on the primitive groups of class ten

On the Primitive Groups of Class TenAuthor(s): W. A. ManningSource: American Journal of Mathematics, Vol. 28, No. 3 (Jul., 1906), pp. 226-236Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2370062 .

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On the Primitive Groups of Class Tlen.

By W. A. MANNING.

Let sl, 2 .-.. be a complete set of conjugate substitutions of prime order (p) in a primitive group G. They generate a transitive group. A certain number (x) of these conjugates may generate an intransitive subgroup II. We wish to prove, before taking up the groups of class 10, that a substitution can always be found in the series sl, S2, . . . which connects the letters of any given set of intransitivity of J1 with the letters of some other set.

Suppose that the theorem is not true. Take a particular transitive set of I1-- ISI 52, s.. in the letters a,, a2, .... Choose from the series sj, .... a substitution SA+1 which connects a1, .... with new letters, anid which, of all the substitutions connecting the letters a,, .. .. with new letters a, has the fewest new letters in the cycles with al, .... There may be a number of such substitutions. Select that one which connects the fewest new letters d with the letters b. Now consider 12 = I1 , S,+ 1 If no substitution of the series s1, . . .. connects the extended set a1, .... with another transitive set of 12, we take 8A+2 as we did

sA +1 and continue in this way until we have an intransitive subgroup 1." and a substitution Se which connects the extended set al, .... with some other set of

le It is now essential to consider closely the substitution s._ by which we pass fromle -2 to I_j. Let the letters of the first set be a,, a2, .... ak, and let the remaining letters of 4p2, which may or may not form a single transitive set, be denoted by b1, b2, . . . .; we shall speak of the letters a and the letters b.

Letters a, new to 4-2, are connected with letters a by se1_. Since s8l1 is of prime order, any power of s1-I connects a's and a's. If se_- has two a's in any one cycle, x can be chosen so that in s"_. these two new letters are adjacent. Then unless s -1 replaces all the k letters a by a's, one of the generators of the group se -$ I4-2 s-1 will connect a's and a's and displace fewer a's than does

se-1, contrary to hypothesis. Suppose that s replaces every a by an a, but- has two a's in the same cycle with two a's, thus:

8_ 1 (a, a, * . - --a2 a2 ....)



MANNING: On the Primitive Groups of Class Ten. 2 27

We can choose y so that SY =1(a1a2 .** ala

and proceed as before. It is now evident that if one cycle of Se-1 displaces p - 1 letters a, all the cycles containing a letter a displace o - 1 a's. In this case

S8_= (a, a,at2/ a..O(lP-l) I /1 a(r - 1) .... t(-) (a >-? and all the substitutions of the series si, S2, ..., not in Ie_2, and which connect a's and new letters, are of this form. Since G is a primitive group there is in it some substitution t which replaces an a by an a, and an a by some other letter w. Consider the group t' 1e2t. One of its generators a connects an a and a letter w. This w cannot be a letter b. Then

a =(a' c' c"l .... c(P-1)

is of the same type as se-1, with the k letters a found in k different cycles. This gives

t =aa' ..a"'c ..ba"..

Then tIe42t' leaves at least one a fixed and has a generator, similar to s,, which connects letters a with other letters. We can at once conclude that se-1 has not two letters a in any cycle, and further that when ]p = 2, se,- replaces an a by an a. We may write

Se= a,, **

where 3 is connected with the b's by Se , and t is arbitrary. If Se_]- has two letters a, as a, aIL +1t adjacent in one of its cycles,

5e-1 Se 5e-1 a,IL+, b *.

Or if Se -1 leaves an a fixed, we can make a, that letter and have

s- is s -=a b.

Hence the theorem as stated is true. All the primitive groups of class 10 which contain a substitution of degree

10 and order 5 are known.* It remains to determine those which do not contain such a substitution. It is convenient to determine first the diedral groups of class 10 which are generated by two substitutions of the form

s= a, a2 . bl 2 - c1 c2. d1 d2 . e1 es.

