the primitive groups of class fourteen

The Primitive Groups of Class FourteenAuthor(s): W. A. ManningSource: American Journal of Mathematics, Vol. 51, No. 4 (Oct., 1929), pp. 619-652Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2370587 .

Accessed: 03/12/2014 21:05

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

The Johns Hopkins University Press is collaborating with JSTOR to digitize, preserve and extend access toAmerican Journal of Mathematics.

http://www.jstor.org

This content downloaded from 128.235.251.160 on Wed, 3 Dec 2014 21:05:07 PMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=jhup

http://www.jstor.org/stable/2370587?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


The Primitive Groups of Class Fourteen. BY W. A. MANNING.

1. No primitive group of class 14 contains a substitution of degree 14 and order 7.* Then the primitive groups of class 14 can be studied to the best advantage under two heads: (a) those in which the positive subgroup is of class > 15, to be determined in ?? 3-36 of this paper, (b) all others, deferred to ?? 37-48. It will be shown that there are but four groups of the first sort, and one of the second:

(1) The simply transitive primitive group according to which the symmetric group of degree 9 permutes its 36 transpositions.

(2) The simnply transitive primitive group of degree 49 and order 2(7!) 2 isomorphic to {ab, bcdefg, a#, fy&Cgr aa b/ 3 cy* dA * eE fg* gr}.

(3) The triply transitive group of degree 22 and order 22 * 21 * 20 * 96 which occurs inr Mathieu's quintuply transitive group of degree 24 as a transitive constituent of the largest subgroup in which the subgroup that leaves two letters fixed is invariant.

(4) The doubly transitive subgroup of (3) of degree 21 and order 21 * 20 * 96.

(5) The doubly transitive group of degree 21 and order 21 [20 288 which occurs in the quintuply transitive group of degree 24 as a transitive constituent of the largest subgroup in which the subgroup that leaves threle letters fixed is invariant.

The positive subgroup of (5) is a doubly transitive group of class 15, which was omitted from the author's published list of the primitive groups of class 15.t The error in that paper is corrected in ?? 39-48. A direct proof of the existence of the quintuply transitive group of degree 24 is given in ? 46.

2. In this investigation the following theorem is of constant use. It is, however, only a special case of a more general theory.:

Let certain substitutions of prime order gen,erate an intvariant subgroup of a primitive group G. If H is an intransitive subgroup generated by some

* Transactions of the American Mathemaitical Society, Vol. 4 (1903), p. 351. t American Journal of Mathematics, Vol. 39 (1917), p. 281. t Transactions of the American Mathematical Society, Vol. 12 (1911), p. 378.

619



620 MANNING: The Primitive Groups of Class Fourteen.

of these substitutions, there is in G a substitution similar to one of the given generators which replaces a- letter of an arbitrarily chosen transitive constituent of H by a letter of another constituent of H.

Occasional use will be made of one of Jordan's theorems, which may be stated as follows: *

Let G be a transitive group generated by certain substitutions of prime order p, some of which generate an intransitive subgroup H. e Let 'there be in G a substitution similar to one of the given generators of G which replaces a letter of an arbitrarily chosen transitive constituent A of H by a letter of some other constituent of H. Of such substitutions let t, t1, * * be those that displace a minimum number of letters new to H. Then if every power of some one of them, say t, replaces a letter of A by a letter of A, there is at most one letter new to H in any cycle of t.

The last condition -of this theorem is certainly satisfied if the generating substitutions are of order 2 and if the degree of A exceeds the number of cycles in t.

The primitive groups of degree less than 21 are known and among them there are no primitive groups of class 14.t

3. If the positive subgroup of the primitive group G is of class > 15, a substitution s2 of order 2 and degree 14 will with

Si - ala2 * blb2 * clc2 * d1d2 * ele2 * flf2 * glg2

generate an Abelian group or one of the following non-Abelian diedral groups:

Do ~fs{l, S2 a1b* cx1ax C2d2 d1/3w d2#2 * ely1 e2/y2 D'o= {sl, S2 albw cid* eif, glal g2a2 /81,82 YlY2}, Di {sl, s2 a1a2 b1c* diel f1' * f2X2 gl/8* g2/32},

2 {sl, s2 alb1 c1d, e1al 622 f 1 3 f232 Y1Y2}, D3 {sl, S2 ala2 b1b3 clc3 d1d3 ele3 ffJ3 glg3jn D4 {Sl, s2 ala3 b1b3 clc3 d1d3 e1e3 flf3 *glg3j D5 {sl, S2 - alb* a2c1 b2c2 d1d3 ele3 flf3 *glg3j D6= {sl, s2 = albw a2c1 b2d, cAd2 elf, gll g2X21n D7 - {s2, S2 a1b* a2c1 b2d1 eflf e2g1 f2a'l g2a2f,

The accuracy of this list can be checked by the construction of the positive substitutions of degree > 15 and < 22 : which can occur in G. D'0 is the same group as D0, with si and S2 transposed.

* C. Jordan, Crelle's Journal, Vol. 79 (1874), p. 249. t See references, Amnerican Journal of Mathomatics, Vol. 35 (1913), p. 229. t Transactions of the American Mathematical Society, Vol. 18 (1917), p. 473.



MANNING: The Primitive Groups- af Class Fourteen. 621

Since no invariant subgroup of G can be Abelian unless it is regular, some two substitutions of a complete set of conjugate substitutions of degree 14 must generate one of these 8 diedral groups. Our immediate object is to show that G can not contain D1, P3, D6, or D7. They will be studied in the order D7, D3,, D, D1, D5, Do, D2, R4. A complete list of the possible Abelian groups generated by two substitutions of degree 14 might be given at this point, but there is nothing to be gained thereby. All such groups will be found in ?? 29-33.

4. The last group of this list, D7, is of order 16 and has only two transitive constituents. It will be shown that it is not a subgroup of any primitive group of the sort we are studying. The following substitutions of D7 will be needed:

S= ala2 blb2 * clc2 dld2 ele2 f1f2 g192, S2-a, b 1a2c1 b2dl* elf,, e2g1 f2(l g2(2,

ti = a2dl b2C c2d2 * eig1* e2fl * f2(2 g2(1l

t2 alb2 * a2bl cld2 c2d flg2 f2gl 1l2

There is a substitution S3 of degree 14 that replaces a1 by a letter of the second constituent and has at most one letter new to D7 in any cycle. UJnder transformation by t1 and e1e2 fflgl f2g2 cclcc2, the four transpositions (a1e,), (a1e2), (a1f1), and (a1g1) are conjugate. So also are (a1f2), (a1g2), (a1a1), and (ala2). For S3 we have to try only (a1e,). . . and (alf2) * . ' -

Since we are not interested in primitive groups of degree < 21j9 it is well to know if a transitive group of degree 16 or 18 of which D7 is a subgroup can be multiply imprimitive.f It is easy to verify the following statements:

The only passible system of imprimitivity of two letters including a1 of any imprimitive group of degree 16 or 18 of which D7 is a subgroup is alb1.

rThe only possible system of imprimitivity of four letters including a1 of any imprimitive group of degree 16 of which D7 is a subgroup is alb1c2d2.

No systems of three or nine letters are possible in an imprimitive group of degree 18 of which D7 is a subgroup.

Then D7 is a subgroup of a primitive group of degree < 21 unless S3 displaces at least four letters new to P87.

Neither {s1, S3}, {S2, S3}, {tlp SI}, or {t2, S3} is Abelian or of the, form P6 or D7. Nor can the first two be D3 or D4.

* See references, American Journal of Matheimatics, Vol. 35 (1913), p. 229. t C. Jordan, Liouville's Journal, Ser. 2, Vol. 16 (1871), p. 383; Marggraff, Dis-

sertation, Ueber primitive Gruppen mit transitive* Untergruppen geringereu Grades, Giessen, 1889; Manning, Primitive Groups, 1921, p. 92.




5. Consider first the substitution

S3 =(ale,) (/3l -) (/2 -) (/3 -) (34 )

Let both {sl, S3} and {S2, S3} be octic. Now S3 (a,el) (a2) (b,) . . so that {t2, s3} is octic and S3 == (ale,) (b2/,l) (b,)* , which is impossible when

{s1 S,S} is octic. Let {sl, s3} be octic and {S2, S3} such a group as D5. Now S3 =

(a,e1) (b,x) (fjy) (,l-) * (a2) (e2), where (xy) is (b2d1), (f2ccl), or (g2cc2). If {sl, s3} is Di,, s3 (ale,) (Cl/,) (c2,82) (a2) (e2) , * * which is impossible with s2. Therefore {s1, s3} is D0 or D2, and S3 -=(ale,) (bid,) (f,b2) or (a1e,) (flccl) (b,2)* , both of which are impossible; or else xy-g2cc2, in which case e2 and g1 are fixed, impossible with s2.

If {s1, S3} is D5 and {S2, S3} is octic, transformatioln by

ale, a2f, ble2* b2gl* clf2 c2cc,l. dg2. d2a2 shifts this case back to the preceding. Let {s1, S3} and {S2, s3} be D5. It is not possible to have, S3

(ale,) (a2x) (e2y) -= (ae,) (blu) (f,v).

6. Consider second S3-= (alf2). Let {s1, S3} and {S2, s3} be octic. Since now s3 fixes a2 and b,, {t2, S3} iS

D0 or D2. Therefore S3 = (alf2) (a2) (b,) (b2) (fl) (gl) (ccal) , and s3 =1

(a2f3l) (a,1) . Thus S3 has no cycle new to s8, and in consequence {S1, S3} is D0. Then S3 fixes g2, so that {t2, S3} is impossible.

Let {sl, S3} be octic and {S2, S3} of order 6. Now S3

(a1f2) (b,x) (aly) (a2) (fl')* , where the cycle (xy) of s2 is (b2di), (e2gl), or (g2cc2). M3oreover, x = g1, el, or e2, and y = di, d2, cl, or C2, for otherwise (bix) (ccly) * can be transformed into (ale,) * * * . We cannot have s3- (a1f2) (b1d,) (acb2) * * *, and neither (alf2) (blg2) (a1a2)* nor (alf2) (bla2)

(alg2) . * is possible. Let {sl, S3} be D5 and {S2, S3} octic. As before, S3 (alf2) (a2X) (fly)

( ,) (c1) , where (xy) is (clc2), (did2), (ele2), or (glg2), and x = el, e2, or g2, y + b1, c1, or di. The three forms of S3, (alf2) (a2Cl) (flc2) *

(a1f2) (a2di) (fid2) . and (alf2) (a2gl) (flg2) . * are all impossible. Finally, {sl, s3} and {S2, S3} cannot both be such groups as D5.

'Therefore D7 can be strucic from our list.

7. Consider a group G in which D3 is a subgroup. If G exists there is in it a substitution s3 ==(albi). . Now s3 is not (a1b1) (a2b2) . . . Nor is s3 -(albi) (a2cl) (b2c2) > or (a1b1) (a2c2) . Then {S1,S3} and {s2,s3} must be octic groups. If s3 fixes a2, it fixes also b2 and b3, so that {s2s1s2, S3}

is an impossibility. If s3 displaces a2, {sl, s3} and {S2, s3} are both D6, S3-



MANNING: The Primitive Groups of Class Fourteen. 623

(a,b,) (a2c,) (b2d) (c2d) (b3 -) d * , and {Sls2sl, S3} is Do, D'o, D1, or D2. Hence s3 (albl)(a2cl) (b2d1) (c2d2)(b3 -)(c3) - . But if {s2,s3} is D6, s3 should displace c3. This consideration removes D3 from our list.

8. We can also rid our list of D6. Its substitutions of degree 14 are

s5 = la2 b1b2 c1c2 d1d2 e1e2 flf2 *g12f S2= alb1 a2c1 b2d1 cAd2 e1f, g1x g2cc2, t=- a1d, a2d2* b1cl b2c2 elf2 e2fl cc1lc2, t2 =al2 a2b2 b1d2 * c1d, e2f2 g1lc2 g2(Xl

All the substitutions of the group {o = e1flf e2f2, ol glg2 c cla2, u2-

b1c2. b2c1* e1e2 f1f2 * 1c2, 03 - a2bl.b2cl * c2d1.elgl e2al fl9g2f2a2} transform D3 into itself. There is then but one substitution S3 to be investigated: (a1e,) - . Upon this substitution may be imposed the condition that it introduces a minimum number of new letters to D6, and therefore at most one new letter to any cycle.

