Introduction

In this section we discuss shortly the problem of finding efficient presentations for the twenty simple groups listed on page 1, A comprehensive discussion including the recent progress in investigating the efficiency of finite simple groups and related groups will appear in a forthcoming survey, Campbell, Robertson and Williams [11].

In 1983 Campbell and Robertson commenced the investigation of the efficiency of all the thirteen simple groups of order up to 10^ excluding PSL(2,p”), n ^ 2. They were successful in finding efficient presentations for all but and PSU(3,3), see [4]. This work used as a starting point the minimal permutation generators given in [35] and the presentations satisfied by these permutations [16]. The method employed to make these presentations efficient was explained in 3.5 of this thesis. Three years later P.E. Kenne showed that and PSU(3,3) are also efficient [32]. To obtain efficient presentations for these groups he used a method which we described in 3.5.

In [37] E.F. Robertson suggested extending the work of [4] to the remaining seven simple groups PSU(3,5), Ji, A9, PSL(3,5), M22, J2 and PSp(4,4). A result of

A

himself and Campbell [6] showed that that J2 , and consequently J2, are efficient.

We will show that the three simple groups PSU(3,5), and M22 are efficient. Taking these results together with the efficient presentation for A9, [9], among the twenty simple groups up to 10^ only the efficiency of PSL(3,5) and PSp(4,4) remain undecided. Note that each of these groups has trivial multiplier and an efficient presentation for each of them requires an equal number of generators and relations.

It is also worth noting that, at the present time, all the simple groups of order up to 178

IQti with the above exceptions have been shown to be efficient [11].

5.2 Method

Suppose G is a moderately large "concrete" group, say a permutation or matrix group, generated by two of its elements x and y with x an involution. In order to be able to obtain an efficient presentation for G we may use methods similar to those of Kenne but since we are dealing with a group of large order we are unable to follow his method of computing presentations on many randomly chosen pairs of generators using Cannon's algorithm. We therefore try to choose a set of generating pairs (x,y) for G with x an involution and y having a specified order. This allows us a more efficient method of finding relations and also simplifies the final stages of reduction to an efficient presentation. In particular if G is a finite simple group < 10^, such generating pairs always exist. Now PERM ( or a similar matrix program) can be used to find a presentation for one of the groups G that we are examining. We then attempt to reduce the number of relations using the method described in 3.5.

Notg.Suppose G is a finite permutation group. The following CAYLEY program tries to determine whether G has a generating pair (x,y) with x an involution and y, xy having specified orders nj and n2. Whenever G is a large group the program can be modifed so that a small set of random elements y of a specific order for which G=<x,y> are sought. > gorder = order (g) ;

>s = null ; >t= null;

> FOR EACH X IN classes (g) DO

> IF order (X ) EQ 2 THEN > s = s JOIN [x] ; > END; > IF order (x) EQ nl THEN > t= t JOIN class (gpc) > END ; > END; > FOR EACH X IN s DO > FOR EACH y IN t DO

> IF order (x*y) EQ n2 THEN

> IF order (<x, y > ) EQ gorder THEN

> PRINT X , y ;

> BREAK ;

> END ; > END ; > END; >END;

The following examples illustrate how the method works in practice. Example 1. Find a deficiency zero presentation for M ^.

The Mathieu group M ^ has a unique pair of minimal generating permutations a and b of degree 11:

a=(l,10)(2,8)(3,ll)(5,7), b=(l,4,7,6)(2,l 1,10,9).

We choose an involution x and an element y of order 5 with <x,y>=Mn. Take x=a and y=abab'lab2 where y is a representative of the single conjugacy class of elements of order 5 chosen so that the product xy has order 11. PERM gives

M i i = < x , y 1 x 2 = y 5 = ( x y ) l l = x y ^ x y x y 2 x y 2 x y x y 2 x y 2 x y 2 x y ~ 2 = l >

and (xy)ll=l is easily shown to be redundant using TC. Now

H=<x,y I x2y5=l, [x-l,y-2][x,y][x,y2][x“l,yl]=y2xy>

is a stem extension of M ^ since H is perfect and x^, being both a power of x and y, is central. But M(Mu)=l so x2=l and H=Mn.

Example 2. Find an efficient presentation for A5XA5,

The group A5 is generated by the permutations (1,3)(4,5) and (1,4,2) so A5xA5=<a,b,c,d> where

a=(l,3)(4,5), b=(l,4,2), c=(6,8)(9,10), d=(6,9,7).

We choose the involution x=ac and look for an element y of order 5 with <x,y>=A^xA3. Using CAYLEY, we find y=(l,3,4,5,2)(6,9,8,10,7). Now PERM gives

A5xA5=<x,y I x2=y5=(xy) 15=(xy 1 )2(xy)4(xy-1 )^xy2= 1 > which proves that A5XA5 is efficient for M(A3xAg)=C2xC2.