*Transactions of the American Mathematical Society, Vol. 4 (1903), p. 351.

30



228 MANNING: On the Primitive Groups of Class Ten.

For brevity D shall be any diedral rotation group. Only one of these groups D is transitive. It is a primitive group of degree 11 and order 22, not contained in a larger primitive group of class 10 of the type we are considering. We neg- lect it enitirely in what follows.

Let sl and 82 be the two generators of a D. The product 81 s2 = S is a positive substitution.

If the degree of S is less than the degree of {s1, S2}, s, and 82 have one or more cycles in common. They cannot have three cycles in common for then S would be of degree 8 at most. If s, and s2 have two cycles in common, we have the group

D1= s,= a, a2 . b6 C1C2 . d12 * el e2, 82 a1a2 6blb2 al1a2 * .1 2 * )l . 2

If s1 and 82 have just one cycle in common, S is of degree 12, 13, .... 16. Neg-

lecting the two letters common to s1 and s2 we have to do with a positive intransitive D of class 8 and degree 12, 13, . .. . or 16. S, if of degree 12, is a

regular substitution of order 2, 3, or 6. There are two groups:

D2= I s2 = ala2. blC1 1 b2 C2 . a, a2 *1 2

D_ s81, 82 a1 a, . b6 11. cyl - d1. . e1 elE.

Corresponding to S6 = 1, the D of class 8 cannot be constructed. A substitution of degree 13 contained in G is necessarily regular. If S is of degree 14 and regular, it is of order 7 and there is no group D. If S is not regular, it is of order 4 and 181, s2t has three octic constituents. These cannot be so arranged as to give, with the other two constituents, a D generated by two negative substitutions of degree 10. Again no isomorphism can be set up when S is of degree 15. When S is of degree 16 we have

D4- s81, 82 =a a2 * ala2 * f(312 *l - 2 *1 71

We can now assume that S is of the same degree as s1, s42. If S is of degree 12, it is regular and is of order 2, 3, or 6. We have at

once D5=- 11, s2=a,b . a2b2 cld c2d2 *a a2

D6- _ sly,t2=_ a, b, a2 cl - b2 C2 dj a, - eig .l t, 83 =;

D7- , s2 a, b6. 6cl2. del . d2al e2a2 S, 5=1;

D8_ 1 s2=a,b1 . a2al. b2a2. Cldl* d2e1 .



MANNING: On the Primitive Grou?ps of Cla8ss Ten. 229

The product. S cannot be of degree 13. If S is of degree 14, it must be non- regular, and is then of order 4. We have

1-4 8 s82- a1b, * .cl . c2a2 - d1a3 - d,2a4d 54 1;

Dlo:-S {, 82- a1 b . c1 d, . e6 al - e2 a2 - a3a4-.

If S is of degree 15:

DI, - s81, S2- a, oa1 . b, g, . cl y, * dt1 3l - el El }

If S is of degree 16:

D12- 1 82 al b= * a2 b2 . al a2 1 P2 * yl 72-

The degrees 17, 18, 19 give no groups D. When S displaces 20 letters, we have,

D13= _ 1i, 82= a, a2 *l1 2 *Yl 72 *1 32 *

Note that D7 and D8 contain D5 and D1, and that D1 and Djo contain D2 and D,,.