If {D6, s3} (= H3) is an imprimitive group of degree 16 or 18, the only system, including a,, of two letters is cal2g1g2, and there can be no system of three letters. Hence if H3 is a transitive group of degree < 20, D6 is a subgroup of a primitive group of degree < 21. That is, if H3 is of degree < 20, it is intransitive.

9. The group {s1, S3} can be Abelian, or of one of the types Do, Do0, D2J D5, or D6.

Let {sl, S3} be Abelian. If S3 and t1 are also commutative, S3-

(a1e,) (a2e2) (d1f2) (d2fl1) . - - . By S2, S3 =(a1e,) (f1d2) (b1c2) (a2e2) (c1) * - Then {t1, S3} is octic. If {t1, S3} is D6, there are just two letters in S3 new to t1, and they replace the letters of one cycle of t1. These new letters are not

gl or g2. Therefore S3 -* (a1e1) (a2e2) (a1,lS1) (2232) - - - . The only other letter new to S2 in S3 is now f2, so that {S2, S3} can only be D1 or D5, and neither case is possible. If {t1, S3} is D'o, D1, or D2, S3 has four or six letters new to t1. Now {S2, S3} is octic because S3 fixes f, and therefore S3 = (ale,) (a2e2) (cl

(b1) (b2) (d1) (d2) (fl) (f2)- * *, and this is impossible. Let {S1, S3} be Do. The substitution S3 (ale,) (a2) (e2) * * * has six let-

ters new to s, in six cycles, of which at least four are /3i, /2, /3, /4. If S3

displaces a, and a2, {t1l S3} is DoP D1, D2, or D5. If it is D5, {s2, S3} is impossible; and if it is Do, D1, or 02, S3 fixes a2, e2, d1, d2, fl, f2, and therefore S3 =(a1e,) (b1,p) (b2132) (c1/3) (c2/.4)+Xlg) (x2g) * * , absurd with S2* If S3

displaces ?x, and fixes aX2. {tl S3} is D2 and S3 = (ale,) (g1ftl) (g2/32) (xlx) * * ,

where x is new to S2 and to t2 and is a letter of si, an impossibility. If S3 fixes 9




, and a2, it displaces six letters / in six cycles, so that {ti, s3} and {82, S3}

are also Do. This is impossible. Let {si, S3} be Dio. Now S3 has six letters new to s, and four of these

new letters are in two cycles. Therefore s3 = (alei) (ao,#3) (a232) (/3X) (/4y)

* **, where (xy) is a cycle of si. We note that {tl, S3} is octic and is not De. Therefore S3 fixes a2, e2, di, f2, and (xy) is (blb2), (clC2), or (glg2). Because of S2, (xy) is not (bib2). If (xy) is (clc2), t1 requires that b, and b2 be displaced, and this si does not permit. If (xy) is (glg2), {S2, S3} is octic and therefore S3 fixes fi and f2, whereas s& requires that fi or f2 be displaced.

Let {Sl, S3} be Di. Here S3 has a cycle in common with s, and has four letters new to sl. If S3 displaces a it is of the form (ale) (ahg) (a2g) - * . For if H3 iS of degree < 20, it is intransitive. This form of S3 iS inconsistent with s2 or t2. Let S3 replace gi by g2. Thus S3 -= (aiei) (gig2) (3l - ) (/2 -)

(/3 -) (/4 -) - - . With S2, t1 and t2, S3 generates octic groups and therefore fixes fi and f2, but should displace a letter from every cycle of si. Let S3 (ale,) (gl,/1) (g2/32) - - . This is commutative with ti and replaces d, by f2. Hence if S has one cycle in common with si, it has two.

Let {Sl, S3} be D2. Now S3 'has six letters new to si, in five cycles. If S3 = (ale,) ((131) (x2 -) (P2 -) (/3 -) (/4 -) *

- - , the groups {tl, S3},

{S2, S3} and {t2, S3} are octic, so that S3 =(alei) ((xi/,) (gX2) (g/82) * , impossible with s2 and t2. If S3 = (alei) (a,#,) (/2 -) (/3 -) (/4 -) (/35 -)

* . . {t-1 s3} is Ds; but (alel) (ol3li) (di - ) (f2-) . . . is absurd if the blanks are to be filled from tl. It would require eight cycles in S3. Then S3 "-(ale,) (ala2) (/3l -) (P2 --) (/3 ) '(/4 -)

. . and S3 + (glg2)

Therefore S3 is not commutative with S2 or t2. Both {S2, S3} and {tl, S3} are octic. Hence S3 fixes fi, f2, gi and g2

10. Let {Si, S3} be D5. Now S3 =(alei) (a2X) (e2y) (/3i -) (/2-)

where (xy) is a cycle of si. InI S3 there are four letters new to si. If one of them is al, H3 iS intransitive and therefore S3 (Xlg) ((X) (g) .

(ale,) (dixi) (f2yl) [(xlyl) is a cycle of ta], which is impossible. Hence S3 -(a) (a2) (ale,) (a2X) (e2y) (/1 -) (32 -) (/3 -) (,4 -). Since {tl, S3}

is Do or D2, S3 fixes d, and f2, and then, because of si, must replace d2 and fu by letters /3. But 53'- (eial) (f/l3i) . is impossible with S2.

Let {Sl, S3} be D6. H3 is intransitive. Then 3 =- (alei) (gl,81) (g2/32) * or (gl') (g2x) , where x is new to si. But (gla) (g2f3) (a) is

impossible, as is (gla) (g2a) . . Now S3 (ale,) (gl3l) (g232) (a() (X2)

(dif2) . Since {S2, S3) and {t2, S3} are of order 6, S3 ale, e1a2f1 b1c, djf2 d2e2 gl131 g2132.




The group H3 = {D6, S3} is unique.' It is transformed into itself by glg2c21-2 /l3,81/2 and by t3 = cle2-c2el.blf2.b2fl-d1d2.a1/32ca2i1, so that for s4

two substitutions, (alg1).- and (alai)- , are to be studied. Now (a1g1). . * can only be the transform of S3 above by the substitution a3-

a2bl b2cl c2dl e1g, e21 l flg2 f2a2 of the group whose substitutions transform D6 into itself: O93S33 = -a1g1 a2b2. bg2 c2c`2ad22al-ex .fly, where x and y are letters new to D6. This is impossible with S3. The other substitution, (a1ca1) *, is the transform of S3 by c`2o3 =r392S392U3 a11. bla2 1 c1d, c2g2* d2g1 e1x fly. This too is inconsistent with S3.

This leaves in our list of diedral groups D0, D'0, D1, D2, D4 and D,.

11. It is also possible to reject the group D1. The four negative substitutions of order 2 in D1 are

si- ala2 b1b2 clc2 d1d2 * ele2 flf2 qlg2, S2 = a1a2 b1c* d1e, f 1 * f2a2* gl/l *22, t= a1a2 b2c2 d2e2 f2al fla2 g2ftl gl/22,

t2 ala2 blc2 Nb2c d1e2 d2el a1a2 /31/32

All the substitutions of the group

{b1b2*c1c2 d1d2 * e1e2.f1f2.g1g2, bcl d1e, f1l l-f2ca2 * g11 * g2/32,

b1d, b2d2 c1el*c2e2, bifi.b2c(I.cif2.c2c22 dig1.d2flle1g2 e2/.2, bic1.b2c2, d1e,*d2e2, f1f2 1ac2, glg2/3tl/2, f1gllf2g2 x1/l1 x232}

transform D1 into itself. Then for S3 only the substitution (albl). . * need be studied.

If S3 is commutative with s5, {S2, s3} is not one of our groups D0, D'o, D1, 12, D4, or D5. If {sl, s3} is octic, S3 fixes a2, b2, c1 and c2 and then {tl, s3} is not a possible subgroup of G. Hence {Sl, S3} is D5: s3 =(a1b1) (a, -) (b2-) * , and (t1s3)4= (S2S3)2 1. Indeed, {t1, s3} is D1. Therefore

s3 -a1b * a2c1 * b2c2 * d2y1 * e2y2 * f1x1 * g1f or s83 a1b * a2c1 * b2c2 * d2y1 * e2y2 f*13 gi1a

The only other substitutions of degree 14 which replace a1 by b1 are the transforms of S3 and S'3 by flf2 alX2 glg2 /31/32 and flf2 g1g2 -1o2 /31/32. The sets of transitivity of H3 are ala2b1b2clc2, d1d2ele2yly2, f1f2cloX2, g1g2/31/32, and of H'3 are ala2b1b2clc2, did2e1e2yly2, flf2glg2cXlc2/ll,/2. Both H3 and H'3 are invariant under the substitutions of the group {die,.d2e2 Y1Y2, b1c2 b2c1 d1d2- e1e2 f1f2 g1g2, f1g1 f2g2 1/l l2fl2, a1a2 b1c1 d1e el fl l2a2g1ft1 g2/32}. Hence for s4 we have to discuss only the three substitutions (a1di) , (aly1) and (a1f1) .



626 MANNING: The Primitir e Groups of Class Fourteen.

It will be convenient to use w to represent s3 and st3 indifferently.

12. Let s4 = (afi).- - - . Take the eight substitutions (a,b,) * - which are possible with {s., s2} and transform them by b1f1 b2(XI c1f2. c2 22 d1g,. d2,81. e1g2. e2/32. Transform again by die,* d2e2 7l72 and by b1c2 b2c1 d1d2 ee2 f1f2 g1g2. There remain for further consideration two substitutions having the first cycles a-if a2f2 /3plr2 ,SlXl fl2x2 and the last two cycles b1b2. d1d2 or b1d2 b2d1. Of the letters xl and x2 we know as yet merely that they are new to D1. Now S4 = (f1al) (aa12)- - * is not possible with S3. The group {s3,, S4} is D,, so that S4 = alf1. a2f2 b1d2. b2d1. a1a2 /3ly1 1272. In the primitive group G there is no other substitution (a1f1) * similar to s,. This means that no G containing this H4 is doubly transitive, for if so there would be four substitutions with (a1fl) in common, conjugate to sl, S2, t1 and t2, respectively. Since { H'3, S4} is transitive of degree 20, and since there is no simply transitive primitive group of degree 20, this S4 --(a,fl) -. can be dropped.

Let S4 = (a1d,)- . Now there are six substitutions S4 with aid, a2el d2e2. b2x1. c2x2 in common and with the last two cycles f1ia1 gl, f gal-g2p2, f2c22g2f2, flfl3- g11, f1f2 -g21, or f2#2 g22. This S4 is commutative with 8i3'81- a1c2 a2b2- b1cl d1 y e1y2 f2,3l1g2a1. Thus we have uniquely s.1 aid, a2el.d2e2. b2y2'c2yY1.f1ft2.g2c2i. The group {H3, S4} is impossible. Of H4 = {H'3, S4} it can at least be said that it has two sets of transitivity. Trans- formation by substitutions of D1 and by sg3t1s'3 = b1c, b2c2 f1g2 f2gl c21)2. a2#1 "1-Y2 shows that there is but one substitution to try for S5: (a1f1)-* and it has already been rejected.

Let S4 = (ay') - * - , with the strong condition that in no substitution of G, similar to s1, is a1 (or a2) replaced by d1, d2, e1, or e2. Let S4 fix a2. If WS4 84W, S4 (aly) (b1d2) * - , and the transform of t1wt1 = (a2b1) by 84 is (a2d2) . . . . If {w, S4} iS octic, S4 = (a1y1) (a2) (b1) (b2 -) (cl-)

(c2) --, and is inconsistent with t2. If {w, S4} is of order 6, 84

(a1y1) (b1 -) (d2 -) (cl -) (a2) (b2) (di) (c2) * * , clearly impossible with t1, Now the only letter by which S4 can replace a2 is 72 84 =

(a1y1) (a2Y2) - - - * If WS4 = S1W, S4 =(a17y) (a272) (b1d2) (b2) . - . Then S48W10'484 (b2Y2). . . and the transform of this by s1ws1 is (a2e1) . If {W, 84} is octic, S4 =(a1,/) (a272) (b1) - , and W84WS4W' (a1d2) *

Then the only mon-Abelian diedral groups left are Do, D'o, D2, D4 and D'5.

13. Consider A5, in which

s= a1a2 b1b2 * c1c2 d1d2 e1e2 f1f2 g1g2, S2= a1b* a2c1 b2c2 d1d3 e1e3 flf3 g1g3, t a2b2 a1c2 *bc* d2d3 e2e3 f2f3 g2g2.