Note. The above presentations for Mjj and A5XA5 are quite different presentations to those found by Kenne in [32] and [30]. In particular, the latter presentation can be

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reduced to a deficiency zero presentation for A5XA5 as follows: Let

G=<x,y I x2y5=l, (x" 1 y ^)2(xy)4(x" ^y 1 )^xy2= 1 >. 180

Then G is a perfect group of order 14400. Now a simple check using TC verifies that the relations [x,(xy)i5]==[y^(xy)l5]-i hold in G. Therefore, (xy)l5 is a central element of G. But, G/<x2,(xy)l5> = A5XA5 so G is a stem extension of A5XA5 . Hence G is a

A A

homomorphic image of A5XA5. Now, since G and A5XA5 have the same order, G = A A

A5XA5.

Note that on adding the relation (xy)l5=l to the presentation for G we find an efficient presentation for A5XA5. Also, adding the relation y5=l to the same presentation gives a perfect group of order 7200 which is denoted by P4 in [40] ( in fact, P4 is the

A A

factor group (A$xA3)/C2 ).

5.3 Efficiency of PSU(3,5), and #

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In this section we give 2-generator 3-relation presentations for the simple group PSU(3,5) of order 126000 and the Mathieu group M22 of order 443520. We also give a 2-generator 2-relation presentation for the Janko group Jj order 175560. This proves that these three simple groups are efficient.

5.3.1 PSU(3,5) is efficient.

Proof. Since M(PSU(3,5)) is cyclic of order 3 we seek a 2-generator 3-relation presentation. We start with the unique minimal generating pair (a, b) with a2=b4=l given in [6]. PSU(3,5) has 4 conjugacy classes of order 5 and, after some experimenting, we choose y=(ab)2ab"l(ab)2 of order 5 and the involution x=a so that xy has order 10.

PERM finds the presentation

PSU(3,5)=<x,y Ix2=y5=(xy)10=[x,y2]4=r=l>

where i^xy-2xy2x(yxy-2xyxy)2. Coset enumeration shows that

j

PSU(3,5)=<X,y Ix2=(xy)10=r=l, [ x ,y 2 ] 4 = y 5 > . |

Let H be the stem extension of PSU(3,5) |

H=<x,y Ix2=(xy)10, r=l, [x,y2]4=y5>, *

Now, coset enumeration shows that IH:<(xy)3>f=12600 so, by 3.5.1 (i), we see that H = PSU(3,5). This gives an efficient presentation as required.

5.3.2 The Janko group Jj has deficiency zero.

Proof. M(Ji) is trivial so every stem extension of is isomorphic to J^. We seek a 2-generator 2-relation presentation starting from the presentation 15.20 for Jj given in

[Q:

Jj=r <a,b I a2=b3 =q7—gl— where r=abab'^ s=(ab)3(ab"l)3abab'i and q=ab'^(ab)2.

Let

H= <a,b I a2=b3=q7, r^=s2> •

Now coset enumeration verifies that H=<r,s> so is central in H and thus H, being perfect, is a stem extension of Jj. Hence H = J^.

Now rewrite r^=s2 as u=l where

u=ab'iab(abab"^)3ab‘lab(ab’^)3abab"l(ab)3 and consider

K=<a,b I a2=b3=l, u=q7>.

Coset enumeration shows, with considerable difficulty, that lK:<b>l=58520 so K ~ Jj. Finally let

L=<a,b I a2b3=l, (a"^b'Hab)2)4(a-lb)2a‘^b’l(ab)3ab"^ab)2(a‘ib"^)2=l>.

But L is perfect, aP- is central in L and L/<a2> = K = Jj. Hence L is a stem extension of

Jj so L = Jj.

5.3.3 The Mathieu group M22 is efficient.

Proof. M(M22) is cyclic of order 12 so we seek a 2-generator 3-relation presentation. Start with the minimal generating pair of permutations given in 18.3 of [35] but rather than use the presentation 18.3 of [6] we use PERM to give the presentation

M22—<â,b I a2=b4=(ab)H=s^=r=l>

where s=abab2 and r=(ab)2(ab-^)2ab2(ab)2ab"^ab(ab2)2. Now coset enumeration shows tiiat

G=<a,b I a2=(ab)H=i, s^=b4, r=b4> is isomorphic to M22. Now let H be the stem extension of G given by

H=<a,b I a2=(ab)H, s'^=b4, r=b4>. Coset enumeration verifies that

IH:<ab>l=IH:<(ab)2>l=IH:<(ab)3>|=IG:<ab>l, so, by 3.5.1 (ii), we obtain H = M22 as required.