The D7, and hence D8, which is the same group, can be thrown out very quickly. It has two transitive systems of 6 letters each. There is a substitution 83 of the series s1, S2, which connects these two systems and brings in at most 4 new letters, one in a cycle,* so that E-s8, s2, s, is a transitive group of degree not greater than 16. We can write S3 SO that it has a cycle new to 81, is not commutative with it and has more than 4 letters new to 81. Hence E is of degree 15 at most. If E is of degree 12, it is of order 24 and the only substitutions of degree 10 found in it are the 6 belonging to s1, s2. If E is of degree 13, its positive subgroup is of class 12, and contains just one subgroup of order 13, which must then be invariant in the group E. But s, cannot transform a group of order 13 into itself. If E is of degree 14, the positive subgroup is of class 12 and of order 14 * 6. By Sylow's theorem this subgroup has a character. istic cyclic subgroup of order 7, which is then invariant in E. Again sl cannot transform a group of order 7 into itself. If E is of degree 15, the positive subgroup is transitive of class 12, and contains no substitutions of degree 13 or 14; hence the subgroup leaving one letter fixed leaves three fixed. Now a group of order 15 * 6 has a single invariant subgroup of order 5. Inr E this is impossible. We can now strike D7 and D8 from our list.

*Transactions, &c., 1. C.




The next to go is the group D3. The tlhree substitutions of degree 10 in D3 are

s= a a2 . b1 b2 . cl c2. d, d2 . eje2, S2= a, a2 . b el* d, * .,E

83= a1a2 . b2 . c2 * d2 . e e.

According to the theorem established in the beginning of this article, there is a substitution s, similar to s,, in the group G, which connects the set a1 a2 with another set of D3. We write s = a1 b1 .... If s leaves a2 fixed we have s3 559 883=

a1b1 . a2 y ... . Then putting = a1 b1 . a2y.... , we see first that y cannot be new to sl. For now that D7 and D8 are thrown out, a group of the type {s, s-

is not found in the list D1, .... , D13. For the samie reason y must belong to 82

and not to 83. Hence y = c1, d1, or e1. Without loss of generalitv we can say that y = c1. Now s,1 s 9 and 82,2 y 8are both D6. Hence from {s1, s 9

s = alb1 . a2 C1 - b2 C2 ....

But (b2 c2) is a cycle new to s2 and on that account {s2,5s cannot be D6. Hence D3 can be dropped from our list.

D6 is next in order of difficulty. The three substitutions of order 2 are

81 = a, a2 . b1 b2 . Cl C2 . d1 d2 . e1 e2,

2= a, b . a2C . b2C2 . da1 * e, a2,

83 = a2 b2 . a1 c2. b1 c . d2 a1 . e2 a2.

There is a substitution s similar to s,, which connects the set a1 .... with the set d1 d2 a1, and brings in at most 4 new letters, one to a cycle. Suppose first that E=- D6, '4 is transitive. If E is of degree 12, it has 4 subgroups of order 3 and is an imprimitive group isomorphic to the symmetric group of degree 4. In (abed) all we get a transitive representation of class 1 0 on 12 letters only by taking as a new element in the multiplication table of the group a subgroup 1, (ab). This substitution (ab) is contained in one subgroup of order 4, in two of order 6, in one of order 8, and in one of order 12. Hence by MARGGRAFF'S extension of a theorem* due to JORDAN, this group E, if conitained in a primitive group, is contained in a doubly transitive group of degree 13, of order 13, 1 2 2.

*MARGGRAFF, Dissertation, Ueber primitive Gruppen mit transitiven Untergruppen geringeren Grades. Giessen (1889). This point is made clear in a paper soon to appear in the Transactions of the American Mathematical Society under the title On Multiply Transitive Groups, by the present writer.



MANNING: On the Primitive Groups of Class Ten. 231

This G13 would by Sylow's tlheorem have to have an invariant subgroup of order 13. This is here impossible. It is clear that E cannot be of degree 13, 14, 15, or 16, if we keep in mind the reasoning in the case of D7.

Now s does not connect mnore than two systems of D6. By ineans of the

conjoin of the regular constituent of D6 we can write without loss of generality

s= a, a,.

Looking over the non-Abelian groups in thie list DI, ...., we see that s can

have 2, 4, or 5 letters new to 8s.