The group D5 is invariant under {o- = b1c2 b2c1 d1d2 e1e2-f1f21*9gg2, Oa2

a2b1 b2c1 d2d3 e2e3 f2f3 g2g3} so that we have to study only the one type of substitution, S3 (a1d1) . If {sl, S3} is an octic group, 1S3S1S3O(r=

(a1d,) (a2d2) , and if {S2, S3} is octic, o1off2S3S2S3q2o1 = (a1di) (a2d2) If (S2S83)2 1, o2S3o92 -(a1d,) (a2d2). . It is not possible to have all the groups {S1, S3}, {52, S3}, {tl. s3} of order 6, as is seen by considering one of the transitive constituents of degree 3. If (SlS3)3 = (s2s3)3 = (t1s3)4 = 1, s3 =(aid,) (c2a) (b2) (a2e2) (e3) * , and the transform of the cycles (a2b2) (e2e3) of t1 by s3s01r241(d1el) (d2e2) (d3e3) is (aid1) (a2d2). Then in all casess3 (aid,) (a2d2) . . If (s2s3)4= 1, s3 (aid,) (a2d2) (b1) (d3) and S3472O1 transforms S2- (a1b) (d1d3) (a2c1) . * into (d2a2) (a1d,) (d3-)

which with S2 generates such a group as D5. Then

S3 a1di a2d2 b1e, b2e2. d3e3 f1f2 g3g4 or s3= a1d, a2d2 b1e, b2e2. d3e3 f3f4 g3g4.

14. We take up the first of these two substitutions and study the group H3. The substitutions of degree 14 in H3 are

Si = aia2 b1b2 clc2 d1d2 e1e2 f1f2 glg2, S2 = a1b1 a2c1 b2c2 d1d3 e1e3 f1f3 glg3, t= a2b2 a1c2 b1c1 d2d3 e2e3 f2f3 g293, S3 = a1d, a2d2 b1e* b2e2 d3e3 f1f2 g3g4, t2 b1d3 c1d2 aie3 c2e2 d1e, f2f3 91g44, t3 b2d3 C2d1 a2e3 cle* d2e2 ff3* g2g4.

The sets of transitivity are a1a2b1b2c1c2d1d2d3e1e2e3, f1f2f3, and g1g2g3g4. H3 is invariant under a = a2c2. b2c1. d2e2* d1e3* d3e l flf3 g1g3. There are then four types of substitutions that need be used for S4: (alfl) * * * , (alf2) . . .

(a1g1) . , and (a1g2) . . If S1S4 is of order 4, S4S1S4 is (a1f2) (a2f1) . -,

(a1fl) (a2f2) * * - , (a1g2) (a2g1) * * * , or (a1g1) (a2g2) , and may be used

for S4. The products S184 and S284 are not both of order 2. Hence the suLbstitu- tion S4 may be determined subject to the following conditions:

(a) (s1s4) 2- (S2S4) 1,.

(b) (S1S4) 3 (s2s4) 2 1, (C) (S1S4) 3 (s2s4)4 1,

(d) (S1S4)2 =(S2S4)4 = S4 =(alf2) . . . or (a1g2)-

(e) (s1s4) 3= (s2s4) 4 1, s4 (a1f2) , or (a1g2)

To justify the limitation in (d) and (e), consider s4 ~==(a1f1) (b1) (f3). The transform t = t3s4s2s4t3 = (alf1) (blf3) * * * , and we may assume that ts, is not of order 4. For if it were of order 4, isit = (a1f2) (a2fl) * could be




used for s4, under (a), or under (d) as stated. If s,===(a,gi) (b&) (g3) we use t = (TS4S2S4o' = (a,g,) (b,g3) * * * and again choose another substitution (alg1) (a2g2). for S4 if tsl is of order 4. It will be seen that not all the substitutions S4 determined subject to the above conditions have to be used for S4, but these five cases certainly include all the necessary forms of S4. The following 19 substitutions result immediately from this classification:

S4 1 alf, a2f2 big, b2g2 dld2 f3g'3 e3 n S4 2 alf2 a2fl clel c2e2 d1d2 e3f3 g3c S4 3 aig* a2g2 bif, b2f2 ele2 f3g3 d3c ,

S4 4 alg2 a2gl cid c2d2 ele2 d3g3 f3 c, S4 5 alf, a2d, blf3 cld3 d2f2 glg3 e2c ,

S4 6 = aig* a2el b1g3 cle3 d2g4 *e2g2 flf3, S4 7= aig1 a2el blg3 cle3 d2g4 e2g2 f2 c, S4 8 = alf2 a2d* b2c1 c1g3 d2fl d3gl *el2 S4 9 aig1 a2e2 blf3 c2cl * dl(22 elg2 flg3,

S4 10 alf2 a2fl blg2 b2gl d1d2 d3(21 g4(22 S4 11 alg2 a2gl blf2 b2fl ele2 e3c1 g4(22 S4 12 alg2 a2e2 bll cl22 d1f3 elgl *f2g`4 S4 13 alg2 a2el blg4 c2d3 d2g3 e2gl f2 a'

Cases (c) and (e) permit also

S414 = a1g2 * a2el b2c(1 clf3 dl(2 e2gl e3fl1

transformed by r inlto alg2 b2f1 clacl C2d3 dlf3 d2g3 e3a2, which with si generates an octic group; and

S4 15 = Ut3 S4 10 t3a alf2 a2d2 blg4 * (x dif* e1g3 g2(2, S4 16 = Ut3 S4 1 t3a- = alf1 a2d2 b1g3 c2(X dlf2 e1g4 f3gl, S4 17 5f S4 2 a0 alf2 a2d2 b2d3 c2f3 dif* e2e3 * (2 , S418 a* S4 4 =- alg2 a2e2 b2e3 c2g3 d2d3 elgl flc, S4 19 = S1S3 S4 5 S38f alf2 a2d* ba * C2e3 d2fl e2f3 g2g4.

The following relations reduce the number of the substitutions we shall have to test for S4:

sls3tlS4 1 t2S4 1 tlS8Sl = 0-S4 40-,

t3S4 8 t2S4 8 t3 S4 1,

t3S4 3 t2S4 3 t3 S4 4, S3818254 9 t3S4 9 S2S4 9 t3S4 9 S2S1S3 S4 6n

tlS4 11 S2S4 11 tl S4 1,

S3S4 12 t3S4 12 S3 = US 4 2 O-,

S4 4 S3S4 4 S4 6,

S4 10 t2S4 10 S2S4 10 t2S4 10 = S4 5,

Thus if S4 6 were fully discussed, S4 4, S4 8 S`4 11, S4 1. S4 9, and S4 3 would require no further attention, as whatever primitive groups they might lQad to would be revealed in following up completely the implications of S4 6. The substitutions then which we shall use for S4 are S4 7, S4 6, S4 2n S4 5, and S4 13*



MANNING: The Primitive, Groups of Class Fourteen. 629

15. No primitive group of the sort under discussion contains H4 (H3, S4 }. For sinice H4 is invariant under the substitutions of {t3, t4}, where t4 t1S4 7 t1 = ble2 b2e1.c1g2-c2g1 d3g4 e3g3.f3f4, it is sufficient to put S5 = (alf) . or (aif2) . . Now when in s5 a, is replaced by f, or f2, s5 can be so taken as to come under one of the cases (a), (b), * * of ? 14. But none of the nine substitutions available are consistent with S4 7.

16. Let S4 6 be our S4. To the six substitutions of order 2 and degree 14 in H3 we add

S4 aig* a2e* b1g3 cie3 d2g4 e2g2 fif3, t4 bie2 b2e* c1g2 c2gl d3g4 e3g3 f1f2, t5 aie2 a2g2 b2g3 C2e3 d1g4 eig* f2f3,

t6 a1d2 a2d* Clgl C2g2 d3g3 e3g4 flf22, t7 a2g3 b1d2 b2g2 c1d3 dig, e1g4 f2f3, t8 a1g3 big, b2d* C2d3 d2g2 e2g4 f1f3.

This group H4 is transformed into itself by cr (of ? 14) and by 'r a2bi *b2c1* digi. d2g3 d3g2- e2e3 f2f3.

The group H3 has the invariant subgroup

1), ss3 a1d2 a2d* bie2 b2e, clc2 d3e3 g1g2 g9g4, s2t3 alb* a2e* b2d* cie3 c2d3 d2e2 g1g3 g294, tlt2 = ale2 a2b2 b1d2 c1d3 c2e3 die, g1g4 g2g3.

This axial group is transformed into itself by

s8s2 =aicib2*a2b1c2- d1d2d3* eie2e3 * f1f2f3*g1g2g3,

and s8 transforms the tetraedral group {S1S3, s2t3, S182} into itself and thereby generates the group {S1S3. s2t3, SlS2,Sl} = {Sl, S2, S3} = H3, of order 24. The 16 letters of the larger constituent of H4 fall into four systems of imprimi- tively of four letters each in just three ways: a1a2e3g3, d1d2d3g4, b1c1elg, b2c2e2g2; aib1d2e2, a2b2d1ei, clc2d3e3, glg2g3g4; or aic2d1g1, a2c1d2g2, b1b2d3g3,

eie2e3g4. There are no other systems of imprimitivity in this constituent, not even systems of two letters each. Now

S2S4 = a1g3 a2e3 big1*b2c2.cie1- dd3 e2g2 d2g4, s1t4 aia2 * biei b2e2- C1g1- C2g2* d1d2* d3g4. e3g3

and the transitive elementary group of order 16

{s1s3, s2t3, S2S4, Slt4}

is invariant in H4. By the adjunction of S182, a group of order 48 is generated. The group {S1S3, S2t3, S284, s1t4, 812, S1} = H4 and is of order 96. There can not be more than 316/4 substitutions of degree 12 and order 2 in the transi-




tive constituent of order 96. The 12 substitutions s&, s2, * , t8 are conjugate in H4. To extend this H4 we have only to consider the single type of substitution (a1f1) * *, on account of t3 and t4, which fix a1. Nor need we go outside the cases (a), (b), * * * of ,? 14. Since TS4 5r = S4 I it is sufficient to use S4 1 for s5. It will be convenient however to put

S5 - t6S4 1 t6 = aia2 bc1. b2c2. d1f d2f2dd3f3.f4g4

The group {H3, so} has a symmetric-5 constituent in the letters glg2g3g4f4. In the other constituent of degree 15 there is an octic subgroup {s1, s2t3}

which fixes f3. This includes s23s1s2t3 = 83, and {si, s2t3, S5} includes (s5s2t3)3 S23 (S5t3) 3 S2, so that {S1, S2, S3, S5} coincides with {S1, s2t3, S5}.

If now the symmetric group of order 120 is represented as the group according to which its 15 octic subgroups are permuted, we have the above constituent of degree 15. Let the 15 octic groups in (12345) all be characterized by an operator of order 4:

a1 1254, a2 - 1245, b1 = 2345, b2 = 1345, c1 = 2354, C2 1354, d1i1253, d2= 1235, d3 =1325, et 2435, e2 1435, e3= 1425, f,= 1243, f2 1432, f3= 1324;

then 12, 13-24, and 45 give rise to

a1a2 * b1b2 c1c2 d1d2 e1e2 flf2, a1b1 a2e1 b2d1 c1e3 c2d3 2e2,

and a1a2 blec1 b2c2 d1f* d2f2 d3f3,

respectively. Hence {Sl, S2, 83, S5} is exactly of order 120 and is a simple' isomorphism between its two constituents. We next note that H,5 has an invariant subgroup of order 16 that leaves every system of imprimitivity of lour letters fixed. It is generated by

s1t4 = a1a2 ble, b2e2 clgl c2g2 d1d2 e3g3 d3g4, S2S4 = aig3 a2e3 b1ga b2c2 cle* d1d3 d2g4 e292,

S5s1t4s5 aja2 big, b2g2 c*le C2e2 e3g3 f1f2 * f3f4, S02W5 55 aje3 a2g3 ble* b2c2 c1gi e2g2 *ff3 f2f4.

The 15 substitutions of order 2 are similar and with the identity constitute an Abelian group. This group has no substitution other than the identity in common with {s1, s2t3, S5,} and therefore the two groups together generate a group of order 1920. This clearly includes S2, S3, and s4. H4, of order 96, is its subgroup that leaves one letter fixed. The only systems of imprimitivity permitted by H5 are ala2e3g3, d1d2d3g4, b1c1elg, b2c2e2g2, and flf2f3f4, so that if H5 is a subgroup of a primitive group of higher degree, it is in a multiply transitive group of degree 21. It is known that there is no primitive group of degree 20 and class 14.




