systems, except when in S 6 and #,, the transitive constituents of degree 27 are primitive groups, the order of G 8 is a power of 3. The order of Cr 217 would then be 31.7.3% with the exception just mentioned. From the argu" ment of §3, it readily follows that, if a > 3, G 211 would contain a subgroup of order > 3 a-2 .7, which order contains a factor = 1 mod 3. If the order of the subgroup were greater than this number, such a subgroup would lead to a tran- sitive representation of the simple group on less than 217 letters. Hence the subgroup, if G m exists, is of order 3 a_2 .7. The simple group would then occur 31 7 3" as a transitive group of degree ' '_ s = 279. When represented on 279 letters the group would be primitive and the subgroup which leaves a given letter fixed would be of order 7.3 a_2 . The degrees of all transitive constituents would be multiples of 7. But 7 is not a divisor of 278. It remains to consider the case where G s has for some of its transitive con- stituents primitive groups of degree 27. In this case all the transitive constitu- ents are either primitive groups of degree 27 or two of them are imprimitive groups of degree 81. In the former case, by Cor., Theor. 6, §4, the order of G 311 would be 7.31.3M3 (a^J> 4). But, by Sylow's theorem, a group of this order contains no more than 63 subgroups of order 31. But this is impossible, since the simple group of odd order would occur of degree 63. In the latter case, we shall also show that the order G g cannot exceed that of one of the transitive con- stituents of degree 27. Let T' be the transitive constituent of degree 27 distinct from T. Let H g be the invariant subgroup of G s corresponding to identity in T, where T is so selected that its order is not greater than that of T'. Then, by Theor. 9, 27 of the subgroups H Sl , E Si , H„ al (defined in §2) are trans- formed by G, according to T'. These 27 subgroups can contain no elements con- tained in T', or H s would have some transitive constituents of degree 13, which is impossible. Hence, these 27 subgroups generate a subgroup contained in the invariant subgroup of G s corresponding to identity in T. But this is impos- 28 Rietz: On Primitive Groups of Odd Order. sible, since, by hypothesis, the order of T is equal to or greater than the order of T. 221 G Sj, = 11, 11, ..... 11. £ 4 = 15, 15, .... , 15, 25, 25, 25, 25. $= 11, 11, 11, 55, 55. St = 15, 15, 25, 25, 25, 25, 45, 45. $3 = 55, 55, 55, 55. If G, has systems S lt the order of G m is, by Cor. 2, Theor. 3, 17.13.11.6" ( 1). But a group of this order cannot be simple, since it is the product of distinct primes* If G, has any of the other systems, the order of G 221 is 17.13.5".3MF. Letp= 17 in Cor. 2, §1, and it follows that G m does not exist. 326 G aS x = any set of systems in which some systems contain 7 letters. S z = 21, 21, . . . . , 21, 49, 49. S 3 = 63, 63, 49, 49. If G s has some transitive constituents of degree 7, the order of 6P 5 is of the form 5 8 .3 a .7 (Cor. 2, Theor. 3). But, by Sylow's theorem, a group of this order cannot contain more than 175 subgroups of order 3 a . If G s has systems S 2 or S 3 , the order of G™ is of the form 513". 7". Let p = 7 in Cor., §3, and it follows that /3 = 1. Then as above G m would contain no more than 175 subgroups of order 3*. Since there is no simple group of odd composite order of degree 175 the group G* 225 does not exist. G* 3 \ Sj, = 15, 15, 25, 25, 25. S s = 45, 45, 45, 45, 25, 25. S z = 15, 15, 25, 25, 75, 75. S t = 23, 23, , 23. $ 3 = 15, 15, .... , 15, 25, 25. S, = 115, 115. Si = 15, 15, . . . . , 15, 25, 25, 45, 45. If G s has some transitive constituent of degree 15, the order of G s is not divisible by 3 8 (Theor. 4). Then G m would be of order 3 2 .7.11.5 a and would contain an invariant subgroup of index 3 2 .f If G s has systems S t or S 1t the * Frobenius, loc. cit., p. 337. tBurnside, loc. cit., pp. 260, 262. Rietz : On Primitive Groups of Odd Order. 29 order of CP n would be of the form 3.7.11". 23".5 1 '; but a group of this order has an invariant subgroup of index 3. If G, has systems S s , the constituents of degree 25 are the primitive group of order 75. As this group is of class 24, we have, by Cor. 3, Theor. 3, that the order of G t is not divisible by 5 3 . The group G m would then be of order 3 a .5 2 .7.11. By Sylow's theorem, a group of this order which does not contain less than 231 subgroups of order 3 a , contains 385 such subgroups. A simple group of this order would then occur as a primitive group of degree 385. The subgroup G x , leaving a given letter fixed, would then be of order 5.3". The only systems of intransitivity of this subgroup G 1 , which are not excluded by the con- ditions stated on p. 15, are 81, 81, 81, 81, 15, 15, 15, 15. In order that the transitive constituents of degree 81 contain in their orders the factor 5, they must be primitive groups of degree 81 of order 81.5. Consider the invariant subgroup H of G x corresponding to identity in a transitive constituent of degree 81. It follows from Theor. 6 that the degree of H cannot exceed 60. As Gi could contain no other similar subgroup, the order of G x would be 81.5. The primitive group of degree 385 of order 3 4 .5 2 .7.11 then contains 324.77 substi- tutions of order 5 of degree less than 385. By Sylow's theorem, a group of order 3 4 .5 2 .7.11 cannot contain more than 11.81 subgroups of order 5 Z . As this number of subgroups contains less than 324.77 substitutions, we have arrived at an absurdity. 235 G # = 117,117. #=13, 13 , 13,39, 39. #=39,39,... ,39. #=13, 13, .... , 13. # = 13, 13, , 13, 39, 39, 39, 39. # = any set of systems of which some contain 9 letters. The order of G 23 * would be of the form 5.47. 3M3". Let^ = 47 in Cor. 2, §1, and it follows that G 33 * does not exist. This completes the determination of the primitive groups of odd order of degree less than 243. The results may be summarized as follows : Aside from the invariant subgroups of the metacyclic groups there are only ten primitive groups of odd order of degree less than 243. The following list of 30 Rietz : On Primitive Groups of Odd Order. numbers gives the orders of these groups, the first factor as they are written being the degree of the group of the given order:* 25.3, 27.13, 27.39, 81.5, 121.3, 121.15, 125.31, 125.93, 169.7, 169.21. Bach of these groups is solvable. The following theorem may then be stated : Theorem 1 3. — A simple group of odd composite order cannot be of degree less than 243. Cornell University. * For the first four groups, see Burnside, Proc. Lond. Math. Soc, Vol. 33, pp. 178-185. Cornell University Library On primitive groups of odd order / 3 1924 032 189 692 olm.anx