If s has just two letters new to s, Js s} is D0 and

s =a a, . eix x . . (a,2) (e),

with all the 6 letters b, c, d displaced. We now arrive at a contradiction. If

ei=el, by is, sL, x e, and by J8,, S}, x =e; there is a simnilar contradiction if e, = e2.

Next let s have just 4 letters new to s1. Here

b b cd s=alal * a21 * e X2 c X3 bd'

or s =aa1 ao . a2 x, * el x2 * e2 x3 ced

Comparison with Sa and S3 shows that these forms for s have to be rejected because of the number of new letters.

If s has 5 letters new to s9,

s = al a . bx, . Cx2 . dX3 . ex4.

Clearly x4 * a2. Then neither s2, s8 nor iS3, st is possible. This clears D6 out of the way.

Dg, and with it D1o, goes out quickly. There are 4 substitutions in Dg of order 2:

s= a, a2 * b, b2 cl c2 * d, d2 . el e2, s2=a, b, . cl a,. c2a2 .d a3 . d2a4

S3r=a, b2 . a2b, . a,a2 . a3a4 * e,e4 s4 =2 b2 * c2al * c,a2 * d2a3 .dia4.

The substitution s may have two forms:

(1) s =e1 a, ...., (2) s =el ...e

Consider s = ea. .. The groups {s,, s and S3, s} are Dg or D10. In either

case s = a, e * - . (a2) (e2) (b2).




If 1s1, st is D9, tS3, SI is also D9, and s= a,e . c1cd. c2&'. docx"'. d2cIV with only one letter, el, new to 82, s, Is is not in our list. If s,1 st is DIo

,q al e, * bix*

where x is a c or d. But jS3, SI is also D1o, and x is an a. Consider the second case, s = e, c1.... If ss1 =1s, we have uniquely:

s = clel . c2e2 . a, a. * 23 1 - a4 g21-

The group Is,, 82, s8 has three sets of intransitivity. We now have, going one

step further, either (1) s' el d1 . e2d2 . b, b2 ....

an impossibility on account of 1s, s'8, or

(2) s' el1g ....

Now s' must either add new letters to the set a. ... or connect d1.... and a1 .

In neither case can s8 =elg .... be constructed. Hence D9 and D1o can be

dropped from our list. Only one non-Abelian D is left. It is DII. If a substitution s of the series

81, .... connects two cycles of another substitution 8' of the series, s and s' are

comnm utative.

We now take up D2. It is generated by

s= ala2 . b1b2 . clc2 . d1d2 * e, e2,

and 82 = ala2 bic . b2c2 *1 2 *131132.

There are two substitutions, (1) s = a b . .. ., (2) 8 = a, d, .. . to be considered

in extending D2. Let s = ab .... Neither ssi = 8 nor82 s= 9 is possible. If s =a,d. , then ss8=s,s. But 182, ald . a2d2 . .. is impossible. Hence G never includes D2.

The generators of D5 are sl and

82= a, b1 . a2 b2 c,d, c 2d2 -a, a2

There are again only two substitutions s to be examined. The first,

8 = ala . blaG2 *e 911

we reject at once. But the second,

S = a1cl . a2C2 . b1d1 * b2d2 * . 2




requires that the next group 181 82, s, 81t be considered. Now s' must be a,a, .... , an impossibility. D5 is rejected. We throw out D1 also. In it

=2 a, a2 . bl b2 . a, C2 * I12 * Yl )2

Obviously we cannot have s = a,cl . a2c2 .... If

s =a bl . a2 b2 . k 82 *1 E2 * '2

we pass step by step by means of the substitutions

St a, l 1.b, S2 . Cl al * d, 13 * el yl,

S/= ale, . b5 3 . cald .d . eyl s' =al b1 . cal d/3 . * ey, (a= a1 or a2, &C.)

to an intransitive group 1s1, .... s"'%, with which we must stop. Then D1 may be rejected.