17. There should now be sought a substitution S6 = (af4)* which with H5 will generate a doubly transitive group of degree 21. Any one of the substitutions s1, s2, , t8 generates with s6 an Abelian or an octic group. Since s6 certainly displaces one of the 16 letters a,, a2, , g4, it may be assumed that g4 is in 86. Because {s5, S6 = (xf4) (g.4 -) } is octic, the letter that follows g4 in 56 iS el, e2, e3, g1 g2, or g3, and after transformation, if necessary, by substitutions of H2 we may make it e3 or g3. But H5 is transformed into itself by a1a2 b1c1 b2c2 e1gl e2g2 e3g3, and therefore s6 (af4) (g4e3)- * * if we like. Suppose then that S4S6 is of order 4; s6 iS commutative with s, and t6, and in consequence fixes the 8 letters c1c2dld2ala2glg2. Then S456 is of order 2 and

s6 = a1gl.a2g2 c1d2 c2d1 e1e2. e3g4. f42.

It will presently be shown that H6 exists; that is, that H5 is the subgroup of H6 that leaves one letter fixed. We proceed on the assumption that H6 exists and is of order 2120-96.

Each subgroup of H6 that leaves two letters fixed, as H4, has one and only one regular subgroup of degree and order 16. Then there are just 21 of these subgroups of order 16 in H6 and each is invariant in a subgroup of order 1920, one constituent of which is the symmetric group of degree 5. Since we have laid down the condition that G contains no substitution of degree 15 and no substitution of degree 14 and order 7, the largest subgroup of G in which the regular subgroup of degree 16, Abelian of type (1, 1, 1, 1), is invariant has no substitutions other than the 16 of the regular group on the 16 letters only of the latter group. Since the constituent in the other five letters is already a symmetric group, and since the subgroup of G that leaves two letters fixed can have but one regular elementary subgroup of order 16, G and H6 are identical.

18. Now we seek to find a triply transitive group of degree 22 of which H6 is a subgroup. Obviously 87 is not (/3X) (f4g4) * * , because of S6. Hence H6 is not a subgroup of a quadruply transitive group with the present limita- tions. Since H4 has a transitive constituent of degree 16, we can, after proper transformation, put S7- (pac) (f4 -) (g4 ) * . . * Since {S6, S7} iS

octic, the letter by which S7 replaces f4 iS fixed by 56 and since we may use for -S7 any one of its transforms by substitutions of H2~ it is enough to consider (/ca) (f4f3) . . only. We find

S7 pa- f4f3 -g4d3- e3g3 .f1f2. did2 ala2.

There exists a quintuply transitive group of degree 24 and of class 16,




discovered by Mathieu * and our groups H,6 and H7 are concealed in it as transitive constituents of certain intransitive subgroups. In fact H7 is the triply transitive constituent of that largest subgroup of Mathieu's group in which the subgroup that leaves two letters fixed is invariant. The generating substitutions of the quintuply transitive group are

Z1 =(4, 19) (15, 12) (16, 7) (10, 8) (18, 9) (6, 13) (11, 20) (21, 22), Z2 (4, 16) (19, 7) (15, 10) (12, 8) (18, 11) (9, 20) (6, 21) (13, 22), Z3 (4, 18) (19, 9) (15, 6) (12, 13) (16, 11) (7, 20) (10, 21) (8, 22), Z4= (4, 12) (19, 15) (16, 8) (7, 10) (18, 13) (9, 6) (11, 22) (20, 21), Y =(3, 14, 17) (7, 19, 21) (13, 15, 12) (4, 10, 11) (9, 8, 16)

(20, 18, 22), U (2, 16, 9, 6, 8) (4, 3, 12, 13, 18) (10, 11, 22, 7, 17)

(20, 15, 14, 19, 21), B (1, 2, 4, 8, 16, 9, 18, 13, 3, 6, 12) (5, 10, 20, 17, 11, 22, 21, 19,

15, 7, 14), A =(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,

20, 21, 22), X= (0, oo) (1, 22) (2, 11) (3, 15) (4, 17) (5, 9) (6, 19) (7, 13)

(8, 20) (10, 16) (12, 21) (14, 18).

If now the substitutions A-2Z1A2, Y-'A-2Z1A2Y, U-'Y-1A_2Z1A2YU, U-'Z1UiY-'A-2Z1A2YU-'Z1U, A-3Z4A3, Z1Z4A-l9BZ1B-1Al9Z4Z1, (B-2 XA-2Z1_

A2XB2Z3) -1U-lZlZ2Z3Z4A-4Z2Z3A4ZlZ2Z3Z4U (B-2XA-2ZlA2XB2Z3), are transformed by 2g4. 3d3 4e3 5a 6g 7g3 8a 9e2 10f2 11c2 12f, 13b2 14d2 15a2 16f3.. 17d d1.18e.-19f4-20cv-21g2-22bi. oofl, they go into Sl, 82, S3 , 4, S5) S6, s7, respectively, multiplied by the transposition (0, 1).

19. With H3 as before, let S4 2 be our s4:

S4 alf2 a2fl cle1 c2e2* d1d2. e3f 3 g3g9, s3s4s3 =t4- a1a2 . b1cl b2c2* d1f1. d2f2 d3f3 . g4g5,

slt4 b1c2. b2cll df2 d2fl.d3f3 e,le2*glg2*g4g5.

For s5 we have to try only (a1g1) and (a1g3) --. If s5= (alg3) (b1) (g1) , s5s2s5 = (alg1) . If s5 = (alg1) (a2) (g2) . .

slt4s5sls5slt4 - (alg) (a2g2) . If st5 (alg1) (b1) (g3) S , S5S2S5

(a1g3) (b1g,) . Then s5 is S4 3, S4 9, (a1g3) (b1g,) , or (a1g3)(bl -)(g1 . But S43 and S49 depend (?14) on S46. If S= (a1g3) (blgl) , we fid at once the unique substitution

s = a1g3 big, b2d1 c2d3 d2g2 .e2g4.flf3 = S183S4 6 S3S1.

UInless gi in s5 = (a1g3) (b, -) (gi -) . * is followed by cl or el, s5 can be

* Mathieu, Liouville's Jomrnal, Ser. 2, Vol. 18 (1873), p. 25.




transformed by substitutions of H4 into (a1gi) * . Both letters are impossible.

20. LetS43bes4: S4 = alf * b1f3 * a2d * c1d3 *d2f2 e2f4 *glg3,

s4t1S4 = t4 = b1d2. b2d1* c1f2. c2f1. d3f3 - e3f4 * g`192,

t2s4t2 = t5 = a2e1 * b1d2 *c1f3 c2f4 d3f2 * e3f *g3g4.

Only (a1g1) ... need be used for s5. If s5=-(agi) (b1) (g3) -

t5s1s3t4s5s2s5t4s1s3t53 (a1g1) (bg3) . If s5 (a1g1) (a2) (g2)

t4s5s1s5t4 = (alg) (a2g2). - . Thus we are restricted to the five cases of ?14 but none of those substitutions give us anything new.

21. Let S4 1,3 be our S4.

S4 = a1g2.a2ei.b1g4.c2d3 d2g3.e2gl f2f4, S384S3 = t4 a2g4- b1d2 b2g1 c2e3 d1g2- e1g3 f1f4.

Therefore we have only to try two types (a1 fl) and (a1f2) . . . for S5. All the substitutions S5 come under cases (a), (b), * of ? 14, and they have all been discussed.

22. Referring to ? 13, we take up the study of the other group H3, of which the substitutions of order 2 and degree 14 are si, s2, t1,

S3 = aid, a2d2* ble* b2e2 d3e3 f3f4 *gg4, t2 = aLe3 b1d3 * c1d2 c2e2 die, f1f4 *gg4, t3 = a2e3 b2d3* c1e1 c2d1 d2e2 f2f4 g2g4.

This group H3 is invariant under cr = a2c2 b2c1. die3. d2e2. d3e1 fjf3 g1g3 and f1gl f2g2.f3g3.f4g4. Then 84 is either (a1g1) . . . or (a1g2) * - . . If we use the transformation by a as in ? 14, we have the same five cases as there and by the same brief process find

S4 = a1gi a2g2.b1ifi.b2f2.d3f4.e3g4-f3g3.

It is true that two other substitutions seem possible with the present H3; they are

s4= a1g1 a2e1 b1g3 c1e3 d2g4, e2g2 f1f. 3

S4" = aig* a2e* b1g3 cie3 d2g4* e292 f2f4.

But a2b1ib2c1. dig, d2g3 .d 3g2. e2e3 f2f3 transforms {S2, S1, S4'} into {Si, S2,

a1d,-a2d2.biei.b2e2.d3e3 fif2.g3g4}, the H3 of ? 14, while the latter group is transformed into {t3, t2, 84"} by alcdlee3gid3.a2eeg4g3b1d2-b2e2g2 - fl4f3.

Now H4 is a transitive group of degree 20. If it is a subgroup of a primitive group, that group is multiply transitive, and in it there is a substitution



634 MANNING: The Primitive, Groups of Class Fourteen.

(alg2) of order 2 and degree 14. We have just seen that there is not such a substitution under our five cases. Then if s = (a1g2) * * exists, {s, SJ} is octic and ssis - s4. But ssis and si have no cycle in common, so that {s, s&} can not be octic. The substitution s = (aig2) . * therefore does not exist.

This comnpletes the discussion of D5, the diedral group of order 6 with one regular constituent.

23. In the group Do there are four substitutions of degree 14:

S1 a1a2 bib2 clc2 d1d2 ele2 fif2 glg2, S2 = a1b1 Clal C2a2 d1i1* d2/32 ely1* e2y2, ti a1b2 a2b, c1a2 /31/32 * ly2 f1f2 g192, t2 a2b2 dCla2 *C2al dif2 d2fl elr2 * e2-y1.

All the substitutions of {flf2, glg2, f1g1 f2g2, C1C2 la2, did2 ,81,82, ele2 *71Y2,

cid, * c2d2* c cll - a2/32, cle* c2e2 -alYl*2Y2} transform Do into itself. Then S3 iS

either (aici). or (aifi) . Let sI==(aici) . . . If Si3 =- S381, {S2, S3} is octic, s3 =(a,cl) (a2c2)

(bi) (b2) * * , and (tis3)3 -=1. Therefore S3'-(acl) (f -) (f2). . is

impossible. If S283 =S3s2, s3 ==(ac1,) (bial) (a2) (c2) (c2). . . and {t2, S3} is of order 6. Then S3 (b281) . and replaces d2 or /3' by a letter new to t2 and in S2. This is impossible.

Now {S1, S3} and {S2, SI} are both octic; {t2, S3} is octic because S3 fixes c2 and c,. Hence S3 should replace a-, by a letter fixed by t2 and displaced by s2, and this is impossible because S3 fixes b1.

Let S3; (alfi) . . Neither si nor t1 can be commutative with S3. The groups {s1, s3} and {t1, s3} are octic. Let (S253)3 = 1. Now s3 can be transformed into (a1f1) (c1x) (d1y) (e1z) (a2) (bi) (N2) (f2) (acl) (P1) (y7) . . . , where x, y, z are new to S2, so that we may put x = gi or 81. But (aifi) (c1g1) (g2). is inconsistent with t1. So too is (a1f1) (C181) (dl82) (el83) * . Let (s2s3)4-= 1. Immediately, S3 (a'f') (big,) (a2) (N2) (f2) (g2). . If

{Sl, S3} 'is D2, S3 fixes both letters of one of the three cycles (ClC2), (d1d2),

(e1e2), and by proper transformation it can be made (eie2). Thus S3

(a1lf) (big,) (ClXl) (C2X2) (diyl) (cd2y2);(ZlZ2) (a2) (b2) (ei) (e2) (f2) (g2) * * where Xi, x2, are new to si. If Xl =- a2, 13l, or /2, S3 fixes c2, di, or d2. If x1 1 -y1, y2 is replaced by di, d2, or C2; that is, by d1 or C2. If by C2, S3

(di,i) (d2f2) * or (d,P2) (d2f1) *. And S3 = (af1i) (big-) (ciyi) (d-y2) (a2) (b2) (ei) (e2) (f2) (g2) (c1) (a2) (f3l) (/32) * is impossible. Then finally S3 = (aifi) (big,) (cl8l) (C282) (d,83) (d284) * . . But {S2, S3} iS octic. If {Si, S3} is DOI, S3 = (nifi) (blgl) (a2) (b2) (f2) (g2) . . . unites two of the three cycles (clc2), (did2), (ele2). The two fixed letters can be made c2 and d2.




Therefore s3'== (aifi) (b1g,) (cidi) (eixi) (e2x2) (a2) (b2) (f2) (g2) (c1) (J81) . * - and is commutative with t2, replacing a2 by /2. The letters x1 and x2 must be new to s, and in t2, so that s3 =(elyl)(e2-y2)* or (ely2)(e2yl).

which, however, would make it commutative with s2.

The proof is complete that G does not include D0o or Dt'

24. The group D2 has the four substitutions:

s=- aia2 b1b2 clc2 d1d2 eie2 fJf2 g192, S2 a1b* c1d* e1al e2a2 fu/3i f2/32 7172, t= a1b2 a2b1 c1d2 c2d1 * 1a2 /31/32 g1g2 t2 a2b2 c2d2 ela2 e2ac1 f12 f2/31 *7172-

It is transformned into itself by {Sl(gig2), S2 (y'ly2), alb .a2b2, cid, C2d2,

e1e2 1a2, flf2 ,81,82, alcl a2c2*bl1b2d2, elfi.e2f2 .c,81 /ia2f2, ale, a2cXl*ble2 b2a2

clfl c2/ldlf2 d22 glyl g2y2}. There are therefore but the three following types to try for s3:

(gal) . , (g1y)** , (glel)*

Let s3 = (gial) . It is commutative with neither si nor t1. Then it fixes b2. If b1 is also fixed, S3 (ailgi) (ylX) (y2)* *, where x is new to S2, and is a letter of t2. After transformation by cid,* c2d2, S3 = (aigi) (y1c2) - - .

Because of s8 and S2, S3 ==(algi) ('lc2) (di) , * * , which is impossible with th. Then S3 ==(a1g1) (bix) (a2) (g2) (b2) * * *, where x is new to s2 and in si, and therefore can be made c2. Because of t1, d1 is fixed, and then because of si, d2 is also fixed. Thus {t2, S3} is neither of our two possible types of non- Abelian diedral groups. The primitive group G contains no substitution of degree 14 that replaces a1, a2, bi, b2, cl, C2, di, d2 by gi or g2; nor el, e2, fl, f2, al, X2 /3l, 12 by yl or 72.

25. Let S3 = (g,y,) . If S3 fixes g2 or Y2, we can transform it ilito (g1y1) (g2) . Now 3)= 1 and after another transformation s3 (giyi) (a, -) (g2) (a2) (b2) * and this means that {t2, S3} is octic; therefore S3=~ (glyl) (-y2x) (g2) (a, -) (a2) (b2) * . Here x is new to S2 and t2, and is a letter of si, an absurdity. If S3 = (gl-yl) (g2X) (y2y) * , x is new to s, and tl, and y is new to S2 and t2; that is, x = 81 andy 82. Then S3 is either (gl-yl) (g2-y2) or (gl-yl) (g2k,) (y282) . , and the four diedral groups generated by s3 and sj, S2, ti, and t2 are octic. One of the four letters a,, a2, bl, and b2 is certainly displaced by S8, and by appropriate transformation it can be made a,. If s3 =(a,83) . * it displaces a2,b1, and b2. Not all these letters can be followed by letters new to D2. Then we can say that x in S3 = (a,x) . . . is a letter of D2. Moreover it is not b1, b2, or a2; but after




transformation can be c1, c2, or e1. Now s3=(g1y1)(aic) . * fixes a2, c2, b1, d1, b2 and d2 and is impossible with s2. Let S3 =(aiC2) (a2) (Ce) (b2) (d1)

It must replace b1 by a letter new to S2 and in s,; that is, by d2. It must omit both e1 and e2 or else both f/ and f2. If el and e2 are fixed by S3, we may transform it by e1fl/e2f2 x1ll1 .X2/32. Now S3 ~=-(a1c2) (b1d2) (a2) (cI) (b2) (d1) (f1) (f2) (elX) (e2y) * where x and y are new to si and are in s2. Neither e is followed by an x. We can transform by flf2 ,31,82 if necessary, and write s3 ==(aic2) (b,d2) (el,fl) (e2,82) (a2) (cl) (b2) (di) (fl) (f2) (aX) (X2)

Therefore two substitutions S3 are determined:

S3 1 glyl ̂ g2y2 aic2 b1d2 el,81 e232 8182p

S3 2 glyl * 72282* a1c2 b1d2 e1l3l e2f82.

Now let S3 (glY1) (alei) (a2) (e2) (b,) (ax) (b2x) (a2y) * . . Here x is new to t1 and is in t2, while y is new to t2 and is in t1. Hence y is ce or di and we can make it c1. It follows that S3 replaces C2 by a letter new to s, and fixes d2. Then x is f, or f2. Since transformation by flf2 ,81182 is possible, x = f, Now 83 must replace c2 by a letter new to s, and in t2, but not by 132. There- fore

S3 3 glyll g2y2 ale, b2fl C1l2 c231 8182, or 83 4 gl1yl 9281 728282 ale, b2fl c2/3,.

26. Let S3 ==(glel) . First let SS3 be of order 4. Then S3=

(elgl) (e2) (g2) (,G,81) (X , and x is in si and new to ti, that is, x is or f2. But not both {S2, S3} and {t2, S3} are octic. Now we have s3 (glel) (g2e2). and since {tlp S3} is octic it displaces an a or a /8. In case S3 =(glel) (g2e2) (alX) (a2y) , * * . x and y are new to S2 and to t2. There must be a cycle in S3 new to S2. If S3 =(glel) (g2e2) (a1lk) (a282) (8384) . . . p

it displaces no other letter new to S2, or to t2, and therefore is impossible. Then the new cycle in S3 displaces two of the four letters a2, b2, C2, and d2. Of these six cycles, a2b2 and c2d2 are in t2 and the other four are conjugate. Then

S3 = (gllel) (gOe2) ((X18l) ((X282) (a2C2) (alcl) (bi) (b2) (di) (d2) (PO) (#2)***

or finally, 83 5 =gel g2e2 acl a2C2 f1f2 181 282

Now let S3 = (glel) (g2e2) (/3 -) (/2 -) (7Y -) ((1) (a2) (72) . The letters following /81, /2, and y1 are new to S2 and t2, and both the letters of the cycle (f/f2) are fixed. Because of S2, S3 must displace a, or b1 and also cl or di. Since S3 can be transformed by aibi a2b2 and by c1d, *CA,




S3 6 = glel g2e2 *18 3*282 7183 ala2 clc2, or S3 7 = glel g2e2 * 1 71*282 *183 a1c2 a2Cl.

27. It is now necessary to consider the groups generated by S3 i and D2, i 1, 2, *, T7. We recall that G contains no substitution of degree 14 that replaces g1 or g2 by a1, a2, b1, b2, c1, c2, d1, or d2; nor yl or y2 by e1, e2, fl, f2, al, /32i, or /32.

Consider S3 1. A substitution S4- (glx) *, where x is el, e2,* *, 2,2 is transformed by S3 1 into (yly) * * *, where y is one of the same letters. Then uniquely S4 == (g81) . Because of S3 1 and t3 = a2cl*b2d1 e132 e2j3* gly2* g2-y1 &82, neither (gl8l) (y182) . * nor (gl8l) (y282) * is possible. Then S4 fixes y1, 72 and 82* If S4 (glAl) (g2x) * * * , x is new to si and to tl, but is in S3 1. impossible. Now S4 = (glA) (g2) (yl) (Y2) . . . and (S1S4)3 1. We seek a substitution S5 (g1E). *, where Ec new to H3, replaces in S4 a letter of si. Now t = S3 1 S4S3 1 = (i182) (XE) (g1) (g2) aiid ts5t = (glx) where x is a letter of D2, not g2, yl, or y2. This disposes of S3 1, and a fortiori of S3 2

For s3 3 and S3 4 the eight substitutions s4 which may replace g1 by el, e2, , /2 reduce on transformation by substitutions of D2 to (gle2) * and (g1l2) . . . . Since S3 354S3 3 (and S3 48483 4 likewise) is (yle2) . or (7fl2) . . . p S3 4 goes out at once and S4 in the case of S3 3 can only be (gA,) . .. Just as in the preceding paragraph this, too, is impossible.

We next take up S3 5. There are four transitive constituents in H3. But S4 can replace g1 only by the letters of the constituent flf2,/1,82. Since s2 and f1f2- J3132 transform {D2, S3 5} into itself, there is but one substitution S4 - (g1f1). . . to be considered. Now s1s4 =1= s4s, because of S3 5. Since it was proved that (g,el). * * must be commutative with si, (g1f1) * must also be commutative with s1, contradicting the last remark.

28. Only S3 6 and S3 7 remain. The arguments leading to the rejection of S3 6 will be seen to be equally valid for S3 7.

83 6 glel* g2e2 8181 7183 ala2 ClC2

t3 ' lgl *2g2* fll* f282 *7283 b1a2 dic2, t e1g2 e2gl *281- 182 *7183 b2b* d2di1 t5 (X2gl * lg2 *f281* f182 *7283 a1b2 c1d2.

The substitution f1f2 /31/32 -812 transforms H3 into itself. There are then but the three substitutions, (g183). * ., (g1lf) . ., (g181)* * ., to try for S4. Let S4 (g183) . S3 6 S4S3 6 =(Yl,el) *., which has been shown to be impossible. Let S4- (g1f) )* . It is commutative with si. Then S4= (g1fl) (g2f2)

(al8&) (a282) * if it is commutative with t3 also. Since S3 6 S4 is of order 3,




S4 displaces y, or 8n, not both. Like S3 6 itself two cyeles of S4 are on four of the eight letters a1, a2, *, d2. Therefore one cycle of S4 is (y,e), and S5 can be nothing but (g1e) *. But s48584 - (fi-y,) * * . Suppose that S4 is not commutative with t3. If S4=(glfl)(g2f2)(elx)(e2y). , x and y are in s8, and not in S3 6 or t4, an impossibility. If S4 =(glfl) (g2f2) (at) (a2)

(el) (e2). **, it fixes the letters of two cycles of S2 and is not commutative with s2. Let S4 = (g,,)* *. It may be that S3 6 and s4 are commutative. If so, S4 (g,a,) (ei13 ) . If S4 displaces g2, it does not replace it by e2 (because of s,) nor by a new letter E. But (g8l) (g282) (e,13,) (e2132) (a,) (a2) . . . is impossible with S2. If S4 fixes g2, it fixes also e2, 12, and a2, and s8s4 is of order 3. Now S4 ==(g18l) (alfl) (e11,) (a-) (b -) (c-) (d-) (g2) (e2) (f2) (a2) (132) (82) * if t3S4 = S4t3, and

S4 - (g,81) (e1l31) (a2f2) (e2) (g2) (fl) (al) (12) (82) if t3S4 + S4t3

All the letters of the two sets g,g2eje2ala2 and fMf2,81328182 of H3 are accounted for, so that H4 is certainly intransitive and S5 can replace g, only by a letter e of S4 = (xE) * , x being one of the eight letters a,a2blb2clc2dld2. A substitution of D2 (s' say) transforms (81x)-* into (Six'). *, where x' is fixed by S4. Then 848S4S58S4 = (glx') *, known to be impossible. Thus S4 is not commutative with S3 6, nor with its transforms by S2 or t2 : t3 and t5. Hence finally S4 (gl8l) (e2132) (el) (fl) (f2) -or (g,a1) (el) (e2) (fl) (f2) ...

and these too are impossible. Hence no G contains A2 without also containing D,. In what follows

the. only non-Abelian diedral group is D4.

29. There are 10 Abelian groups generated by two substitutions of degree 14.

'Consider the group generated by

s8 and S2 = alb, a2b2 cid, c2d2. *elfl / e2f2 * .a2.

We may transform tnis group by ala2- b,b2 and by albl.a2b2. Then only (glal). . . and (g,al) - need be considered for S3. Let S3 =(gal) - a .

We have immediately S3 = (glal) (g2a2) (b,) (b2) * * , and since we may transform by cld, c2d2 and by elf. e2f2 if necessary, S3 =i (g,a,) (g2a2) (clx) (c2y)

(e,z) (e2w) (a, -) (b,) (b2) (a2) (dl) (d2) (fl) (f2), where x, y, z and w should be letters of si. Let S3 - (g,)*x **. Uniquely S3 -ga, aA,- b282. C183A d284' e8t; *f286. Next 84 = (ga1i) . and S3483 ==(alx) *, where x is a letter of sl, not g, or g2. But this, as shown above, is not possible.