It is now clear that D4 also may be struck from our list. Only D1l, D12, and D13 are left. Consider D1I. It leads to two primitive groups of class 10. The substitu-

tions of order 2 in D1l are sl, s8 = aa3 . b b3 . cl C3 . dld3 . ele3,

83= a2a3 . b2b3 * C2 C3 * d2d3 * e2 e3.

There must be in s8, .... a fourth substitution S4 non-commutative with two of these three, non-commutative withi s and.82, say. This follows from the fact that Js1, .... t is transitive. There are only two forms we need assume for 84:

s4 ala4 . blb4 . cl c4 . dld4 . ele4 (e1)

s4= a2b3.a3b2.c1c4.d1d4.e,e4. (2)

Consider (2). The group. E=-4D1l,S4 has in all 6 substitutions of degree 10. The two not yet written down are:

s5 = a, b3 . a3b, . c2c4 . d2d4 . e2e4,

=6 a2 b1 . a, b2 * C3 c4 . d3 d4 . e3 e4. There is in E a set of intransitivity of 6 letters, 1 . . , so that we may extend E by means of

S7=alc1 . a2c2 . a3 C3 d4d5, .*e4e or aIC3 . a3 Cl . b2 c4 * d2d5 . e2e5.

These two substitutions are conjugate under

t = a3 62 * C1 C3 * C2 C4 * d1 d3 . d2 d4 . e1 e3. e2 e4, which also transforms s2 into s2, s1 into s,, and 83 into s4. Hence only the first



234 MANNING: On the Prim'itive Groups of Class Ten.

form of S7 need be examined. Beside the substitutions already written down E' - IE, S7} has the following three substitutions similar to s1:

88 alc4 . b3c2 . b2c3 . d1 dF5 . el e,

sg = b1c, . a2 c4 . blC3 . d2d5 . e2e5,

s1o = b2cl . b, c2 . a3 c4 . d3d5 . e3e5 .

In extending E', two substitutions sll are to be considered. The one, s1l = a, d3.... is impossible. The other

s11 -a1d1 . a2 d2 . a3d3 . c4d5 . e4eB,

gives the group E"', in which the other substitutions of order 2 and class 10 are

s2 a1 d4 . b3d2 . b2d3 . cl d . el e6,

S13- a2 d4. b3 d1 . b1d3 . 2 d5 . e,2e6,

814= a3d4 . bld2 . b2d1 . c3Ad . e3 e6,

si5= cld1 . c2d2 . c3d3 . c4d4 . e5 e6. Finally

816 a, el . a2e2 . a3e3 . c4e, - d4e6

is unique. The transitive group 82, 82 1 84 q 87 811 , 816 d is primitive of degree 21, and is isomorphic to the symmetric group on 7 letters. The subgroup leaving one letter fixed is E', E 16l811 = d1 e1 . d2 e2 . d3 e3 . d4 e4 . d5 e; t. This subgroup has two sets of intransitivity. It is of order 2 (5!). Hence the isoniorphism between I E", 816t E"' and (abcdefg) all is simple.

This group E"' cannot be contained in a doubly transitive group, for then we should have a substitution

s-a1 e6 . e1 d4 * d1e4 * b1- c1-,

similar to s8, which cannot exist at the same tinme as S4. This also means that no group containing B"' can have other substitutions similar to 89. Nor can E"' be invariant in a larger group of degree 21.

There remains the first form of 84:

84- a1 a4. bl4 .Cl C4 d1 d4 e1 e4

Only one form for S7 is possible:

=7=a a5 . b1 b5. c0 c5 d1 d5 * ee 05.