Consider s, and 82 = alb, a2b2- cid,c2d2 ele2a,la2.13,l32. Now S3 is



MANNING: The Prim1itive Groups of Class Fourteen. 639

(e1a,) . . or (e1f1) * * . If we try the first, S3 = (e1al) (e2a2) . . . by s, and (elal) (e2b1) . * by S2 If S3 ==(elfl) (e2f2) ,

* {S2, S3} i,S octic. Consider s, and s2 albba2b2.clc2 .c1c22 .31/32 -y12'8182,

si and 82 ala2 *1a2 P1P2 71y72 '8182 ' *lElg2

For S3 both (c1al) . and (c1d1). are impossible.

30. Consider s1 and s2== alb1 a2b2-c1c2 d1d2cXla2 B1pB2y1y2. Both (c1al). and (cie1). are impossible as before. Let S3 == (cld1) (C 22) * *

No S4 replaces cl by one of the other 16 letters of our group D. Then S3 has at least one new cycle (8182). In fact S4 is uniquely determined as S4 = C181 dl82 all'b2E2 .elZl'f1/ , gll Since transformation by alb2 a2bl E1E2

is possible, S5 is (c1E1) or (c11) , (g2) being another cycle of s3. If S5 =(c1El) - -, SlS5Sl ((c2C1) (c1) c , and s4s8s58l84 =(c2a) * .-. We now have s8 = (c1Cl) (d,g2) (C2) (d2) - ** Because the transitive constituent cl.. ..is increased only by the addition of the two letters of a cycle of S3 at this and each succeeding step, the substitutions si, S2, cannot in this case generate a transitive group.

Consider s8 and S2 = ala2. blb2 X1l2 *2il2 717Y2 8l82 * E1E2. The reasoning of the preceding paragraph is again applicable.

31. Consider si and S2 = ala2 * V2 - ClC2 * #1#2 *yly2 *8182. This is the last of the Abelian D groups in which s8 and S2 have a cycle in common. Therefore s3 =(albl) (a2b2) (Cl) (C2) (e1e2) * - .. To follow this we have S4 (alcl) (a2c2) . . or S4 =(ale) (blE2) (ClK) * - '. If S4 =(alcl) (a2c2)- .

S (alcl) (blE2) (ClC3) (a2) (b2) (C2) . . or (a1E2) (b1el) (c2X) (lE3) (a2) (b2cl) In either case the letters of the last four cycles of S5 are letters of s&

and S2. Like S5, S6 will replace a1 by a letter of a transitive constituent of {S38, 84} of degree 3, and so on. But these steps will not lead us to a transitive group. If 84 = (alEl) (b1E2) (ClK) * , it can be completed as follows:

S4= alE,-blE2-clK-di(axel/31 e2yl-.fl8l If S5 -(a1K) * *, SMSl8 lS4 (a2C1)*

already discussed. If SI = (a1gl) (blC2). . ., the last four cycles are again built up from the last four cycles of si and S2. The third cycle is (c1X) or (C2K). It is evident that we are not going to build up a transitive group by continuing with S6, S7,-

32. Consider si and S2 =alb a2b2 * cid, c2d2ala2 * /31 2 * yly2. For 83 we have (e1al) * * *, (e1f1) * * *, or (e1,1) * . It is to be remembered that no two substitutions of degree 14 have a common cycle. In the first case, immediately and uniquely, S3 = ela, e2a2 Ctfi . C2f2. acIXc3 . 13 yl'3. This case can

10




be continued by either (g1al). or (g1y1) *. Since H3 is transformed into itself by a1a2 b1b2 e1e2 and c1c2 d1d2. f,f2, we get

S4 1 = gia1 'g2a2. d1f2. d2fl . 1?:4 P134 7:174,

S4 2= gial* g2a2 * dlf2* d2fl * 1l4 #232y3 *72P3,

S4 3 =gly *el2 bla3 - f 12 83 -a2 4 C2,4

Let us take up the three corresponding groups 14 1, 14 2 and 14 3 in turn. Because c1c2- d1d2 -f1f2 transforms 14 1 into itself, only (gcl). * , (g1al) . *,

(gla4) * * need be tried for s5. A short calculation shows tihat all three are impossible. With 14 2 there are four possibilities for S3 : (g1cl) * . ., (g1ax) . . .

(gla4).. and (gl/31). *. The substitution s5 can be completed only in the last case, s5 =gl,8/31 a28 bly3 ely2 clX4 d2. - f2a2. The transitive sets of H5 are c1c2d1d2flf2ac1(X2a3X4 and one other, 8a, * . Only two forms of S6 are to be tried, (8c,) * * and (8ac)- ; or, if we prefer, their transforms, (glx1) * * and (gla4) * *. These two substitutions were inadmissible when proposed for s8. Let us consider 14 ,. To extend this group we have (glal) * *, (gla4) * -, and (g1al) - . Calculation shows that there are but two substitutions s5, one of which is 85 glal72el-bl173 a281 - c1f4 f22/2 d2/3.

With this group H5 only one substitution s6 need be considered because H5 has just two transitive sets, one of which is clc2dld2f1f2,l81/2/83fl4. We may put s6 == (1b4) * and this is at once seen to be impossible. The other s5 gla1.2a27y2aX,y3a2 fid2.f2dl./1'f4 goes out with S4 2.

33. The second possibility for s3, (e1al). * *, is to be studied. Uniquely, s3 = e(21fiP.gly1 a,ASlb282'c1EJ d2E2. The group H3 is on the five sets of letters: ala2blb28182, clC2dld2ElE2, ele2alX2, fif2ftd32, glg2yly2. For S4 we can have only (e,fi) . . or (elf2) . *. Now (elf2) is impossible because it would have to replace g, or yl by one of the letters a,, a2, . -, d2. Then S4 = elf, e2f2 ?l1fl1 2, alb2 2b, 8182 or S4 elf, e2f2 l /3 a2/23 U2 "1lq2 ?1 1202 2

The sets of 114 are ele2f1f2alX2fllf/2, ala2blb28182, glg2y1ly2, . Then

s5 == (elgl) - . , (eig2). . , or, in case S4 =(U2 2)* . . , (e11) - .- We need not have written down (elg2) - . for since (elf2). . . cannot be S4, it cannot be S,. Now s8 = (e1g1) (e2g2) (calyl) (a272) (fl) *. Considering only the first form of S4, it is seen that S, must displace a1 or b2, and by means of SlS2 this letter can be made a,. But S5 =(elgl) (e2g2) (alYl) (a272) (alcl) . . . or (e1g1) (e2g2) (21-yl) (a2y2) (a1d2) - * requires eight cycles. There remains only S4 = elf1 * e2f2. *Pi *2/32 *1glg2 ?q2- 0102. Try SB = e1g1 * e2g2- alyl * 272-l2 -

T1h'3'0103s For s6 we have only (e1t1) . . . and (elC2) . . . If S6 =

(eelgl) (f1C2) (g,C3) (ax) . .*, it fixes ,, x2., /3l, /25 yl and 72, while x should be in S2 and not in si, an impossibility. If S6 = (elt2) (fltl) . .*, it fixes




g1, 21, ,22 Pi and /2; therefore it must replace yi by a letter of one of the first four cycles of S2, which takes us back to the case of S3 (e1al). Finally try S5 ==(e1g1) (f1l2) *. This is impossible with si andc s2.

The third and last possibility for S3 is to be studied: S3 (e1f,) (e2f2) The substitutions S3, S4, * * * can replace e1 only by g1, g2, or a letter 8 new to H2, and any one of these substitutions is commutative with S2. Hence no one of them adds a new letter to 2122, 31/2, or Y1Y2, and therefore the group {81 S2, 8 S3,- * * } is not transitive.

This disposes of the group

{Sl, S2 alb1 .a2b2 'e 1d-. *C2d2o*112/3t1f82 -y1Y2}-

Only two Abelian D groups, one of degree 24 and one of degree 28, and only one non-Abelian D group, of order 3 and degree 21, are left.

34. In the group D4 there are the following three substitutions:

S1 aia2.b1b2.c1c2 d1d2 eie2 flf2 glg2, s2 a1a3ia.b1b3 * cic3* did3 e,1e3.* flf3.*g1g3, t a2a3 b2b3*C2C3 *d2d3 *e2e3 *f 2f.3*g2g3.

There is at least one substitution s3, similar to si, non-commutative with at least two of the above three substitutions. We assume without loss of gen- erality that s3 is not commutative with si and S2. Since this substitution s3 may or may not connect the transitive sets of D4, two cases arise, as {t, s3} is of degree 24 or 28.

35. Case I:

s3 a2b3.a3b2.clc4.dld4.ele4.flf4.glg4. In H3 are also t1 a1b3a3bl c2c4. d2d4.e2e4.f2f4.g2g4,

t2 aib2 * a2bl c3c4. d3d4. e3e4 f 3f 4 g3g4.

We may extend H3 by S4 alcl.a2C2*a3C3.d4d5.e4e5.f4f5.g4g5 or by a1c3* a3c1 * b2c4- d2d5- e2e5 f2f. 5g2g5. These two substitutions are conjugate under a2b2 clc3 c2c4 d1d3. d2d4. eie3 e2e4 flf3 f2f4 glg3 g2g4, which transforms si, s8, S3 into t2, S2, s., respectively. Then we use the first only. In H4 there are also the substitutions

t3 alC4b3c2 b2c3. d1d5. e1e5 f1f5 g1g5, t4 b3c1 a2c4 b1c3. d2d5 .e2e5. f2f5 .g2g5, to b2c1 blc2 a3c4. d3d5. e3e5 * f3fs5.*g3g5.

If s, replaces a1 by d3, s5 a1d3 a3dlb2d4.c2x e2y f2z.g2w, where x is new to H4, y, z and w are new to H3, but displaced by S4. Then s$=

a1d3 a3d1* b2d4* c2 * e2f5 *f2e5 *g2-, an impossibility. Then




s5 a1d3.a2d2 a3d3.C4d5 * e4e6 * f4f6*g4g6.

This s5 is unique because (a1d4) . * and (aid2) . can be transformed into (a1d,) ... or (a1d3) ... by a substitution of H4 that fixes a1. Next,

S6 ale, ct 2e2 *a3e3 * C4e5* d4e6 .f4f79g4g9, St ajfl*a2f2*a3f3 * c4f5 d4f6.e4f7 * g4g8,

s8 ajgj*a2g2-a3g3 - c4g5 *d4g'6.e4g7.f4g8.

The groups H6, H7, H8 are unique. H8 is simply isomorphic to the symmetric group of degree 9. In fact it is the group according to which that symmetric group permutes its 36 transpositions. Because of the impossibility of a substitution (a1d3)- * * of degree 14, H8 is not a subgroup of a double transitive group of the same or higher degree. H8 contains all the permissible substitutions similar to s, on the letters of H8 only, except perhaps (a1b1)* (a1c2). - and (a1c3). * . Now t4 and t5 reduce (a1c2) ... and (a1c3) to (a1b1) * . This last substitution (a1bI) (a2b2) (a3b3) . fixes all the other letters of H3, and therefore displaces d5, e5, f5 and g5, because of S4. In the same way s5 demands e6, f6, g6, S6 calls for f7, g7, and S7 for g8. Hence 18 is invariant in any larger primitive group G, if one exists. But H8 is a complete group.

36. Case II.

S3 a1a4. b1b4 clc4. d1d4. e1e4f1f4 g1g4.

The case in which s, (i 3, 4,. *) is non-commutative with the substitutions of such a group as D4 and connects two of its sets of transitivity has been fully considered. Then since we know that there is a substitution similar to si which replaces a, by a letter of another constituent of Hi (i 3, 4 ) we write down immediately and uniquely:

S4 a1b* a2b2 a3b3 a4b4 a5b5 a6b6 a7b7, s5 alc, a2C2 a3C3 a4C4 *a5c5.5 a6c6 a7c7, S6 aid, a2d2 a3d3 a4d4 a5d5 a6d6 ad7, S7- ale, a2e2 a3e3 a4e4 a5e5 a6e6 a7e7, s8 alf* a2f2 a3f3 a4f4 a5f5 a6f6 a7f7, sg alg1 a2g2 a3g3 a4g4 a5g5 a6g6 a797, s=o aja5 b1b5 c1c5 d1d5 eje5 ff5* g1g5, s=l aja6 blb6 c1c6 d1d6 eje6 flf6 g1g6, s12 = ala7 *bb7 clc7 d1d7 e.e7 f1f7 g9g7.

This imprimitive group H12 of degree 49 and order (7!)2 is a subgroup of one and only one primitive group G, and G is of degree 49 and order 2( !) 2. A substitution of G that transforms H12 into itself is




a2b1 a3cj.a4dj a5e, a6f. a7g91b3c2.b4d2.b5e2 * b6f2*b7g2 * c4d3. c*5e3. C6f3- C7g3 d5e4* d6f4. d7g4* e6f5* e7g5 f7g6.

37. At this point we remove the restriction that the positive subgroup of G is of class > 15. But G contains no substitution of degree 15 and order 5, for the primitive groups of class 15 containing such a substitution are of degree 16 or 17. The substitutions of degree 15 of G are all of order 3 and generate an invariant, and therefore transitive, subgroup H of G of degree > 20. For the determination of H we have precisely the conditions of our former study of the primitive groups of class, 15.* The transitive groups H of class 15 (with no substitution of degree 15 arid order 5), generated by substitutions of degree 15 and order 3 were found to be:

(1) A simply transitive primitive group of degree 21 simply isomorphic to the alternating group of degree 7.

(2) An imprimitive group of degree 25 simply isomorphic to the direct product of two alternating groups of degree 5.

But there was a serious oversight in that study of the groups of cl-ass 15 (as will be shown in ? 39) because of which another such group is to be taken into account:

(3) The positive subgroup of the larger transitive constituent of that subgroup of Mathieu's quintuply transitive group of degree 24 in which the subgroup that fixes three letters is invariant.

38. It will now be shown that no G of class 14 contains either (1) or (2).

Suppose the group H of degree 21 is a subgroup of G. Now G contains a substitution of degree 14 and order 2 which transforms H into itself and therefore with H generates a primitive group of degree 21 and order 7!. It is easy to show tha-t this G is of class < 14. Let G1 and H1 be the subgroups of G and H, respectively, which fix one letter. The subgroup H1 is known. It is isomorphic to the symmetric group of degree 5, a complete group. Then G1 is a direct product of H1 and a subgroup of G1 of order 2. Of the two transitive constituents of degree 10 of H1, one is primitive and the other is imprimitive, so that G1 is also intransitive, with one constituent, at least, primitive. This primitive constituent of G1 can have no invariant substitution except the identity, and therefore remains of order 120 in G1, as in H1. The invariant substitution of G1 is therefore of degree 10 on the letters of the other transitive constituent.

* American Journal of Mathematios, Vol. 39 (1917), p. 281.




As to (2), the imprimitive group of degree 25 and order 3600, it has been proved * that the only primitive groups of degree 25 in which it is invariant are of class 10 or 15.

39. In order to conclude this investigation of the primitive groups of class 14, it is necessary to correct the error made by the author in line 20, page 284, of the paper on primitive groups of class 15. The statement " nor is H an imprimitive group of degree 16, for then its order is of lnecessity 48, and this case has been examined " is false. The conclusion referred to is in the second and third lines from the bottom of page 283, " the primitive groups unider consideration do not contain a transitive group of degree 16 and order 48 ', quite true because the groups in question are generated by two substitutions of degree 15 and order 3. But the reasoning is not applicable to the groups of the next page, which have three, instead of two, genierators of degree 15 and order 3. This paper was written up completely as the research progressed. It was laid aside for some weeks at a poilnt between these two statements and on resuming the task the last quoted result was misinterpreted.

The lines quoted from page 284 are to be replaced by ?? 40-48 below.

40. The group D is tetraedral, with one transitive constituent of degree 12 and onle of degree 4:

D = {si, S2 = alb2c3 b1a2d3 cCd2a3 d1c2b3.e1e3x}.

For the larger conistituent, its conjoin might be used, but it is conjugate to the given regular group.

The group H = {D, s3} is transitive. Suppose it to be of degree 16. It is imprimitive. It is not in a primitive group of lower degree than 21, anld then only by virtue of having 5 systems of imprimitivity of four letters each with a letter (x, say) in common. Each of the systems to which x belongs is left fixed by si, and these 5 systems can only be xeie2e3, xa1a2a3, xb b2b3, xc1c2c3, xdld2d3. The group {sl, S3} is of order 12 and like D has transitive constituents of degrees 12 and 4. The letter following x in S3 is one of the 12 letters of the large constituent of D, and by transformation we can make it d2. Then s3 =(did3x) (d2) (el -) (e2 -) (e3 -) . .. The 20 systems of imprimitivity of H are completely determined by D. They are

xala2a3, Xblb2b3, XClC2C3, xdld2d3, xele2e3, eib2d3ci, eia2c3dA, e1d2b3a1, eic2a3bi, d1a1blc1, e3c3b1d2, e3d3alc2, e3a3d1b2, e3b3c1a2, d2a2b2c2, e2d1c2b3, e2c1d2a3, e2b1a2d3, e2alb2c3, d3a3b3c3.

* American Joutrnal of Mathematics, Vol. 39 (1917), pp. 308, 309.



MANNING: The Prim,itive Groups of Class Fourteen. 645

It may be well to note that after the first 17 systems are written down from D, the remaining system d1, has no letter except d1 in common with any of the preceding systems, and therefore is d1a,b,c,. Likewise we have the systems d2a2b2c2 and d3a3b3c3. Now s3 replaces the system xa1a2a3 by e2d1c2b3, the latter by e,b2d3c1, and similar statements can be made for the systems xblb2b3, , xc1c2c3, . Then

S3 = (d,d3X) (a,b3e,) (c1a3e2) (bIc3e3) (a2c2b4,

completely determined by the 20 systems of imprimitivity of H. 41. This transitive group H has an elementary subgroup E of order 16

generated by el S12S2 ale a2b2 b di c2d2 a3d3. b3c3*elx .e2e3,

di S12S3 ale2 a2b3 blel1*b2c3 cie'3*c2a3* dix d2d3, e3= 82812 albl*a2d2 b2c2c1d a3c3 b3d3 *e3X.e1e2, d,3 838S2 -a1b2 a2cl blc2 b3e3 c3e2*e1a3.d1d2 d3x.

As a notation for the 15 substitutions of the regular group E we use the letter which follows x in that substitution. Now s1 transforms {e1, e3} and {d,, d3} each into itself. Moreover, {E, sJ} = {Sl, S2, S3} = H. Then E is invariant in H and H is of order 48. H has 16 conjugate subgroups of order 3.

That subgroup of the group of isomorphisms of the elementary group E, in which s, is invariant, is the transitive group

=s {s,a,ctb1-a2b2ct3b3'cldi c2d2.c3d3, a,el. a2c2 a3c3 bid,, b2d2* b3d3, ale, a2e2 a3e3 b1d3. b2d,.b3d2, a,bl * a2b3 * a3b2* c2c3 * d2d3 * ele2}

of order 360. These generators transform {E, s,} into itself. The group of isomorphisns of the elementary group of order 16 is isomorphic to the alternating-8 group, in which {(abc)} is invariant in a subgroup of order 360, and { (abc) (def)} is invariant in a subgroup of order 36, so that the order of I, is right.

42. In the imprimitive group of degree 20 towards which we are working there are 5 systems of four letters each permuted according to the alternating-5 group. We can take for the substitution which with H generates this imprimitive group (call it H1) a substitution of E and transform it into a substitution t which fixes an arbitrary one of the 5 systems- and transforms the new system yly2y3y4 into the system x - -. Trans- formation of H and t by substitutions of I, permits us to write

t=(y4x) (y1e1) (y2e2) (y3e3) (a,-) (b1-) (ci-) (di-) (as) (b3) (c3) (d3).




The four blanks in t are to be filled by a2, b2, c0 and d2 in some order as yet unknown. Now ts1t == (y1y_y3) (x) (e1) (e2) (e3) (y,) . . . is commutative with s1, and therefore one of its cycles is the inverse of one of the cycles of s1. The same is true if for s1 we put s2- (a01b2c3) (b1a2d3) (d1c2b3) (c1d2a3) (ele.3x), (alc2d3) (bld2c3) (c1a2b3) (d1b2a3) (elxe2), or (a1d2b3) (blc2a3) (c1b2d3) (d1a2c3) (e2xe3), all of D. Then one cycle of t is ala2, b1b2, clc2, or d1d2. TTansformation by the substitution albi, a2b2 a3b3 cid1' c2d2 c3d3, aid, a2d2 a3d3 bclc b2c2 b3c3, or aict 'a2C2 ' a3C3 * bid, * b2d2 b3d3 of Is en- able us to fix upon b1b2 as this cycle. The reason for this choice is that it seems to be the most convenient when we comq later to connect our groups with subgrouLps, of the 4uintuply transitive group of degree 24. From s2 we see that t has either (d1c2) or (c1d2) ; from e1s1, either (alc2) or (cla2) ; froM b3s1, either (d1a2) or (a1d2). If t= (a1c2) * - *, the other cycles are determined by these considerations. Thus t may be

blb2 a1c2 c c1d2 d1a2* yiel* y2e2* y3e3 WY

But if this t is multiplied by the substitution b3 of E the product is of degree 12. Then

t a a1d2* b1b2 * c1a2 * d1c2 * ely1 * e2y2 * e3y3 * Xy4

uniquely.

43. Consider the imprimitive group {E, t}. It has the same 5 systems of four letters each as Hi, and permuted according to the alternating group. There is a subgroup A l= {el, e3, telt, teat} in {E, t}, an elementary group of order 16 with 5 regular axial constituents, obviously invariant in {E, t}. Another subgroup of {E, t} is r = {a2, a3, t}, of which one transitive constituent is the simple group { (b3e3) (d1c2), (b3d1) (c2e2), (d1c2) (e2y2) } of order 60. The other transitive constituent is of degree 15 and order 60, and because of the systems of imprimitivity of {E, t}, r is not a direct product. Since A and r are permutable groups and have only the identity in

common, {Ar, } is of order 960. Now {A, r} = {e1, e3, te1t, te3t, a2, as, t} ; {E, t}-

Because s12Es1 E-and s12tsl c1ta2tcl, {E, t} is an invariant subgroup of H1 = {E, sl, t}, and the latter is therefore of order 2880. The head of H1 is the intransitive subgroup As - {A, V} of order 48, where V = tsits- aidl1cl* a2d2C2 * a3d3c3 * ele2e3 * Yly2Y3.

The order of the subgroup F of H1 that fixes Y4 is 144. F has two transitive constituents, one on Y1Y2Y3 and the other on the remaining 16 letters.




44. If H1 is the subgroup that fixes one letter of the doubly transitive group if2, there exists a substitution T = (yiz) which transforms H into H. T can be assumed to be!a transform of one of the substitutions of E. If T displaces x, its transform by one of the substitutions of E will fix x. Since THT = H, T replaces ino letter (of H by a y,, y2, y3, or Y4.

Then T- (yiz) (x) . . . transforms s& into s1 or St2, and 6 of its 8 cycles come from Is; the only cycle unaccounted for must be (yy), with subscripts to be determined. The substitutions of order 2 of Is which permute four cycles of s, fall into two sets of conjugates under the substitutions of As that fix x (the head of H1 was called As), with the three that fix the cycle (ele2e3) of s, in one set, and the 12 that displace (e1e2e3) in the other set. For example, one substitution of As is te3s1ts1 = V2- b1c1dr b2c2d2 b3c3d3 ele2e8 Yl.3Y4. Another is V = ab1dl 2b2d2 a3b3d8 e1e2e3 * y2y4y3, the transform of slts1t by telt. Since all the substitutions of As that fix x are commutative with sl, and transform E into E, we need try for T 'only two of these 15 possible substitutions of Is. Similarly, of the 10 substitutions of IS which permute only two cycles of si, the 6 which fix (ele2e3) are conjugate under the substitutions of As which fix x, and the four which displace (ele2e8) *are conjugate under the same subgroup of the head. There are then four possibilities for T:-

To= a1e, a2e2 a3e3 b1d3 b2d1 b3d2 yy YZ, T=- ab, * a2b2* a3b83* c1d1 c2d2* c83d., yy* yz, T2- a1e3 *a2e2 -a8el b2b8 cc3* d1d2 yy yz, T3 =- a1bl a2b3 3b2 c2C3 d2d3 * ee2* yy * yz.

If te2t= a1b1 *c1d1 a2b2 * c2d2 a3b3 -c3d3 yly3 y2y4 of A is multiplied by T1, the product is of degree ? 5 and is not the identity. If te2t is mnulti- plied by T3, the product is (cld) (z, with a square not the identity and yet of degree ? 5. Multiply T2 by the substitution VT of As; the product is

(alelb3d2) (a2e3bldl) (a3e2b2d3) (C2c3) (zy) (yj) (y2y4y3).