Continuing as before we get the unique set of generating operators, s1, 82, $4, 87,

and St a,b1 * a2 b2 . a3 b3 * a4 b4 . a5 b,

2- ai1c* a2 2 a3C3 * a 4C4 . a5 c, S3/ = a, d, . a2 d2 . a3 d3 * a4 d4 . aA d5, S4 = a, el . a2 e2 * a3 e3. a4 e4 . a5 e5

These give an imprimitive group Hof order (5!)2, and of degree 52. It is not possible to form another substitution s similar to a1. Hence H is invariant in any primitive group of class 10 containing it. Adjoining the substitution

t =a2bl a3cl . a4dl . a5el . b3c2 . b4d2 . b5e2 . c4d3 . cd e4 ,

we have a primitive group G= IH, tt of class 10, degree 25, and order 2(5 !)2. For all integral values of k greater than 2 there exists a primnitive group G

of degree ld and class 2 k with an invariant subgroup H of the same type as H12.

Thus the imprimitive group H is generated by

So = (a, a2) (bl b2) (cl c2) . U(112l A) (klk 2), 83 = (a1 a3) (bl b3) (el c3) . . . . . . . . (.jj3) (kl k3),

*~~~~ ~ ~ . .* * . . . . . . . . . . . . . . . . . . . .

Sk= (a, ak) (bl bk) (cl CO.).* .... (jl1k) (k1 kk), Gb= (a, b1) (a2 b2) (a3 b3) . (aj bj) (ak bk),

(a, cl) (a2 c2) (a3 c3). . (ajcj) (ak CA),

ak = (a, kl) (a2 A2) (a3 k3) . ..... (aj k,) (a kkk).

If now we adjoin to H the substitution

t = (a,) (a2 bl) (a3 cl) (a4 dl) .... (a3 jl) (ak kl)

(b2) (bs c2) (b4d2) .. . .(bj A) (bkk2)

(c3) (c 3). . . . (ci j3) (Ck k3)

(1j) (jkkj)

the group G H, t is primitive since the subgroup

1 S2, 831 . . . . Sj, lb I ce * * )vt0

of order 2[(k - 1)!]2 and omitting kk, is maximal when k is greater than 2. This group G is the only transitive group in which H is invariant.

31



236 MANNING: On the Primitive Groups of ClasS Ten.

If a second transitive group (G') exists in which H is invariant, there is a primitive group G"I G, G' t of order greater than 2(k!)2 in which H is invariant. Let GI, H1, .... be the subgroups of G, H, .... respectively, which leave one letter fixed. Since systems of imprimitivity of H can be chosen in only two ways, G" is simply transitive.* Then GUt is intransitive, and since it includes GI, the two sets of intransitivity of the latter are also the sets of intransitivity of GUs. Now the larger set of intransitivity of G1 is exactly that group of the system we are considering which corresponds to the value k - 1 of k. Hence if the theorem is true for one value of k, it is true for the next succeeding value. But it certainly' holds for k = 3. Therefore the theorem is true in general.

The transitive group generated by the complete set of conjugates sl, .... cannot be Abelian. Then.our task is completed. There are three primitive groups of class 10 which do not contain a substitution of order 5 on 10 letters: the diedral G."; the G" isomorphic to the symmetric group on 7 letters.; and the new group GU!,". We recall the four groups found in a previous paper: t the metacyclic a- G".-o, the Mathieu 4 GU2.11.lo, and their positive subgroups. Two of these groups, the G21 and the G25, are omitted by JORDAN j and one group is in his list for class 10 which does not belong there. All of these groups belong to infinite systemis. The system to which G21 belongs is that of the symmetric-k

group written on k(k - 1) letters. These groups are of class 2k - 4 (not 2 2k - 2 as JORDAN has it) and are primitive if k is greater than 4.

PARISI 19 Jan., 1905.

* On the Primitive Groups of Class Eiight. Bulletin of the American Mathematical Society (1904), 2nd Ser., v. 10, p. 286. An invariant imprimitive subgroup of a multiply transitive group is at least three-fold imprimitive..

t Transactions of the American Mathematical Society, 1. c.

I Comptes Rendus. Vol. 75 (1872), p. 1754.



on the primitive groups of class ten

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