The product of (zy) (yy) by (y2y4ly), where the undetermined subscripts are 1, 2, 3, or 4, is a substitution of the alternating-5 group and therefore not a cycle of four letters. It is of order 2, 3, or 5, and the degree of the second or fourth power of the partial product is ? 12, and is not the identity.

In To the subscripts of the y's are to be determined. Consider VYTo - (a1d3e3) (a2d1et) (a3d2e2) (b1b2b83) (y24yy3) (yy) (yz). The partial product of

(y2y4y3) by (yy) (yz) must be of order 3 and therefore To fixes yl. IV2To= (a1ela2e2a3e3) (biclb2c2b3c3) (did2d3) * . The product of (YysY4) by (yy) (yz) (yr) must be of order 2 and degree 4, hence is (y1_4) (z-), so that



648 MIANNING: The Primitive Groups of Class Fourteen.

To= Y3Y4 y2z ale, a2e2 * a3e3* bld3 b2dI bsd2.

After transformation by te2t, we have finally the unique substitution

T = y4z * YlY2* ble, * b2e2 * b3e3* alc3* a2CI1 a3C2.

45. It must next be proved that H11 is indeed the subgroup of H2

{iH, T} that fixes z. Now

H1= Fs + FstFs + F.te2tF.,

where F8 is the subgroup of H1 that fixes y4. We have to show that TFsT = FS, and that (tT)3 and (te2tT)3 are substitutions of Fs. Now F, {H, V = s1ts&t}. But THT = -, as we know, and TIVT 811V2. Also,

tT xzy4* b1e2yI * b2e1y2 * b3e3y3 * Ad2C8 asC2dA and te2tT = y4y3z,* ala2a3* b1d1el, b2d2e2 * b3d3e3 * c1c3c2,

so that (tT)3 = 1 and (te2tT)3 = 1. Then 12 exists. Since there is no positive doubly transitive group of order 21*20 144 of class < 15, H2 is certainly of class 15, an essential point not previously proved.

Since s1T Ts1, the doubly transitive group {E, t, T} is invariant in 12. Now P= {E, Y} where

Y - tclta2te2 = ala2e2 b1b2X, c1c2el dld2e3 * a3d3c3 * Yly3Y2,

is the subgroup of {E, t} that fixes y4, anid

{E, t}=F+FtF+Fte2tF;

while TET=- E, and TYT = c,Y2. We have as above (tT)3 - (te2tT)3 1, so that {E, t, T} is of order 21 20 * 48.

46. In this connection it is easy to prove the existence of Mathieu's quintuply transitive group of degree 24.* We shall show that the group {E, t, T} whose existence has just been proved is the subgroup that leaves three letters of Mathieu's group fixed, and that H2 is a subgroup of the transitive constituent of the largest subgroup of Mathieu's G24 in which {E, t, T} (or say G21) is invariant.

The generating substitutions of G24 given by Mathieu are in ? 18 above. Since the connection with the immediately preceding work must be made clear, we now replace Mathieu's letters xi, or rather the subscripts i (i oo,

* Mathieu, 'iousville's Jownl, Vol. 18 (1873), p. 25. Miller, Messenger of Mathemoatics, Vol. 27 (1897), p. 187; Bulletin de la Soceite

Matthe6matique de France, Vol. 28 (1900), p. 206. De Seguier, Groupes de substitutions (1912), pp. 156-163.




O, 1,* * *, 22), by cc, ,O W, Y4, Y3, a1, z, b3, b1, c1, e1, a2, e2, d2, e3, Y2, d1, c2, ys, a3, b2, c3, x, d3, respectively. Then Mathieu's {Z1, Z2, Z3, Z4} is precisely our group E, for Z1 = d3, Z2 = b3, Z3 = a2, Z4 = C3, and his Y is our Y. Moreover,

U y4c2ejb3c1w y3d2e3a3al * yja2e2d3bl * y2b2XC3d, - (b2Yb2te1)3,

a substitution of {E, t}. And t is in {E, Y, U}, for t = b2Y2b2U2e1. Thus it is proved that {E, t} = {E, Y, U}.

Mathieu gives no substitution that with G20 generates G21, but the substitution YBa3B-'Y-1 fixes oo, 0 and w, and is our T. From Mathieu's set of generators we also get

T P-1B-7c2B7Y zw YZY2 a2d3 a3b2 b1d1 b3c2 c3d2 e2e3,

T'llt A-2d3A2b3 -wo* YlY2, ajc2 l 2c1 a3c3 b1b2 d1d2 eje2,

T"' = Ye2TA-'XT"'XA.Te2tY-le2 - 0? c) YlY2 * aict * a2c3 a3C2 * b2b3 d2d3 * e2e3.

Since wuo oo aja2a3 b1b2b3 c1c2c3 . d1d2d3 eje2e3,

T"T' =zwo a1b3c2 a2c1d3 *a3d2br b2djc3* eje3e2, TT y4z, ajc3d2 a2d3cI * a3e2b3 * bid,el b2c2e3,

the existence of G24 will be proved as soon as it is shown that

T'{E, t}T'== {E, t}, T"{E, t, T}T" = {E, t, T3, T"'{E, t, T, T'}T"`= (F, t, T, T'}.

This is established as follows:

TT'E= T"ET" - El T'tT' = T'"tT"' tb3Y2b3, T"tT" ==t, T"TT" -T"f'TT"' == T., T"'T'T"= - T'.

In the course of the above proof of the existence of the quintuply transitive group G24 we found a substitution T"T" of order 3 which transforms {E, t, T} into itself, and which except for the cycle (wo oo ) is our si. The transitive constituent of degree 21 of {G21, T"'T"} is H2.

It should be noted that G24 is of class 16.

47. It remains to be seen if H2 is a subgroup of a larger primitive group of class 15. Suppose it a subgroup of a group G of the same degree. Because of H, G is not triply transitive, and G1 (the subgroup that fixes z) has the sam'e 5 systems of imprimitiv'ity as H1. The transitive constituents




of the head of G1 cannot be symmetric groups, for then the class of G would be ? 10. If the 5 alternating-4 groups of the head are cormbined to make its order > 48, there must be more substitutions of order 2 and of degree 16 or 20, and this is impossible because such a substitution would have at least four cycles in common with a substitution of A and the resulting product would be of degree -' 12. The head of G1 is A, of order 48. Then the group in the systems is the symmetric group of order 120.

In H1 there are 96 subgroups of order 5, and in H1, {U}), for example, is invariant in the subgroup { U, C3, e2V1e2} of order 30. According to Sylow's Theorem, G1 has 96 subgroups of order 5 and each is invariant in a subgroup I. of order 60. Suppose that IX has substitutions which transform U into U2 without permutation of cycles. One such substitution (Q) fixes y4 and therefore as a substitution of the imprimitive group G1 permutes YI, Y2 and y3 among themselves, that is, fixes y., y2 and y3. Therefore

Q - c2e1c1b3 d2e3a1a3 * a2e2b1d3 * b2xd1c3.

Now b2Q = ala2c1d2blc2 * a3b3d3 eje2e3 b2d1, and (b2Q)3 is of degree 8. Then all the substitutions of IX not in the invariant subgroup {U, C3} permute cycles of U according to a group of order 6, the symmetric-3 group. A substitution of IX which permutes two cycles of U cannot be commutative with U, for then it or its fifth power would be of degree 10. The known subgroup {U, e12V1e2} of IX shows that the cycle (yi * ) of U is fixed by I. Then a substitution which permutes two cycles of U is WT (Y2Y3) (y4) (y1) . . . I with the property of transforming U into U4 or U2. No one of the 30 substitutions of IX which permute two cycles of U is commutative with U. Now C3 transforms U into its inverse. If W does the same, Wc3 is commutative with U, and therefore W is in {U, C3, e2Vle2}. Then T-1UW-= U2. The subgroup {U, W} has one transitive constituent of degree 10 and two of degree 5 each. The transitive constituent of degree 10 is completely determined by U and the given cycle (Y2Y3) of W, and W is completely determined as to the other two constituents because yi and 'y4 are fixed by it. Thus

= c2e1c1b3 * a2e2b,d3 * Y2Y3 * d2saM3 * b2e3dla3.

Since W2 = C3, W3EW E, W3S1W a2S1, W3tW alta1, W3TW =- VTV2, {H2, W} is a doubly transitive group of order 21 * 20 288. There is no doubly transitive group of this order among the groups of class < 14. Now

Td2WY2 - a1C2 * a2cl * a3G3* b1b2 * CIC2 * e1e2 * yly2,

so that {(12, Wj} is actually of class 14. There is therefore no group of class 15 and degree 21 of which our H2 is a subgroup.



MANNING: The Primitive' Groups of Class Fourteen. 651

The substitution Vd2WV2 above is (wo)T", so that {H2, W} is the transitive constituent of degree 21 of the largest subgroup {E, t, T, T", T"'} of G24 in which {E, t, T} is invariant. We have showtn that {'H2, W} is the only primitive group of class 14 in which H2 is invariant.

48. Can H2 be the subgroup that leaves one letter fiXed of a triply transitive group I., of degree 22? If HE3 exists, there is a substitution T= (zwu) (y4) ... of order 2 and degree 16 in H13 which transforms H1 {E, Si, t} into itself. Now H, has only one system of imprimitivity of four letters including Y4, Y1Y2Y3Y4, and therefore T1 permutes the three letters yi, Y2 and Y3 among themselves. T1 can be chosen- to fix x. This T, transforms E into E and {sl} into {sl}. Then T1 has 6 cycles from Is and may now be written (zw) (y1y2) (y4) (y3) (X) .- - . Since T1 respects the systems of four letters each of H,, only three substitutions of order 2 of Is are available for T1: a1b1 cid1 a2b2 * c2d2 a3b3 c3d3, a1c1 b1d, a2c2 * b2d2. a3c3 * b3d3 and aid,, bic, * a2d2 * b2c2* a3d3 * b3c3. But these three substitutions occur in substitutions of A, which, multiplied by T1, would lower the class of 113 to 4 or 6. There is no group 113 of degree 22 with H2 as a subgroup of index 22.

49. At line 32 of page 288 of the paper on Class 15 all the partitions of the degree of the subgroup F have been rejected except two: 12, 9 and 9, 12. There is a slip in the discussion of these two cases which may be corrected by noting that the conditions of the following theorem are ful- filled. By the notation G(x) is meant the subgroup of G that leaves fixed the letter x, and by G(Q) (a) the subgroup of G(x) that fixes a.

Let G be a transitive group whose G(x) has just two transitive constituents, A on the m + h lett,ers a, a, . . and B of a lo,wler degree mr Letl G(x) (a), of degree *> 2i, have no constituent of degree < h, but at least one transitive constituent of degree mh Then G is imtprimtive.

The theorem is true if m < 3< * Then it is assumed that m ? 3. Let bi, b2, * * *, bm be the letters of B.

Suppose the theorem false. Then G(x) is a maximal subgroup of G. Since A is the only constituent of G (x) of its degree, it is paired with itself and G contains a substitution S (xa) . . . t which transposes G (x) and G (a) and transforms G (x) (a) into itself. S cannot permute the m letters

* Miller, Proceedings of the London Mathematical Society, Vol. 38 (1897), p. 533. f Burnside, Proceedlings of the London Mathemnatical Society, Vol. 33 (1901), p.

1162.




of B among themselves for then G (a) and G (x) would generate an intransitive group, a subgroup of G, contrary to the assumption that G (x) is maximal. If S transforms the given transitive constituent of degree m of G (x) (a) into itself, its letters are certainly a1, a2, . . ., am of the constituent A. All the other h - 1 letters a'. * * of A are fixed by G (x) (a), for if displaced they would occur in transitive constituents of degree < h, contrary to hypothesis. Then because the degree of G(x) (a) is _ 2n, it displaces all the letters of B anld permutes them among themselves, which is impossible.

But G(x) (a) may have two transitive constituents of degree m, transposed by S. Then one is on b1, b2, . . ., bm and the other is on a1, a2, * * , am. Now h > 2 for if h - 1, A is doubly transitive, which is impossible when G is primitive.* The other letters of G(x) are as before all fixed by G(x) (a), so that G(x) (a) G(x) ('). In this case both G(a) and G(a') have transitive constituents of degree m on a,, a2, * a.. The group generated by G(a) and G(a') is intransitive. But G(a), a conjugate of G(x), is a maximal subgroup of G.

Hence G (x) cannot be a maximal subgroup of G, and G is therefore imprimitive.

STANFORD UNIVEBSITY, CALIFORNIA.

* Transactions of the American Mathematical Society, Vol. 20 (1919), p. 66, Th. 16.



the primitive groups of class fourteen

Documents